Authors David Joyner Minh Van Nguyen Nathann Cohen
License GFDL-1.3-no-invariants-or-later
Algorithmic Graph Theory
David Joyner, Minh Van Nguyen, Nathann Cohen
Version 0.7-r1946
2012 February 25
Copyright © 2010–2012 David Joyner hwdjoyner@gmail.comi
Copyright © 2009–2012 Minh Van Nguyen hmvngu.name@gmail.comi
Copyright © 2010 Nathann Cohen hnathann.cohen@gmail.comi
Permission is granted to copy, distribute and/or modify this document under the terms
of the GNU Free Documentation License, Version 1.3 or any later version published by
the Free Software Foundation; with no Invariant Sections, no Front-Cover Texts, and no
Back-Cover Texts. A copy of the license is included in the section entitled “GNU Free
Documentation License”.
The latest version of the book is available from its website at
http://code.google.com/p/graphbook/
Edition
Version 0.7-r1946
2012 February 25
Contents
Acknowledgments iv
1 Introduction to graph theory 1
1.1 Graphs and digraphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
1.2 Subgraphs and other graph types . . . . . . . . . . . . . . . . . . . . . . 11
1.3 Representing graphs as matrices . . . . . . . . . . . . . . . . . . . . . . . 18
1.4 Isomorphic graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
1.5 New graphs from old . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
1.6 Common applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37
1.7 Application: finite automata . . . . . . . . . . . . . . . . . . . . . . . . . 39
1.8 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46
2 Graph algorithms 53
2.1 Representing graphs in a computer . . . . . . . . . . . . . . . . . . . . . 54
2.2 Graph searching . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60
2.3 Weights and distances . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71
2.4 Dijkstra’s algorithm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74
2.5 Bellman-Ford algorithm . . . . . . . . . . . . . . . . . . . . . . . . . . . 77
2.6 Floyd-Roy-Warshall algorithm . . . . . . . . . . . . . . . . . . . . . . . . 78
2.7 Johnson’s algorithm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85
2.8 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87
3 Trees and forests 105
3.1 Definitions and examples . . . . . . . . . . . . . . . . . . . . . . . . . . . 105
3.2 Properties of trees . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112
3.3 Minimum spanning trees . . . . . . . . . . . . . . . . . . . . . . . . . . . 116
3.4 Binary trees . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127
3.5 Huffman codes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134
3.6 Tree traversals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140
3.7 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145
4 Tree data structures 153
4.1 Priority queues . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154
4.2 Binary heaps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155
4.3 Binomial heaps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165
4.4 Binary search trees . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173
4.5 AVL trees . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 180
4.6 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 190
i
ii Contents
5 Distance and connectivity 197
5.1 Paths and distance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 197
5.2 Vertex and edge connectivity . . . . . . . . . . . . . . . . . . . . . . . . . 202
5.3 Menger’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 207
5.4 Whitney’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 210
5.5 Centrality of a vertex . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211
5.6 Network reliability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213
5.7 The spectrum of a graph . . . . . . . . . . . . . . . . . . . . . . . . . . . 213
5.8 Expander graphs and Ramanujan graphs . . . . . . . . . . . . . . . . . . 219
5.9 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 222
6 Optimal graph traversals 226
6.1 Eulerian graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 226
6.2 Hamiltonian graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 226
6.3 The Chinese Postman Problem . . . . . . . . . . . . . . . . . . . . . . . 227
6.4 The Traveling Salesman Problem . . . . . . . . . . . . . . . . . . . . . . 227
7 Planar graphs 228
7.1 Planarity and Euler’s Formula . . . . . . . . . . . . . . . . . . . . . . . . 228
7.2 Kuratowski’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 228
7.3 Planarity algorithms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 230
8 Graph coloring 231
8.1 Vertex coloring . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 231
8.2 Edge coloring . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 232
8.3 Applications of graph coloring . . . . . . . . . . . . . . . . . . . . . . . . 232
9 Network flows 234
9.1 Flows and cuts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 234
9.2 Chip firing games . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 239
9.3 Ford-Fulkerson theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 245
9.4 Edmonds and Karp’s algorithm . . . . . . . . . . . . . . . . . . . . . . . 250
9.5 Goldberg and Tarjan’s algorithm . . . . . . . . . . . . . . . . . . . . . . 250
10 Random graphs 251
10.1 Network statistics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 251
10.2 Binomial random graph model . . . . . . . . . . . . . . . . . . . . . . . . 255
10.3 Erdős-Rényi model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 262
10.4 Small-world networks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 264
10.5 Scale-free networks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 269
10.6 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 274
11 Graph problems and their LP formulations 280
11.1 Maximum average degree . . . . . . . . . . . . . . . . . . . . . . . . . . . 280
11.2 Traveling Salesman Problem . . . . . . . . . . . . . . . . . . . . . . . . . 281
11.3 Edge-disjoint spanning trees . . . . . . . . . . . . . . . . . . . . . . . . . 283
11.4 Steiner tree . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 284
11.5 Linear arboricity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 286
11.6 H-minor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 288
Contents iii
A Asymptotic growth 292
B GNU Free Documentation License 293
1. APPLICABILITY AND DEFINITIONS . . . . . . . . . . . . . . . . . . . . 293
2. VERBATIM COPYING . . . . . . . . . . . . . . . . . . . . . . . . . . . . 295
3. COPYING IN QUANTITY . . . . . . . . . . . . . . . . . . . . . . . . . . . 295
4. MODIFICATIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 296
5. COMBINING DOCUMENTS . . . . . . . . . . . . . . . . . . . . . . . . . 297
6. COLLECTIONS OF DOCUMENTS . . . . . . . . . . . . . . . . . . . . . . 298
7. AGGREGATION WITH INDEPENDENT WORKS . . . . . . . . . . . . . 298
8. TRANSLATION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 298
9. TERMINATION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 299
10. FUTURE REVISIONS OF THIS LICENSE . . . . . . . . . . . . . . . . . 299
11. RELICENSING . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 299
ADDENDUM: How to use this License for your documents . . . . . . . . . . . 300
Bibliography 301
Index 310
Acknowledgments
Fidel Barrera-Cruz: reported typos in Chapter 3. See changeset 101. Suggested
making a note about disregarding the direction of edges in undirected graphs. See
changeset 277.
Daniel Black: reported a typo in Chapter 1. See changeset 61.
Kevin Brintnall: reported typos in the definition of iadj(v) ∩ oadj(v); see change-
sets 240 and 242. Solution to Example 1.14(2); see changeset 246.
John Costella: helped to clarify the idea that the adjacency matrix of a bipar-
tite graph can be permuted to obtain a block diagonal matrix. See page 20 and
revisions 1865 and 1869.
Aaron Dutle: reported a typo in Figure 1.15. See changeset 125.
Péter L. Erdős (http://www.renyi.hu/∼elp) for informing us of the reference [78] on
the Havel-Hakimi theorem for directed graphs.
Noel Markham: reported a typo in Algorithm 2.5. See changeset 131 and Issue 2.
Caroline Melles: clarify definitions of various graph types (weighted graphs, multi-
graphs, and weighted multigraphs); clarify definitions of degree, isolated vertices,
and pendant and using the butterfly graph with 5 vertices (see Figure 1.10) to
illustrate these definitions; clarify definitions of trails, closed paths, and cycles;
see changeset 448. Some rearrangements of materials in Chapter 1 to make the
reading flow better and a few additions of missing definitions; see changeset 584.
Clarifications about unweighted and weighted degree of a vertex in a multigraph;
notational convention about a graph being simple unless otherwise stated; an exam-
ple on graph minor; see changeset 617. Reported a missing edge in Figure 1.5(b);
see changeset 1945.
Pravin Paratey: simplify the sentence formation in the definition of digraphs; see
changeset 714 and Issue 7.
Henrique Rennó: pointed out the ambiguity in the definition of weighted multi-
graphs; see changeset 1936. Reported typos; see changeset 1938.
The world map in Figure 2.16 was adapted from an SVG image file from Wikipedia.
The original SVG file was accessed on 2010-10-01 at http://en.wikipedia.org/wiki/
File:WorldmapwlocationwNEDw50m.svg.
iv
Chapter 1
Introduction to graph theory
— Spiked Math, http://spikedmath.com/120.html
Our journey into graph theory starts with a puzzle that was solved over 250 years ago by
Leonhard Euler (1707–1783). The Pregel River flowed through the town of Königsberg,
which is present day Kaliningrad in Russia. Two islands protruded from the river.
On either side of the mainland, two bridges joined one side of the mainland with one
island and a third bridge joined the same side of the mainland with the other island. A
bridge connected the two islands. In total, seven bridges connected the two islands with
both sides of the mainland. A popular exercise among the citizens of Königsberg was
determining if it was possible to cross each bridge exactly once during a single walk. For
historical perspectives on this puzzle and Euler’s solution, see Gribkovskaia et al. [94]
and Hopkins and Wilson [108].
To visualize this puzzle in a slightly different way, consider Figure 1.1. Imagine that
points a and c are either sides of the mainland, with points b and d being the two islands.
Place the tip of your pencil on any of the points a, b, c, d. Can you trace all the lines
in the figure exactly once, without lifting your pencil? Known as the seven bridges of
Königsberg puzzle, Euler solved this problem in 1735 and with his solution he laid the
foundation of what is now known as graph theory.
1
2 Chapter 1. Introduction to graph theory
c
b d
a
Figure 1.1: The seven bridges of Königsberg puzzle.
1.1 Graphs and digraphs
When I use a word, it means just what I choose it to mean—neither more nor less.
— Humpty Dumpty in Lewis Carroll’s Through the Looking Glass
The word “graph” is commonly understood to mean a visual representation of a dataset,
such as a bar chart, a histogram, a scatterplot, or a plot of a function. Examples of such
graphs are shown in Figure 1.2.
1
0.5
0.5
f (x)
0 0
y
−0.5
−0.5
−1
−6 −4 −2 0 2 4 6 −4 −2 0 2 4
x x
(a) Plots of functions. (b) A scatterplot.
Figure 1.2: Visual representations of datasets as plots.
This book is not about graphs in the sense of plots of functions or datasets. Rather,
our focus is on combinatorial graphs or graphs for short. A graph in the combinatorial
sense is a collection of discrete interconnected elements, an example of which is shown
in Figure 1.1. How can we elaborate on this brief description of combinatorial graphs?
To paraphrase what Felix Klein said about curves,1 it is easy to define a graph until we
realize the countless number of exceptions. There are directed graphs, weighted graphs,
multigraphs, simple graphs, and so on. Where do we begin?
Notation If S is a set, let S (n) denote the set of unordered n-tuples (with possible
repetition). We shall sometimes refer to an unordered n-tuple as an n-set.
1
“Everyone knows what a curve is, until he has studied enough mathematics to become confused
through the countless number of possible exceptions.”
1.1. Graphs and digraphs 3
We start by calling a “graph” what some call an “unweighted, undirected graph without
multiple edges.”
Definition 1.1. A graph G = (V, E) is an ordered pair of finite sets. Elements of V are
called vertices or nodes, and elements of E ⊆ V (2) are called edges or arcs. We refer to
V as the vertex set of G, with E being the edge set. The cardinality of V is called the
order of G, and |E| is called the size of G. We usually disregard any direction of the
edges and consider (u, v) and (v, u) as one and the same edge in G. In that case, G is
referred to as an undirected graph.
One can label a graph by attaching labels to its vertices. If (v1 , v2 ) ∈ E is an edge of
a graph G = (V, E), we say that v1 and v2 are adjacent vertices. For ease of notation, we
write the edge (v1 , v2 ) as v1 v2 . The edge v1 v2 is also said to be incident with the vertices
v1 and v2 .
a
e b
d c
Figure 1.3: A house graph.
Example 1.2. Consider the graph in Figure 1.3.
1. List the vertex and edge sets of the graph.
2. For each vertex, list all vertices that are adjacent to it.
3. Which vertex or vertices have the largest number of adjacent vertices? Similarly,
which vertex or vertices have the smallest number of adjacent vertices?
4. If all edges of the graph are removed, is the resulting figure still a graph? Why or
why not?
5. If all vertices of the graph are removed, is the resulting figure still a graph? Why
or why not?
Solution. (1) Let G = (V, E) denote the graph in Figure 1.3. Then the vertex set of G
is V = {a, b, c, d, e}. The edge set of G is given by
E = {ab, ae, ba, bc, be, cb, cd, dc, de, ed, eb, ea}. (1.1)
We can also use Sage to construct the graph G and list its vertex and edge sets:
4 Chapter 1. Introduction to graph theory
sage : G = Graph ({ " a " :[ " b " ," e " ] , " b " :[ " a " ," c " ," e " ] , " c " :[ " b " ," d " ] ,
... " d " :[ " c " ," e " ] , " e " :[ " a " ," b " ," d " ]})
sage : G
Graph on 5 vertices
sage : G . vertices ()
[ ’a ’ , ’b ’ , ’c ’ , ’d ’ , ’e ’]
sage : G . edges ( labels = False )
[( ’a ’ , ’b ’) , ( ’a ’ , ’e ’) , ( ’b ’ , ’e ’) , ( ’c ’ , ’b ’) , ( ’c ’ , ’d ’) , ( ’e ’ , ’d ’ )]
The graph G is undirected, meaning that we do not impose direction on any edges.
Without any direction on the edges, the edge ab is the same as the edge ba. That is why
G.edges() returns six edges instead of the 12 edges listed in (1.1).
(2) Let adj(v) be the set of all vertices that are adjacent to v. Then we have
adj(a) = {b, e}
adj(b) = {a, c, e}
adj(c) = {b, d}
adj(d) = {c, e}
adj(e) = {a, b, d}.
The vertices adjacent to v are also referred to as its neighbors. We can use the function
G.neighbors() to list all the neighbors of each vertex.
sage : G . neighbors ( " a " )
[ ’b ’ , ’e ’]
sage : G . neighbors ( " b " )
[ ’a ’ , ’c ’ , ’e ’]
sage : G . neighbors ( " c " )
[ ’b ’ , ’d ’]
sage : G . neighbors ( " d " )
[ ’c ’ , ’e ’]
sage : G . neighbors ( " e " )
[ ’a ’ , ’b ’ , ’d ’]
(3) Taking the cardinalities of the above five sets, we get |adj(a)| = |adj(c)| =
|adj(d)| = 2 and |adj(b)| = |adj(e)| = 3. Thus a, c and d have the smallest number
of adjacent vertices, while b and e have the largest number of adjacent vertices.
(4) If all the edges in G are removed, the result is still a graph, although one without
any edges. By definition, the edge set of any graph is a subset of V (2) . Removing all
edges of G leaves us with the empty set ∅, which is a subset of every set.
(5) Say we remove all of the vertices from the graph in Figure 1.3 and in the process
all edges are removed as well. The result is that both of the vertex and edge sets are
empty. This is a special graph known as the empty or null graph.
Example 1.3. Consider the illustration in Figure 1.4. Does Figure 1.4 represent a
graph? Why or why not?
Solution. If V = {a, b, c} and E = {aa, bc}, it is clear that E ⊆ V (2) . Then (V, E) is a
graph. The edge aa is called a self-loop of the graph. In general, any edge of the form
vv is a self-loop.
In Figure 1.3, the edges ae and ea represent one and the same edge. If we do not
consider the direction of the edges in the graph of Figure 1.3, then the graph has six
edges. However, if the direction of each edge is taken into account, then there are 12 edges
as listed in (1.1). The following definition captures the situation where the direction of
the edges are taken into account.
A directed edge is an edge such that one vertex incident with it is designated as
the head vertex and the other incident vertex is designated as the tail vertex. In this
1.1. Graphs and digraphs 5
a
c b
Figure 1.4: A figure with a self-loop.
situation, we may assume that the set of edges is a subset of the ordered pairs V × V .
A directed edge uv is said to be directed from its tail u to its head v. A directed graph
or digraph G is a graph each of whose edges is directed. The indegree id(v) of a vertex
v ∈ V (G) counts the number of edges such that v is the head of those edges. The
outdegree od(v) of a vertex v ∈ V (G) is the number of edges such that v is the tail of
those edges. The degree deg(v) of a vertex v of a digraph is the sum of the indegree and
the outdegree of v.
Let G be a graph without self-loops and multiple edges. It is important to distinguish
a graph G as being directed or undirected. If G is undirected and uv ∈ E(G), then uv
and vu represent the same edge. In case G is a digraph, then uv and vu are different
directed edges. For a digraph G = (V, E) and a vertex v ∈ V , all the neighbors of v
in G are contained in adj(v), i.e. the set of all neighbors of v. Just as we distinguish
between indegree and outdegree for a vertex in a digraph, we also distinguish between in-
neighbors and out-neighbors. The set of in-neighbors iadj(v) ⊆ adj(v) of v ∈ V consists
of all those vertices that contribute to the indegree of v. Similarly, the set of out-neighbors
oadj(v) ⊆ adj(v) of v ∈ V are those vertices that contribute to the outdegree of v. Then
iadj(v) ∩ oadj(v) = {u | uv ∈ E and vu ∈ E}
and adj(v) = iadj(v) ∪ oadj(v).
1.1.1 Multigraphs
This subsection presents a larger class of graphs. For simplicity of presentation, in this
book we shall assume usually that a graph is not a multigraph. In other words, when you
read a property of graphs later in the book, it will be assumed (unless stated explicitly
otherwise) that the graph is not a multigraph. However, as multigraphs and weighted
graphs are very important in many applications, we will try to keep them in the back
of our mind. When appropriate, we will add as a remark how an interesting property of
“ordinary” graphs extends to the multigraph or weighted graph case.
An important class of graphs consist of those graphs having multiple edges between
pairs of vertices. A multigraph is a graph in which there are multiple edges between a
pair of vertices. A multi-undirected graph is a multigraph that is undirected. Similarly,
a multidigraph is a directed multigraph.
Example 1.4. Sage can compute with and plot multigraphs, or multidigraphs, having
loops.
sage : G = Graph ({0:{0: ’ e0 ’ ,1: ’ e1 ’ ,2: ’ e2 ’ ,3: ’ e3 ’} , 2:{5: ’ e4 ’ }})
sage : G . show ( vertex_labels = True , edge_labels = True , graph_border = True )
sage : H = DiGraph ({0:{0: " e0 " }} , Loops = True )
sage : H . add_edges ([(0 ,1 , ’ e1 ’) , (0 ,2 , ’ e2 ’) , (0 ,2 , ’ e3 ’) , (1 ,2 , ’ e4 ’) , (1 ,0 , ’ e5 ’ )])
sage : H . show ( vertex_labels = True , edge_labels = True , graph_border = True )
6 Chapter 1. Introduction to graph theory
5
e4
2
2 e3
e4 e2
e2
e5
0 e0
1 0
e3 e1
3 e0 1 e1
(a) (b)
Figure 1.5: A graph G and digraph H with a loop and multi-edges.
These graphs are plotted in Figure 1.5.
As we indicated above, a graph may have “weighted” edges.
Definition 1.5. A weighted graph is a graph G = (V, E) where each set V and E is a
pair consisting of a vertex and a real number called the weight.
The illustration in Figure 1.1 is actually a multigraph, a graph with multiple edges,
called the Königsberg graph.
Definition 1.6. For a weighted multigraph G, we are given:
A finite set V whose elements are pairs (v, wv ), where v is called a vertex and
wv ∈ R is the vertex weight. (Sometimes, the pair (v, wv ) is called the vertex.)
A finite set E whose elements are weighted edges. We do not necessarily assume
that E ⊆ V (2) , where V (2) is the set of unordered pairs of vertices.2 Each weighted
edge can be represented as a 3-tuple of the form (we , u, v), where (u, v) is the edge
in question and we ∈ R is the edge weight.
An incidence function
i : E → V (2) . (1.2)
Such a multigraph is denoted G = (V, E, i). An orientation on G is a function
h:E→V (1.3)
where h(e) ∈ i(e) for all e ∈ E. The element v = h(e) is called the head of i(e). If G has
no self-loops, then i(e) is a set having exactly two elements denoted i(e) = {h(e), t(e)}.
The element v = t(e) is called the tail of i(e). For self-loops, we set t(e) = h(e). A
multigraph with an orientation can therefore be described as the 4-tuple (V, E, i, h).
In other words, G = (V, E, i, h) is a multidigraph. Figure 1.6 illustrates a weighted
multigraph.
2
However, we always assume that E ⊆ R × V (2) , where the R-component is called the weight of the
edge.
1.1. Graphs and digraphs 7
1
v3 v4
2
3 1
1
1 2 3 v5
3 3
6
v2 v1
1
Figure 1.6: An example of a weighted multigraph.
The vertex degree of a weighted multigraph must be defined. There is a weighted
degree and an unweighted degree. Let G be a graph as in Definition 1.6. The unweighted
indegree of a vertex v ∈ V counts the edges going into v:
X
deg+ (v) = 1.
e∈E
h(e)=v
The unweighted outdegree of a vertex v ∈ V counts the edges going out of v:
X
deg− (v) = 1.
e∈E
v∈i(e)={v,v 0 }
h(e)=v 0
The unweighted degree deg(v) of a vertex v of a weighted multigraph is the sum of the
unweighted indegree and the unweighted outdegree of v:
deg(v) = deg+ (v) + deg− (v). (1.4)
Loops are counted twice.
Similarly, there is the set of in-neighbors
iadj(v) = {w ∈ V | for some e ∈ E, i(e) = {v, w}, h(e) = v}
and the set of out-neighbors
oadj(v) = {w ∈ V | for some e ∈ E, i(e) = {v, w}, h(e) = w}.
Define the adjacency of v to be the union of these:
adj(v) = iadj(v) ∪ oadj(v). (1.5)
It is clear that deg+ (v) = | iadj(v)| and deg− (v) = | oadj(v)|.
The weighted indegree of a vertex v ∈ V counts the weights of edges going into v:
X
wdeg + (v) = wv .
e∈E
h(e)=v
8 Chapter 1. Introduction to graph theory
The weighted outdegree of a vertex v ∈ V counts the weights of edges going out of v:
X
wdeg − (v) = wv .
e∈E
v∈i(e)={v,v 0 }
h(e)=v 0
The weighted degree of a vertex of a weighted multigraph is the sum of the weighted
indegree and the weighted outdegree of that vertex,
wdeg(v) = wdeg + (v) + wdeg − (v).
In other words, it is the sum of the weights of the edges incident to that vertex, regarding
the graph as an undirected weighted graph. Unweighted degrees are a special case of
weighted degrees. For unweighted degrees, we merely set each edge weight to unity.
Definition 1.7. Let G = (V, E, h) be an unweighted multidigraph. The line graph of G,
denoted L(G), is the multidigraph whose vertices are the edges of G and whose edges are
(e, e0 ) where h(e) = t(e0 ) (for e, e0 ∈ E). A similar definition holds if G is undirected.
For example, the line graph of the cyclic graph is itself.
1.1.2 Simple graphs
Our life is frittered away by detail. . . . Simplify, simplify. Instead of three meals a day, if
it be necessary eat but one; instead of a hundred dishes, five; and reduce other things in
proportion.
— Henry David Thoreau, Walden, 1854, Chapter 2: Where I Lived, and What I Lived For
A simple graph is a graph with no self-loops and no multiple edges. Figure 1.7 illustrates
a simple graph and its digraph version, together with a multidigraph version of the
Königsberg graph. The edges of a digraph can be visually represented as directed arrows,
similar to the digraph in Figure 1.7(b) and the multidigraph in Figure 1.7(c). The digraph
in Figure 1.7(b) has the vertex set {a, b, c} and the edge set {ab, bc, ca}. There is an arrow
from vertex a to vertex b, hence ab is in the edge set. However, there is no arrow from
b to a, so ba is not in the edge set of the graph in Figure 1.7(b). The family Sh(n) of
Shannon multigraphs is illustrated in Figure 1.8 for integers 2 ≤ n ≤ 7. These graphs
are named after Claude E. Shannon (1916–2001) and are sometimes used when studying
edge colorings. Each Shannon multigraph consists of three vertices, giving rise to a total
of three distinct unordered pairs. Two of these pairs are connected by bn/2c edges and
the third pair of vertices is connected by b(n + 1)/2c edges.
Notational convention Unless stated otherwise, all graphs are simple graphs in the
remainder of this book.
Definition 1.8. For any vertex v in a graph G = (V, E), the cardinality of adj(v) (as
in 1.5) is called the degree of v and written as deg(v) = | adj(v)|. The degree of v counts
the number of vertices in G that are adjacent to v. If deg(v) = 0, then v is not incident
to any edge and we say that v is an isolated vertex. If G has no loops and deg(v) = 1,
then v is called a pendant.
1.1. Graphs and digraphs 9
c
a a b d
c b c b a
(a) Simple graph. (b) Digraph. (c) Multidigraph.
Figure 1.7: A simple graph, its digraph version, and a multidigraph.
(a) Sh(2) (b) Sh(3) (c) Sh(4)
(d) Sh(5) (e) Sh(6) (f) Sh(7)
Figure 1.8: The family of Shannon multigraphs Sh(n) for n = 2, . . . , 7.
10 Chapter 1. Introduction to graph theory
Some examples would put the above definition in concrete terms. Consider again
the graph in Figure 1.4. Note that no vertices are isolated. Even though vertex a is
not incident to any vertex other than a itself, note that deg(a) = 2 and so by definition
a is not isolated. Furthermore, each of b and c is a pendant. For the house graph in
Figure 1.3, we have deg(b) = 3. For the graph in Figure 1.7(b), we have deg(b) = 2.
If V 6= ∅ and E = ∅, then G is a graph consisting entirely of isolated vertices. From
Example 1.2 we know that the vertices a, c, d in Figure 1.3 have the smallest degree in
the graph of that figure, while b, e have the largest degree.
The minimum degree among all vertices in G is denoted δ(G), whereas the maximum
degree is written as ∆(G). Thus, if G denotes the graph in Figure 1.3 then we have
δ(G) = 2 and ∆(G) = 3. In the following Sage session, we construct the digraph in
Figure 1.7(b) and compute its maximum and minimum number of degrees.
sage : G = DiGraph ({ " a " : " b " , " b " : " c " , " c " : " a " })
sage : G
Digraph on 3 vertices
sage : G . degree ( " a " )
2
sage : G . degree ( " b " )
2
sage : G . degree ( " c " )
2
So for the graph G in Figure 1.7, we have δ(G) = ∆(G) = 2.
The graph G in Figure 1.7 has the special property that its minimum degree is the
same as its maximum degree, i.e. δ(G) = ∆(G). Graphs with this property are referred
to as regular . An r-regular graph is a regular graph each of whose vertices has degree r.
For instance, G is a 2-regular graph. The following result, due to Euler, counts the total
number of degrees in any graph.
P
Theorem 1.9. Euler 1736. If G = (V, E) is a graph, then v∈V deg(v) = 2|E|.
Proof. Each edge e = v1 v2 ∈ E is incident with two vertices, so e is counted twice
towards the total sum of degrees. The first time, we count e towards the degree of vertex
v1 and the second time we count e towards the degree of v2 .
Theorem 1.9 is sometimes called the “handshaking lemma,” due to its interpretation
as in the following story. Suppose you go into a room. Suppose there are n people in the
room (including yourself) and some people shake hands with others and some do not.
Create the graph with n vertices, where each vertex is associated with a different person.
Draw an edge between two people if they shook hands. The degree of a vertex is the
number of times that person has shaken hands (we assume that there are no multiple
edges, i.e. that no two people shake hands twice). The theorem above simply says that
the total number of handshakes is even. This is “obvious” when you look at it this way
since each handshake is counted twice (A shaking B’s hand is counted, and B shaking A’s
hand is counted as well, since the sum in the theorem is over all vertices). To interpret
Theorem 1.9 in a slightly different way within the context of the same room of people,
there is an even number of people who shook hands with an odd number of other people.
This consequence of Theorem 1.9 is recorded in the following corollary.
Corollary 1.10. A graph G = (V, E) contains an even number of vertices with odd
degrees.
Proof. Partition V into two disjoint subsets: Ve is the subset of V that contains only
vertices with even degrees; and Vo is the subset of V with only vertices of odd degrees.
1.2. Subgraphs and other graph types 11
That is, V = Ve ∪ Vo and Ve ∩ Vo = ∅. From Theorem 1.9, we have
X X X
deg(v) = deg(v) + deg(v) = 2|E|
v∈V v∈Ve v∈Vo
which can be re-arranged as
X X X
deg(v) = deg(v) − deg(v).
v∈Vo v∈V v∈Ve
P P
As v∈V deg(v) and v∈Ve deg(v) are both even, their difference is also even.
As E ⊆ V (2) , then E can be the empty set, in which case the total degree of G =
(V, E) is zero. Where E 6= ∅, then the total degree of G is greater than zero. By
Theorem 1.9, the total degree of G is nonnegative and even. This result is an immediate
consequence of Theorem 1.9 and is captured in the following corollary.
Corollary 1.11. If G is a graph, then the sum of its vertex degrees is nonnegative and
even.
If G = (V, E) is an r-regular graph with n vertices and m edges, it is clear by definition
of r-regular graphs that the total degree of G is rn. By Theorem 1.9 we have 2m = rn
and therefore m = rn/2. This result is captured in the following corollary.
Corollary 1.12. If G = (V, E) is an r-regular graph having n vertices and m edges,
then m = rn/2.
1.2 Subgraphs and other graph types
We now consider several common types of graphs. Along the way, we also present basic
properties of graphs that could be used to distinguish different types of graphs.
Let G be a multigraph as in Definition 1.6, with vertex set V (G) and edge set E(G).
Consider a graph H such that V (H) ⊆ V (G) and E(H) ⊆ E(G). Furthermore, if
e ∈ E(H) and i(e) = {u, v}, then u, v ∈ V (H). Under these conditions, H is called a
subgraph of G.
1.2.1 Walks, trails, and paths
I like long walks, especially when they are taken by people who annoy me.
— Noel Coward
If u and v are two vertices in a graph G, a u-v walk is an alternating sequence of vertices
and edges starting with u and ending at v. Consecutive vertices and edges are incident.
Formally, a walk W of length n ≥ 0 can be defined as
W : v0 , e1 , v1 , e2 , v2 , . . . , vn−1 , en , vn
where each edge ei = vi−1 vi and the length n refers to the number of (not necessarily
distinct) edges in the walk. The vertex v0 is the starting vertex of the walk and vn is
the end vertex, so we refer to W as a v0 -vn walk. The trivial walk is the walk of length
n = 0 in which the start and end vertices are one and the same vertex. If the graph has
12 Chapter 1. Introduction to graph theory
no multiple edges then, for brevity, we omit the edges in a walk and usually write the
walk as the following sequence of vertices:
W : v0 , v1 , v2 , . . . , vn−1 , vn .
For the graph in Figure 1.9, an example of a walk is an a-e walk: a, b, c, b, e. In other
words, we start at vertex a and travel to vertex b. From b, we go to c and then back to
b again. Then we end our journey at e. Notice that consecutive vertices in a walk are
adjacent to each other. One can think of vertices as destinations and edges as footpaths,
say. We are allowed to have repeated vertices and edges in a walk. The number of edges
in a walk is called its length. For instance, the walk a, b, c, b, e has length 4.
a b
c
g
d
f e
Figure 1.9: Walking along a graph.
A trail is a walk with no repeating edges. For example, the a-b walk a, b, c, d, f, g, b in
Figure 1.9 is a trail. It does not contain any repeated edges, but it contains one repeated
vertex, i.e. b. Nothing in the definition of a trail restricts a trail from having repeated
vertices. A walk with no repeating vertices, except possibly the first and last, is called a
path. Without any repeating vertices, a path cannot have repeating edges, hence a path
is also a trail.
Proposition 1.13. Let G = (V, E) be a simple (di)graph of order n = |V |. Any path in
G has length at most n − 1.
Proof. Let V = {v1 , v2 , . . . , vn } be the vertex set of G. Without loss of generality, we can
assume that each pair of vertices in the digraph G is connected by an edge, giving a total
of n2 possible edges for E = V × V . We can remove self-loops from E, which now leaves
us with an edge set E1 that consists of n2 − n edges. Start our path from any vertex,
say, v1 . To construct a path of length 1, choose an edge v1 vj1 ∈ E1 such that vj1 ∈ / {v1 }.
Remove from E1 all v1 vk such that vj1 6= vk . This results in a reduced edge set E2 of
n2 − n − (n − 2) elements and we now have the path P1 : v1 , vj1 of length 1. Repeat the
same process for vj1 vj2 ∈ E2 to obtain a reduced edge set E3 of n2 − n − 2(n − 2) elements
and a path P2 : v1 , vj1 , vj2 of length 2. In general, let Pr : v1 , vj1 , vj2 , . . . , vjr be a path of
length r < n and let Er+1 be our reduced edge set of n2 − n − r(n − 2) elements. Repeat
the above process until we have constructed a path Pn−1 : v1 , vj1 , vj2 , . . . , vjn−1 of length
n − 1 with reduced edge set En of n2 − n − (n − 1)(n − 2) elements. Adding another
vertex to Pn−1 means going back to a vertex that was previously visited, because Pn−1
already contains all vertices of V .
1.2. Subgraphs and other graph types 13
A walk of length n ≥ 3 whose start and end vertices are the same is called a closed
walk . A trail of length n ≥ 3 whose start and end vertices are the same is called a closed
trail . A path of length n ≥ 3 whose start and end vertices are the same is called a closed
path or a cycle (with apologies for slightly abusing terminology).3 For example, the
walk a, b, c, e, a in Figure 1.9 is a closed path. A path whose length is odd is called odd ,
otherwise it is referred to as even. Thus the walk a, b, e, a in Figure 1.9 is a cycle. It is
easy to see that if you remove any edge from a cycle, then the resulting walk contains no
closed walks. An Euler subgraph of a graph G is either a cycle or an edge-disjoint union
of cycles in G. An example of a closed walk which is not a cycle is given in Figure 1.10.
0 4
2
1 3
Figure 1.10: Butterfly graph with 5 vertices.
The length of the shortest cycle in a graph is called the girth of the graph. By
convention, an acyclic graph is said to have infinite girth.
Example 1.14. Consider the graph in Figure 1.9.
1. Find two distinct walks that are not trails and determine their lengths.
2. Find two distinct trails that are not paths and determine their lengths.
3. Find two distinct paths and determine their lengths.
4. Find a closed trail that is not a cycle.
5. Find a closed walk C which has an edge e such that C − e contains a cycle.
Solution. (1) Here are two distinct walks that are not trails: w1 : g, b, e, a, b, e and
w2 : f, d, c, e, f, d. The length of walk w1 is 5 and the length of walk w2 is also 5.
(2) Here are two distinct trails that are not paths: t1 : a, b, c, e, b and t2 : b, e, f, d, c, e.
The length of trail t1 is 4 and the length of trail t2 is 5.
(3) Here are two distinct paths: p1 : a, b, c, d, f, e and p2 : g, b, a, e, f, d. The length of
path p1 is 5 and the length of path p2 is also 5.
(4) Here is a closed trail that is not a cycle: d, c, e, b, a, e, f, d.
(5) Left to the reader.
Theorem 1.15. Every u-v walk in a graph contains a u-v path.
Proof. A walk of length n = 0 is the trivial path. So assume that W is a u-v walk of
length n > 0 in a graph G:
W : u = v0 , v1 , . . . , vn = v.
It is possible that a vertex in W is assigned two different labels. If W has no repeated
vertices, then W is already a path. Otherwise W has at least one repeated vertex. Let
3
A cycle in a graph is sometimes also called a “circuit”. Since that terminology unfortunately
conflicts with the closely related notion of a circuit of a matroid, we do not use it here.
14 Chapter 1. Introduction to graph theory
0 ≤ i, j ≤ n be two distinct integers with i < j such that vi = vj . Deleting the vertices
vi , vi+1 , . . . , vj−1 from W results in a u-v walk W1 whose length is less than n. If W1 is
a path, then we are done. Otherwise we repeat the above process to obtain a u-v walk
shorter than W1 . As W is a finite sequence, we only need to apply the above process a
finite number of times to arrive at a u-v path.
A graph is said to be connected if for every pair of distinct vertices u, v there is a
u-v path joining them. A graph that is not connected is referred to as disconnected .
The empty graph is disconnected and so is any nonempty graph with an isolated vertex.
However, the graph in Figure 1.7 is connected. A geodesic path or shortest path between
two distinct vertices u, v of a graph is a u-v path of minimum length. A nonempty graph
may have several shortest paths between some distinct pair of vertices. For the graph
in Figure 1.9, both a, b, c and a, e, c are geodesic paths between a and c. Let H be a
connected subgraph of a graph G such that H is not a proper subgraph of any connected
subgraph of G. Then H is said to be a component of G. We also say that H is a maximal
connected subgraph of G. Any connected graph is its own component. The number of
connected components of a graph G will be denoted ω(G).
The following is an immediate consequence of Corollary 1.10.
Proposition 1.16. Suppose that exactly two vertices of a graph have odd degree. Then
those two vertices are connected by a path.
Proof. Let G be a graph all of whose vertices are of even degree, except for u and v. Let
C be a component of G containing u. By Corollary 1.10, C also contains v, the only
remaining vertex of odd degree. As u and v belong to the same component, they are
connected by a path.
Example 1.17. Determine whether or not the graph in Figure 1.9 is connected. Find a
shortest path from g to d.
Solution. In the following Sage session, we first construct the graph in Figure 1.9 and
use the method is_connected() to determine whether or not the graph is connected.
Finally, we use the method shortest_path() to find a geodesic path between g and d.
sage : g = Graph ({ " a " :[ " b " ," e " ] , " b " :[ " a " ," g " ," e " ," c " ] , \
... " c " :[ " b " ," e " ," d " ] , " d " :[ " c " ," f " ] , " e " :[ " f " ," a " ," b " ," c " ] , \
... " f " :[ " g " ," d " ," e " ] , " g " :[ " b " ," f " ]})
sage : g . is_connected ()
True
sage : g . shortest_path ( " g " , " d " )
[ ’g ’ , ’f ’ , ’d ’]
This shows that g, f, d is a shortest path from g to d. In fact, any other g-d path has
length greater than 2, so we can say that g, f, d is the shortest path between g and d.
Remark 1.18. We will explain Dijkstra’s algorithm in Chapter 2, which gives one of
the best algorithms for finding shortest paths between two vertices in a connected graph.
What is very remarkable is that, at the present state of knowledge, finding the shortest
path from a vertex v to a particular (but arbitrarily given) vertex w appears to be as
hard as finding the shortest path from a vertex v to all other vertices in the graph!
Trees are a special type of graphs that are used in modelling structures that have
some form of hierarchy. For example, the hierarchy within an organization can be drawn
as a tree structure, similar to the family tree in Figure 1.11. Formally, a tree is an
undirected graph that is connected and has no cycles. If one vertex of a tree is specially
designated as the root vertex , then the tree is called a rooted tree. Chapter 3 covers trees
in more details.
1.2. Subgraphs and other graph types 15
grandma
mum uncle aunt
me sister brother cousin1 cousin2
Figure 1.11: A family tree.
1.2.2 Subgraphs, complete and bipartite graphs
Let G be a graph with vertex set V (G) and edge set E(G). Suppose we have a graph
H such that V (H) ⊆ V (G) and E(H) ⊆ E(G). Furthermore, suppose the incidence
function i of G, when restricted to E(H), has image in V (H)(2) . Then H is a subgraph
of G. In this situation, G is referred to as a supergraph of H.
Starting from G, one can obtain its subgraph H by deleting edges and/or vertices
from G. Note that when a vertex v is removed from G, then all edges incident with
v are also removed. If V (H) = V (G), then H is called a spanning subgraph of G. In
Figure 1.12, let G be the left-hand side graph and let H be the right-hand side graph.
Then it is clear that H is a spanning subgraph of G. To obtain a spanning subgraph
from a given graph, we delete edges from the given graph.
(a) (b)
Figure 1.12: A graph and one of its subgraphs.
We now consider several standard classes of graphs. The complete graph Kn on n
vertices is a graph such that any two distinct vertices are adjacent. As |V (Kn )| = n,
then |E(Kn )| is equivalent to the total number of 2-combinations from a set of n objects:
n n(n − 1)
|E(Kn )| = = . (1.6)
2 2
Thus for any simple graph G with n vertices, its total number of edges |E(G)| is bounded
above by
n(n − 1)
|E(G)| ≤ . (1.7)
2
Figure 1.13 shows complete graphs each of whose total number of vertices is bounded by
1 ≤ n ≤ 5. The complete graph K1 has one vertex with no edges. It is also called the
trivial graph.
The following result is an application of inequality (1.7).
16 Chapter 1. Introduction to graph theory
(a) K5 (b) K4 (c) K3 (d) K2 (e) K1
Figure 1.13: Complete graphs Kn for 1 ≤ n ≤ 5.
Theorem 1.19. Let G be a simple graph with n vertices and k components. Then G
has at most 21 (n − k)(n − k + 1) edges.
Proof. If ni is the number of vertices in component i, then ni > 0 and it can be shown (see
the proof of Lemma 2.1 in [84, pp.21–22]) that
X X 2 X
n2i ≤ ni − (k − 1) 2 ni − k . (1.8)
(This
P result holds true for any nonempty but finite set of positive integers.) Note that
ni = n and by (1.7) each component i has at most 21 ni (ni − 1) edges. Apply (1.8) to
get
X ni (ni − 1) 1X 2 1X
= ni − ni
2 2 2
1 1
≤ (n2 − 2nk + k 2 + 2n − k) − n
2 2
(n − k)(n − k + 1)
=
2
as required.
The cycle graph on n ≥ 3 vertices, denoted Cn , is the connected 2-regular graph on n
vertices. Each vertex in Cn has degree exactly 2 and Cn is connected. Figure 1.14 shows
cycles graphs Cn where 3 ≤ n ≤ 6. The path graph on n ≥ 1 vertices is denoted Pn . For
n = 1, 2 we have P1 = K1 and P2 = K2 . Where n ≥ 3, then Pn is a spanning subgraph
of Cn obtained by deleting one edge.
(a) C6 (b) C5 (c) C4 (d) C3
Figure 1.14: Cycle graphs Cn for 3 ≤ n ≤ 6.
A bipartite graph G is a graph with at least two vertices such that V (G) can be split
into two disjoint subsets V1 and V2 , both nonempty. Every edge uv ∈ E(G) is such that
1.2. Subgraphs and other graph types 17
u ∈ V1 and v ∈ V2 , or v ∈ V1 and u ∈ V2 . See Kalman [122] for an application of bipartite
graphs to the problem of allocating satellites to radio stations.
Theorem 1.20. A graph is bipartite if and only if it has no odd cycles.
Proof. Necessity (=⇒): Assume G to be bipartite. Traversing each edge involves going
from one side of the bipartition to the other. For a walk to be closed, it must have
even length in order to return to the side of the bipartition from which the walk started.
Thus, any cycle in G must have even length.
Sufficiency (⇐=): Assume G = (V, E) has order n ≥ 2 and no odd cycles. If G is
connected, choose any vertex u ∈ V and define a partition of V thus:
X = {x ∈ V | d(u, x) is even},
Y = {y ∈ V | d(u, y) is odd}
where d(u, v) denotes the distance (or length of the shortest path) from u to v. If (X, Y )
is a bipartition of G, then we are done. Otherwise, (X, Y ) is not a bipartition of G.
Then one of X and Y has two vertices v, w joined by an edge e. Let P1 be a shortest
u-v path and P2 be a shortest u-w path. By definition of X and Y , both P1 and P2 have
even lengths or both have odd lengths. From u, let x be the last vertex common to both
P1 and P2 . The subpath u-x of P1 and u-x of P2 have equal length. That is, the subpath
x-v of P1 and x-w of P2 both have even or odd lengths. Construct a cycle C from the
paths x-v and x-w, and the edge e joining v and w. Since x-v and x-w both have even
or odd lengths, the cycle C has odd length, contradicting our hypothesis that G has no
odd cycles. Hence, (X, Y ) is a bipartition of G.
Finally, if G is disconnected, each of its components has no odd cycles. Repeat the
above argument for each component to conclude that G is bipartite.
The complete bipartite graph Km,n is the bipartite graph whose vertex set is parti-
tioned into two nonempty disjoint sets V1 and V2 with |V1 | = m and |V2 | = n. Any
vertex in V1 is adjacent to each vertex in V2 , and any two distinct vertices in Vi are not
adjacent to each other. If m = n, then Kn,n is n-regular. Where m = 1 then K1,n is
called the star graph. Figure 1.15 shows a bipartite graph together with the complete
bipartite graphs K4,3 and K3,3 , and the star graph K1,4 .
(a) bipartite (b) K4,3 (c) K3,3 (d) K1,4
Figure 1.15: Bipartite, complete bipartite, and star graphs.
As an example of K3,3 , suppose that there are 3 boys and 3 girls dancing in a room.
The boys and girls naturally partition the set of all people in the room. Construct a
18 Chapter 1. Introduction to graph theory
graph having 6 vertices, each vertex corresponding to a person in the room, and draw
an edge form one vertex to another if the two people dance together. If each girl dances
three times, once with with each of the three boys, then the resulting graph is K3,3 .
1.3 Representing graphs as matrices
Neo: What is the Matrix?
Morpheus: Unfortunately, no one can be told what the Matrix is. You have to see it for
yourself.
— From the movie The Matrix, 1999
An m × n matrix A can be represented as
a11 a12 · · · a1n
a21 a22 · · · a2n
A=
. . . . . . . . . . . . . . . . . . . .
am1 am2 · · · amn
The positive integers m and n are the row and column dimensions of A, respectively.
The entry in row i column j is denoted aij . Where the dimensions of A are clear from
context, A is also written as A = [aij ].
Representing a graph as a matrix is very inefficient in some cases and not so in
other cases. Imagine you walk into a large room full of people and you consider the
“handshaking graph” discussed in connection with Theorem 1.9. If not many people
shake hands in the room, it is a waste of time recording all the handshakes and also all
the “non-handshakes.” This is basically what the adjacency matrix does. In this kind
of “sparse graph” situation, it would be much easier to simply record the handshakes as
a Python dictionary.4 This section requires some concepts and techniques from linear
algebra, especially matrix theory. See introductory texts on linear algebra and matrix
theory [20] for coverage of such concepts and techniques.
1.3.1 Adjacency matrix
Let G be an undirected graph with vertices V = {v1 , . . . , vn } and edge set E. The
adjacency matrix of G is the n × n matrix A = [aij ] defined by
(
1, if vi vj ∈ E,
aij =
0, otherwise.
The adjacency matrix of G is also written as A(G). As G is an undirected graph, then
A is a symmetric matrix. That is, A is a square matrix such that aij = aji .
Now let G be a directed graph with vertices V = {v1 , . . . , vn } and edge set E. The
(0, −1, 1)-adjacency matrix of G is the n × n matrix A = [aij ] defined by
1, if vi vj ∈ E,
aij = −1, if vj vi ∈ E,
0, otherwise.
4
A Python dictionary is basically an indexed set. See the reference manual at http://www.python.org
for further details.
1.3. Representing graphs as matrices 19
3 f
6 2 d e
5 1 b c
4 a
(a) (b)
Figure 1.16: What are the adjacency matrices of these graphs?
Example 1.21. Compute the adjacency matrices of the graphs in Figure 1.16.
Solution. Define the graphs in Figure 1.16 using DiGraph and Graph. Then call the
method adjacency_matrix().
sage : G1 = DiGraph ({1:[2] , 2:[1] , 3:[2 ,6] , 4:[1 ,5] , 5:[6] , 6:[5]})
sage : G2 = Graph ({ " a " :[ " b " ," c " ] , " b " :[ " a " ," d " ] , " c " :[ " a " ," e " ] , \
... " d " :[ " b " ," f " ] , " e " :[ " c " ," f " ] , " f " :[ " d " ," e " ]})
sage : m1 = G1 . adjacency_matrix (); m1
[0 1 0 0 0 0]
[1 0 0 0 0 0]
[0 1 0 0 0 1]
[1 0 0 0 1 0]
[0 0 0 0 0 1]
[0 0 0 0 1 0]
sage : m2 = G2 . adjacency_matrix (); m2
[0 1 1 0 0 0]
[1 0 0 1 0 0]
[1 0 0 0 1 0]
[0 1 0 0 0 1]
[0 0 1 0 0 1]
[0 0 0 1 1 0]
sage : m1 . is_symmetric ()
False
sage : m2 . is_symmetric ()
True
In general, the adjacency matrix of a digraph is not symmetric, while that of an undi-
rected graph is symmetric.
More generally, if G is an undirected multigraph with edge eij = vi vj having mul-
tiplicity wij , or a weighted graph with edge eij = vi vj having weight wij , then we can
define the (weighted) adjacency matrix A = [aij ] by
(
wij , if vi vj ∈ E,
aij =
0, otherwise.
For example, Sage allows you to easily compute a weighted adjacency matrix.
sage : G = Graph ( sparse = True , weighted = True )
sage : G . add_edges ([(0 ,1 ,1) , (1 ,2 ,2) , (0 ,2 ,3) , (0 ,3 ,4)])
sage : M = G . w e i g h t e d _ a d ja c e n c y _ m a t r i x (); M
[0 1 3 4]
[1 0 2 0]
[3 2 0 0]
[4 0 0 0]
20 Chapter 1. Introduction to graph theory
Bipartite case
Suppose G = (V, E) is an undirected bipartite graph with n = |V | vertices. Any ad-
jacency matrix A of G is symmetric and we assume that it is indexed from zero up to
n − 1, inclusive. Then there exists a permutation π of the index set {0, 1, . . . , n − 1} such
that the matrix A0 = [aπ(i)π(j) ] is also an adjacency matrix for G and has the form
0 0 B
A = (1.9)
BT 0
where 0 is a zero matrix. The matrix B is called a reduced adjacency matrix or a bi-
adjacency matrix (the literature also uses the terms “transfer matrix” or the ambiguous
term “adjacency matrix”). In fact, it is known [10, p.16] that any undirected graph is
bipartite if and only if there is a permutation π on {0, 1, . . . , n − 1} such that A0 (G) =
[aπ(i)π(j) ] can be written as in (1.9).
Tanner graphs
If H is an m × n (0, 1)-matrix, then the Tanner graph of H is the bipartite graph
G = (V, E) whose set of vertices V = V1 ∪V2 is partitioned into two sets: V1 corresponding
to the m rows of H and V2 corresponding to the n columns of H. For any i, j with
1 ≤ i ≤ m and 1 ≤ j ≤ n, there is an edge ij ∈ E if and only if the (i, j)-th entry of
H is 1. This matrix H is sometimes called the reduced adjacency matrix or the check
matrix of the Tanner graph. Tanner graphs are used in the theory of error-correcting
codes. For example, Sage allows you to easily compute such a bipartite graph from its
matrix.
sage : H = Matrix ([(1 ,1 ,1 ,0 ,0) , (0 ,0 ,1 ,0 ,1) , (1 ,0 ,0 ,1 ,1)])
sage : B = BipartiteGraph ( H )
sage : B . r ed u c e d_ a d j ac e n c y_ m a t ri x ()
[1 1 1 0 0]
[0 0 1 0 1]
[1 0 0 1 1]
sage : B . plot ( graph_border = True )
The corresponding graph is similar to that in Figure 1.17.
1 1
2
3 2
4
5 3
Figure 1.17: A Tanner graph.
Theorem 1.22. Let A be the adjacency matrix of a graph G with vertex set V =
{v1 , v2 , . . . , vp }. For each positive integer n, the ij-th entry of An counts the number
of vi -vj walks of length n in G.
1.3. Representing graphs as matrices 21
Proof. We shall prove by induction on n. For the base case n = 1, the ij-th entry of
A1 counts the number of walks of length 1 from vi to vj . This is obvious because A1 is
merely the adjacency matrix A.
Suppose for induction that for some positive integer k ≥ 1, the ij-th entry of Ak
counts the number of walks of length k from vi to vj . We need to show that the ij-th
entry of Ak+1 counts the number of vi -vj walks of length k + 1. Let A = [aij ], Ak = [bij ],
and Ak+1 = [cij ]. Since Ak+1 = AAk , then
p
X
cij = air brj
r=1
for i, j = 1, 2, . . . , p. Note that air is the number of edges from vi to vr , and brj is the
number of vr -vj walks of length k. Any edge from vi to vr can be joined with any vr -vj
walk to create a walk vi , vr , . . . , vj of length k + 1. Then for each r = 1, 2, . . . , p, the
value air brj counts the number of vi -vj walks of length k + 1 with vr being the second
vertex in the walk. Thus cij counts the total number of vi -vj walks of length k + 1.
1.3.2 Incidence matrix
The relationship between edges and vertices provides a very strong constraint on the
data structure, much like the relationship between points and blocks in a combinatorial
design or points and lines in a finite plane geometry. This incidence structure gives rise
to another way to describe a graph using a matrix.
Let G be a digraph with edge set E = {e1 , . . . , em } and vertex set V = {v1 , . . . , vn }.
The incidence matrix of G is the n × m matrix B = [bij ] defined by
−1, if vi is the tail of ej ,
1, if vi is the head of ej ,
bij = (1.10)
2, if ej is a self-loop at vi ,
0, otherwise.
Each column of B corresponds to an edge and each row corresponds to a vertex. The
definition of incidence matrix of a digraph as contained in expression (1.10) is applicable
to digraphs with self-loops as well as multidigraphs.
For the undirected case, let G be an undirected graph with edge set E = {e1 , . . . , em }
and vertex set V = {v1 , . . . , vn }. The unoriented incidence matrix of G is the n × m
matrix B = [bij ] defined by
1, if vi is incident to ej ,
bij = 2, if ej is a self-loop at vi ,
0, otherwise.
An orientation of an undirected graph G is an assignment of direction to each edge of G.
The oriented incidence matrix of G is defined similarly to the case where G is a digraph:
it is the incidence matrix of any orientation of G. For each column of B, we have 1 as
an entry in the row corresponding to one vertex of the edge under consideration and −1
as an entry in the row corresponding to the other vertex. Similarly, bij = 2 if ej is a
self-loop at vi .
22 Chapter 1. Introduction to graph theory
1.3.3 Laplacian matrix
The degree matrix of a graph G = (V, E) is an n × n diagonal matrix D whose i-th
diagonal entry is the degree of the i-th vertex in V . The Laplacian matrix L of G is the
difference between the degree matrix and the adjacency matrix:
L = D − A.
In other words, for an undirected unweighted simple graph, L = [`ij ] is given by
−1, if i 6= j and vi vj ∈ E,
`ij = di , if i = j,
0, otherwise,
where di = deg(vi ) is the degree of vertex vi .
Sage allows you to compute the Laplacian matrix of a graph:
sage : G = Graph ({1:[2 ,4] , 2:[1 ,4] , 3:[2 ,6] , 4:[1 ,3] , 5:[4 ,2] , 6:[3 ,1]})
sage : G . laplacian_matrix ()
[ 3 -1 0 -1 0 -1]
[ -1 4 -1 -1 -1 0]
[ 0 -1 3 -1 0 -1]
[ -1 -1 -1 4 -1 0]
[ 0 -1 0 -1 2 0]
[ -1 0 -1 0 0 2]
There are many remarkable properties of the Laplacian matrix. It shall be discussed
further in Chapter 5.
1.3.4 Distance matrix
Recall that the distance (or geodesic distance) d(v, w) between two vertices v, w ∈ V in a
connected graph G = (V, E) is the number of edges in a shortest path connecting them.
The n × n matrix [d(vi , vj )] is the distance matrix of G. Sage helps you to compute the
distance matrix of a graph:
sage : G = Graph ({1:[2 ,4] , 2:[1 ,4] , 3:[2 ,6] , 4:[1 ,3] , 5:[4 ,2] , 6:[3 ,1]})
sage : d = [[ G . distance (i , j ) for i in range (1 ,7)] for j in range (1 ,7)]
sage : matrix ( d )
[0 1 2 1 2 1]
[1 0 1 1 1 2]
[2 1 0 1 2 1]
[1 1 1 0 1 2]
[2 1 2 1 0 3]
[1 2 1 2 3 0]
The distance matrix is an important quantity which allows one to better understand
the “connectivity” of a graph. Distance and connectivity will be discussed in more detail
in Chapters 5 and 10.
1.4 Isomorphic graphs
Determining whether or not two graphs are, in some sense, the “same” is a hard but
important problem. Two graphs G and H are isomorphic if there is a bijection f :
V (G) → V (H) such that whenever uv ∈ E(G) then f (u)f (v) ∈ E(H). The function f
is an isomorphism between G and H. Otherwise, G and H are non-isomorphic. If G
and H are isomorphic, we write G ∼
= H.
1.4. Isomorphic graphs 23
(a) (b)
Figure 1.18: Two representations of the Franklin graph.
e f 1 2 e f
c d 3 4 c d
a b 5 6 a b
(a) C6 (b) G1 (c) G2
Figure 1.19: Isomorphic and nonisomorphic graphs.
A graph G is isomorphic to a graph H if these two graphs can be labelled in such a
way that if u and v are adjacent in G, then their counterparts in V (H) are also adjacent
in H. To determine whether or not two graphs are isomorphic is to determine if they are
structurally equivalent. Graphs G and H may be drawn differently so that they seem
different. However, if G ∼ = H then the isomorphism f : V (G) → V (H) shows that both
of these graphs are fundamentally the same. In particular, the order and size of G are
equal to those of H, the isomorphism f preserves adjacencies, and deg(v) = deg(f (v)) for
all v ∈ G. Since f preserves adjacencies, then adjacencies along a given geodesic path are
preserved as well. That is, if v1 , v2 , v3 , . . . , vk is a shortest path between v1 , vk ∈ V (G),
then f (v1 ), f (v2 ), f (v3 ), . . . , f (vk ) is a geodesic path between f (v1 ), f (vk ) ∈ V (H). For
example, the two graphs in Figure 1.18 are isomorphic to each other.
Example 1.23. Consider the graphs in Figure 1.19. Which pair of graphs are isomor-
phic, and which two graphs are non-isomorphic?
Solution. If G is a Sage graph, one can use the method G.is_isomorphic() to determine
whether or not the graph G is isomorphic to another graph. The following Sage session
illustrates how to use G.is_isomorphic().
sage : C6 = Graph ({ " a " :[ " b " ," c " ] , " b " :[ " a " ," d " ] , " c " :[ " a " ," e " ] , \
... " d " :[ " b " ," f " ] , " e " :[ " c " ," f " ] , " f " :[ " d " ," e " ]})
sage : G1 = Graph ({1:[2 ,4] , 2:[1 ,3] , 3:[2 ,6] , 4:[1 ,5] , \
... 5:[4 ,6] , 6:[3 ,5]})
sage : G2 = Graph ({ " a " :[ " d " ," e " ] , " b " :[ " c " ," f " ] , " c " :[ " b " ," f " ] , \
... " d " :[ " a " ," e " ] , " e " :[ " a " ," d " ] , " f " :[ " b " ," c " ]})
sage : C6 . is_isomorphic ( G1 )
True
sage : C6 . is_isomorphic ( G2 )
False
sage : G1 . is_isomorphic ( G2 )
24 Chapter 1. Introduction to graph theory
False
Thus, for the graphs C6 , G1 and G2 in Figure 1.19, C6 and G1 are isomorphic, but G1
and G2 are not isomorphic.
An important notion in graph theory is the idea of an “invariant”. An invariant is
an object f = f (G) associated to a graph G which has the property
G∼
= H =⇒ f (G) = f (H).
For example, the number of vertices of a graph, f (G) = |V (G)|, is an invariant.
1.4.1 Adjacency matrices
Two n × n matrices A1 and A2 are permutation equivalent if there is a permutation
matrix P such that A1 = P A2 P −1 . In other words, A1 is the same as A2 after a suitable
re-ordering of the rows and a corresponding re-ordering of the columns. This notion of
permutation equivalence is an equivalence relation.
To show that two undirected graphs are isomorphic depends on the following result.
Theorem 1.24. Consider two directed or undirected graphs G1 and G2 with respective
adjacency matrices A1 and A2 . Then G1 and G2 are isomorphic if and only if A1 is
permutation equivalent to A2 .
This says that the permutation equivalence class of the adjacency matrix is an in-
variant.
Define an ordering on the set of n×n (0, 1)-matrices as follows: we say A1 < A2 if the
list of entries of A1 is less than or equal to the list of entries of A2 in the lexicographical
ordering. Here, the list of entries of a (0, 1)-matrix is obtained by concatenating the
entries of the matrix, row-by-row. For example,
1 1 1 1
< .
0 1 1 1
Algorithm 1.1 is an immediate consequence of Theorem 1.24. The lexicographically
maximal element of the permutation equivalence class of the adjacency matrix of G is
called the canonical label of G. Thus, to check if two undirected graphs are isomorphic,
we simply check if their canonical labels are equal. This idea for graph isomorphism
checking is presented in Algorithm 1.1.
1.4.2 Degree sequence
Let G be a graph with n vertices. The degree sequence of G is the ordered n-tuple of the
vertex degrees of G arranged in non-increasing order.
The degree sequence of G may contain the same degrees, repeated as often as they
occur. For example, the degree sequence of C6 is 2, 2, 2, 2, 2, 2 and the degree sequence
of the house graph in Figure 1.3 is 3, 3, 2, 2, 2. If n ≥ 3 then the cycle graph Cn has the
degree sequence
2, 2, 2, . . . , 2 .
| {z }
n copies of 2
1.4. Isomorphic graphs 25
Algorithm 1.1: Computing graph isomorphism using canonical labels.
Input: Two undirected simple graphs G1 and G2 , each having n vertices.
Output: True if G1 ∼
= G2 ; False otherwise.
1 for i ← 1, 2 do
2 Ai ← adjacency matrix of Gi
3 pi ← permutation equivalence class of Ai
4 A0i ← lexicographically maximal element of pi
5 if A01 = A02 then
6 return True
7 return False
The path Pn , for n ≥ 3, has the degree sequence
2, 2, 2, . . . , 2, 1, 1 .
| {z }
n−2 copies of 2
For positive integer values of n and m, the complete graph Kn has the degree sequence
n − 1, n − 1, n − 1, . . . , n − 1
| {z }
n copies of n−1
and the complete bipartite graph Km,n has the degree sequence
n, n, n, . . . , n, m, m, m, . . . , m .
| {z }| {z }
m copies of n n copies of m
Let S be a non-increasing sequence of non-negative integers. Then S is said to be
graphical if it is the degree sequence of some graph. If G is a graph with degree sequence
S, we say that G realizes S.
Let S = (d1 , d2 , . . . , dn ) be a graphical sequence,
P i.e. di ≥ dj for all i ≤ j such that
1 ≤ i, j ≤ n. From Corollary 1.11 we see that di ∈S di = 2k for some integer k ≥ 0. In
other words, the sum of a graphical sequence is nonnegative and even. In 1961, Erdős
and Gallai [75] used this observation as part of a theorem that provides necessary and
sufficient conditions for a sequence to be realized by a simple graph. The result is stated
in Theorem 1.25, but the original paper of Erdős and Gallai [75] does not provide an
algorithm to construct a simple graph with a given degree sequence. For a simple graph
that has a degree sequence with repeated elements, e.g. the degree sequences of Cn ,
Pn , Kn , and Km,n , it is redundant to verify inequality (1.11) for repeated elements of
that sequence. In 2003, Tripathi and Vijay [195] showed that one only needs to verify
inequality (1.11) for as many times as there are distinct terms in S.
Theorem 1.25. Erdős & Gallai 1961 [75]. Let d = (d1 , d2 , . . . , dn ) be a sequence
positive integers such that di ≥ di+1 . Then d is realized by a simple graph if and only
of P
if i di is even and
Xk Xn
di ≤ k(k + 1) + min{k, di } (1.11)
i=1 j=k+1
for all 1 ≤ k ≤ n − 1.
26 Chapter 1. Introduction to graph theory
As noted above, Theorem 1.25 is an existence result showing that something exists
without providing a construction of the object under consideration. Havel [102] and
Hakimi [99, 100] independently provided an algorithmic approach that allows for con-
structing a simple graph with a given degree sequence. See Sierksma and Hoogeveen [183]
for a coverage of seven criteria for a sequence of integers to be graphic. See Erdős et al. [78]
for an extension of the Havel-Hakimi theorem to digraphs.
Theorem 1.26. Havel 1955 [102] & Hakimi 1962–3 [99, 100]. Consider the
non-increasing sequence S1 = (d1 , d2 , . . . , dn ) of nonnegative integers, where n ≥ 2 and
d1 ≥ 1. Then S1 is graphical if and only if the sequence
S2 = (d2 − 1, d3 − 1, . . . , dd1 +1 − 1, dd1 +2 , . . . , dn )
is graphical.
Proof. Suppose S2 is graphical. Let G2 = (V2 , E2 ) be a graph of order n − 1 with vertex
set V2 = {v2 , v3 , . . . , vn } such that
(
di − 1, if 2 ≤ i ≤ d1 + 1,
deg(vi ) =
di , if d1 + 2 ≤ i ≤ n.
Construct a new graph G1 with degree sequence S1 as follows. Add another vertex v1
to V2 and add to E2 the edges v1 vi for 2 ≤ i ≤ d1 + 1. It is clear that deg(v1 ) = d1 and
deg(vi ) = di for 2 ≤ i ≤ n. Thus G1 has the degree sequence S1 .
On the other hand, suppose S1 is graphical and let G1 be a graph with degree sequence
S1 such that
(i) The graph G1 has the vertex set V (G1 ) = {v1 , v2 , . . . , vn } and deg(vi ) = di for
i = 1, . . . , n.
(ii) The degree sum of all vertices adjacent to v1 is a maximum.
To obtain a contradiction, suppose v1 is not adjacent to vertices having degrees
d2 , d3 , . . . , dd1 +1 .
Then there exist vertices vi and vj with dj > di such that v1 vi ∈ E(G1 ) but v1 vj 6∈ E(G1 ).
As dj > di , there is a vertex vk such that vj vk ∈ E(G1 ) but vi vk 6∈ E(G1 ). Replacing the
edges v1 vi and vj vk with v1 vj and vi vk , respectively, results in a new graph H whose degree
sequence is S1 . However, the graph H is such that the degree sum of vertices adjacent to
v1 is greater than the corresponding degree sum in G1 , contradicting property (ii) in our
choice of G1 . Consequently, v1 is adjacent to d1 other vertices of largest degree. Then
S2 is graphical because G1 − v1 has degree sequence S2 .
The proof of Theorem 1.26 can be adapted into an algorithm to determine whether
or not a sequence of nonnegative integers can be realized by a simple graph. If G is
a simple graph, the degree of any vertex in V (G) cannot exceed the order of G. By
the handshaking lemma (Theorem 1.9), the sum of all terms in the sequence cannot be
odd. Once the sequence passes these two preliminary tests, we then adapt the proof of
Theorem 1.26 to successively reduce the original sequence to a smaller sequence. These
ideas are summarized in Algorithm 1.2.
1.4. Isomorphic graphs 27
Algorithm 1.2: Havel-Hakimi test for sequences realizable by simple graphs.
Input: A nonincreasing sequence S = (d1 , d2 , . . . , dn ) of nonnegative integers,
where n ≥ 2.
Output: True if S is realizable by a simple graph; False otherwise.
P
1 if i di is odd then
2 return False
3 while True do
4 if min(S) < 0 then
5 return False
6 if max(S) = 0 then
7 return True
8 if max(S) > length(S) − 1 then
9 return False
10 S ← (d2 − 1, d3 − 1, . . . , dd1 +1 − 1, dd1 +2 , . . . , dlength(S) )
11 sort S in nonincreasing order
We now show that Algorithm 1.2 determines whether or not a sequence of integers
is realizable by a simple graph. Our input is a sequence S = (d1 , d2 , . . . , dn ) arranged
in non-increasing order, where each di ≥ 0. The first test as contained in the if block,
otherwise known as a conditional, on line 1 uses the handshaking lemma (Theorem 1.9).
During the first run of the while loop, the conditional on line 4 ensures that the sequence
S only consists of nonnegative integers. At the conditional on line 6, we know that S
is arranged in non-increasing order and has nonnegative integers. If this conditional
holds true, then S is a sequence of zeros and it is realizable by a graph with only isolated
vertices. Such a graph is simple by definition. The conditional on line 8 uses the following
property of simple graphs: If G is a simple graph, then the degree of each vertex of G
is less than the order of G. By the time we reach line 10, we know that S has n terms,
max(S) > 0, and 0 ≤ di ≤ n − 1 for all i = 1, 2, . . . , n. After applying line 10, S is now a
sequence of n − 1 terms with max(S) > 0 and 0 ≤ di ≤ n − 2 for all i = 1, 2, . . . , n − 1. In
general, after k rounds of the while loop, S is a sequence of n − k terms with max(S) > 0
and 0 ≤ di ≤ n − k − 1 for all i = 1, 2, . . . , n − k. And after n − 1 rounds of the while
loop, the resulting sequence has one term whose value is zero. In other words, eventually
Algorithm 1.2 produces a sequence with a negative term or a sequence of zeros.
1.4.3 Invariants revisited
In some cases, one can distinguish non-isomorphic graphs by considering graph invariants.
For instance, the graphs C6 and G1 in Figure 1.19 are isomorphic so they have the same
number of vertices and edges. Also, G1 and G2 in Figure 1.19 are non-isomorphic because
the former is connected, while the latter is not connected. To prove that two graphs
are non-isomorphic, one could show that they have different values for a given graph
invariant. The following list contains some items to check off when showing that two
graphs are non-isomorphic:
1. the number of vertices,
2. the number of edges,
28 Chapter 1. Introduction to graph theory
3. the degree sequence,
4. the length of a geodesic path,
5. the length of the longest path,
6. the number of connected components of a graph.
1.5 New graphs from old
This section provides a brief survey of operations on graphs to obtain new graphs from
old graphs. Such graph operations include unions, products, edge addition, edge deletion,
vertex addition, and vertex deletion. Several of these are briefly described below.
1.5.1 Union, intersection, and join
The disjoint union of graphs is defined as follows. For two graphs G1 = (V1 , E1 ) and
G2 = (V2 , E2 ) with disjoint vertex sets, their disjoint union is the graph
G1 ∪ G2 = (V1 ∪ V2 , E1 ∪ E2 ).
For example, Figure 1.20 shows the vertex disjoint union of the complete bipartite graph
K1,5 with the wheel graph W4 . The adjacency matrix A of the disjoint union of two
graphs G1 and G2 is the diagonal block matrix obtained from the adjacency matrices A1
and A2 , respectively. Namely,
A1 0
A= .
0 A2
Sage can compute graph unions, as the following example shows.
sage : G1 = Graph ({1:[2 ,4] , 2:[1 ,3] , 3:[2 ,6] , 4:[1 ,5] , 5:[4 ,6] , 6:[3 ,5]})
sage : G2 = Graph ({7:[8 ,10] , 8:[7 ,10] , 9:[8 ,12] , 10:[7 ,9] , 11:[10 ,8] , 12:[9 ,7]})
sage : G1u2 = G1 . union ( G2 )
sage : G1u2 . adjacency_matrix ()
[0 1 0 1 0 0 0 0 0 0 0 0]
[1 0 1 0 0 0 0 0 0 0 0 0]
[0 1 0 0 0 1 0 0 0 0 0 0]
[1 0 0 0 1 0 0 0 0 0 0 0]
[0 0 0 1 0 1 0 0 0 0 0 0]
[0 0 1 0 1 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 1 0 1 0 1]
[0 0 0 0 0 0 1 0 1 1 1 0]
[0 0 0 0 0 0 0 1 0 1 0 1]
[0 0 0 0 0 0 1 1 1 0 1 0]
[0 0 0 0 0 0 0 1 0 1 0 0]
[0 0 0 0 0 0 1 0 1 0 0 0]
In the case where V1 = V2 , then G1 ∪ G2 is simply the graph consisting of all edges in G1
or in G2 . In general, the union of two graphs G1 = (V1 , E1 ) and G2 = (V2 , E2 ) is defined
as
G1 ∪ G2 = (V1 ∪ V2 , E1 ∪ E2 )
where V1 ⊆ V2 , V2 ⊆ V1 , V1 = V2 , or V1 ∩ V2 = ∅. Figure 1.21(c) illustrates the graph
union where one vertex set is a proper subset of the other. If G1 , G2 , . . . , Gn are the
components
S of a graph G, then G is obtained by the disjoint union of its components,
i.e. G = Gi .
1.5. New graphs from old 29
Figure 1.20: The vertex disjoint union K1,5 ∪ W4 .
3 3
3 3
1 2 1 2
4 4
4 4
1 2 5 6 5 6 1 2
(a) G1 (b) G2 (c) G1 ∪ G2 (d) G1 ∩ G2
Figure 1.21: The union and intersection of graphs with overlapping vertex sets.
The intersection of graphs is defined as follows. For two graphs G1 = (V1 , E1 ) and
G2 = (V2 , E2 ), their intersection is the graph
G1 ∩ G2 = (V1 ∩ V2 , E1 ∩ E2 ).
Figure 1.21(d) illustrates the intersection of two graphs whose vertex sets overlap.
The symmetric difference of graphs is defined as follows. For two graphs G1 = (V1 , E1 )
and G2 = (V2 , E2 ), their symmetric difference is the graph
G1 ∆G2 = (V, E)
where V = V1 ∆V2 and the edge set is given by
E = (E1 ∆E2 )\{uv | u ∈ V1 ∩ V2 or v ∈ V1 ∩ V2 }.
Recall that the symmetric difference of two sets S1 and S2 is defined by
S1 ∆S2 = {x ∈ S1 ∪ S2 | x ∈
/ S1 ∩ S2 }.
In the case where V1 = V2 , then G1 ∆G2 is simply the empty graph. See Figure 1.22 for
an illustration of the symmetric difference of two graphs.
The join of two disjoint graphs G1 and G2 , denoted G1 +G2 , is their graph union, with
each vertex of one graph connecting to each vertex of the other graph. For example, the
join of the cycle graph Cn−1 with a single vertex graph is the wheel graph Wn . Figure 1.23
shows various wheel graphs.
1.5.2 Edge or vertex deletion/insertion
Vertex deletion subgraph
If G = (V, E) is any graph with at least 2 vertices, then the vertex deletion subgraph is
the subgraph obtained from G by deleting a vertex v ∈ V and also all the edges incident
to that vertex. The vertex deletion subgraph of G is sometimes denoted G − {v}. Sage
can compute vertex deletions, as the following example shows.
30 Chapter 1. Introduction to graph theory
5
5
3
9 7
7 9 3
4
1 2 1 2 4
(a) G1 (b) G2 (c) G1 ∆G2
Figure 1.22: The symmetric difference of graphs.
(a) W4 (b) W5 (c) W6
(d) W7 (e) W8 (f) W9
Figure 1.23: The wheel graphs Wn for n = 4, . . . , 9.
1.5. New graphs from old 31
sage : G = Graph ({1:[2 ,4] , 2:[1 ,4] , 3:[2 ,6] , 4:[1 ,3] , 5:[4 ,2] , 6:[3 ,1]})
sage : G . vertices ()
[1 , 2 , 3 , 4 , 5 , 6]
sage : E1 = Set ( G . edges ( labels = False )); E1
{(1 , 2) , (4 , 5) , (1 , 4) , (2 , 3) , (3 , 6) , (1 , 6) , (2 , 5) , (3 , 4) , (2 , 4)}
sage : E4 = Set ( G . edges_incident ( vertices =[4] , labels = False )); E4
{(4 , 5) , (3 , 4) , (2 , 4) , (1 , 4)}
sage : G . delete_vertex (4)
sage : G . vertices ()
[1 , 2 , 3 , 5 , 6]
sage : E2 = Set ( G . edges ( labels = False )); E2
{(1 , 2) , (1 , 6) , (2 , 5) , (2 , 3) , (3 , 6)}
sage : E1 . difference ( E2 ) == E4
True
Figure 1.24 presents a sequence of subgraphs obtained by repeatedly deleting vertices.
As the figure shows, when a vertex is deleted from a graph, all edges incident on that
vertex are deleted as well.
c c
b
a e a e
d d
(a) G (b) G − {b}
c c
e
d d
(c) G − {a, b} (d) G − {a, b, e}
(e) G − {a, b, c, d, e}
Figure 1.24: Obtaining subgraphs via repeated vertex deletion.
Edge deletion subgraph
If G = (V, E) is any graph with at least 1 edge, then the edge deletion subgraph is the
subgraph obtained from G by deleting an edge e ∈ E, but not the vertices incident to
that edge. The edge deletion subgraph of G is sometimes denoted G − {e}. Sage can
compute edge deletions, as the following example shows.
sage : G = Graph ({1:[2 ,4] , 2:[1 ,4] , 3:[2 ,6] , 4:[1 ,3] , 5:[4 ,2] , 6:[3 ,1]})
sage : E1 = Set ( G . edges ( labels = False )); E1
{(1 , 2) , (4 , 5) , (1 , 4) , (2 , 3) , (3 , 6) , (1 , 6) , (2 , 5) , (3 , 4) , (2 , 4)}
32 Chapter 1. Introduction to graph theory
sage : V1 = G . vertices (); V1
[1 , 2 , 3 , 4 , 5 , 6]
sage : E14 = Set ([(1 ,4)]); E14
{(1 , 4)}
sage : G . delete_edge ([1 ,4])
sage : E2 = Set ( G . edges ( labels = False )); E2
{(1 , 2) , (4 , 5) , (2 , 3) , (3 , 6) , (1 , 6) , (2 , 5) , (3 , 4) , (2 , 4)}
sage : E1 . difference ( E2 ) == E14
True
Figure 1.25 shows a sequence of graphs resulting from edge deletion. Unlike vertex
deletion, when an edge is deleted the vertices incident on that edge are left intact.
b b b
c a c a c a
(a) G (b) G − {ac} (c) G − {ab, ac, bc}
Figure 1.25: Obtaining subgraphs via repeated edge deletion.
Vertex cut, cut vertex, or cutpoint
A vertex cut (or separating set) of a connected graph G = (V, E) is a subset W ⊆ V
such that the vertex deletion subgraph G − W is disconnected. In fact, if v1 , v2 ∈ V are
two non-adjacent vertices, then you can ask for a vertex cut W for which v1 , v2 belong
to different components of G − W . Sage’s vertex_cut method allows you to compute a
minimal cut having this property. For many connected graphs, the removal of a single
vertex is sufficient for the graph to be disconnected (see Figure 1.25(c)).
Edge cut, cut edge, or bridge
If deleting a single, specific edge would disconnect a graph G, that edge is called a
bridge. More generally, the edge cut (or disconnecting set or seg) of a connected graph
G = (V, E) is a set of edges F ⊆ E whose removal yields an edge deletion subgraph
G − F that is disconnected. A minimal edge cut is called a cut set or a bond . In fact, if
v1 , v2 ∈ V are two vertices, then you can ask for an edge cut F for which v1 , v2 belong
to different components of G − F . Sage’s edge_cut method allows you to compute a
minimal cut having this property. For example, any of the three edges in Figure 1.25(c)
qualifies as a bridge and those three edges form an edge cut for the graph in question.
Theorem 1.27. Let G be a connected graph. An edge e ∈ E(G) is a bridge of G if and
only if e does not lie on a cycle of G.
Proof. First, assume that e = uv is a bridge of G. Suppose for contradiction that e lies
on a cycle
C : u, v, w1 , w2 , . . . , wk , u.
Then G − e contains a u-v path u, wk , . . . , w2 , w1 , v. Let u1 , v1 be any two vertices in
G − e. By hypothesis, G is connected so there is a u1 -v1 path P in G. If e does not lie
1.5. New graphs from old 33
on P , then P is also a path in G − e so that u1 , v1 are connected, which contradicts our
assumption of e being a bridge. On the other hand, if e lies on P , then express P as
u1 , . . . , u, v, . . . , v1 or u1 , . . . , v, u, . . . , v1 .
Now
u1 , . . . , u, wk , . . . , w2 , w1 , v, . . . , v1 or u1 , . . . , v, w1 , w2 , . . . , wk , u, . . . , v1
respectively is a u1 -v1 walk in G − e. By Theorem 1.15, G − e contains a u1 -v1 path,
which contradicts our assumption about e being a bridge.
Conversely, let e = uv be an edge that does not lie on any cycles of G. If G − e has no
u-v paths, then we are done. Otherwise, assume for contradiction that G − e has a u-v
path P . Then P with uv produces a cycle in G. This cycle contains e, in contradiction
of our assumption that e does not lie on any cycles of G.
Edge contraction
An edge contraction is an operation which, like edge deletion, removes an edge from a
graph. However, unlike edge deletion, edge contraction also merges together the two
vertices the edge used to connect. For a graph G = (V, E) and an edge uv = e ∈ E, the
edge contraction G/e is the graph obtained as follows:
1. Delete the vertices u, v from G.
2. In place of u, v is a new vertex ve .
3. The vertex ve is adjacent to vertices that were adjacent to u, v, or both u and v.
The vertex set of G/e = (V 0 , E 0 ) is defined as V 0 = V \{u, v} ∪ {ve } and its edge set is
E 0 = wx ∈ E | {w, x} ∩ {u, v} = ∅ ∪ ve w | uw ∈ E\{e} or vw ∈ E\{e} .
Make the substitutions
E1 = wx ∈ E | {w, x} ∩ {u, v} = ∅
E2 = ve w | uw ∈ E\{e} or vw ∈ E\{e} .
Let G be the wheel graph W6 in Figure 1.26(a) and consider the edge contraction G/ab,
where ab is the gray colored edge in that figure. Then the edge set E1 denotes all those
edges in G each of which is not incident on a, b, or both a and b. These are precisely
those edges that are colored red. The edge set E2 means that we consider those edges in
G each of which is incident on exactly one of a or b, but not both. The blue colored edges
in Figure 1.26(a) are precisely those edges that E2 suggests for consideration. The result
of the edge contraction G/ab is the wheel graph W5 in Figure 1.26(b). Figures 1.26(a)
to 1.26(f) present a sequence of edge contractions that starts with W6 and repeatedly
contracts it to the trivial graph K1 .
34 Chapter 1. Introduction to graph theory
d
d
d
f
e c
f e c
f
a b vab = g e vcg = h
(a) G1 (b) G2 = G1 /ab (c) G3 = G2 /cg
f
e vdh = i e vf i = j vej
(d) G4 = G3 /dh (e) G5 = G4 /f i (f) G6 = G5 /ej
Figure 1.26: Contracting the wheel graph W6 to the trivial graph K1 .
1.5.3 Complements
The complement of a simple graph has the same vertices, but exactly those edges that
are not in the original graph. In other words, if Gc = (V, E c ) is the complement of
G = (V, E), then two distinct vertices v, w ∈ V are adjacent in Gc if and only if they are
not adjacent in G. We also write the complement of G as G. The sum of the adjacency
matrix of G and that of Gc is the matrix with 1’s everywhere, except for 0’s on the
main diagonal. A simple graph that is isomorphic to its complement is called a self-
complementary graph. Let H be a subgraph of G. The relative complement of G and H
is the edge deletion subgraph G − E(H). That is, we delete from G all edges in H. Sage
can compute edge complements, as the following example shows.
sage : G = Graph ({1:[2 ,4] , 2:[1 ,4] , 3:[2 ,6] , 4:[1 ,3] , 5:[4 ,2] , 6:[3 ,1]})
sage : Gc = G . complement ()
sage : EG = Set ( G . edges ( labels = False )); EG
{(1 , 2) , (4 , 5) , (1 , 4) , (2 , 3) , (3 , 6) , (1 , 6) , (2 , 5) , (3 , 4) , (2 , 4)}
sage : EGc = Set ( Gc . edges ( labels = False )); EGc
{(1 , 5) , (2 , 6) , (4 , 6) , (1 , 3) , (5 , 6) , (3 , 5)}
sage : EG . difference ( EGc ) == EG
True
sage : EGc . difference ( EG ) == EGc
True
sage : EG . intersection ( EGc )
{}
Theorem 1.28. If G = (V, E) is self-complementary, then the order of G is |V | = 4k
or |V | = 4k + 1 for some nonnegative integer k. Furthermore, if n = |V | is the order of
G, then the size of G is |E| = n(n − 1)/4.
Proof. Let G be a self-complementary graph of order n. Each of G and Gc contains half
the number of edges in Kn . From (1.6), we have
1 n(n − 1) n(n − 1)
|E(G)| = |E(Gc )| = · = .
2 2 4
1.5. New graphs from old 35
Then n | n(n − 1), with one of n and n − 1 being even and the other odd. If n is even,
n − 1 is odd so gcd(4, n − 1) = 1, hence by [182, Theorem 1.9] we have 4 | n and so
n = 4k for some nonnegative k ∈ Z. If n − 1 is even, use a similar argument to conclude
that n = 4k + 1 for some nonnegative k ∈ Z.
Theorem 1.29. A graph and its complement cannot be both disconnected.
Proof. If G is connected, then we are done. Without loss of generality, assume that G
is disconnected and let G be the complement of G. Let u, v be vertices in G. If u, v
are in different components of G, then they are adjacent in G. If both u, v belong to
some component Ci of G, let w be a vertex in a different component Cj of G. Then u, w
are adjacent in G, and similarly for v and w. That is, u and v are connected in G and
therefore G is connected.
1.5.4 Cartesian product
The Cartesian product GH of graphs G and H is a graph such that the vertex set of
GH is the Cartesian product
V (GH) = V (G) × V (H).
Any two vertices (u, u0 ) and (v, v 0 ) are adjacent in GH if and only if either
1. u = v and u0 is adjacent with v 0 in H; or
2. u0 = v 0 and u is adjacent with v in G.
The vertex set of GH is V (GH) and the edge set of GH is
E(GH) = V (G) × E(H) ∪ E(G) × V (H) .
Sage can compute Cartesian products, as the following example shows.
sage : Z = graphs . CompleteGraph (2); len ( Z . vertices ()); len ( Z . edges ())
2
1
sage : C = graphs . CycleGraph (5); len ( C . vertices ()); len ( C . edges ())
5
5
sage : P = C . cartesian_product ( Z ); len ( P . vertices ()); len ( P . edges ())
10
15
The path graph Pn is a tree with n vertices V = {v1 , v2 , . . . , vn } and edges E =
{(vi , vi+1 ) | 1 ≤ i ≤ n − 1}. In this case, deg(v1 ) = deg(vn ) = 1 and deg(vi ) = 2 for
1 < i < n. The path graph Pn can be obtained from the cycle graph Cn by deleting
one edge of Cn . The ladder graph Ln is the Cartesian product of path graphs, i.e.
Ln = Pn P1 .
The Cartesian product of two graphs G1 and G2 can be visualized as follows. Let V1 =
{u1 , u2 , . . . , um } and V2 = {v1 , v2 , . . . , vn } be the vertex sets of G1 and G2 , respectively.
Let H1 , H2 , . . . , Hn be n copies of G1 . Place each Hi at the location of vi in G2 . Then
ui ∈ V (Hj ) is adjacent to ui ∈ V (Hk ) if and only if vjk ∈ E(G2 ). See Figure 1.27 for an
illustration of obtaining the Cartesian product of K3 and P3 .
The hypercube graph Qn is the n-regular graph having vertex set
V = (1 , . . . , n ) | i ∈ {0, 1}
of cardinality 2n . That is, each vertex of Qn is a bit string of length n. Two vertices
v, w ∈ V are connected by an edge if and only if v and w differ in exactly one coordinate.5
5
In other words, the Hamming distance between v and w is equal to 1.
36 Chapter 1. Introduction to graph theory
(a) K3 (b) P3 (c) K3 P3
Figure 1.27: The Cartesian product of K3 and P3 .
The Cartesian product of n edge graphs K2 is a hypercube:
(K2 )n = Qn .
Figure 1.28 illustrates the hypercube graphs Qn for n = 1, . . . , 4.
(a) Q1 (b) Q2 (c) Q3 (d) Q4
Figure 1.28: Hypercube graphs Qn for n = 1, . . . , 4.
Example 1.30. The Cartesian product of two hypercube graphs is another hypercube,
i.e. Qi Qj = Qi+j .
Another family of graphs that can be constructed via Cartesian product is the mesh.
Such a graph is also referred to as grid or lattice. The 2-mesh is denoted M (m, n) and
is defined as the Cartesian product M (m, n) = Pm Pn . Similarly, the 3-mesh is defined
as M (k, m, n) = Pk Pm Pn . In general, for a sequence a1 , a2 , . . . , an of n > 0 positive
integers, the n-mesh is given by
M (a1 , a2 , . . . , an ) = Pa1 Pa2 · · · Pan
where the 1-mesh is simply the path graph M (k) = Pk for some positive integer k.
Figure 1.29(a) illustrates the 2-mesh M (3, 4) = P3 P4 , while the 3-mesh M (3, 2, 3) =
P3 P2 P3 is presented in Figure 1.29(b).
1.6. Common applications 37
(a) M (3, 4) (b) M (3, 2, 3)
Figure 1.29: The 2-mesh M (3, 4) and the 3-mesh M (3, 2, 3).
1.5.5 Graph minors
A graph H is called a minor of a graph G if H is isomorphic to a graph obtained by a
sequence of edge contractions on a subgraph of G. The order in which a sequence of such
contractions is performed on G does not affect the resulting graph H. A graph minor is
not in general a subgraph. However, if G1 is a minor of G2 and G2 is a minor of G3 , then
G1 is a minor of G3 . Therefore, the relation “being a minor of” is a partial ordering on
the set of graphs. For example, the graph in Figure 1.26(c) is a minor of the graph in
Figure 1.26(a).
The following non-intuitive fact about graph minors was proven by Neil Robertson
and Paul Seymour in a series of 20 papers spanning 1983 to 2004. This result is known
by various names including the Robertson-Seymour theorem, the graph minor theorem,
or Wagner’s conjecture (named after Klaus Wagner).
Theorem 1.31. Robertson & Seymour 1983–2004. If an infinite list G1 , G2 , . . .
of finite graphs is given, then there always exist two indices i < j such that Gi is a minor
of Gj .
Many classes of graphs can be characterized by forbidden minors: a graph belongs
to the class if and only if it does not have a minor from a certain specified list. We shall
see examples of this in Chapter 7.
1.6 Common applications
(a) 2,4,4-trimethylheptane (b) naphthalene
Figure 1.30: Two molecular graphs.
Graph theory, and especially undirected graphs, is used in chemistry to study the struc-
ture of molecules. The graph theoretical representation of the structure of a molecule
38 Chapter 1. Introduction to graph theory
is called a molecular graph; two such examples are illustrated in Figure 1.30. Below we
list a few common problems arising in applications of graph theory. See Foulds [84] and
Walther [200] for surveys of applications of graph theory in science, engineering, social
sciences, economics, and operation research.
If the edge weights are all nonnegative, find a “cheapest” closed path which contains
all the vertices. This is related to the famous traveling salesman problem and is
further discussed in Chapters 2 and 6.
Find a walk that visits each vertex, but contains as few edges as possible and
contains no cycles. This type of problem is related to spanning trees and is discussed
in further details in Chapter 3.
Determine which vertices are “more central” than others. This is connected with
various applications to social network analysis and is covered in more details in
Chapters 5 and 10. An example of a social network is shown in Figure 1.31, which
illustrates the marriage ties among Renaissance Florentine families [35]. Note that
one family has been removed because its inclusion would create a disconnected
graph.
Pazzi Salviati Barbadori
Ginori Albizzi Medici Ridolfi
Acciaiuol Strozzi Castellan
Lambertes Tornabuon
Guadagni Bischeri Peruzzi
Figure 1.31: Marriage ties among Renaissance Florentine families.
A planar graph is a graph that can be drawn on the plane in such a way that its
edges intersect only at their endpoints. Can a graph be drawn entirely in the plane,
with no crossing edges? In other words, is a given graph planar? This problem is
important for designing computer chips and wiring diagrams. Further discussion
is contained in Chapter 7.
Can you label or color all the vertices of a graph in such a way that no adjacent
vertices have the same color? If so, this is called a vertex coloring. Can you label
or color all the edges of a graph in such a way that no incident edges have the same
color? If so, this is called an edge coloring. Figure 1.32(a) shows a vertex coloring
of the wheel graph W4 using two colors; Figure 1.32(b) shows a vertex coloring
of the Petersen graph using three colors. Graph coloring has several remarkable
applications, one of which is to scheduling of jobs relying on a shared resource.
This is discussed further in Chapter 8.
1.7. Application: finite automata 39
(a) (b)
Figure 1.32: Vertex coloring with two and three colors.
In some fields, such as operations research, a directed graph with nonnegative edge
weights is called a network , the vertices are called nodes, the edges are called arcs,
and the weight on an edge is called its capacity. A network flow must satisfy
the restriction that the amount of flow into a node equals the amount of flow out
of it, except when it is a source node, which has more outgoing flow, or a sink
node, which has more incoming flow. The flow along an edge must not exceed the
capacity. What is the maximum flow on a network and how to you find it? This
problem, which has many industrial applications, is discussed in Chapter 9.
1.7 Application: finite automata
In theoretical computer science, automata are used as idealized mathematical models
of computation. The studies of computability (i.e. what can be computed) and com-
plexity (i.e. the time and space requirements of a computation) are based on automata
theory to provide precise mathematical models of computers. For an intuitive appreci-
ation of automata, consider a vending machine that dispenses food or beverages. We
insert a fixed amount of money into the vending machine and make our choice of food
or beverage by pressing buttons that correspond to our choice. If the amount of money
inserted is sufficient to cover the cost of our choice of food or beverage, the machine
dispenses the item of our choice. Otherwise we need to insert more money until the
required amount is reached and then make our selection again. Embodied in the above
vending machine example are notions of input (money), machine states (has a selection
been made? has sufficient money been inserted?), state transition (move from money
insertion state to food/beverage selection state), and output (dispense item of choice).
In the above vending machine example, we should note that the vending machine
only accepts a finite number of objects as legitimate input. The vending machine can
accept dollar bills and coins of a fixed variety of denominations and belonging to a specific
locale, e.g. Australia. Thus we say that the vending machine has finite input and the
automaton that models the vending machine is referred to as a finite automaton having
a finite alphabet.
40 Chapter 1. Introduction to graph theory
δ 20¢ 50¢
0¢ 20¢ 50¢
20¢ 40¢ 70¢
40¢ 60¢ 90¢
50¢ 70¢ ≥ $1
60¢ 80¢ ≥ $1
70¢ 90¢ ≥ $1
80¢ ≥ $1 ≥ $1
90¢ ≥ $1 ≥ $1
≥ $1 ≥ $1 ≥ $1
Table 1.1: Transition table of a simple vending machine.
1.7.1 Automaton and language
Before presenting a precise definition of finite automata, we take a detour to describe
notations associated with valid input to finite automata. Let Σ be a nonempty finite
alphabet. By Σ∗ we mean the set of all finite strings over Σ. Each element of Σ∗ is a
string or word of finite length whose components are elements of Σ. That is, if w ∈ Σ∗
then w = w1 w2 · · · wn for some integer n ≥ 0 and each wi ∈ Σ. It follows that Σ ⊆ Σ∗ .
We also consider the empty string ε as a valid string over Σ. The string ε is sometimes
called the null string.
Definition 1.32. Finite automata. Let Q and Σ be nonempty finite sets. A finite
automaton is a 5-tuple A = (Q, Σ, δ, q0 , F ) where
1. Q is a finite set of states.
2. Σ is a finite set of input alphabet.
3. δ : Q × Σ → Q is the transition function.
4. q0 ∈ Q is the start or initial state.
5. F ⊆ Q is the set of accepting or final states.
For each possible combination of state and input symbol, the transition function δ
specifies exactly one subsequent or next state. The finite automaton A must have at least
one initial state, but this lower bound does not necessarily apply to its set of final states.
It is possible that the set of final states be empty, in which case A has no accepting
states.
Example 1.33. Figure 1.33 illustrates a finite-automaton representation of a basic vend-
ing machine. The initial state is depicted as a circle with an arrow pointing to it, with
no other state at the tail of the arrow. The final state is shown as a circle with two
concentric rings. We can consider the visual representation in Figure 1.33, also called
a state diagram, as a multidigraph where each vertex is a state and each directed edge
is a transition from one state to another. The state diagram can also be represented in
tabular form as shown in Table 1.1.
Let q1 , q2 ∈ Q. The finite-state automaton A is said to be a deterministic finite-state
automaton (DFA) if for all (q, a) ∈ Q × Σ, the mappings (q, a) 7→ q1 and (q, a) 7→ q2
imply that q1 = q2 . Furthermore, for each state q ∈ Q and each input symbol a ∈ Σ, we
have (q, a) 7→ q 0 for some q 0 ∈ Q. In other words |δ(q, a)| = 1.
1.7. Application: finite automata 41
20¢ 20¢
20¢ 40¢ 60¢
20¢
20¢
50¢ 50¢
80¢
20¢ 50¢
0¢ 70¢ 90¢
20¢
20¢ 20¢
50¢
50¢
50¢ 50¢
50¢ ≥ $1
50¢
20¢, 50¢
Figure 1.33: State diagram of a simple vending machine.
Definition 1.34. Nondeterministic finite-state automata. A nondeterministic
finite-state automaton (NFA) is a 5-tuple A = (Q, Σ, δ, Q0 , F ) where
1. Q is a finite set of states.
2. Σ is a finite set of input alphabet.
3. δ is a transition function defined by δ : Q × Σ → 2Q , where 2Q is the power set of
Q.
4. Q0 ⊆ Q is a set of initial states.
5. F ⊆ Q is a set of accepting or final states.
Intuitively, A is said to be an NFA if there exist some (q, a) ∈ Q × Σ and q1 , q2 ∈ Q
such that the transitions (q, a) 7→ q1 and (q, a) 7→ q2 imply q1 6= q2 . That is, correspond-
ing to each state/input pair is a multitude of subsequent states. Note the contrast to
DFA, where it is required that each state/input pair has at most one subsequent state.
Example 1.35. Let A = (Q, Σ, δ, q0 , F ) be defined by Q = {1, 2}, Σ = {a, b}, q0 = 1,
F = {2} and the transition function δ given by
δ(1, a) = 1, δ(1, b) = 2, δ(2, a) = 2, δ(2, b) = 2.
Figure 1.34 shows a digraph representation of A. It is easily verifiable by definition that
A is indeed a DFA.
a Σ
1 2
b
Figure 1.34: A deterministic finite-state automaton.
42 Chapter 1. Introduction to graph theory
Example 1.36. Let A = (Q, Σ, δ, Q0 , F ) be defined by Q = {1, 2}, Σ = {a, b}, Q0 = {1},
F = {2}, and the transition function δ given by
δ(1, a) = 1, δ(1, a) = 2, δ(2, a) = 2, δ(2, b) = 2.
Figure 1.35 shows a digraph representation of A. Note that δ(1, a) = 1 and δ(1, a) = 2.
It follows by definition that A is an NFA.
a Σ
1 2
a
Figure 1.35: A nondeterministic finite-state automaton.
We can inductively define a transition function δ̂ of a DFA A = (Q, Σ, δ, q0 , F ) oper-
ating on finite strings over Σ. That is,
δ̂ : Q × Σ∗ → Q. (1.12)
Let q ∈ Q and let s = s1 s2 · · · sn ∈ Σ∗ . In the case of the empty string, define δ̂(q, ε) = q.
When i = 1, we have δ̂(q, s1 ) = δ(q, s1 ). For 1 < i ≤ n, define
δ̂(q, s1 s2 · · · si ) = δ̂ δ̂(q, s1 s2 · · · si−1 ), si .
For convenience, we write δ(q, s) instead of δ̂(q, s). Where δ(q0 , s) ∈ F , we say that the
string s is accepted by A. Any subset of Σ∗ is said to be a language over Σ. The language
L of A is the set of all finite strings accepted by A, i.e.
L(A) = {s ∈ Σ∗ | δ(q0 , s) ∈ F } .
The special language L(A) is also referred to as a regular language. Referring back to
example 1.35, any string accepted by A has zero or more a, followed by exactly one b,
and finally zero or more occurrences of a or b. We can describe this language using the
regular expression a∗ b(a|b)∗ .
For NFAs, we can similarly define a transition function operating on finite strings.
Each input is a string over Σ and the transition function δ̂ returns a subset of Q. Formally,
our transition function for NFAs operating on finite strings is the map
δ̂ : Q × Σ∗ → 2Q .
Let q ∈ Q and let w = xa, where x ∈ Σ∗ and a ∈ Σ. The input symbol a can be
interpreted as being the very last symbol in the string w. Then x is interpreted as being
the substring of w excluding the symbol a. In the case of the empty string, we have
δ̂(q, ε) = {q}. For the inductive case, assume that δ̂(q, x) = {p1 , p2 , . . . , pk } where each
pi ∈ Q. Then δ̂(q, w) is defined by
δ̂(q, w) = δ̂ δ̂(q, x), a
= δ̂ ({p1 , p2 , . . . , pk }, a)
k
[
= δ(pi , a).
i=1
It may happen that for some state pi , there are no transitions from pi with input a. We
cater for this possibility by writing δ(pi , a) = ∅.
1.7. Application: finite automata 43
1.7.2 Simulating NFAs using DFAs
Any NFA can be simulated by a DFA. One way of accomplishing this is to allow the DFA
to keep track of all the states that the NFA can be in after reading an input symbol.
The formal proof depends on this construction of an equivalent DFA and then showing
that the language of the DFA is the same as that of the NFA.
Theorem 1.37. Determinize an NFA. If A is a nondeterministic finite-state au-
tomaton, then there exists a deterministic finite-state automaton A0 such that L(A) =
L(A0 ).
Proof. Let the NFA A be defined by A = (Q, Σ, δ, Q0 , F ) and define a DFA A0 =
(Q0 , Σ, δ 0 , q00 , F 0 ) as follows. The state space of A0 is the power set of Q, i.e. Q0 = 2Q .
The accepting state space F 0 of A0 is a subset of Q0 , where each f ∈ F 0 is a set containing
at least an accepting state of A. In symbols, we write F 0 ⊆ Q0 where
F 0 = {q ∈ Q0 | p ∈ F for some p ∈ q} .
Denote each element q ∈ Q0 by q = [q1 , q2 , . . . , qi ] where q1 , q2 , . . . , qi ∈ Q. Thus the
initial state of A0 is q00 = [Q0 ]. Now define the transition function δ 0 by
δ 0 ([q1 , q2 , . . . , qi ], s) = [p1 , p2 , . . . , pj ]
i
[ (1.13)
⇐⇒ δ ({q1 , q2 , . . . , qi }, s) = δ(qk , s) = {p1 , p2 , . . . , pj }.
k=1
For any input string w, we now show by induction on the length of w that
δ 0 (q00 , w) = [q1 , q2 , . . . , qi ]
(1.14)
⇐⇒ δ(Q0 , w) = {q1 , q2 , . . . , qi }.
For the basis step, let |w| = 0 so that w = ε. Then it is clear that
δ 0 (q00 , w) = δ 0 (q00 , ε) = [q00 ]
⇐⇒ δ(Q0 , w) = δ(Q0 , ε) = Q0 .
Next, assume for induction that statement (1.14) holds for all strings of length less than
or equal to m > 0. Let w be a string of length m and let a ∈ Σ so that |wa| = m + 1.
Then δ 0 (q00 , wa) = δ 0 δ 0 (q00 , w), a . By our inductive hypothesis, we have
δ 0 (q00 , w) = [p1 , p2 , . . . , pj ]
⇐⇒ δ(Q0 , w) = {p1 , p2 , . . . , pj }
and applying (1.14) we get
δ 0 ([p1 , p2 , . . . , pj ], a) = [r1 , r2 , . . . , rk ]
⇐⇒ δ ({p1 , p2 , . . . , pj }, a) = {r1 , r2 , . . . , rk }.
Hence
δ 0 (q00 , wa) = [r1 , r2 , . . . , rk ]
⇐⇒ δ(Q0 , wa) = {r1 , r2 , . . . , rk }
which establishes that statement (1.14) holds for all finite strings over Σ. Finally,
δ 0 (q00 , w) ∈ F 0 if and only if there is some p ∈ δ(Q0 , w) such that p ∈ F . Therefore
L(A) = L(A0 ).
44 Chapter 1. Introduction to graph theory
Theorem 1.37 tells us that any NFA corresponds to some DFA that accepts the
same language. For this reason, the theorem is said to provide us with a procedure
for determinizing NFAs. The actual procedure itself is contained in the proof of the
theorem, although it must be noted that the procedure is inefficient since it potentially
yields transitions from states that are unreachable from the initial state. If q ∈ Q0 is
a state of A0 that is unreachable from q00 , then there are no input strings w such that
δ 0 (q00 , w) = q. Such unreachable states are redundant insofar as they do not affect L(A0 ).
Another inefficiency of the procedure in the proof of Theorem 1.37 is the problem of
state space explosion. As Q0 = 2Q is the power set of Q, the resulting DFA can potentially
have exponentially more states than the NFA it is simulating. In the worse case, each
element of Q0 is a state of the resulting DFA that is reachable from q00 = [Q0 ]. The
best-case scenario is when each state of the DFA is a singleton, hence the DFA has the
same number of states as its corresponding NFA. However, according to the procedure
in the proof of Theorem 1.37, we generate all the possible 2n states of the DFA, where
n = |Q|. After considering all the transitions whose starting states are singletons, we
then consider all transitions starting from each of the remaining 2n − n elements in Q0 .
In the best-case, none of those remaining 2n − n states are reachable from q00 , hence it is
redundant to generate transitions starting at each of those 2n − n states. Example 1.38
concretizes our discussion.
Example 1.38. Use the procedure in Theorem 1.37 to determinize the NFA in Fig-
ure 1.36.
a b
a
1 2
c c
3
Figure 1.36: An NFA with 3 states and 3 input symbols.
Solution. The NFA A = (Q, Σ, δ, q0 , F ) in Figure 1.36 has the states Q = {1, 2, 3}, the
initial state q0 = 1, the final state set F = {3}, and the input alphabet Σ = {a, b, c}.
Its transitions are contained in Table 1.2. To determinize A, we construct a DFA A0 =
δ a b c
1 {1, 2} ∅ {3}
2 ∅ {2} {3}
3 ∅ ∅ ∅
Table 1.2: Transition table for the NFA in Figure 1.36.
(Q0 , Σ, δ 0 , q00 , F 0 ). As Q0 is the power set of Q, then all the possible states of A0 are
contained in Q0 = {∅, [1], [2], [3], [1, 2], [1, 3], [2, 3], [1, 2, 3]}. The alphabet of A0 is the
same as the alphabet of A, namely Σ. The initial state of A0 is q00 = [q0 ] = [1]. All
the possible accepting states of A0 are contained in F 0 = {[3], [1, 3], [2, 3], [1, 2, 3]}. Next,
1.7. Application: finite automata 45
we apply (1.13) to construct all the possible transitions of A0 . These transitions are
contained in Table 1.3. Using those transitions, we obtain the digraph representation in
Figure 1.37, from which it is clear that the states [1], [2], [3], and [1, 2] are the only states
δ0 a b c
[1] [1, 2] ∅ [3]
[2] ∅ [2] [3]
[3] ∅ ∅ ∅
[1, 2] [1, 2] [2] [3]
[1, 3] [1, 2] ∅ [3]
[2, 3] ∅ [2] [3]
[1, 2, 3] [1, 2] [2] [3]
Table 1.3: Transition table of a deterministic version of the NFA in Figure 1.36.
reachable from the initial state q00 = [1]. The remaining states [1, 3], [2, 3], and [1, 2, 3]
are not reachable from q00 = [1]. In other words, starting at q00 = [1] there are no input
strings that would result in a transition to any of [1, 3], [2, 3], and [1, 2, 3]. Therefore these
states, and the transitions starting from them, can be deleted from Figure 1.37 without
affecting the language of A0 . Figure 1.38 shows an equivalent DFA with redundant states
removed.
1
a c
a
c
1, 2 3
b c c
a
b 2 c 1, 2, 3
c
a
b
b
2, 3 1, 3
Figure 1.37: A DFA accepting the same language as the NFA in Figure 1.36.
46 Chapter 1. Introduction to graph theory
1
a c
c
a 1, 2 b 3
b c
2
Figure 1.38: A DFA equivalent to that in Figure 1.37, with redundant states removed.
1.8 Problems
A problem left to itself dries up or goes rotten. But fertilize a problem with a solution—
you’ll hatch out dozens.
— N. F. Simpson, A Resounding Tinkle, 1958
1.1. For each graph in Figure 1.7, do the following:
(a) Construct the graph using Sage.
(b) Find its adjacency matrix.
(c) Find its node and edge sets.
(d) How many nodes and edges are in the graph?
(e) If applicable, find all of each node’s in-coming and out-going edges. Hence
find the node’s indegree and outdegree.
Alice Bob
Carol
Figure 1.39: Graph representation of a social network.
1.2. In the friendship network of Figure 1.39, Carol is a mutual friend of Alice and Bob.
How many possible ways are there to remove exactly one edge such that, in the
resulting network, Carol is no longer a mutual friend of Alice and Bob?
1.3. The routing network of German cities in Figure 1.40 shows that each pair of distinct
cities are connected by a flight path. The weight of each edge is the flight distance
in kilometers between the two corresponding cities. In particular, there is a flight
path connecting Karlsruhe and Stuttgart. What is the shortest route between
Karlsruhe and Stuttgart? Suppose we can remove at least one edge from this
network. How many possible ways are there to remove edges such that, in the
resulting network, Karlsruhe is no longer connected to Stuttgart via a flight path?
1.4. Let D = (V, E) be a digraph of size q. Show that
X X
id(v) = od(v) = q.
v∈V v∈V
1.8. Problems 47
Karlsruhe
197 54
Augsburg Mannheim
57 72
Munich 383 Kassel 145 Frankfurt
149 97
Nuremberg 90 Würzburg
157 154
Stuttgart Erfurt
Figure 1.40: Graph representation of a routing network.
48 Chapter 1. Introduction to graph theory
1.5. If G is a simple graph of order n > 0, show that deg(v) < n for all v ∈ V (G).
1.6. Let G be a graph of order n and size m. Then G is called an overfull graph if
m > ∆(G) · bn/2c. If m = ∆(G) · bn/2c + 1, then G is said to be just overfull.
It can be shown that overfull graphs have odd order. Equivalently, let G be of
odd order. We can define G to be overfull if m > ∆(G) · (n − 1)/2, and G is just
overfull if m = ∆(G) · (n − 1)/2 + 1. Find an overfull graph and a graph that is
just overfull. Some basic results on overfull graphs are presented in [56].
1.7. Fix a positive integer n and denote by Γ(n) the number of simple graphs on n
vertices. Show that n
Γ(n) = 2( 2 ) = 2n(n−1)/2 .
1.8. Let G be an undirected graph whose unoriented incidence matrix is Mu and whose
oriented incidence matrix is Mo .
(a) Show that the sum of the entries in any row of Mu is the degree of the
corresponding vertex.
(b) Show that the sum of the entries in any column of Mu is equal to 2.
(c) If G has no self-loops, show that each column of Mo sums to zero.
1.9. Let G be a loopless digraph and let M be its incidence matrix.
(a) If r is a row of M , show that the number of occurrences of −1 in r counts
the outdegree of the vertex corresponding to r. Show that the number of
occurrences of 1 in r counts the indegree of the vertex corresponding to r.
(b) Show that each column of M sums to 0.
1.10. Let G be a digraph and let M be its incidence matrix. For any row r of M , let m
be the frequency of −1 in r, let p be the frequency of 1 in r, and let t be twice the
frequency of 2 in r. If v is the vertex corresponding to r, show that the degree of
v is deg(v) = m + p + t.
1.11. Let G be an undirected graph without self-loops and let M and its oriented in-
cidence matrix. Show that the Laplacian matrix L of G satisfies L = M × M T ,
where M T is the transpose of M .
1.12. Let J1 denote the incidence matrix of G1 and let J2 denote the incidence matrix of
G2 . Find matrix theoretic criteria on J1 and J2 which hold if and only if G1 ∼
= G2 .
In other words, find the analog of Theorem 1.24 for incidence matrices.
1.13. Show that the complement of an edgeless graph is a complete graph.
1.14. Let GH be the Cartesian product of two graphs G and H. Show that |E(GH)| =
|V (G)| · |E(H)| + |E(G)| · |V (H)|.
1.15. In 1751, Leonhard Euler posed a problem to Christian Goldbach, a problem that
now bears the name “Euler’s polygon division problem”. Given a plane convex
polygon having n sides, how many ways are there to divide the polygon into tri-
angles using only diagonals? For our purposes, we consider only regular polygons
having n sides for n ≥ 3 and any two diagonals must not cross each other. For
1.8. Problems 49
Figure 1.41: Euler’s polygon division problem for the hexagon.
example, the triangle is a regular 3-gon, the square a regular 4-gon, the pentagon
a regular 5-gon, etc. In the case of the hexagon considered as the cycle graph C6 ,
there are 14 ways to divide it into triangles, as shown in Figure 1.41, resulting in
14 graphs. However, of those 14 graphs only 3 are nonisomorphic to each other.
(a) What is the number of ways to divide a pentagon into triangles using only
diagonals? List all such divisions. If each of the resulting so divided pentagons
is considered a graph, how many of those graphs are nonisomorphic to each
other?
(b) Repeat the above exercise for the heptagon.
(c) Let En be the number of ways to divide an n-gon into triangles using only
diagonals. For n ≥ 1, the Catalan numbers Cn are defined as
1 2n
Cn = .
n+1 n
Dörrie [68, pp.21–27] showed that En is related to the Catalan numbers via
the equation En = Cn−1 . Show that
1 2n + 2
Cn = .
4n + 2 n + 1
For k ≥ 2, show that the Catalan numbers satisfy the recurrence relation
4k − 2
Ck = Ck−1 .
k+1
1.16. A graph is said to be planar if it can be drawn on the plane in such a way that
no two edges cross each other. For example, the complete graph Kn is planar for
n = 1, 2, 3, 4, but K5 is not planar (see Figure 1.13). Draw a planar version of K4 as
presented in Figure 1.13(b). Is the graph in Figure 1.9 planar? For n = 1, 2, . . . , 5,
enumerate all simple nonisomorphic graphs on n vertices that are planar; only work
with undirected graphs.
50 Chapter 1. Introduction to graph theory
1.17. If n ≥ 3, show that the join of Cn and K1 is the wheel graph Wn+1 . In other words,
show that Cn + K1 = Wn+1 .
1.18. A common technique for generating “random” numbers is the linear congruential
method, a generalization of the Lehmer generator [140] introduced in 1949. First,
we choose four integers:
m, modulus, 0<m
a, multiplier, 0≤a<m
c, increment, 0≤c<m
X0 , seed, 0 ≤ X0 < m
where the value X0 is also referred to as the starting value. Then iterate the
relation
Xn+1 = (aXn + c) mod m, n≥0
and halt when the relation produces the seed X0 or when it produces an integer
Xk such that Xk = Xi for some 0 ≤ i < k. The resulting sequence
S = (X0 , X1 , . . . , Xn )
is called a linear congruential sequence. Define a graph theoretic representation of
S as follows: let the vertex set be V = {X0 , X1 , . . . , Xn } and let the edge set be
E = {Xi Xi+1 | 0 ≤ i < n}. The resulting graph G = (V, E) is called the linear
congruential graph of the linear congruential sequence S. See chapter 3 of [128] for
other techniques for generating “random” numbers.
(a) Compute the linear congruential sequences Si with the following parameters:
(i) S1 : m = 10, a = c = X0 = 7
(ii) S2 : m = 10, a = 5, c = 7, X0 = 0
(iii) S3 : m = 10, a = 3, c = 7, X0 = 2
(iv) S4 : m = 10, a = 2, c = 5, X0 = 3
(b) Let Gi be the linear congruential graph of Si . Draw each of the graphs Gi .
Draw the graph resulting from the union
[
Gi .
i
(c) Let m, a, c, and X0 be the parameters of a linear congruential sequence where
(i) c is relatively prime to m;
(ii) b = a − 1 is a multiple of p for each prime p that divides m; and
(iii) 4 divides b if 4 divides m.
Show that the corresponding linear congruential graph is the wheel graph Wm
on m vertices.
1.19. We want to generate a random bipartite graph whose first and second partitions
have n1 and n2 vertices, respectively. Describe and present pseudocode to generate
the required random bipartite graph. What is the worst-case runtime of your
algorithm? Modify your algorithm to account for a third parameter m that specifies
the number of edges in the resulting bipartite graph.
1.8. Problems 51
1.20. Describe and present pseudocode to generate a random regular graph. What is the
worst-case runtime of your algorithm?
1.21. The Cantor-Schröder-Bernstein theorem states that if A, B are sets and we have an
injection f : A → B and an injection g : B → A, then there is a bijection between A
and B, thus proving that A and B have the same cardinality. Here we use bipartite
graphs and other graph theoretic concepts to prove the Cantor-Schröder-Bernstein
theorem. The full proof can be found in [213].
(a) Is it possible for A and B to be bipartitions of V and yet satisfy A ∩ B 6= ∅?
(b) Now assume that A ∩ B = ∅ and define a bipartite graph G = (V, E) with A
and B being the two partitions of V , where for any x ∈ A and y ∈ B we have
xy ∈ E if and only if either f (x) = y or g(y) = x. Show that deg(v) = 1 or
deg(v) = 2 for each v ∈ V .
(c) Let C be a component of G and let A0 ⊆ A and B 0 ⊆ B contain all vertices
in the component C. Show that |A0 | = |B 0 |.
1.22. Fermat’s little theorem states that if p is prime and a is an integer not divisible
by p, then p divides ap − a. Here we cast the problem within the context of graph
theory and prove it using graph theoretic concepts. The full proof can be found
in [103, 213].
(a) Let G = (V, E) be a graph with V being the set of all sequences (a1 , a2 , . . . , ap )
of integers 1 ≤ ai ≤ a and aj 6= ak for some j 6= k. Show that G has ap − a
vertices.
(b) Define the edge set of G as follows. If u, v ∈ V such that u = (u1 , u2 , . . . , up )
and v = (up , u1 , . . . , up−1 ), then uv ∈ E. Show that each component of G is a
cycle of length p.
(c) Show that G has (ap − a)/p components.
1.23. For the finite automaton in Figure 1.33, identify the following:
(a) The states set Q.
(b) The alphabet set Σ.
(c) The transition function δ : Q × Σ → Q.
(d) The initial state q0 ∈ Q.
(e) The set of final states F ⊆ Q.
1.24. The cycle graph Cn is a 2-regular graph. If 2 < r < n/2, unlike the cycle graph
there are various realizations of an r-regular graph; see Figure 1.42 for the case of
r = 3 and n = 10. The k-circulant graph on n vertices can be considered as an
intermediate graph between Cn and a k-regular graph. Let k and n be positive
integers satisfying k < n/2 with k being even. Suppose G = (V, E) is a simple
undirected graph with vertex set V = {0, 1, . . . , n − 1}. Define the edge set of G
as follows. Each i ∈ V is incident with each of i + j mod n and i − j mod n for
j ∈ {1, 2, . . . , k/2}. With the latter edge set, G is said to be a k-circulant graph, a
type of graphs used in constructing small-world networks (see section 10.4). Refer
to Figure 1.43 for examples of k-circulant graphs.
52 Chapter 1. Introduction to graph theory
3 2 3 2 3 2
4 1 4 1 4 1
5 0 5 0 5 0
6 9 6 9 6 9
7 8 7 8 7 8
(a) (b) (c)
Figure 1.42: Various 3-regular graphs on 10 vertices.
(a) k = 4 (b) k = 6 (c) k = 8
Figure 1.43: Various k-circulant graphs for k = 4, 6, 8.
(a) Describe and provide pseudocode of an algorithm to construct a k-circulant
graph on n vertices.
(b) Show that the cycle graph Cn is 2-circulant.
(c) Show that the sum of all degrees of a k-circulant graph on n vertices is nk.
(d) Show that a k-circulant graph is k-regular.
(e) Let C be the collection of all k-regular graphs on n vertices. If each k-regular
graph from C is equally likely to be chosen, what is the probability that a
k-circulant graph be chosen from C?
Chapter 2
Graph algorithms
— Randall Munroe, xkcd, http://xkcd.com/518/
Graph algorithms have many applications. Suppose you are a salesman with a product
you would like to sell in several cities. To determine the cheapest travel route from city-
to-city, you must effectively search a graph having weighted edges for the “cheapest”
route visiting each city once. Each vertex denotes a city you must visit and each edge
has a weight indicating either the distance from one city to another or the cost to travel
from one city to another.
Shortest path algorithms are some of the most important algorithms in algorithmic
graph theory. In this chapter, we first examine several common graph traversal algo-
rithms and some basic data structures underlying these algorithms. A data structure is
a combination of methods for structuring a collection of data (e.g. vertices and edges)
and protocols for accessing the data. We then consider a number of common shortest
path algorithms, which rely in one way or another on graph traversal techniques and
basic data structures for organizing and managing vertices and edges.
53
54 Chapter 2. Graph algorithms
2.1 Representing graphs in a computer
To err is human but to really foul things up requires a computer.
— Anonymous, Farmers’ Almanac for 1978, “Capsules of Wisdom”
In section 1.3, we discussed how to use matrices for representing graphs and digraphs. If
A = [aij ] is an m×n matrix, the adjacency matrix representation of a graph would require
representing all the mn entries of A. Alternative graph representations exist that are
much more efficient than representing all entries of a matrix. The graph representation
used can be influenced by the size of a graph or the purpose of the representation. Sec-
tion 2.1.1 discusses the adjacency list representation that can result in less storage space
requirement than the adjacency matrix representation. The graph6 format discussed in
section 2.1.3 provides a compact means of storing graphs for archival purposes.
2.1.1 Adjacency lists
A list is a sequence of objects. Unlike sets, a list may contain multiple copies of the same
object. Each object in a list is referred to as an element of the list. A list L of n ≥ 0
elements is written as L = [a1 , a2 , . . . , an ], where the i-th element ai can be indexed
as L[i]. In case n = 0, the list L = [ ] is referred to as the empty list. Two lists are
equivalent if they both contain the same elements at exactly the same positions.
Define the adjacency lists of a graph as follows. Let G be a graph with vertex set
V = {v1 , v2 , . . . , vn }. Assign to each vertex vi a list Li containing all the vertices that
are adjacent to vi . The list Li associated with vi is referred to as the adjacency list of
vi . Then Li = [ ] if and only if vi is an isolated vertex. We say that Li is the adjacency
list of vi because any permutation of the elements of Li results in a list that contains
the same vertices adjacent to vi . We are mainly concerned with the neighbors of vi , but
disregard the position where each neighbor is located in Li . If each adjacency list Li
contains si elements where 0 ≤ si ≤ n, we say that Li has length P si . The adjacency
list representation of the graph G requires that we represent i si = 2 · |E(G)| ≤ n2
elements in a computer’s memory, since each edge appears twice in the adjacency list
representation. An adjacency list is explicit about which vertices are adjacent to a vertex
and implicit about which vertices are not adjacent to that same vertex. Without knowing
the graph G, given the adjacency lists L1 , L2 , . . . , Ln , we can reconstruct G. For example,
Figure 2.1 shows a graph and its adjacency list representation.
4 3
7
8 6 L1 = [2, 8] L5 = [6, 8]
L2 = [1, 6] L6 = [2, 5, 8]
5
L3 = [4] L7 = [ ]
1 2 L4 = [3] L8 = [1, 5, 6]
Figure 2.1: A graph and its adjacency lists.
Example 2.1. The Kneser graph with parameters (n, k), also known as the (n, k)-Kneser
graph, is the graph whose vertices are all the k-subsets of {1, 2, . . . , n}. Furthermore, two
2.1. Representing graphs in a computer 55
vertices are adjacent if their corresponding sets are disjoint. Draw the (5, 2)-Kneser
graph and find its order and adjacency lists. In general, if n and k are positive, what is
the order of the (n, k)-Kneser graph?
Solution. The (5, 2)-Kneser graph is the graph whose vertices are the 2-subsets
{1, 2}, {1, 3}, {1, 4}, {1, 5}, {2, 3}, {2, 4}, {2, 5}, {3, 4}, {3, 5}, {4, 5}
of {1, 2, 3, 4, 5}. That is, each vertex of the (5, 2)-Kneser graph
is5×4
a 2-combination of the
5
set {1, 2, 3, 4, 5} and therefore the graph itself has order 2 = 2! = 10. The edges of
this graph are
({1, 3}, {2, 4}), ({2, 4}, {1, 5}), ({2, 4}, {3, 5}), ({1, 3}, {4, 5}), ({1, 3}, {2, 5})
({3, 5}, {1, 4}), ({3, 5}, {1, 2}), ({1, 4}, {2, 3}), ({1, 4}, {2, 5}), ({4, 5}, {2, 3})
({4, 5}, {1, 2}), ({1, 5}, {2, 3}), ({1, 5}, {3, 4}), ({3, 4}, {1, 2}), ({3, 4}, {2, 5})
from which we obtain the following adjacency lists:
L{1,2} = [{3, 4}, {3, 5}, {4, 5}], L{1,3} = [{2, 4}, {2, 5}, {4, 5}],
L{1,4} = [{2, 3}, {3, 5}, {2, 5}], L{1,5} = [{2, 4}, {3, 4}, {2, 3}],
L{2,3} = [{1, 5}, {1, 4}, {4, 5}], L{2,4} = [{1, 3}, {1, 5}, {3, 5}],
L{2,5} = [{1, 3}, {3, 4}, {1, 4}], L{3,4} = [{1, 2}, {1, 5}, {2, 5}],
L{3,5} = [{2, 4}, {1, 2}, {1, 4}], L{4,5} = [{1, 3}, {1, 2}, {2, 3}].
The (5, 2)-Kneser graph itself is shown in Figure 2.2. Using Sage, we have
sage : K = graphs . KneserGraph (5 , 2); K
Kneser graph with parameters 5 ,2: Graph on 10 vertices
sage : for v in K . vertices ():
... print (v , K . neighbors ( v ))
...
({4 , 5} , [{1 , 3} , {1 , 2} , {2 , 3}])
({1 , 3} , [{2 , 4} , {2 , 5} , {4 , 5}])
({2 , 5} , [{1 , 3} , {3 , 4} , {1 , 4}])
({2 , 3} , [{1 , 5} , {1 , 4} , {4 , 5}])
({3 , 4} , [{1 , 2} , {1 , 5} , {2 , 5}])
({3 , 5} , [{2 , 4} , {1 , 2} , {1 , 4}])
({1 , 4} , [{2 , 3} , {3 , 5} , {2 , 5}])
({1 , 5} , [{2 , 4} , {3 , 4} , {2 , 3}])
({1 , 2} , [{3 , 4} , {3 , 5} , {4 , 5}])
({2 , 4} , [{1 , 3} , {1 , 5} , {3 , 5}])
If n and k are positive integers, then the (n, k)-Kneser graph has
n n(n − 1) · · · (n − k + 1)
=
k k!
vertices.
We can categorize a graph G = (V, E) as dense or sparse based upon its size. A dense
2 2
graph has size |E| that is close to |V | , i.e. |E| = Ω |V | , in which case it is feasible to
2
represent G as an adjacency matrix. The size of a sparse graph is much less than |V | ,
i.e. |E| = Ω |V | , which renders the adjacency matrix representation as unsuitable. For
a sparse graph, an adjacency list representation can require less storage space than an
adjacency matrix representation of the same graph.
56 Chapter 2. Graph algorithms
{3, 4}
{1, 2}
{1, 5} {2, 5}
{2, 3}
{3, 5} {4, 5}
{1, 4}
{2, 4} {1, 3}
Figure 2.2: The (5, 2)-Kneser graph.
2.1.2 Edge lists
Lists can also be used to store the edges of a graph. To create an edge list L for a graph
G, if uv is an edge of G then we let uv or the ordered pair (u, v) be an element of L. In
general, let
v0 v1 , v2 v3 , . . . , vk vk+1
be all the edges of G, where k is even. Then the edge list of G is given by
L = [v0 v1 , v2 v3 , . . . , vk vk+1 ].
In some cases, it is desirable to have the edges of G be in contiguous list representation.
If the edge list L of G is as given above, the contiguous edge list representation of the
edges of G is
[v0 , v1 , v2 , v3 , . . . , vk , vk+1 ].
That is, if 0 ≤ i ≤ k is even then vi vi+1 is an edge of G.
2.1.3 The graph6 format
The graph formats graph6 and sparse6 were developed by Brendan McKay [149] at
The Australian National University as a compact way to represent graphs. These two
formats use bit vectors and printable characters of the American Standard Code for
Information Interchange (ASCII) encoding scheme. The 64 printable ASCII characters
used in graph6 and sparse6 are those ASCII characters with decimal codes from 63 to
126, inclusive, as shown in Table 2.1. This section shall only cover the graph6 format.
For full specification on both of the graph6 and sparse6 formats, see McKay [149].
Bit vectors
Before discussing how graph6 and sparse6 represent graphs using printable ASCII char-
acters, we first present encoding schemes used by these two formats. A bit vector is, as
2.1. Representing graphs in a computer 57
binary decimal glyph binary decimal glyph
0111111 63 ? 1011111 95 _
1000000 64 @ 1100000 96 ‘
1000001 65 A 1100001 97 a
1000010 66 B 1100010 98 b
1000011 67 C 1100011 99 c
1000100 68 D 1100100 100 d
1000101 69 E 1100101 101 e
1000110 70 F 1100110 102 f
1000111 71 G 1100111 103 g
1001000 72 H 1101000 104 h
1001001 73 I 1101001 105 i
1001010 74 J 1101010 106 j
1001011 75 K 1101011 107 k
1001100 76 L 1101100 108 l
1001101 77 M 1101101 109 m
1001110 78 N 1101110 110 n
1001111 79 O 1101111 111 o
1010000 80 P 1110000 112 p
1010001 81 Q 1110001 113 q
1010010 82 R 1110010 114 r
1010011 83 S 1110011 115 s
1010100 84 T 1110100 116 t
1010101 85 U 1110101 117 u
1010110 86 V 1110110 118 v
1010111 87 W 1110111 119 w
1011000 88 X 1111000 120 x
1011001 89 Y 1111001 121 y
1011010 90 Z 1111010 122 z
1011011 91 [ 1111011 123 {
1011100 92 \ 1111100 124 |
1011101 93 ] 1111101 125 }
1011110 94 ^ 1111110 126 ~
Table 2.1: ASCII printable characters used by graph6 and sparse6.
58 Chapter 2. Graph algorithms
its name suggests, a vector whose elements are 1’s and 0’s. It can be represented as a list
of bits, e.g. E can be represented as the ASCII bit vector [1, 0, 0, 0, 1, 0, 1]. For brevity,
we write a bit vector in a compact form such as 1000101. The length of a bit vector
is its number of bits. The most significant bit of a bit vector v is the bit position with
the largest value among all the bit positions in v. Similarly, the least significant bit is
the bit position in v having the least value among all the bit positions in v. The least
significant bit of v is usually called the parity bit because when v is interpreted as an
integer the parity bit determines whether the integer is even or odd. Reading 1000101
from left to right, the first bit 1 is the most significant bit, followed by the second bit 0
which is the second most significant bit, and so on all the way down to the seventh bit
1 which is the least significant bit.
The order in which we process the bits of a bit vector
v = bn−1 bn−2 · · · b0 (2.1)
is referred to as endianness. Processing v in big-endian order means that we first process
the most significant bit of v, followed by the second most significant bit, and so on all the
way down to the least significant bit of v. Thus, in big-endian order we read the bits bi of
v from left to right in increasing order of powers of 2. Table 2.2 illustrates the big-endian
interpretation of the ASCII binary representation of E. Little-endian order means that
we first process the least significant bit, followed by the second least significant bit, and
so on all the way up to the most significant bit. In little-endian order, the bits bi are read
from right to left in increasing order of powers of 2. Table 2.3 illustrates the little-endian
interpretation of the ASCII binary representation of E. In his novel Gulliver’s Travels
first published in 1726, Jonathan Swift used the terms big- and little-endian in satirizing
politicians who squabbled over whether to break an egg at the big end or the little end.
Danny Cohen [59, 60] first used the terms in 1980 as an April fool’s joke in the context
of computer architecture.
Suppose the bit vector (2.1) is read in big-endian order. To determine the integer
representation of v, multiply each bit value by its corresponding position value, then add
up all the results. Thus, if v is read in big-endian order, the integer representation of v
is obtained by evaluating the polynomial
n−1
X
p(x) = xi bi = xn−1 bn−1 + xn−2 bn−2 + · · · + xb1 + b0 . (2.2)
i=0
at x = 2. See problem 2.2 for discussion of an efficient method to compute the integer
representation of a bit vector.
position 0 1 2 3 4 5 6
bit value 1 0 0 0 1 0 1
position value 20 21 22 23 24 25 26
Table 2.2: Big-endian order of the ASCII binary code of E.
In graph6 and sparse6 formats, the length of a bit vector must be a multiple of 6.
Suppose v is a bit vector of length k such that 6 - k. To transform v into a bit vector
having length a multiple of 6, let r = k mod 6 be the remainder upon dividing k by 6,
and pad 6 − r zeros to the right of v.
2.1. Representing graphs in a computer 59
position 0 1 2 3 4 5 6
bit value 1 0 0 0 1 0 1
position value 26 25 24 23 22 21 20
Table 2.3: Little-endian order of the ASCII binary code of E.
Suppose v = b1 b2 · · · bk is a bit vector of length k, where 6 | k. We split v into k/6
bit vectors vi , each of length 6. For 0 ≤ i ≤ k/6, the i-th bit vector is given by
vi = b6i−5 b6i−4 b6i−3 b6i−2 b6i−1 b6i .
Consider each vi as the big-endian binary representation of a positive integer. Use (2.2)
to obtain the integer representation Ni of each vi . Then add 63 to each Ni to obtain Ni0
and store Ni0 in one byte of memory. That is, each Ni0 can be represented as a bit vector
of length 8. Thus the required number of bytes to store v is dk/6e. Let Bi be the byte
representation of Ni0 so that
R(v) = B1 B2 · · · Bdk/6e (2.3)
denotes the representation of v as a sequence of dk/6e bytes.
We now discuss how to encode an integer n in the range 0 ≤ n ≤ 236 − 1 using (2.3)
and denote such an encoding of n as N (n). Let v be the big-endian binary representation
of n. Then N (n) is given by
n + 63, if 0 ≤ n ≤ 62,
N (n) = 126 R(v), if 63 ≤ n ≤ 258047, (2.4)
126 126 R(v), if 258048 ≤ n ≤ 236 − 1.
Note that n + 63 requires one byte of storage memory, while 126 R(v) and 126 126 R(v)
require 4 and 8 bytes, respectively.
The graph6 format
The graph6 format is used to represent simple, undirected graphs of order from 0 to
236 − 1, inclusive. Let G be a simple, undirected graph of order 0 ≤ n ≤ 236 − 1. If
n = 0, then G is represented in graph6 format as “?”. Suppose n > 0. Let M = [aij ]
be the adjacency matrix of G. Consider the upper triangle of M , excluding the main
diagonal, and write that upper triangle as the bit vector
v = a0,1 a0,2 a1,2 a0,3 a1,3 a2,3 · · · a0,i a1,i · · · ai−1,i · · · a0,n a1,n · · · an−1,n
|{z} | {z } | {z } | {z } | {z }
c1 c2 c3 ci cn
where ci denotes the entries a0,i a1,i · · · ai−1,i in column i of M . Then the graph6 repre-
sentation of G is N (n)R(v), where R(v) and N (n) are as in (2.3) and (2.4), respectively.
That is, N (n) encodes the order of G and R(v) encodes the edges of G.
60 Chapter 2. Graph algorithms
2.2 Graph searching
Errors, like straws, upon the surface flow;
He who would search for pearls must dive below.
— John Dryden, All for Love, 1678
This section discusses two fundamental algorithms for graph traversal: breadth-first
search and depth-first search. The word “search” used in describing these two algorithms
is rather misleading. It would be more accurate to describe them as algorithms for
constructing trees using the adjacency information of a given graph. However, the names
“breadth-first search” and “depth-first search” are entrenched in literature on graph
theory and computer science. From hereon, we use these two names as given above,
bearing in mind their intended purposes.
2.2.1 Breadth-first search
Breadth-first search (BFS) is a strategy for running through the vertices of a graph. It
was presented by Moore [157] in 1959 within the context of traversing mazes. Lee [139]
independently discovered the same algorithm in 1961 in his work on routing wires on
circuit boards. In the physics literature, BFS is also known as a “burning algorithm” in
view of the analogy of a fire burning and spreading through an area, a piece of paper,
fabric, etc.
The basic BFS algorithm can be described as follows. Starting from a given vertex
v of a graph G, we first explore the neighborhood of v by visiting all vertices that are
adjacent to v. We then apply the same strategy to each of the neighbors of v. The
strategy of exploring the neighborhood of a vertex is applied to all vertices of G. The
result is a tree rooted at v and this tree is a subgraph of G. Algorithm 2.1 presents a
general template for the BFS strategy. The tree resulting from the BFS algorithm is
called a breadth-first search tree.
Algorithm 2.1: A general breadth-first search template.
Input: A directed or undirected graph G = (V, E) of order n > 0. A vertex s
from which to start the search. The vertices are numbered from 1 to
n = |V |, i.e. V = {1, 2, . . . , n}.
Output: A list D of distances of all vertices from s. A tree T rooted at s.
1 Q ← [s] /* queue of nodes to visit */
2 D ← [∞, ∞, . . . , ∞] /* n copies of ∞ */
3 D[s] ← 0
4 T ← []
5 while length(Q) > 0 do
6 v ← dequeue(Q)
7 for each w ∈ adj(v) do
8 if D[w] = ∞ then
9 D[w] ← D[v] + 1
10 enqueue(Q, w)
11 append(T, vw)
12 return (D, T )
2.2. Graph searching 61
The breadth-first search algorithm makes use of a special type of list called a queue.
This is analogous to a queue of people waiting in line to be served. A person may enter
the queue by joining the rear of the queue. The person who is in the queue the longest
amount of time is served first, followed by the person who has waited the second longest
time, and so on. Formally, a queue Q is a list of elements. At any time, we only have
access to the first element of Q, known as the front or start of the queue. We insert
a new element into Q by appending the new element to the rear or end of the queue.
The operation of removing the front of Q is referred to as dequeue, while the operation
of appending to the rear of Q is called enqueue. That is, a queue implements a first-in
first-out (FIFO) protocol for adding and removing elements. As with lists, the length of
a queue is its total number of elements.
1 2 1 2
3 3
7 7
4 4
6 5 6 5
(a) Original undirected graph. (b) First iteration of while loop.
1 2 1 2
3 3
7 7
4 4
6 5 6 5
(c) Second iteration of while loop. (d) Third iteration of while loop.
1
1 2
2 5
3
7
3 7 6
4
6 5 4
(e) Fourth iteration of while loop. (f) Final BFS tree.
Figure 2.3: Breadth-first search tree for an undirected graph.
Note that the BFS Algorithm 2.1 works on both undirected and directed graphs. For
an undirected graph, line 7 means that we explore all the neighbors of vertex v, i.e. the
set adj(v) of vertices adjacent to v. In the case of a digraph, we replace “w ∈ adj(v)”
on line 7 with “w ∈ oadj(v)” because we only want to explore all vertices that are out-
neighbors of v. The algorithm returns two lists D and T . The list T contains a subset
62 Chapter 2. Graph algorithms
1 2 1 2
3 3
7 7
4 4
6 5 6 5
(a) Original digraph. (b) First iteration of while loop.
1 2 1 2
3 3
7 7
4 4
6 5 6 5
(c) Second iteration of while loop. (d) Third iteration of while loop.
4
1 2
3 6
3
7
2 5 7
4
6 5
1
(e) Fourth iteration of while loop. (f) Final BFS tree.
Figure 2.4: Breadth-first search tree for a digraph.
2.2. Graph searching 63
of edges in E(G) that make up a tree rooted at the given start vertex s. As trees are
connected graphs without cycles, we may take the vertices comprising the edges of T to
be the vertex set of the tree. It is clear that T represents a tree by means of a list of
edges, which allows us to identify the tree under consideration as the edge list T . The
list D has the same number of elements as the order of G = (V, E), i.e. length(D) = |V |.
The i-th element D[i] counts the number of edges in T between the vertices s and vi . In
other words, D[i] is the length of the s-vi path in T . It can be shown that D[i] = ∞ if
and only if G is disconnected. After one application of Algorithm 2.1, it may happen that
D[i] = ∞ for at least one vertex vi ∈ V . To traverse those vertices that are unreachable
from s, again we apply Algorithm 2.1 on G with starting vertex vi . Repeat this algorithm
as often as necessary until all vertices of G are visited. The result may be a tree that
contains all the vertices of G or a collection of trees, each of which contains a subset of
V (G). Figures 2.3 and 2.4 present BFS trees resulting from applying Algorithm 2.1 on
an undirected graph and a digraph, respectively.
Theorem 2.2. The worst-case time complexity of Algorithm 2.1 is O(|V | + |E|).
Proof. Without loss of generality, we can assume that G = (V, E) is connected. The
initialization steps in lines 1 to 4 take O(|V |) time. After initialization, all but one
vertex are labelled ∞. Line 8 ensures that each vertex is enqueued at most once and
hence dequeued at most once. Each of enqueuing and dequeuing takes constant time.
The total time devoted to queue operations is O(|V |). The adjacency list of a vertex
is scanned after dequeuing that vertex, so each adjacency list is scanned at most once.
Summing the lengths of the adjacency lists, we have Θ(|E|) and therefore we require
O(|E|) time to scan the adjacency lists. After the adjacency list of a vertex is scanned,
at most k edges are added to the list T , where k is the length of the adjacency list under
consideration. Like queue operations, appending to a list takes constant time, hence we
require O(|E|) time to build the list T . Therefore, BFS runs in O(|V | + |E|) time.
Theorem 2.3. For the list D resulting from Algorithm 2.1, let s be a starting vertex
and let v be a vertex such that D[v] 6= ∞. Then D[v] is the length of any shortest path
from s to v.
Proof. It is clear that D[v] = ∞ if and only if there are no paths from s to v. Let v be
a vertex such that D[v] 6= ∞. As v can be reached from s by a path of length D[v], the
length d(s, v) of any shortest s-v path satisfies d(s, v) ≤ D[v]. Use induction on d(s, v) to
show that equality holds. For the base case s = v, we have d(s, v) = D[v] = 0 since the
trivial path has length zero. Assume for induction that if d(s, v) = k, then d(s, v) = D[v].
Let d(s, u) = k + 1 with the corresponding shortest s-u path being (s, v1 , v2 , . . . , vk , u).
By our induction hypothesis, (s, v1 , v2 , . . . , vk ) is a shortest path from s to vk of length
d(s, vk ) = D[vk ] = k. In other words, D[vk ] < D[u] and the while loop spanning lines 5
to 11 processes vk before processing u. The graph under consideration has the edge vk u.
When examining the adjacency list of vk , BFS reaches u (if u is not reached earlier) and
so D[u] ≤ k + 1. Hence, D[u] = k + 1 and therefore d(s, u) = D[u] = k + 1.
In the proof of Theorem 2.3, we used d(u, v) to denote the length of the shortest path
from u to v. This shortest path length is also known as the distance from u to v, and
will be discussed in further details in section 2.3 and Chapter 5. The diameter diam(G)
of a graph G = (V, E) is defined as
diam(G) = max d(u, v). (2.5)
u,v∈V
u6=v
64 Chapter 2. Graph algorithms
Using the above definition, to find the diameter we first determine the distance between
each pair of distinct vertices, then we compute the maximum of all such distances.
Breadth-first search is a useful technique for finding the diameter: we simply run breadth-
first search from each vertex. An interesting application of the diameter appears in the
small-world phenomenon [126, 155, 205], which contends that a certain special class of
sparse graphs have low diameter.
2.2.2 Depth-first search
— Randall Munroe, xkcd, http://xkcd.com/761/
A depth-first search (DFS) is a graph traversal strategy similar to breadth-first search.
Both BFS and DFS differ in how they explore each vertex. Whereas BFS explores
the neighborhood of a vertex v before moving on to explore the neighborhoods of the
neighbors, DFS explores as deep as possible a path starting at v. One can think of BFS
as exploring the immediate surrounding, while DFS prefers to see what is on the other
side of the hill. In the 19th century, Lucas [146] and Tarry [191] investigated DFS as
a strategy for traversing mazes. Fundamental properties of DFS were discovered in the
early 1970s by Hopcroft and Tarjan [106, 190].
To get an intuitive appreciation for DFS, suppose we have an 8 × 8 chessboard in
front of us. We place a single knight piece on a fixed square of the board, as shown in
Figure 2.5(a). Our objective is to find a sequence of knight moves that visits each and
every square exactly once, while obeying the rules of chess that govern the movement
of the knight piece. Such a sequence of moves, if one exists, is called a knight’s tour .
How do we find such a tour? We could make one knight move after another, recording
each move to ensure that we do not step on a square that is already visited, until we
could not make any more moves. Acknowledging defeat when encountering a dead end,
2.2. Graph searching 65
0Z0Z0Z0Z 0Z0Z0Z0Z
Z0Z0Z0Z0 Z0Z0Z0Z0
0Z0Z0Z0Z 0Z0Z0Z0Z
Z0Z0m0Z0 Z0Z0Z0Z0
0Z0Z0Z0Z 0Z0Z0Z0Z
Z0Z0Z0Z0 Z0Z0Z0Z0
0Z0Z0Z0Z 0Z0Z0Z0Z
Z0Z0Z0Z0 Z0Z0Z0Z0
(a) The knight’s initial position. (b) A knight’s tour.
(c) Graph representation of tour.
Figure 2.5: The knight’s tour from a given starting position.
66 Chapter 2. Graph algorithms
it might make sense to backtrack a few moves and try again, hoping we would not get
stuck. If we fail again, we try backtracking a few more moves and traverse yet another
path, hoping to make further progress. Repeat this strategy until a tour is found or until
we have exhausted all possible moves. The above strategy for finding a knight’s tour
is an example of depth-first search, sometimes called backtracking. Figure 2.5(b) shows
a knight’s tour with the starting position as shown in Figure 2.5(a); and Figure 2.5(c)
is a graph representation of this tour. The black-filled nodes indicate the endpoints
of the tour. A more interesting question is: What is the number of knight’s tours
on an 8 × 8 chessboard? Loebbing and Wegener [144] announced in 1996 that this
number is 33,439,123,484,294. The answer was later corrected by McKay [150] to be
13,267,364,410,532. See [73] for a discussion of the knight’s tour and its relationship to
mathematics.
Algorithm 2.2: A general depth-first search template.
Input: A directed or undirected graph G = (V, E) of order n > 0. A vertex s
from which to start the search. The vertices are numbered from 1 to
n = |V |, i.e. V = {1, 2, . . . , n}.
Output: A list D of distances of all vertices from s. A tree T rooted at s.
1 S ← [s] /* stack of nodes to visit */
2 D ← [∞, ∞, . . . , ∞] /* n copies of ∞ */
3 D[s] ← 0
4 T ← []
5 while length(S) > 0 do
6 v ← pop(S)
7 for each w ∈ adj(v) do
8 if D[w] = ∞ then
9 D[w] ← D[v] + 1
10 push(S, w)
11 append(T, vw)
12 return (D, T )
Algorithm 2.2 formalizes the above description of depth-first search. The tree re-
sulting from applying DFS on a graph is called a depth-first search tree. The general
structure of this algorithm bears close resemblance to Algorithm 2.1. A significant dif-
ference is that instead of using a queue to structure and organize vertices to be visited,
DFS uses another special type of list called a stack . To understand how elements of a
stack are organized, we use the analogy of a stack of cards. A new card is added to
the stack by placing it on top of the stack. Any time we want to remove a card, we
are only allowed to remove the top-most card that is on the top of the stack. A list
L = [a1 , a2 , . . . , ak ] of k elements is a stack when we impose the same rules for element
insertion and removal. The top and bottom of the stack are L[k] and L[1], respectively.
The operation of removing the top element of the stack is referred to as popping the
element off the stack. Inserting an element into the stack is called pushing the element
onto the stack. In other words, a stack implements a last-in first-out (LIFO) protocol
for element insertion and removal, in contrast to the FIFO policy of a queue. We also
use the term length to refer to the number of elements in the stack.
The depth-first search Algorithm 2.2 can be analyzed similar to how we analyzed
2.2. Graph searching 67
1 2 1 2
3 3
7 7
4 4
6 5 6 5
(a) Original undirected graph. (b) First iteration of while loop.
1 2 1 2
3 3
7 7
4 4
6 5 6 5
(c) Second iteration of while loop. (d) Third iteration of while loop.
1
2 5
3 6
7 4
(e) Final DFS tree.
Figure 2.6: Depth-first search tree for an undirected graph.
68 Chapter 2. Graph algorithms
1 2 1 2
3 3
7 7
4 4
6 5 6 5
(a) Original digraph. (b) First iteration of while loop.
1 2 1 2
3 3
7 7
4 4
6 5 6 5
(c) Second iteration of while loop. (d) Third iteration of while loop.
4
1 2
3 6
3
7
5 7
4
6 5
1 2
(e) Fourth iteration of while loop. (f) Final DFS tree.
Figure 2.7: Depth-first search tree for a digraph.
2.2. Graph searching 69
Algorithm 2.3. Just as BFS is applicable to both directed and undirected graphs, we
can also have undirected graphs and digraphs as input to DFS. For the case of an
undirected graph, line 7 of Algorithm 2.2 considers all vertices adjacent to the current
vertex v. In case the input graph is directed, we replace “w ∈ adj(v)” on line 7 with
“w ∈ oadj(v)” to signify that we only want to consider the out-neighbors of v. If any
neighbors (respectively, out-neighbors) of v are labelled as ∞, we know that we have
not explored any paths starting from any of those vertices. So we label each of those
unexplored vertices with a positive integer and push them onto the stack S, where
they will wait for later processing. We also record the paths leading from v to each of
those unvisited neighbors, i.e. the edges vw for each vertex w ∈ adj(v) (respectively,
w ∈ oadj(v)) are appended to the list T . The test on line 8 ensures that we do not
push onto S any vertices on the path that lead to v. When we resume another round of
the while loop that starts on line 5, the previous vertex v have been popped off S and
the neighbors (respectively, out-neighbors) of v have been pushed onto S. For example,
in step 2 of Figure 2.6, vertex 5 is considered in DFS (in contrast to the vertex 2 in
step 2 of the BFS in the graph in Figure 2.3) because DFS is organized by the LIFO
protocol (in contrast to the FIFO protocol of BFS). To explore a path starting at v,
we choose any unexplored neighbors of v by popping an element off S and repeat the
for loop starting on line 7. Repeat the DFS algorithm as often as required in order to
traverse all vertices of the input graph. The output of DFS consists of two lists D and
T : T is a tree rooted at the starting vertex s; and each D[i] counts the length of the s-vi
path in T . Figures 2.6 and 2.7 show the DFS trees resulting from running Algorithm 2.2
on an undirected graph and a digraph, respectively. The worst-case time complexity of
DFS can be analyzed using an argument similar to that in Theorem 2.2. Arguing along
the same lines as in the proof of Theorem 2.3, we can also show that the list D returned
by DFS contains lengths of any shortest paths in the tree T from the starting vertex s
to any other vertex in T (but not necessarily for shortest paths in the original graph G).
0
5
1 4
6 9
7 8
2 3
Figure 2.8: The Petersen graph.
Example 2.4. In 1898, Julius Petersen published [169] a graph that now bears his name:
the Petersen graph shown in Figure 2.8. Compare the search trees resulting from running
breadth- and depth-first searches on the Petersen graph with starting vertex 0.
Solution. The Petersen graph in Figure 2.8 can be constructed and searched as follows.
70 Chapter 2. Graph algorithms
sage : g = graphs . PetersenGraph (); g
Petersen graph : Graph on 10 vertices
sage : list ( g . breadth_first_search (0))
[0 , 1 , 4 , 5 , 2 , 6 , 3 , 9 , 7 , 8]
sage : list ( g . depth_first_search (0))
[0 , 5 , 8 , 6 , 9 , 7 , 2 , 3 , 4 , 1]
From the above Sage session, we see that starting from vertex 0 breadth-first search
yields the edge list
[01, 04, 05, 12, 16, 43, 49, 57, 58]
and depth-first search produces the corresponding edge list
[05, 58, 86, 69, 97, 72, 23, 34, 01].
Our results are illustrated in Figure 2.9.
0 0
5 5
1 4 1 4
6 9 6 9
7 8 7 8
2 3 2 3
(a) Breadth-first search. (b) Depth-first search.
Figure 2.9: Traversing the Petersen graph starting from vertex 0.
2.2.3 Connectivity of a graph
Both BFS and DFS can be used to determine if an undirected graph is connected. Let
G = (V, E) be an undirected graph of order n > 0 and let s be an arbitrary vertex
of G. We initialize a counter c ← 1 to mean that we are starting our exploration at
s, hence we have already visited one vertex, i.e. s. We apply either BFS or DFS,
treating G and s as input to any of these algorithms. Each time we visit a vertex that
was previously unvisited, we increment the counter c. At the end of the algorithm, we
compare c with n. If c = n, we know that we have visited all vertices of G and conclude
that G is connected. Otherwise, we conclude that G is disconnected. This procedure is
summarized in Algorithm 2.3.
Note that Algorithm 2.3 uses the BFS template of Algorithm 2.1, with some minor
changes. Instead of initializing the list D with n = |V | copies of ∞, we use n copies of
0. Each time we have visited a vertex w, we make the assignment D[w] ← 1, instead
of incrementing the value D[v] of w’s parent vertex and Passign that value to D[w]. At
the end of the while loop, we have the equality c = d∈D d. The value of this sum
could be used in the test starting from line 12. However, the value of the counter c
is incremented immediately after we have visited an unvisited vertex. An advantage is
2.3. Weights and distances 71
Algorithm 2.3: Determining whether an undirected graph is connected.
Input: An undirected graph G = (V, E) of order n > 0. A vertex s from which to
start the search. The vertices are numbered from 1 to n = |V |,
i.e. V = {1, 2, . . . , n}.
Output: True if G is connected; False otherwise.
1 Q ← [s] /* queue of nodes to visit */
2 D ← [0, 0, . . . , 0] /* n copies of 0 */
3 D[s] ← 1
4 c←1
5 while length(Q) > 0 do
6 v ← dequeue(Q)
7 for each w ∈ adj(v) do
8 if D[w] = 0 then
9 D[w] ← 1
10 c←c+1
11 enqueue(Q, w)
12 if c = |V | then
13 return True
14 return False
that we do not need to perform a separate summation outside of the while loop. To
use the DFS template for determining graph connectivity, we simply replace the queue
implementation in Algorithm 2.3 with a stack implementation (see problem 2.20).
2.3 Weights and distances
In Chapter 1, we briefly mentioned some applications of weighted graphs, but we did
not define the concept of weighted graphs. A graph is said to be weighted when we
assign a numeric label or weight to each of its edges. Depending on the application,
we can let the vertices represent physical locations and interpret the weight of an edge
as the distance separating two adjacent vertices. There might be a cost involved in
traveling from a vertex to one of its neighbors, in which case the weight assigned to the
corresponding edge can represent such a cost. The concept of weighted digraphs can be
similarly defined. When no explicit weights are assigned to the edges of an undirected
graph or digraph, it is usually convenient to consider each edge as having a weight of
one or unit weight.
Based on the concept of weighted graphs, we now define what it means for a path
to be a shortest path. Let G = (V, E) be a (di)graph with nonnegative edge weights
w(e) ∈ R for each edge e ∈ E. The length or distance d(P ) of a u-v path P from u ∈ V
to v ∈ V is the sum of the edge weights for edges in P . Denote by d(u, v) the smallest
value of d(P ) for all paths P from u to v. When we regard edge weights as physical
distances, a u-v path that realizes d(u, v) is sometimes called a shortest path from u to v.
The above definitions of distance and shortest path also apply to graphs with negative
edge weights. Unless otherwise specified, where the weight of an edge is not explicitly
given, we usually consider the edge to have unit weight.
The distance function d on a graph with nonnegative edge weights is known as a
72 Chapter 2. Graph algorithms
metric function. Intuitively, the distance between two physical locations is greater than
zero. When these two locations coincide, i.e. they are one and the same location, the
distance separating them is zero. Regardless of whether we are measuring the distance
from location a to b or from b to a, we would obtain the same distance. Imagine now
a third location c. The distance from a to b plus the distance from b to c is greater
than or equal to the distance from a to c. The latter principle is known as the triangle
inequality. In summary, given three vertices u, v, w in a graph G, the distance function
d on G satisfies the following property.
Lemma 2.5. Path distance as metric function. Let G = (V, E) be a graph with
weight function w : E → R. Define a distance function d : V × V → R given by
(
∞, if there are no paths from u to v,
d(u, v) =
min{w(W ) | W is a u-v walk}, otherwise.
Then d is a metric on V if it satisfies the following properties:
1. Nonnegativity: d(u, v) ≥ 0 with d(u, v) = 0 if and only if u = v.
2. Symmetry: d(u, v) = d(v, u).
3. Triangle inequality: d(u, v) + d(v, w) ≥ d(u, w).
The pair (V, d) is called a metric space, where the word “metric” refers to the distance
function d. Any graphs we consider are assumed to have finite sets of vertices. For this
reason, (V, d) is also known as a finite metric space. The distance matrix D = [d(vi , vj )]
of a connected graph is the distance matrix of its finite metric space. The topic of metric
space is covered in further details in topology texts such as Runde [176] and Shirali and
Vasudeva [181]. See Buckley and Harary [47] for an in-depth coverage of the distance
concept in graph theory.
Many different algorithms exist for computing a shortest path in a weighted graph.
Some only work if the graph has no negative weight cycles. Some assume that there is a
single start or source vertex. Some compute the shortest paths from any vertex to any
other and also detect if the graph has a negative weight cycle. No matter what algorithm
is used for the special case of nonnegative weights, the length of the shortest path can
neither equal nor exceed the order of the graph.
Lemma 2.6. Fix a vertex v in a connected graph G = (V, E) of order n = |V |. If there
are no negative weight cycles in G, then there exists a shortest path from v to any other
vertex w ∈ V that uses at most n − 1 edges.
Proof. Suppose that G contains no negative weight cycles. Observe that at most n − 1
edges are required to construct a path from v to any vertex w (Proposition 1.13). Let P
denote such a path:
P : v0 = v, v1 , v2 , . . . , vk = w.
Since G has no negative weight cycles, the weight of P is no less than the weight of
P 0 , where P 0 is the same as P except that all cycles have been removed. Thus, we can
remove all cycles from P and obtain a v-w path P 0 of lower weight. Since the final path
is acyclic, it must have no more than n − 1 edges.
2.3. Weights and distances 73
Algorithm 2.4: A template for shortest path algorithms.
Input: A weighted graph or digraph G = (V, E), where the vertices are numbered
as V = {1, 2, . . . , n}. A starting vertex s.
Output: A list D of distances from s to all other vertices. A list P of parent
vertices such that P [v] is the parent of v.
1 D ← [∞, ∞, . . . , ∞] /* n copies of ∞ */
2 C ← list of candidate vertices to visit
3 while length(C) > 0 do
4 select v ∈ C
5 C ← remove(C, v)
6 for each u ∈ adj(v) do
7 if D[u] > D[v] + w(vu) then
8 D[u] ← D[v] + w(vu)
9 P [u] ← v
10 if u ∈/ C then
11 add u to C
12 return (D, P )
Having defined weights and distances, we are now ready to discuss shortest path
algorithms for weighted graphs. The breadth-first search Algorithm 2.1 can be applied
where each edge has unit weight. Moving on to the general case of graphs with positive
edge weights, algorithms for determining shortest paths in such graphs can be classified
as weight-setting or weight-correcting [87]. A weight-setting method traverses a graph
and assigns weights that, once assigned, remain unchanged for the duration of the al-
gorithm. Weight-setting algorithms cannot deal with negative weights. On the other
hand, a weight-correcting method is able to change the value of a weight many times
while traversing a graph. In contrast to a weight-setting algorithm, a weight-correcting
algorithm is able to deal with negative weights, provided that the weight sum of any
cycle is nonnegative. The term negative cycle refers to the weight sum s of a cycle such
that s < 0. Some algorithms halt upon detecting a negative cycle; examples of such
algorithms include the Bellman-Ford and Johnson’s algorithms.
Algorithm 2.4 is a general template for many shortest path algorithms. With a tweak
here and there, one could modify it to suit the problem at hand. Note that w(vu) is the
weight of the edge vu. If the input graph is undirected, line 6 considers all the neighbors
of v. For digraphs, we are interested in out-neighbors of v and accordingly we replace
“u ∈ adj(v)” in line 6 with “u ∈ oadj(v)”. The general flow of Algorithm 2.4 follows the
same pattern as depth-first and breadth-first searches.
74 Chapter 2. Graph algorithms
2.4 Dijkstra’s algorithm
— Randall Munroe, xkcd, http://xkcd.com/342/
Dijkstra’s algorithm [65], discovered by E. W. Dijkstra in 1959, is a graph search algo-
rithm that solves the single-source shortest path problem for a graph with nonnegative
edge weights. The algorithm is a generalization of breadth-first search. Imagine that
the vertices of a weighted graph represent cities and edge weights represent distances
between pairs of cities connected by a direct road. Dijkstra’s algorithm can be used to
find a shortest route from a fixed city to any other city.
Let G = (V, E) be a (di)graph with nonnegative edge weights. Fix a start or source
vertex s ∈ V . Dijkstra’s Algorithm 2.5 performs a number of steps, basically one step
for each vertex in V . First, we initialize a list D with n copies of ∞ and then assign 0 to
D[s]. The purpose of the symbol ∞ is to denote the largest possible value. The list D is
to store the distances of all shortest paths from s to any other vertices in G, where we
take the distance of s to itself to be zero. The list P of parent vertices is initially empty
and the queue Q is initialized to all vertices in G. We now consider each vertex in Q,
removing any vertex after we have visited it. The while loop starting on line 5 runs until
we have visited all vertices. Line 6 chooses which vertex to visit, preferring a vertex v
whose distance value D[v] from s is minimal. After we have determined such a vertex v,
we remove it from the queue Q to signify that we have visited v. The for loop starting
on line 8 adjusts the distance values of each neighbor u of v such that u is also in Q. If
G is directed, we only consider out-neighbors of v that are also in Q. The conditional
starting on line 9 is where the adjustment takes place. The expression D[v] + w(vu)
sums the distance from s to v and the distance from v to u. If this total sum is less than
the distance D[u] from s to u, we assign this lesser distance to D[u] and let v be the
parent vertex of u. In this way, we are choosing a neighbor vertex that results in minimal
distance from s. Each pass through the while loop decreases the number of elements in
Q by one without adding any elements to Q. Eventually, we would exit the while loop
and the algorithm returns the lists D and P .
Example 2.7. Apply Dijkstra’s algorithm to the graph in Figure 2.10(a), with starting
vertex v1 .
Solution. Dijkstra’s Algorithm 2.5 applied to the graph in Figure 2.10(a) yields the
sequence of intermediary graphs shown in Figure 2.10, culminating in the final shortest
paths graph of Figure 2.10(f) and Table 2.4. For any column vi in the table, each 2-tuple
represents the distance and parent vertex of vi . As we move along the graph, processing
vertices according to Dijkstra’s algorithm, the distance and parent vertex of a column
2.4. Dijkstra’s algorithm 75
Algorithm 2.5: A general template for Dijkstra’s algorithm.
Input: An undirected or directed graph G = (V, E) that is weighted and has no
self-loops. The order of G is n > 0. A vertex s ∈ V from which to start
the search. Vertices are numbered from 1 to n, i.e. V = {1, 2, . . . , n}.
Output: A list D of distances such that D[v] is the distance of a shortest path
from s to v. A list P of vertex parents such that P [v] is the parent of v,
i.e. v is adjacent from P [v].
1 D ← [∞, ∞, . . . , ∞] /* n copies of ∞ */
2 D[s] ← 0
3 P ← []
4 Q←V /* list of nodes to visit */
5 while length(Q) > 0 do
6 find v ∈ Q such that D[v] is minimal
7 Q ← remove(Q, v)
8 for each u ∈ adj(v) ∩ Q do
9 if D[u] > D[v] + w(vu) then
10 D[u] ← D[v] + w(vu)
11 P [u] ← v
12 return (D, P )
2 2
v2 v4 v2 v4
10 1 7 10 1 7
8 8
4 9 4 9
v1 v3 v5 v1 v3 v5
3 2 3 2
(a) Original digraph. (b) First iteration of while loop.
2 2
v2 v4 v2 v4
10 1 7 10 1 7
8 8
4 9 4 9
v1 v3 v5 v1 v3 v5
3 2 3 2
(c) Second iteration of while loop. (d) Third iteration of while loop.
2 2
v2 v4 v2 v4
10 1 7
8
4 9 4
v1 v3 v5 v1 v3 v5
3 2 3 2
(e) Fourth iteration of while loop. (f) Final shortest paths graph.
Figure 2.10: Searching a weighted digraph using Dijkstra’s algorithm.
76 Chapter 2. Graph algorithms
v1 v2 v3 v4 v5
(0, −) (∞, −) (∞, −) (∞, −) (∞, −)
(10, v1 ) (3, v1 ) (11, v3 ) (5, v3 )
(7, v3 ) (9, v2 )
Table 2.4: Stepping through Dijkstra’s algorithm.
are updated. The underlined 2-tuple represents the final distance and parent vertex
produced by Dijkstra’s algorithm. From Table 2.4, we have the following shortest paths
and distances:
v1 -v2 : v1 , v3 , v2 d(v1 , v2 ) = 7
v1 -v3 : v1 , v3 d(v1 , v3 ) = 3
v1 -v4 : v1 , v3 , v2 , v4 d(v1 , v4 ) = 9
v1 -v5 : v1 , v3 , v5 d(v1 , v5 ) = 5
Intermediary vertices for a u-v path are obtained by starting from v and work backward
using the parent of v, then the parent of the parent, and so on.
Dijkstra’s algorithm is an example of a greedy algorithm. Whenever it tries to find the
next vertex, it chooses only that vertex that minimizes the total weight so far. Greedy
algorithms may not produce the best possible result. However, as the following theorem
shows, Dijkstra’s algorithm does indeed produce shortest paths.
Theorem 2.8. Correctness of Algorithm 2.5. Let G = (V, E) be a weighted
(di)graph with a nonnegative weight function w. When Dijkstra’s algorithm is applied to
G with source vertex s ∈ V , the algorithm terminates with D[u] = d(s, u) for all u ∈ V .
Furthermore, if D[v] 6= ∞ and v 6= s, then s = u1 , u2 , . . . , uk = v is a shortest s-v path
such that ui−1 = P [ui ] for i = 2, 3, . . . , k.
Proof. If G is disconnected, then any v ∈ V that cannot be reached from s has distance
D[v] = ∞ upon algorithm termination. Hence, it suffices to consider the case where G
is connected. Let V = {s = v1 , v2 , . . . , vn } and use induction on i to show that after
visiting vi we have
D[v] = d(s, v) for all v ∈ Vi = {v1 , v2 , . . . , vi }. (2.6)
For i = 1, equality holds. Assume for induction that (2.6) holds for some 1 ≤ i ≤ n − 1,
so that now our task is to show that (2.6) holds for i + 1. To verify D[vi+1 ] = d(s, vi+1 ),
note that by our inductive hypothesis,
D[vi+1 ] = min {d(s, v) + w(vu) | v ∈ Vi and u ∈ adj(v) ∩ (Q\Vi )}
and respectively
D[vi+1 ] = min {d(s, v) + w(vu) | v ∈ Vi and u ∈ oadj(v) ∩ (Q\Vi )}
if G is directed. Therefore, D[vi+1 ] = d(s, vi+1 ).
Let v ∈ V such that D[v] 6= ∞ and v 6= s. We now construct an s-v path. When
Algorithm 2.5 terminates, we have D[v] = D[v1 ] + w(v1 v), where P [v] = v1 and d(s, v) =
d(s, v1 ) + w(v1 v). This means that v1 is the second-to-last vertex in a shortest s-v path.
Repeated application of this process using the parent list P , we eventually produce a
shortest s-v path s = vm , vm−1 , . . . , v1 , v, where P [vi ] = vi+1 for i = 1, 2, . . . , m − 1.
2.5. Bellman-Ford algorithm 77
To analyze the worst case time complexity of Algorithm 2.5, note that initializing D
takes O(n + 1) and initializing Q takes O(n), for a total of O(n) devoted to initialization.
Each extraction of a vertex v with minimal D[v] requires O(n) since we search through
the entire list Q to determine the minimum value, for a total of O(n2 ). Each insertion
into D requires constant time and the same holds for insertion into P . Thus, insertion
into D and P takes O(|E| + |E|) = O(|E|), which require at most O(n) time. In the
worst case, Dijkstra’s Algorithm 2.5 has running time O(n2 + n) = O(n2 ).
Can we improve the run time of Dijkstra’s algorithm? The time complexity of Dijk-
stra’s algorithm depends on its implementation. With a simple list implementation as
presented in Algorithm 2.5, we have a worst case time complexity of O(n2 ), where n is
the order of the graph under consideration. Let m be the size of the graph. Table 2.5
presents time complexities of Dijkstra’s algorithm for various implementations. Out of
all the four implementations in this table, the heap implementations are much more
efficient than the list implementation presented in Algorithm 2.5. A heap is a type of
tree, a topic which will be covered in Chapter 3. Of all the heap implementations in
Table 2.5, the Fibonacci heap implementation [86] yields the best runtime. Chapter 4
discusses how to use trees for efficient implementations of priority queues via heaps.
Implementation Time complexity
list O(n2 )
binary heap O (n + m) ln n
k-ary heap O (kn + m) ln n
ln k
Fibonacci heap O(n ln n + m)
Table 2.5: Implementation specific worst-case time complexity of Dijkstra’s algorithm.
2.5 Bellman-Ford algorithm
— Randall Munroe, xkcd, http://xkcd.com/69/
A disadvantage of Dijkstra’s Algorithm 2.5 is that it cannot handle graphs with negative
edge weights. The Bellman-Ford algorithm computes single-source shortest paths in
a weighted graph or digraph, where some of the edge weights may be negative. This
algorithm is a modification of the one published in 1957 by Richard E. Bellman [22] and
that by Lester Randolph Ford, Jr. [83] in 1956. Shimbel [180] independently discovered
the same method in 1955, and Moore [157] in 1959. In contrast to the “greedy” approach
that Dijkstra’s algorithm takes, i.e. searching for the “cheapest” path, the Bellman-Ford
algorithm searches over all edges and keeps track of the shortest one found as it searches.
The Bellman-Ford Algorithm 2.6 runs in time O(mn), where m and n are the size
and order of an input graph, respectively. To see this, note that the initialization on
78 Chapter 2. Graph algorithms
Algorithm 2.6: The Bellman-Ford algorithm.
Input: An undirected or directed graph G = (V, E) that is weighted and has no
self-loops. Negative edge weights are allowed. The order of G is n > 0. A
vertex s ∈ V from which to start the search. Vertices are numbered from
1 to n, i.e. V = {1, 2, . . . , n}.
Output: A list D of distances such that D[v] is the distance of a shortest path
from s to v. A list P of vertex parents such that P [v] is the parent of v,
i.e. v is adjacent from P [v]. If G has negative-weight cycles, then return
False. Otherwise, return D and P .
1 D ← [∞, ∞, . . . , ∞] /* n copies of ∞ */
2 D[s] ← 0
3 P ← []
4 for i ← 1, 2, . . . , n − 1 do
5 for each edge uv ∈ E do
6 if D[v] > D[u] + w(uv) then
7 D[v] ← D[u] + w(uv)
8 P [v] ← u
9 for each edge uv ∈ E do
10 if D[v] > D[u] + w(uv) then
11 return False
12 return (D, P )
lines 1 to 3 takes O(n). Each of the n − 1 rounds of the for loop starting on line 4 takes
O(m), for a total of O(mn) time. Finally, the for loop starting on line 9 takes O(m).
The loop starting on line 4 performs at most n − 1 updates of the distance D[v] of
each head of an edge. Many graphs have sizes that are less then n − 1, resulting in
a number of redundant rounds of updates. To avoid such redundancy, we could add
an extra check in the outer loop spanning lines 4 to 8 to immediately terminate that
outer loop after any round that did not result in an update of any D[v]. Algorithm 2.7
presents a modification of the Bellman-Ford Algorithm 2.6 that avoids redundant rounds
of updates.
2.6 Floyd-Roy-Warshall algorithm
The shortest distance between two points is not a very interesting journey.
— R. Goldberg
Let D be a weighted digraph of order n and size m. Dijkstra’s Algorithm 2.5 and
the Bellman-Ford Algorithm 2.6 can be used to determine shortest paths from a single
source vertex to all other vertices of D. To determine a shortest path between each pair
of distinct vertices in D, we repeatedly apply either of these algorithms to each vertex
of D. Such repeated application of Dijkstra’s and the Bellman-Ford algorithms results
in algorithms that run in time O(n3 ) and O(n2 m), respectively.
The Floyd-Roy-Warshall algorithm (FRW), or the Floyd-Warshall algorithm, is an
algorithm for finding shortest paths in a weighted, directed graph. Like the Bellman-
Ford algorithm, it allows for negative edge weights and detects a negative weight cycle
if one exists. Assuming that there are no negative weight cycles, a single execution of
2.6. Floyd-Roy-Warshall algorithm 79
Algorithm 2.7: The Bellman-Ford algorithm with checks for redundant updates.
Input: An undirected or directed graph G = (V, E) that is weighted and has no
self-loops. Negative edge weights are allowed. The order of G is n > 0. A
vertex s ∈ V from which to start the search. Vertices are numbered from
1 to n, i.e. V = {1, 2, . . . , n}.
Output: A list D of distances such that D[v] is the distance of a shortest path
from s to v. A list P of vertex parents such that P [v] is the parent of v,
i.e. v is adjacent from P [v]. If G has negative-weight cycles, then return
False. Otherwise, return D and P .
1 D ← [∞, ∞, . . . , ∞] /* n copies of ∞ */
2 D[s] ← 0
3 P ← []
4 for i ← 1, 2, . . . , n − 1 do
5 updated ← False
6 for each edge uv ∈ E do
7 if D[v] > D[u] + w(uv) then
8 D[v] ← D[u] + w(uv)
9 P [v] ← u
10 updated ← True
11 if updated = False then
12 exit the loop
13 for each edge uv ∈ E do
14 if D[v] > D[u] + w(uv) then
15 return False
16 return (D, P )
80 Chapter 2. Graph algorithms
the FRW algorithm will find the shortest paths between all pairs of vertices. It was
discovered independently by Bernard Roy [175] in 1959, Robert Floyd [82] in 1962, and
by Stephen Warshall [201] in 1962.
In some sense, the FRW algorithm is an example of dynamic programming, which
allows one to break the computation into simpler steps using some sort of recursive
procedure. The rough idea is as follows. Temporarily label the vertices of a weighted
digraph G as V = {1, 2, . . . , n} with n = |V (G)|. Let W = [w(i, j)] be the weight matrix
of G where
w(ij), if ij ∈ E(G),
w(i, j) = 0, if i = j, (2.7)
∞, otherwise.
Let Pk (i, j) be a shortest path from i to j such that its intermediate vertices are in
{1, 2, . . . , k}. Let Dk (i, j) be the weight (or distance) of Pk (i, j). If no shortest i-j
paths exist, define Pk (i, j) = ∞ and Dk (i, j) = ∞ for all k ∈ {1, 2, . . . , n}. If k = 0,
then P0 (i, j) : i, j since no intermediate vertices are allowed in the path and hence
D0 (i, j) = w(i, j). In other words, if i and j are adjacent, a shortest i-j path is the
edge ij itself and the weight of this path is simply the weight of ij. Now consider
Pk (i, j) for k > 0. Either Pk (i, j) passes through k or it does not. If k is not on the
path Pk (i, j), then the intermediate vertices of Pk (i, j) are in {1, 2, . . . , k − 1}, as are
the vertices of Pk−1 (i, j). In case Pk (i, j) contains the vertex k, then Pk (i, j) traverses
k exactly once by the definition of path. The i-k subpath in Pk (i, j) is a shortest i-k
path whose intermediate vertices are drawn from {1, 2, . . . , k − 1}, which is also the set
of intermediate vertices for the k-j subpath in Pk (i, j). That is, to obtain Pk (i, j), we
take the union of the paths Pk−1 (i, k) and Pk−1 (k, j). We compute the weight Dk (i, j)
of Pk (i, j) using the expression
(
w(i, j), if k = 0,
Dk (i, j) = (2.8)
min{Dk−1 (i, j), Dk−1 (i, k) + Dk−1 (k, j)}, if k > 0.
The key to the Floyd-Roy-Warshall algorithm lies in exploiting expression (2.8). If
n = |V |, then this is a O(n3 ) time algorithm. For comparison, the Bellman-Ford al-
gorithm has complexity O(|V | · |E|), which is O(n3 ) time for dense graphs. However,
Bellman-Ford only yields the shortest paths emanating from a single vertex. To achieve
comparable output, we would need to iterate Bellman-Ford over all vertices, which would
be an O(n4 ) time algorithm for dense graphs. Except possibly for sparse graphs, Floyd-
Roy-Warshall is better than an iterated implementation of Bellman-Ford. Note that
Pk (i, k) = Pk−1 (i, k) and Pk (k, i) = Pk−1 (k, i), consequently Dk (i, k) = Dk−1 (i, k) and
Dk (k, i) = Dk−1 (k, i). This observation allows us to replace Pk (i, j) with P (i, j) for
k = 1, 2, . . . , n. The final results of P (i, j) and D(i, k) are the same as Pn (i, j) and
Dn (i, j), respectively. Algorithm 2.8 summarizes the above discussion into an algorith-
mic presentation.
Like the Bellman-Ford algorithm, the Floyd-Roy-Warshall algorithm can also detect
the presence of negative weight cycles. If G is a weighted digraph without self-loops,
by (2.7) we have D(i, i) = 0 for i = 1, 2, . . . , n. Any path p starting and ending at i
could only improve upon the initial weight of 0 if the weight sum of p is less than zero, i.e.
a negative weight cycle. Upon termination of Algorithm 2.8, if D(i, i) < 0, we conclude
that there is a path starting and ending at i whose weight sum is negative.
2.6. Floyd-Roy-Warshall algorithm 81
Algorithm 2.8: The Floyd-Roy-Warshall algorithm for all-pairs shortest paths.
Input: A weighted digraph G = (V, E) that has no self-loops. Negative edge
weights are allowed. The order of G is n > 0. Vertices are numbered from
1 to n, i.e. V = {1, 2, . . . , n}. The weight matrix W = [w(i, j)] of G as
defined in (2.7).
Output: A matrix P = [aij ] of shortest paths in G. A matrix D = [aij ] of
distances where D[i, j] is the weight (or distance) of a shortest i-j path
in G.
1 n ← |V |
2 P [aij ] ← an n × n zero matrix
3 D[aij ] ← W [w(i, j)]
4 for k ← 1, 2, . . . , n do
5 for i ← 1, 2, . . . , n do
6 for j ← 1, 2, . . . , n do
7 if D[i, j] > D[i, k] + D[k, j] then
8 P [i, j] ← k
9 D[i, j] ← D[i, k] + D[k, j]
10 return (P, D)
Here is an implementation in Sage.
def fl oy d_roy_warshall ( A ):
"""
Shortest paths
INPUT :
- A -- weighted adjacency matrix
OUTPUT :
- dist -- a matrix of distances of shortest paths .
- paths -- a matrix of shortest paths .
"""
G = Graph (A , format = " w e i g h t e d _ a d j a c e n c y _ m a tr i x " )
V = G . vertices ()
E = [( e [0] , e [1]) for e in G . edges ()]
n = len ( V )
dist = [[0]* n for i in range ( n )]
paths = [[ -1]* n for i in range ( n )]
# initialization step
for i in range ( n ):
for j in range ( n ):
if (i , j ) in E :
paths [ i ][ j ] = j
if i == j :
dist [ i ][ j ] = 0
elif A [ i ][ j ] < >0:
dist [ i ][ j ] = A [ i ][ j ]
else :
dist [ i ][ j ] = infinity
# iteratively finding the shortest path
for j in range ( n ):
for i in range ( n ):
if i <> j :
for k in range ( n ):
if k <> j :
if dist [ i ][ k ] > dist [ i ][ j ]+ dist [ j ][ k ]:
paths [ i ][ k ] = V [ j ]
dist [ i ][ k ] = min ( dist [ i ][ k ] , dist [ i ][ j ] + dist [ j ][ k ])
for i in range ( n ):
if dist [ i ][ i ] < 0:
raise ValueError , " A negative edge weight cycle exists . "
82 Chapter 2. Graph algorithms
return dist , matrix ( paths )
Here are some examples.
sage : A = matrix ([[0 ,1 ,2 ,3] , [0 ,0 ,2 ,1] , [ -5 ,0 ,0 ,3] , [1 ,0 ,1 ,0]]); A
sage : f lo yd_roy_warshall ( A )
Traceback ( click to the left of this block for traceback )
...
ValueError : A negative edge weight cycle exists .
The plot of this weighted digraph with four vertices appears in Figure 2.11.
1
3 3 2
2
3 1 2
1 5
0 1
1
Figure 2.11: Demonstrating the Floyd-Roy-Warshall algorithm.
sage : A = matrix ([[0 ,1 ,2 ,3] , [0 ,0 ,2 ,1] , [ -1/2 ,0 ,0 ,3] , [1 ,0 ,1 ,0]]); A
sage : f lo yd_roy_warshall ( A )
([[0 , 1 , 2 , 2] , [3/2 , 0 , 2 , 1] , [ -1/2 , 1/2 , 0 , 3/2] , [1/2 , 3/2 , 1 , 0]] ,
[ -1 1 2 1]
[ 2 -1 2 3]
[ -1 0 -1 1]
[ 2 2 -1 -1])
The plot of this weighted digraph with four vertices appears in Figure 2.12.
Example 2.9. Section 1.6 briefly presented the concept of molecular graphs in chem-
istry. The Wiener number of a molecular graph was first published in 1947 by Harold
Wiener [208] who used it in chemistry to study properties of alkanes. Other applica-
tions [98] of the Wiener number to chemistry are now known. If G = (V, E) is a
connected graph with vertex set V = {v1 , v2 , . . . , vn }, then the Wiener number W of G is
defined by
X
W (G) = d(vi , vj ) (2.9)
i<j
where d(vi , vj ) is the distance from vi to vj . What is the Wiener number of the molecular
graph in Figure 2.13?
Solution. Consider the molecular graph in Figure 2.13 as directed with unit weight.
To compute the Wiener number of this graph, use the Floyd-Roy-Warshall algorithm to
obtain a distance matrix D = [di,j ], where di,j is the distance from vi to vj , and apply the
definition (2.9). The distance matrix resulting from the Floyd-Roy-Warshall algorithm
2.6. Floyd-Roy-Warshall algorithm 83
1
1 2
3
3 1 1 2
1 −0.5
3 2
0
Figure 2.12: Another demonstration of the Floyd-Roy-Warshall algorithm.
Figure 2.13: Molecular graph of 1,1,3-trimethylcyclobutane.
84 Chapter 2. Graph algorithms
is
0 2 1 2 3 2 4
2 0 1 2 3 2 4
1 1 0 1 2 1 3
M =
2 2 1 0 1 2 2
.
3 3 2 1 0 1 1
2 2 1 2 1 0 2
4 4 3 2 1 2 0
Sum all entries in the upper (or lower) triangular of M to obtain the Wiener number
W = 42. Using Sage, we have
sage : G = Graph ({1:[3] , 2:[3] , 3:[4 ,6] , 4:[5] , 6:[5] , 5:[7]})
sage : D = G . sh or t es t _p at h _a l l_ pa i rs ()[0]
sage : M = [ D [ i ][ j ] for i in D for j in D [ i ]]
sage : M = matrix (M , nrows =7 , ncols =7)
sage : W = 0
sage : for i in range ( M . nrows () - 1):
... for j in range ( i +1 , M . ncols ()):
... W += M [i , j ]
sage : W
42
which verifies our computation above. See Gutman et al. [98] for a survey of some results
concerning the Wiener number.
2.6.1 Transitive closure
Consider a digraph G = (V, E) of order n = |V |. The transitive closure of G is defined
as the digraph G∗ = (V, E ∗ ) having the same vertex set as G. However, the edge set
E ∗ of G∗ consists of all edges uv such that there is a u-v path in G and uv ∈ / E. The
∗
transitive closure G answers an important question about G: If u and v are two distinct
vertices of G, are they connected by a path with length ≥ 1?
To compute the transitive closure of G, we let each edge of G be of unit weight and
apply the Floyd-Roy-Warshall Algorithm 2.8 on G. By Proposition 1.13, for any i-j path
in G we have D[i, j] < n, and if there are no paths from i to j in G, we have D[i, j] = ∞.
This procedure for computing transitive closure runs in time O(n3 ).
Modifying the Floyd-Roy-Warshall algorithm slightly, we obtain an algorithm for
computing transitive closure that, in practice, is more efficient than Algorithm 2.8 in
terms of time and space. Instead of using the operations min and + as is the case in the
Floyd-Roy-Warshall algorithm, we replace these operations with the logical operations
∨ (logical OR) and ∧ (logical AND), respectively. For i, j, k = 1, 2, . . . , n, define Tk (i, j) = 1
if there is an i-j path in G with all intermediate vertices belonging to {1, 2, . . . , k}, and
Tk (i, j) = 0 otherwise. Thus, the edge ij belongs to the transitive closure G∗ if and only
if Tk (i, j) = 1. The definition of Tk (i, j) can be cast in the form of a recursive definition
as follows. For k = 0, we have
(
0, if i 6= j and ij ∈
/ E,
T0 (i, j) =
1, if i = j or ij ∈ E
and for k > 0, we have
Tk (i, j) = Tk−1 (i, j) ∨ Tk−1 (i, k) ∧ Tk−1 (k, j) .
We need not use the subscript k at all and instead let T be a boolean matrix such that
T [i, j] = 1 if and only if there is an i-j path in G, and T [i, j] = 0 otherwise. Using
2.7. Johnson’s algorithm 85
the above notations, the Floyd-Roy-Warshall algorithm is translated to Algorithm 2.9
for obtaining the boolean matrix T . We can then use T and the definition of transitive
closure to obtain the edge set E ∗ in the transitive closure G∗ = (V, E ∗ ) of G = (V, E).
A more efficient transitive closure algorithm can be found in the PhD thesis of Esko
Nuutila [165]. See also the method of four Russians [2,9]. The transitive closure algorithm
as presented in Algorithm 2.9 is due to Stephen Warshall [201]. It is a special case of a
more general algorithm in automata theory due to Stephen Kleene [125], called Kleene’s
algorithm.
Algorithm 2.9: Variant of the Floyd-Roy-Warshall algorithm for transitive closure.
Input: A digraph G = (V, E) that has no self-loops. Vertices are numbered from
1 to n, i.e. V = {1, 2, . . . , n}.
Output: The boolean matrix T such that T [i, j] = 1 if and only if there is an i-j
path in G, and T [i, j] = 0 otherwise.
1 n ← |V |
2 T ← adjacency matrix of G
3 for k ← 1, 2, . . . , n do
4 for i ← 1, 2, . . . , n do
5 for j ← 1, 2, . . . , n do
6 T [i, j] ← T [i, j] ∨ T [i, k] ∧ T [k, j]
7 return T
2.7 Johnson’s algorithm
The shortest distance between two points is under construction.
— Noelie Altito
Let G = (V, E) be a sparse digraph with edge weights but no negative cycles. Johnson’s
algorithm [117] finds a shortest path between each pair of vertices in G. First published
in 1977 by Donald B. Johnson, the main insight of Johnson’s algorithm is to combine
the technique of edge reweighting with the Bellman-Ford and Dijkstra’s algorithms. The
Bellman-Ford algorithm is first used to ensure that G has no negative cycles. Next,
we reweight edges in such a manner as to preserve shortest paths. The final stage
makes use of Dijkstra’s algorithm for computing shortest paths between all vertex pairs.
Pseudocode for Johnson’s algorithm is presented in Algorithm 2.10. With a Fibonacci
heap implementation of the minimum-priority queue, the time complexity for sparse
graphs is O(|V |2 log |V | + |V | · |E|), where n = |V | is the number of vertices of the
original graph G.
To prove the correctness of Algorithm 2.10, we need to show that the new set of edge
weights produced by ŵ must satisfy two properties:
1. The reweighted edges preserve shortest paths. That is, let p be a u-v path for
u, v ∈ V . Then p is a shortest weighted path using weight function w if and only
if p is also a shortest weighted path using weight function ŵ.
2. The reweight function ŵ produces nonnegative weights. In other words, if u, v ∈ V
then ŵ(uv) ≥ 0.
86 Chapter 2. Graph algorithms
Algorithm 2.10: Johnson’s algorithm for sparse graphs.
Input: A sparse weighted digraph G = (V, E), where the vertex set is
V = {1, 2, . . . , n}.
Output: If G has negative-weight cycles, then return False. Otherwise, return an
n × n matrix D of shortest-path weights and a list P such that P [v] is a
parent list resulting from running Dijkstra’s algorithm on G with start
vertex v.
1 s ← vertex not in V
2 V 0 ← V ∪ {s}
3 E 0 ← E ∪ {sv | v ∈ V }
4 G0 ← digraph (V 0 , E 0 ) with weight w(sv) = 0 for all v ∈ V
5 if BellmanFord(G0 , w, s) = False then
6 return False
7 d ← distance list returned by BellmanFord(G0 , w, s)
8 for each edge uv ∈ E 0 do
9 ŵ(uv) ← w(uv) + d[u] − d[v]
10 for each u ∈ V do
11 (δ̂, P̂ ) ← distance and parent lists returned by Dijkstra(G, ŵ, u)
12 P [u] ← P̂
13 for each v ∈ V do
14 D[u, v] ← δ̂[v] + d[v] − d[u]
15 return (D, P )
Both of these properties are proved in Lemma 2.10.
Lemma 2.10. Reweighting preserves shortest paths. Let G = (V, E) be a weighted
digraph having weight function w : E → R and let h : V → R be a mapping of vertices
to real numbers. Let ŵ be another weight function for G such that
ŵ(uv) = w(uv) + h(u) − h(v)
for all uv ∈ E. Suppose p : v0 , v1 , . . . , vk is any path in G. Then we have the following
results.
1. The path p is a shortest v0 -vk path with weight function w if and only if it is a
shortest v0 -vk path with weight function ŵ.
2. The graph G has a negative cycle using weight function w if and only if G has a
negative cycle using ŵ.
3. If G has no negative cycles, then ŵ(uv) ≥ 0 for all uv ∈ E.
Proof. Write δ and δ̂ for the shortest path weights derived from w and ŵ, respectively.
To prove part 1, we need to show that w(p) = δ(v0 , vk ) if and only if ŵ(p) = δ̂(v0 , vk ).
2.8. Problems 87
First, note that any v0 -vk path p satisfies ŵ(p) = w(p) + h(v0 ) − h(vk ) because
k
X
ŵ(p) = ŵ(vi−1 vi )
i=1
Xk
= w(vi−1 vi ) + h(vi−1 ) − h(vi )
i=1
k
X k
X
= w(vi−1 vi ) + h(vi−1 ) − h(vi )
i=1 i=1
Xk
= w(vi−1 vi ) + h(v0 ) − h(vk )
i=1
= w(p) + h(v0 ) − h(vk ).
Any v0 -vk path shorter than p and using weight function w is also shorter using ŵ.
Therefore, w(p) = δ(v0 , vk ) if and only if ŵ(p) = δ̂(v0 , vk ).
To prove part 2, consider any cycle c : v0 , v1 , . . . , vk where v0 = vk . Using the proof
of part 1, we have
ŵ(c) = w(c) + h(v0 ) − h(vk )
= w(c)
thus showing that c is a negative cycle using w if and only if it is a negative cycle using
ŵ.
To prove part 3, we construct a new graph G0 = (V 0 , E 0 ) as follows. Consider a vertex
s∈ / V and let V 0 = V ∪ {s} and E 0 = E ∪ {sv | v ∈ V }. Extend the weight function w
to include w(sv) = 0 for all v ∈ V . By construction, s has no incoming edges and any
path in G0 that contains s has s as the source vertex. Thus G0 has no negative cycles if
and only if G has no negative cycles. Define the function h : V → R by v 7→ δ(s, v) with
domain V 0 . By the triangle inequality (see Lemma 2.5),
δ(s, u) + w(uv) ≥ δ(s, v) ⇐⇒ h(u) + w(uv) ≥ h(v)
thereby showing that ŵ(uv) = w(uv) + h(u) − h(v) ≥ 0.
2.8 Problems
I believe that a scientist looking at nonscientific problems is just as dumb as the next guy.
— Richard Feynman
2.1. The Euclidean algorithm is one of the oldest known algorithms. Given two positive
integers a and b with a ≥ b, let a mod b be the remainder obtained upon dividing
a by b. The Euclidean algorithm determines the greatest common divisor gcd(a, b)
of a and b. The procedure is summarized in Algorithm 2.11. Refer to Chabert [52]
for a history of algorithms from ancient to modern times.
(a) Implement Algorithm 2.11 in Sage and use your implementation to compute
the greatest common divisors of various pairs of integers. Use the built-in
Sage command gcd to verify your answer.
88 Chapter 2. Graph algorithms
Algorithm 2.11: The Euclidean algorithm.
Input: Two integers a > 0 and b ≥ 0 with a ≥ b.
Output: The greatest common divisor of a and b.
1 x←a
2 y←b
3 while y 6= 0 do
4 r ← x mod y
5 x←y
6 y←r
7 return x
(b) Modify Algorithm 2.11 to compute the greatest common divisor of any pair
of integers.
2.2. Given a polynomial p(x) = an xn + an−1 xn−1 + · · · + a1 x + a0 of degree n, we can use
Horner’s method [109] to efficiently evaluate p at a specific value x = x0 . Horner’s
method evaluates p(x) by expressing the polynomial as
n
X
p(x) = ai xi = (· · · (an x + an−1 )x + · · · )x + a0
i=0
to obtain Algorithm 2.12.
(a) Compare the runtime of polynomial evaluation using equation (2.2) and Horner’s
method.
(b) Let v be a bit vector read using big-endian order. Write a Sage function that
uses Horner’s method to compute the integer representation of v.
(c) Modify Algorithm 2.12 to evaluate the integer representation of a bit vector
v read using little-endian order. Hence, write a Sage function to convert v to
its integer representation.
Algorithm 2.12: Polynomial evaluation using Horner’s method.
P
Input: A polynomial p(x) = ni=0 ai xi , where an 6= 0 and x0 ∈ R.
Output: An evaluation of p at x = x0 .
1 b ← an
2 for i ← n − 1, n − 2, . . . , 0 do
3 b ← bx0 + ai
4 return b
2.3. Let G = (V, E) be an undirected graph, let s ∈ V , and D is the list of distances
resulting from running Algorithm 2.1 with G and s as input. Show that G is
connected if and only if D[v] is defined for each v ∈ V .
2.4. Show that the worst-case time complexity of depth-first search Algorithm 2.2 is
O(|V | + |E|).
2.8. Problems 89
2.5. Let D be the list of distances returned by Algorithm 2.2, let s be a starting vertex,
and let v be a vertex such that D[v] 6= ∞. Show that D[v] is the length of any
shortest path from s to v.
2.6. Consider the graph in Figure 2.10 as undirected. Run this undirected version
through Dijkstra’s algorithm with starting vertex v1 .
1
v3 v4
2
3 1
1
1 2 3 v5
3
6
v2 v1
1
Figure 2.14: Searching a directed house graph using Dijkstra’s algorithm.
2.7. Consider the graph in Figure 2.14. Choose any vertex as a starting vertex and run
Dijkstra’s algorithm over it. Now consider the undirected version of that digraph
and repeat the exercise.
2.8. For each vertex v of the graph in Figure 2.14, run breadth-first search over that
graph with v as the starting vertex. Repeat the exercise for depth-first search.
Compare the graphs resulting from the above exercises.
2.9. A list data structure can be used to manage vertices and edges. If L is a nonempty
list of vertices of a graph, we may want to know whether the graph contains a
particular vertex. We could search the list L, returning True if L contains the vertex
in question and False otherwise. Linear search is a simple searching algorithm.
Given an object E for which we want to search, we consider each element e of L in
turn and compare E to e. If at any point during our search we found a match, we
halt the algorithm and output an affirmative answer. Otherwise, we have scanned
through all elements of L and each of those elements do not match E. In this
case, linear search reports that E is not in L. Our discussion is summarized in
Algorithm 2.13.
(a) Implement Algorithm 2.13 in Sage and test your implementation using the
graphs presented in the figures of this chapter.
(b) What is the maximum number of comparisons during the running of Algo-
rithm 2.13? What is the average number of comparisons?
(c) Why must the input list L be nonempty?
2.10. Binary search is a much faster searching algorithm than linear search. The binary
search algorithm assumes that its input list is ordered in some manner. For sim-
plicity, we assume that the input list L consists of positive integers. The main idea
of binary search is to partition L into two halves: the left half and the right half.
Our task now is to determine whether the object E of interest is in the left half or
90 Chapter 2. Graph algorithms
Algorithm 2.13: Linear search for lists.
Input: A nonempty list L of vertices or edges. An object E for which we want to
search in L.
Output: True if E is in L; False otherwise.
1 for each e ∈ L do
2 if E = e then
3 return True
4 return False
Algorithm 2.14: Binary search for lists of positive integers.
Input: A nonempty list L of positive integers. Elements of L are sorted in
nondecreasing order. An integer i for which we want to search in L.
Output: True if i is in L; False otherwise.
1 low ← 0
2 high ← |L| − 1
3 while low ≤ high do
4 mid ← b low +2 high c
5 if i = L[mid] then
6 return True
7 if i < L[mid] then
8 high ← mid − 1
9 else
10 low ← mid + 1
11 return False
2.8. Problems 91
the right half, and apply binary search recursively to the half in which E is located.
Algorithm 2.14 provides pseudocode of our discussion of binary search.
(a) Implement Algorithm 2.14 in Sage and test your implementation using the
graphs presented in the figures of this chapter.
(b) What is the worst case runtime of Algorithm 2.14? How does this compare
to the worst case runtime of linear search?
(c) Why must the input list L be sorted in nondecreasing order? Would Algo-
rithm 2.14 work if L is sorted in nonincreasing order? If not, modify Algo-
rithm 2.14 so that it works with an input list that is sorted in nonincreasing
order.
(d) Line 4 of Algorithm 2.14 uses the floor function to compute the index of the
middle value. Would binary search still work if we use the ceiling function
instead of the floor function?
(e) An improvement on the time complexity of binary search is to not blindly use
the middle value of the interval of interest, but to guess more precisely where
the target falls within this interval. Interpolation search uses this heuristic
to improve on the runtime of binary search. Provide an algorithm for inter-
polation search, analyze its worst-case runtime, and compare its theoretical
runtime efficiency with that of binary search (see pages 419–420 in Knuth [129]
and pages 201–202 in Sedgewick [177]).
2.11. Let G be a simple undirected graph having distance matrix D = [d(vi , vj )], where
d(vi , vj ) ∈ R denotes the shortest distance from vi ∈ V (G) to vj ∈ V (G). If
vi = vj , we set d(vi , vj ) = 0. For each pair of distinct vertices (vi , vj ), we have
d(vi , vj ) = d(vj , vi ). The i-j entry of D is also written as di,j and denotes the entry
in row i and column j.
(a) The total distance td(u) of a fixed vertex u ∈ V (G) is the sum of distances
from u to each vertex in G:
X
td(u) = d(u, v).
v∈V (G)
If G is connected, i is the row index of vertex u in the distance matrix D, and
j is the column index of u in D, show that the total distance of u is
X X
td(u) = di,k = dk,j . (2.10)
k k
(b) Let the vertices of G be labeled V = {v1 , v2 , . . . , vn }, where n = |V (G)| is
the order of G. The total distance td(G) of G is obtained by summing all the
d(vi , vj ) for i < j. If G is connected, show that the total distance of G is equal
to the sum of all entries in the upper (or lower) triangular of D:
!
X X 1 XX
td(G) = di,j = di,j = d(u, v) . (2.11)
i<j i>j
2 u∈V v∈V
Hence show that the total distance of G is equal to its Wiener number:
td(G) = W (G).
92 Chapter 2. Graph algorithms
(c) Would equations (2.10) and (2.11) hold if G is not connected or directed?
2.12. The following result is from Yeh and Gutman [214]. Let G1 and G2 be graphs with
orders ni = |V (Gi )| and sizes mi = |E(Gi )|, respectively.
(a) If each of G1 and G2 is connected, show that the Wiener number of the
Cartesian product G1 G2 is
W (G1 G2 ) = n22 · W (G1 ) + n21 · W (G2 ).
(b) If G1 and G2 are arbitrary graphs, show that the Wiener number of the join
G1 + G2 is
W (G1 + G2 ) = n21 − n1 + n22 − n2 + n1 n2 − m1 − m2 .
2.13. The following results originally appeared in Entringer et al. [74] and independently
rediscovered many times since.
(a) If Pn is the path graph on n ≥ 0 vertices, show that the Wiener number of
Pn is W (Pn ) = 61 n(n2 − 1).
(b) If Cn is the cycle graph on n ≥ 0 vertices, show that the Wiener number of
Cn is (1
8
n(n2 − 1), if n is odd,
W (Cn ) =
1 3
8
n, if n is even.
(c) If Kn is the complete graph on n vertices, show that its Wiener number is
W (Kn ) = 21 n(n − 1).
(d) Show that the Wiener number of the complete bipartite graph Km,n is
W (Km,n ) = mn + m(m − 1) + n(n − 1).
2.14. Consider the world map of major capital cities in Figure 2.16.
(a) Run breadth- and depth-first searches over the graph in Figure 2.16 and com-
pare your results.
(b) Convert the graph in Figure 2.16 to a digraph as follows. Let 0 ≤ α ≤ 1 be a
fixed threshold probability and let V = {v1 , . . . , vn } be the vertex set of the
graph. For each edge vi vj , let 0 ≤ p ≤ 1 be its orientation probability and
define the directedness dir(vi , vj ) by
(
vi vj , if p ≤ α,
dir(vi , vj ) =
vj vi , otherwise.
That is, dir(vi , vj ) takes the endpoints of an undirected edge vi vj and returns
a directed version of this edge. The result is either the directed edge vi vj
or the directed edge vj vi . Use the above procedure to convert the graph of
Figure 2.16 to a digraph, and run breadth- and depth-first searches over the
resulting digraph.
2.8. Problems 93
(c) Table 2.6 lists distances in kilometers between the major capital cities shown
in Figure 2.16. Let those distances be the weights of the edges shown in Fig-
ure 2.16. Repeatedly run Dijkstra’s algorithm over this undirected, weighted
graph with each city as the start vertex. Use the procedure from the previous
exercise to obtain a directed version of the graph in Figure 2.16 and repeat
the current exercise with the resulting weighted digraph.
(d) Repeat the previous exercise, but use the Bellman-Ford algorithm instead of
Dijkstra’s algorithm. Compare the results you obtain using these two different
algorithms.
(e) Consider a weighted digraph version of the graph in Figure 2.16. Run the
Floyd-Roy-Warshall and Johnson’s algorithms over the weighted digraph.
Compare the results output by these two algorithms to the results of the
previous exercise.
2.15. Consider a finite square lattice Λ where points on Λ can connect to other points
in their von Neumann or Moore neighborhoods, or other points in the lattice. See
Figures 2.15(a) and 2.15(b) for illustrations of the von Neumann and Moore neigh-
borhoods of a central node, respectively. We treat each point as a node and forbid
each node from connecting to itself. The whole lattice can be considered as a sim-
ple undirected graph Γ. Let n be the maximum number of nodes in any connected
components. For each i ∈ {1, 2, . . . , n}, how many connected components have
exactly n nodes? In other words, what is the density of connected components in
Γ? For example, consider the lattice in Figure 2.15(c). Denote by κ(i) the number
of connected components with i nodes. Then the maximum number of nodes of
any connected components is n = 6, with κ(1) = 2, κ(2) = 2, κ(3) = 2, κ(4) = 1,
κ(5) = 0, and κ(6) = 1.
(a) Von Neumann neighbor- (b) Moore neighborhood. (c) Square lattice with compo-
hood. nents.
Figure 2.15: Component density in a square lattice.
94
Moscow
London
Ottawa Berlin
Beijing
Washington DC Madrid Tokyo
New Delhi
Bangkok
Lima
Brasilia
Pretoria Sydney
Buenos Aires
Figure 2.16: Major capital cities of the world.
Chapter 2. Graph algorithms
Bangkok Beijing Berlin Brasilia Buenos Aires Lima London Madrid Moscow New Delhi Ottawa Pretoria Sydney Tokyo Washington DC
Bangkok 3282 2908 7540
Beijing 3282 5807 3784 2103
Berlin 929 1866 1619 5796
Brasilia 2314 3154 7906 6764
Buenos Aires 2314
Lima 3154
2.8. Problems
London 929 1261 5379 5915
Madrid 1866 1261 7288 8033
Moscow 5807 1619
New Delhi 2908 3784 5796 7288
Ottawa 5379 734
Pretoria 7906 8033
Sydney 7540
Tokyo 2103
Washington DC 6764 5915 734
Table 2.6: Distances in kilometers between major world capital cities.
95
96 Chapter 2. Graph algorithms
2.16. Various efficient search techniques exist that cater for special situations. Some of
these are covered in chapter 6 in Knuth [129] and chapters 14–18 in Sedgewick [177].
Investigate an algorithm for and time complexity of trie search. Hashing techniques
can result in searches that run in O(1) time. Furthermore, hashing has important
applications outside of the searching problem, a case in point being cryptology.
Investigate how hashing can be used to speed up searches. For further information
on hashing and its application to cryptology, see Menezes et al. [152], Stinson [186],
or Trappe and Washington [192].
2.17. In addition to searching, there is the related problem of sorting a list according to
an ordering relation. If the given list L = [e1 , e2 , . . . , en ] consists of real numbers,
we want to order elements in nondecreasing order. Bubble sort is a basic sorting
algorithm that can be used to sort a list of real numbers, indeed any collection of
objects that can be ordered according to an ordering relation. During each pass
through the list L from left to right, we consider ei and its right neighbor ei+1 . If
ei ≤ ei+1 , then we move on to consider ei+1 and its right neighbor ei+2 . If ei > ei+1 ,
then we swap these two values around in the list and then move on to consider ei+1
and its right neighbor ei+2 . Each successive pass pushes to the right end an element
that is the next largest in comparison to the previous largest element pushed to
the right end. Hence the name bubble sort for the algorithm. Algorithm 2.15
summarizes our discussion.
Algorithm 2.15: Bubble sort.
Input: A list L of n > 1 elements that can be ordered using the “less than or
equal to” relation “≤”.
Output: The same list as L, but sorted in nondecreasing order.
1 for i ← n, n − 1, . . . , 2 do
2 for j ← 2, 3, . . . , i do
3 if L[j − 1] > L[j] then
4 swap the values of L[j − 1] and L[j]
5 return L
(a) Analyze the worst-case runtime of Algorithm 2.15.
(b) Modify Algorithm 2.15 so that it sorts elements in nonincreasing order.
(c) Line 4 of Algorithm 2.15 is where elements are actually sorted. Swapping
the values of two elements is such a common task that sometimes we want
to perform the swap as efficiently as possible, i.e. using as few operations
as possible. A common way to swap the values of two elements a and b
is to create a temporary placeholder t and realize the swap as presented in
Algorithm 2.16. Some programming languages allow for swapping the values
of two elements without creating a temporary placeholder for an intermediary
value. Investigate how to realize this swapping method using a programming
language of your choice.
2.18. Selection sort is another simple sorting algorithm that works as follows. Let L =
[e1 , e2 , . . . , en ] be a list of elements that can be ordered according to the relation
“≤”, e.g. the ei can all be real numbers or integers. On the first scan of L
2.8. Problems 97
Algorithm 2.16: Swapping values using a temporary placeholder.
Input: Two objects a and b.
Output: The values of a and b swapped with each other.
1 t←a
2 a←b
3 b←t
from left to right, among the elements L[2], . . . , L[n] we find the smallest element
and exchange it with L[1]. On the second scan, we find the smallest element
among L[3], . . . , L[n] and exchange that smallest element with L[2]. In general,
during the i-th scan we find the smallest element among L[i + 1], . . . , L[n] and
exchange that with L[i]. At the end of the i-th scan, the element L[i] is in its final
position and would not be processed again. When the index reaches i = n, the list
would have been sorted in nondecreasing order. The procedure is summarized in
Algorithm 2.17.
(a) Analyze the worst-case runtime of Algorithm 2.17 and compare your result to
the worst-case runtime of the bubble sort Algorithm 2.15.
(b) Modify Algorithm 2.17 to sort elements in nonincreasing order.
(c) Line 6 of Algorithm 2.17 assumes that among L[i+1], L[i+2], . . . , L[n] there is
a smallest element L[k] such that L[i] > L[k], hence we perform the swap. It is
possible that L[i] < L[k], obviating the need to carry out the value swapping.
Modify Algorithm 2.17 to take account of our discussion.
Algorithm 2.17: Selection sort.
Input: A list L of n > 1 elements that can be ordered using the relation “≤”.
Output: The same list as L, but sorted in nondecreasing order.
1 for i ← 1, 2, . . . , n − 1 do
2 min ← i
3 for j ← i + 1, i + 2, . . . , n do
4 if L[j] < L[min] then
5 min ← j
6 swap the values of L[min] and L[i]
7 return L
2.19. In addition to bubble and selection sort, other algorithms exist whose runtime is
more efficient than these two basic sorting algorithms. Chapter 5 in Knuth [129] de-
scribes various efficient sorting techniques. See also chapters 8–13 in Sedgewick [177].
(a) Investigate and provide pseudocode for insertion sort and compare its runtime
efficiency with that of selection sort. Compare the similarities and differences
between insertion and selection sort.
(b) Shellsort is a variation on insertion sort that can speed up the runtime of
insertion sort. Describe and provide pseudocode for shellsort. Compare the
98 Chapter 2. Graph algorithms
time complexity of shellsort with that of insertion sort. In what ways is
shellsort different from or similar to insertion sort?
(c) The quicksort algorithm due to C. A. R. Hoare [104] was published in 1962.
Describe and provide pseudocode for quicksort and compare its runtime com-
plexity with the other sorting algorithms covered in this chapter.
2.20. Algorithm 2.3 uses breadth-first search to determine the connectivity of an undi-
rected graph. Modify this algorithm to use depth-first search. How can Algo-
rithm 2.3 be used or modified to test the connectivity of a digraph?
2.21. The following problem is known as the river crossing problem. A man, a goat,
a wolf, and a basket of cabbage are all on one side of a river. They have a boat
that could be used to cross to the other side of the river. The boat can only hold
at most two passengers, one of whom must be able to row the boat. One of the
two passengers must be the man and the other passenger can be either the goat,
the wolf, or the basket of cabbage. When crossing the river, if the man leaves the
wolf with the goat, the wolf would prey on the goat. If he leaves the goat with the
basket of cabbage, the goat would eat the cabbage. The objective is to cross the
river in such a way that the wolf has no chance of preying on the goat, nor that
the goat eat the cabbage.
(a) Let M , G, W , and C denote the man, the goat, the wolf, and the basket of
cabbage, respectively. Initially all four are on the left side of the river and none
of them are on the right side. Denote this by the ordered pair (M GW C, w),
which is called the initial state of the problem. When they have all crossed to
the right side of the river, the final state of the problem is (w, M GW C). The
underscore “w” means that neither M , G, W , nor C are on the corresponding
side of the river. List a finite sequence of moves to get from (M GW C, w) to
(w, M GW C). Draw your result as a digraph.
(b) In the digraph Γ obtained from the previous exercise, let each edge of Γ be of
unit weight. Find a shortest path from (M GW C, w) to (w, M GW C).
(c) Rowing from one side of the river to the other side is called a crossing. What
is the minimum number of crossings needed to get from (M GW C, w) to
(w, M GW C)?
2.22. Symbolic computation systems such as Magma, Maple, Mathematica, Maxima,
and Sage are able to read in a symbolic expression such as
(a + b)^2 - (a - b)^2
and determine whether or not the brackets match. A bracket is any of the following
characters:
( ) { } [ ]
A string S of characters is said to be balanced if any left bracket in S has a corre-
sponding right bracket that is also in S. Furthermore, if there are k occurrences
of one type of left bracket, then there must be k occurrences of the corresponding
right bracket. The balanced bracket problem is concerned with determining whether
or not the brackets in S are balanced. Algorithm 2.18 contains a procedure to de-
termine if the brackets in S are balanced, and if so return a list of positive integers
to indicate how the brackets match.
2.8. Problems 99
(a) Implement Algorithm 2.18 in Sage and test your implementation on various
strings containing brackets. Test your implementation on nonempty strings
without any brackets.
(b) Modify Algorithm 2.18 so that it returns True if the brackets of an input
string are balanced, and returns False otherwise.
(c) What is the worst-case runtime of Algorithm 2.18?
Algorithm 2.18: A brackets parser.
Input: A nonempty string S of characters.
Output: A list L of positive integers indicating how the brackets match. If the
brackets are not balanced, return the empty string ε.
1 L ← []
2 T ← empty stack
3 c←1
4 n ← |S|
5 for i ← 0, 1, . . . , n do
6 if S[i + 1] is a left bracket then
7 append(L, c)
8 push (S[i + 1], c) onto T
9 c←c+1
10 if S[i + 1] is a right bracket then
11 if T is empty then
12 return ε
13 (left, d) ← pop(T )
14 if left matches S[i + 1] then
15 append(L, d)
16 else
17 return ε
18 if T is empty then
19 return L
20 return ε
2.23. An arithmetic expression written in the form a + b is said to be in infix notation
because the operator is in between the operands. The same expression can also be
written in reverse Polish notation (or postfix notation) as
ab+
with the operator following its two operands. Given an arithmetic expression A =
e0 e1 · · · en written in reverse Polish notation, we can use the stack data structure
to evaluate the expression. Let P = [e0 , e1 , . . . , en ] be the stack representation of
A, where traversing P from left to right we are moving from the top of the stack
to the bottom of the stack. We call P the Polish stack and the stack E containing
intermediate results the evaluation stack. While P is not empty, pop the Polish
stack and assign the extracted result to x. If x is an operator, we pop the evaluation
stack twice: the result of the first pop is assigned to b and the result of the second
100 Chapter 2. Graph algorithms
pop is assigned to a. Compute the infix expression a x b and push the result onto
E. However, if x is an operand, we push x onto E. Iterate the above process until
P is empty, at which point the top of E contains the evaluation of A. Refer to
Algorithm 2.19 for pseudocode of the above discussion.
(a) Prove the correctness of Algorithm 2.19.
(b) What is the worst-case runtime of Algorithm 2.19?
(c) Modify Algorithm 2.19 to support the exponentiation operator.
Algorithm 2.19: Evaluate arithmetic expressions in reverse Polish notation.
Input: A Polish stack P containing an arithmetic expression in reverse Polish
notation.
Output: An evaluation of the arithmetic expression represented by P .
1 E ← empty stack
2 v ← NULL
3 while P is not empty do
4 x ← pop(P )
5 if x is an operator then
6 b ← pop(E)
7 a ← pop(E)
8 if x is addition operator then
9 v ←a+b
10 else if x is subtraction operator then
11 v ←a−b
12 else if x is multiplication operator then
13 v ←a×b
14 else if x is division operator then
15 v ← a/b
16 else
17 exit algorithm with error
18 push(E, v)
19 else
20 push(E, x)
21 v ← pop(E)
22 return v
2.24. Figure 2.5 provides a knight’s tour for the knight piece with initial position as
in Figure 2.5(a). By rotating the chessboard in Figure 2.5(b) by 90n degrees for
positive integer values of n, we obtain another knight’s tour that, when represented
as a graph, is isomorphic to the graph in Figure 2.5(c).
(a) At the beginning of the 18th century, de Montmort and de Moivre provided
the following strategy [13, p.176] to solve the knight’s tour problem on an
8 × 8 chessboard. Divide the board into an inner 4 × 4 square and an outer
shell of two squares deep, as shown in Figure 2.17(a). Place a knight on a
square in the outer shell and move the knight piece around that shell, always
2.8. Problems 101
in the same direction, so as to visit each square in the outer shell. After that,
move into the inner square and solve the knight’s tour problem for the 4 × 4
case. Apply this strategy to solve the knight’s tour problem with the initial
position as in Figure 2.17(b).
(b) Use the Montmort-Moivre strategy to obtain a knight’s tour, starting at the
position of the black-filled node in the outer shell in Figure 2.5(b).
(c) A re-entrant or closed knight’s tour is a knight’s tour that starts and ends
at the same square. Find re-entrant knight’s tours with initial positions as in
Figure 2.18.
(d) Devise a backtracking algorithm to solve the knight’s tour problem on an n×n
chessboard for n > 3.
0Z0Z0Z0Z 0Z0Z0Z0m
Z0Z0Z0Z0 Z0Z0Z0Z0
0Z0Z0Z0Z 0Z0Z0Z0Z
Z0Z0Z0Z0 Z0Z0Z0Z0
0Z0Z0Z0Z 0Z0Z0Z0Z
Z0Z0Z0Z0 Z0Z0Z0Z0
0Z0Z0Z0Z 0Z0Z0Z0Z
Z0Z0Z0Z0 Z0Z0Z0Z0
(a) A 4 × 4 inner square. (b) Initial position in the outer shell.
Figure 2.17: De Montmort and de Moivre’s solution strategy for the 8 × 8 knight’s tour
problem.
0Z0Z0Z0Z
Z0Z0Z0Z0
0Z0Z0Z 0Z0Z0Z0Z
Z0Z0Z0 Z0m0Z0Z0
0Z0Z0Z 0Z0Z0Z0Z
Z0Z0Z0 Z0Z0Z0Z0
0Z0Z0Z 0Z0Z0Z0Z
m0Z0Z0 Z0Z0Z0Z0
(a) A 6 × 6 chessboard. (b) An 8 × 8 chessboard.
Figure 2.18: Initial positions of re-entrant knight’s tours.
2.25. The n-queens problem is concerned with the placement of n queens on an n × n
chessboard such that no two queens can attack each other. Two queens can attack
102 Chapter 2. Graph algorithms
0ZQZ0Z0Z
Z0Z0Z0ZQ
0Z0L0Z0Z
Z0Z0Z0L0
0L0Z QZ0Z0Z0Z
Z0ZQ Z0Z0ZQZ0
QZ0Z 0L0Z0Z0Z
Z0L0 Z0Z0L0Z0
(a) n = 4 (b) n = 8
Figure 2.19: Solutions of the n-queens problem for n = 4, 8.
each other if they are in the same row, column, diagonal, or antidiagonal of the
chessboard. The trivial case n = 1 is easily solved by placing the one queen in the
only given position. There are no solutions for the cases n = 2, 3. Solutions for the
cases n = 4, 8 are shown in Figure 2.19. Devise a backtracking algorithm to solve
the n-queens problem for the case where n > 3. See Bell and Stevens [21] for a
survey of the n-queens problem and its solutions.
2.26. Hampton Court Palace in England is well-known for its maze of hedges. Figure 2.20
shows a maze and its graph representation; the figure is adapted from page 434 in
Sedgewick [177]. To obtain the graph representation, we use a vertex to represent
an intersection in the maze. An edge joining two vertices represents a path from
one intersection to another.
(a) Suppose the entrance to the maze is represented by the lower-left black-filled
vertex in Figure 2.20(b) and the exit is the upper-right black-filled vertex.
Solve the maze by providing a path from the entrance to the exit.
(b) Repeat the previous exercise for each pair of distinct vertices, letting one
vertex of the pair be the entrance and the other vertex the exit.
(c) What is the diameter of the graph in Figure 2.20(b)?
(d) Investigate algorithms for generating and solving mazes.
2.27. For each of the algorithms below: (i) justify whether or not it can be applied
to multigraphs or multidigraphs; (ii) if not, modify the algorithm so that it is
applicable to multigraphs or multidigraphs.
(a) Breadth-first search Algorithm 2.1.
(b) Depth-first search Algorithm 2.2.
(c) Graph connectivity test Algorithm 2.3.
(d) General shortest path Algorithm 2.4.
(e) Dijkstra’s Algorithm 2.5.
2.8. Problems 103
(a) (b)
Figure 2.20: A maze and its graph representation.
(f) The Bellman-Ford Algorithms 2.6 and 2.7.
(g) The Floyd-Roy-Warshall Algorithm 2.8.
(h) The transitive closure Algorithm 2.9.
(i) Johnson’s Algorithm 2.10.
(a) 2 × 2 (b) 3 × 3 (c) 4 × 4
Figure 2.21: Grid graphs for n = 2, 3, 4.
2.28. Let n be a positive integer. An n × n grid graph is a graph on the Euclidean plane,
where each vertex is an ordered pair from Z × Z. In particular, the vertices are
ordered pairs (i, j) ∈ Z × Z such that
0 ≤ i, j < n. (2.12)
Each vertex (i, j) is adjacent to any of the following vertices provided that ex-
pression (2.12) is satisfied: the vertex (i − 1, j) immediately to its left, the vertex
(i + 1, j) immediately to its right, the vertex (i, j + 1) immediately above it, or
the vertex (i, j − 1) immediately below it. Figure 2.21 illustrates some examples
of grid graphs. The 1 × 1 grid graph is the trivial graph K1 .
(a) Fix a positive integer n > 1. Describe and provide pseudocode of an algorithm
to generate all nonisomorphic n × n grid graphs. What is the worst-case
runtime of your algorithm?
(b) How many n × n grid graphs are there? How many of those graphs are
nonisomorphic to each other?
104 Chapter 2. Graph algorithms
(c) Describe and provide pseudocode of an algorithm to generate a random n × n
grid graph. Analyze the worst-case runtime of your algorithm.
(d) Extend the grid graph by allowing edges to be diagonals. That is, a vertex
(i, j) can also be adjacent to any of the following vertices so long as expres-
sion (2.12) holds: (i − 1, j − 1), (i − 1, j + 1), (i + 1, j + 1), (i + 1, j − 1).
With this extension, repeat the previous exercises.
2.29. Let G = (V, E) be a digraph with integer weight function w : E → Z\{0}, where
either w(e) > 0 or w(e) < 0 for each e ∈ E. Yamada and Kinoshita [211] provide
a divide-and-conquer algorithm to enumerate all the negative cycles in G. Investi-
gate the divide and conquer technique for algorithm design. Describe and provide
pseudocode of the Yamada-Kinoshita algorithm. Analyze its runtime complexity
and prove the correctness of the algorithm.
Chapter 3
Trees and forests
— Randall Munroe, xkcd, http://xkcd.com/71/
In section 1.2.1, we briefly touched upon trees and provided examples of how trees could
be used to model hierarchical structures. This chapter provides an in-depth study of
trees, their properties, and various applications. After defining trees and related con-
cepts in section 3.1, we then present various basic properties of trees in section 3.2.
Each connected graph G has an underlying subgraph called a spanning tree that con-
tains all the vertices of G. Spanning trees are discussed in section 3.3 together with
various common algorithms for finding spanning trees. We then discuss binary trees in
section 3.4, followed by an application of binary trees to coding theory in section 3.5.
Whereas breadth- and depth-first searches are general methods for traversing a graph,
trees require specialized techniques in order to visit their vertices, a topic that is taken
up in section 3.6.
3.1 Definitions and examples
I think that I shall never see
A poem lovely as a tree.
— Joyce Kilmer, Trees and Other Poems, 1914, “Trees”
Recall that a path in a graph G = (V, E) whose start and end vertices are the same is
called a cycle. We say G is acyclic, or a forest, if it has no cycles. In a forest, a vertex
of degree one is called an endpoint or a leaf . Any vertex that is not a leaf is called an
105
106 Chapter 3. Trees and forests
internal vertex. A connected forest is a tree. In other words, a tree is a graph without
cycles and each edge is a bridge. A forest can also be considered as a collection of trees.
A rooted tree T is a tree with a specified root vertex v0 , i.e. exactly one vertex has
been specially designated as the root of T . However, if G is a rooted tree with root
vertex v0 having degree one, then by convention we do not call v0 an endpoint or a leaf.
The depth depth(v) of a vertex v in T is its distance from the root. The height height(T )
of T is the length of a longest path starting from the root vertex, i.e. the height is the
maximum depth among all vertices of T . It follows by definition that depth(v) = 0 if and
only if v is the root of T , height(T ) = 0 if and only if T is the trivial graph, depth(v) ≥ 0
for all v ∈ V (T ), and height(T ) ≤ diam(T ).
The Unix, in particular Linux, filesystem hierarchy can be viewed as a tree (see
Figure 3.1). As shown in Figure 3.1, the root vertex is designated with the forward
slash, which is also referred to as the root directory. Other examples of trees include the
organism classification tree in Figure 3.2, the family tree in Figure 3.3, and the expression
tree in Figure 3.4.
A directed tree is a digraph which would be a tree if the directions on the edges
were ignored. A rooted tree can be regarded as a directed tree since we can imagine an
edge uv for u, v ∈ V being directed from u to v if and only if v is further away from v0
than u is. If uv is an edge in a rooted tree, then we call v a child vertex with parent u.
Directed trees are pervasive in theoretical computer science, as they are useful structures
for describing algorithms and relationships between objects in certain datasets.
/
bin etc home lib opt proc tmp usr ...
anne sam ... bin include local share src ...
acyclic diff dot gc neato ...
Figure 3.1: The Linux filesystem hierarchy.
An ordered tree is a rooted tree for which an ordering is specified for the children of
each vertex. An n-ary tree is a rooted tree for which each vertex that is not a leaf has
at most n children. The case n = 2 are called binary trees. An n-ary tree is said to
be complete if each of its internal vertices has exactly n children and all leaves have the
same depth. A spanning tree of a connected, undirected graph G is a subgraph that is
a tree and containing all vertices of G.
Example 3.1. Consider the 4 × 4 grid graph with 16 vertices and 24 edges. Two
examples of a spanning tree are given in Figure 3.5 by using a darker line shading for its
edges.
Example 3.2. For n = 1, . . . , 6, how many distinct (nonisomorphic) trees are there of
order n? Construct all such trees for each n.
3.1. Definitions and examples 107
organism
plant animal
tree flower invertebrate vetebrate
deciduous evergreen bird mammal
finch rosella sparrow dolphin human whale
Figure 3.2: Classification tree of organisms.
Nikolaus senior
Jacob Nicolaus Johann
Nicolaus I Nicolaus II Daniel Johann II
Johann III Daniel II Jakob II
Figure 3.3: Bernoulli family tree of mathematicians.
+
× × ×
a a 2 a b b b
Figure 3.4: Expression tree for the perfect square a2 + 2ab + b2 .
108 Chapter 3. Trees and forests
(a) (b)
Figure 3.5: Two spanning trees for the 4 × 4 grid graph.
Solution. For n = 1, there is only one tree of order 1, i.e. K1 . The same is true for n = 2
and n = 3, where the required trees are P2 and P3 , respectively (see Figure 3.6). We
have two trees of order n = 4 (see Figure 3.7), three of order n = 5 (see Figure 3.8), and
six of order n = 6 (see Figure 3.9).
(a) n = 1 (b) n = 2 (c) n = 3
Figure 3.6: All distinct trees of order n = 1, 2, 3.
(a) (b)
Figure 3.7: All distinct trees of order n = 4.
Example 3.3. Let T = (V, E) be a tree with vertex set
V = {a, b, c, d, e, f, v, w, x, y, z}
edge set
E = {va, vw, wx, wy, xb, xc, yd, yz, ze, zf }
and root vertex v. Verify that T is a binary tree. Suppose that x is the root of the branch
we want to remove from T . Find all children of x and cut off the branch rooted at x from
T . Is the resulting graph also a binary tree?
3.1. Definitions and examples 109
(a) (b) (c)
Figure 3.8: All distinct trees of order n = 5.
(a) (b) (c) (d) (e)
(f)
Figure 3.9: All distinct trees of order n = 6.
110 Chapter 3. Trees and forests
Solution. We construct the tree T in Sage as follows:
sage : T = DiGraph ({
... " v " :[ " a " ," w " ] , " w " :[ " x " ," y " ] ,
... " x " :[ " c " ," b " ] , " y " :[ " z " ," d " ] ,
... " z " :[ " f " ," e " ]})
sage : for v in T . vertex_iterator ():
... print ( v ) ,
a c b e d f w v y x z
sage : for e in T . edge_iterator ():
... print ( " % s % s " % ( e [0] , e [1])) ,
wy wx va vw yd yz xc xb ze zf
Each vertex in a binary tree has at most 2 children. Use this definition to test whether
or not a graph is a binary tree.
sage : T . is_tree ()
True
sage : def is_bintree1 ( G ):
... for v in G . vertex_iterator ():
... if len ( G . neighbors_out ( v )) > 2:
... return False
... return True
sage : is_bintree1 ( T )
True
Here’s another way to test for binary trees. Let T be an undirected rooted tree. Each
vertex in a binary tree has a maximum degree of 3. If the root vertex is the only vertex
with degree 2, then T is a binary tree. (Problem 3.5 asks you to prove this result.) We
can use this test because the root vertex v of T is the only vertex with two children.
sage : def is_bintree2 ( G ):
... if G . is_tree () and max ( G . degree ()) == 3 and G . degree (). count (2) == 1:
... return True
... return False
sage : is_bintree2 ( T . to_undirected ())
True
As x is the root vertex of the branch we want to cut off from T , we could use breadth-
or depth-first search to determine all the children of x. We then delete x and its children
from T .
sage : T2 = copy ( T )
sage : # using breadth - first search
sage : V = list ( T . breadth_first_search ( " x " )); V
[ ’x ’ , ’c ’ , ’b ’]
sage : T . delete_vertices ( V )
sage : for v in T . vertex_iterator ():
... print ( v ) ,
a e d f w v y z
sage : for e in T . edge_iterator ():
... print ( " % s % s " % ( e [0] , e [1])) ,
wy va vw yd yz ze zf
sage : # using depth - first search
sage : V = list ( T2 . depth_first_search ( " x " )); V
[ ’x ’ , ’b ’ , ’c ’]
sage : T2 . delete_vertices ( V )
sage : for v in T2 . vertex_iterator ():
... print ( v ) ,
a e d f w v y z
sage : for e in T2 . edge_iterator ():
... print ( " % s % s " % ( e [0] , e [1])) ,
wy va vw yd yz ze zf
The resulting graph T is a binary tree because each vertex has at most two children.
sage : T
Digraph on 8 vertices
sage : is_bintree1 ( T )
True
Notice that the test defined in the function is_bintree2 can no longer be used to test
whether or not T is a binary tree, because T now has two vertices, i.e. v and w, each of
3.1. Definitions and examples 111
which has degree 2.
Consider again the organism classification tree in Figure 3.2. We can view the vertex
“organism” as the root of the tree and having two children. The first branch of “or-
ganism” is the subtree rooted at “plant” and its second branch is the subtree rooted
at “animal”. We form the complete tree by joining an edge between “organism” and
“plant”, and an edge between “organism” and “animal”. The subtree rooted at “plant”
can be constructed in the same manner. The first branch of this subtree is the subtree
rooted at “tree” and the second branch is the subtree rooted at “flower”. To construct
the subtree rooted at “plant”, we join an edge between “plant” and “tree”, and an
edge between “plant” and “flower”. The other subtrees of the tree in Figure 3.2 can be
constructed using the above recursive procedure.
In general, the recursive construction in Theorem 3.4 provides an alternative way
to define trees. We say construction because it provides an algorithm to construct a
tree, as opposed to the nonconstructive definition presented earlier in this section, where
we defined the conditions under which a graph qualifies as a tree without presenting a
procedure to construct a tree. Furthermore, we say recursive since a larger tree can be
viewed as being constructed from smaller trees, i.e. join up existing trees to obtain a
new tree. The recursive construction of trees as presented in Theorem 3.4 is illustrated
in Figure 3.10.
Theorem 3.4. Recursive construction of trees. An isolated vertex is a tree. That
single vertex is the root of the tree. Given a collection T1 , T2 , . . . , Tn of n > 0 trees,
construct a new tree as follows:
1. Let T be a tree having exactly the one vertex v, which is the root of T .
2. Let vi be the root of the tree Ti .
3. For i = 1, 2, . . . , n, add the edge vvi to T and add Ti to T . That is, each vi is now
a child of v.
The result is the tree T rooted at v with vertex set
!
[
V (T ) = {v} ∪ V (Ti )
i
and edge set [
E(T ) = {vvi } ∪ E(Ti ) .
i
The following game is a variant of the Shannon switching game, due to Edmonds and
Lehman. We follow the description in Oxley’s survey [166]. Recall that a minimal edge
cut of a graph is also called a bond of the graph. The following two-person game is played
on a connected graph G = (V, E). Two players Alice and Bob alternately tag elements
of E. Alice’s goal is to tag the edges of a spanning tree, while Bob’s goal is to tag the
edges of a bond. If we think of this game in terms of a communication network, then
Bob’s goal is to separate the network into pieces that are no longer connected to each
other, while Alice is aiming to reinforce edges of the network to prevent their destruction.
Each move for Bob consists of destroying one edge, while each move for Alice involves
securing an edge against destruction. The next result characterizes winning strategies
on G. The full proof can be found in Oxley [166]. See Rasmussen [172] for optimization
algorithms for solving similar games.
112 Chapter 3. Trees and forests
v
...
T1 T2 Tn
Figure 3.10: Recursive construction of a tree.
Theorem 3.5. The following statements are equivalent for a connected graph G =
(V, E).
1. Bob plays first and Alice can win against all possible strategies of Bob.
2. The graph G has 2 edge-disjoint spanning trees.
3. For all partitions P of the vertex set V , the number of edges of G that join vertices
in different classes of the partition is at least 2(|P | − 1).
3.2 Properties of trees
All theory, dear friend, is grey, but the golden tree of actual life springs ever green.
— Johann Wolfgang von Goethe, Faust, part 1, 1808
By Theorem 1.27, each edge of a tree is a bridge. Removing any edge of a tree partitions
the tree into two components, each of which is a subtree of the original tree. The following
results provide further basic characterizations of trees.
Theorem 3.6. Any tree T = (V, E) has size |E| = |V | − 1.
Proof. This follows by induction on the number of vertices. By definition, a tree has
no cycles. We need to show that any tree T = (V, E) has size |E| = |V | − 1. For the
base case |V | = 1, there are no edges. Assume for induction that the result holds for
all integers less than or equal to k ≥ 2. Let T = (V, E) be a tree having k + 1 vertices.
Remove an edge from T , but not the vertices it is incident to. This disconnects T into
two components T1 = (V1 , E1 ) and T2 = (V2 , E2 ), where |E| = |E1 | + |E2 | + 1 and
|V | = |V1 | + |V2 | (and possibly one of the Ei is empty). Each Ti is a tree satisfying the
conditions of the induction hypothesis. Therefore,
|E| = |E1 | + |E2 | + 1
= |V1 | − 1 + |V2 | − 1 + 1
= |V | − 1.
as required.
3.2. Properties of trees 113
Corollary 3.7. If T = (V, E) is a graph of order |V | = n, then the following are
equivalent:
1. T is a tree.
2. T contains no cycles and has n − 1 edges.
3. T is connected and has n − 1 edges.
4. Every edge of T is a cut set.
Proof. (1) =⇒ (2): This holds by definition of trees and Theorem 3.6.
(2) =⇒ (3): If T = (V, E) has k connected components then it is a disjoint union
of trees Ti = (Vi , Ei ), i = 1, 2, . . . , k, for some k. By part (2), each of these satisfy
|Ei | = |Vi | − 1
so
k
X
|E| = |Ei |
i=1
k
X
= |Vi | − k
i=1
= |V | − k.
This contradicts part (2) unless k = 1. Therefore, T is connected.
(3) =⇒ (4): If removing an edge e ∈ E leaves T = (V, E) connected then T 0 =
(V, E 0 ) is a tree, where E 0 = E−e. However, this means that |E 0 | = |E|−1 = |V |−1−1 =
|V | − 2, which contradicts part (3). Therefore e is a cut set.
(4) =⇒ (1): From part (2) we know that T has no cycles and from part (3) we
know that T is connected. Conclude by the definition of trees that T is a tree.
Theorem 3.8. Let T = (V, E) be a tree and let u, v ∈ V be distinct vertices. Then T
has exactly one u-v path.
Proof. Suppose for contradiction that
P : v0 = u, v1 , v2 , . . . , vk = v
and
Q : w0 = u, w1 , w2 , . . . , w` = v
are two distinct u-v paths. Then P and Q has a common vertex x, which is possibly
x = u. For some i ≥ 0 and some j ≥ 0 we have vi = x = wj , but vi+1 6= wj+1 . Let
y be the first vertex after x such that y belongs to both P and Q. (It is possible that
y = v.) We now have two distinct x-y paths that have only x and y in common. Taken
together, these two x-y paths result in a cycle, contradicting our hypothesis that T is a
tree. Therefore T has only one u-v path.
Theorem 3.9. If T = (V, E) is a graph then the following are equivalent:
114 Chapter 3. Trees and forests
1. T is a tree.
2. For any new edge e, the join T + e has exactly one cycle.
Proof. (1) =⇒ (2): Let e = uv be a new edge connecting u, v ∈ V . Suppose that
P : v0 = w, v1 , v2 , . . . , vk = w
and
P 0 : v00 = w, v10 , v20 , . . . , v`0 = w
are two cycles in T + e. If either P or P 0 does not contain e, say P does not contain e,
then P is a cycle in T . Let u = v0 and let v = v1 . The edge (v0 = w, v1 ) is a u-v path
and the sequence v = v1 , v2 , . . . , vk = w = u taken in reverse order is another u-v path.
This contradicts Theorem 3.8.
We may now suppose that P and P 0 both contain e. Then P contains a subpath
P0 = P − e (which is not closed) that is the same as P except it lacks the edge from u
to v. Likewise, P 0 contains a subpath P00 = P 0 − e (which is not closed) that is the same
as P 0 except it lacks the edge from u to v. By Theorem 3.8, these u-v paths P0 and P00
must be the same. This forces P and P 0 to be the same, which proves part (2).
(2) =⇒ (1): Part (2) implies that T is acyclic. (Otherwise, it is trivial to make
two cycles by adding an extra edge.) We must show T is connected. Suppose T is
disconnected. Let u be a vertex in one component, T1 say, of T and v a vertex in another
component, T2 say, of T . Adding the edge e = uv does not create a cycle (if it did then
T1 and T2 would not be disjoint), which contradicts part (2).
Taking together the results in this section, we have the following characterizations of
trees.
Theorem 3.10. Basic characterizations of trees. If T = (V, E) is a graph with n
vertices, then the following statements are equivalent:
1. T is a tree.
2. T contains no cycles and has n − 1 edges.
3. T is connected and has n − 1 edges.
4. Every edge of T is a cut set.
5. For any pair of distinct vertices u, v ∈ V , there is exactly one u-v path.
6. For any new edge e, the join T + e has exactly one cycle.
Let G = (V1 , E1 ) be a graph and T = (V2 , E2 ) a subgraph of G that is a tree. As in
part (6) of Theorem 3.10, we see that adding just one edge in E1 − E2 to T will create
a unique cycle in G. Such a cycle is called a fundamental cycle of G. The set of such
fundamental cycles of G depends on T .
The following result essentially says that if a tree has at least one edge, then the tree
has at least two vertices each of which has degree one. In other words, each tree of order
≥ 2 has at least two pendants.
Theorem 3.11. Every nontrivial tree has at least two leaves.
3.2. Properties of trees 115
Proof. Let T be a nontrivial tree of order m and size n. Consider the degree sequence
d1 , d2 , . . . , dm of T where d1 ≤ d2 ≤ · · · ≤ dm . As T is nontrivial and connected, then
m ≥ 2 and di ≥ 1 for i = 1, 2, . . . , m. If T has less than two leaves, then d1 ≥ 1 and
di ≥ 2 for 2 ≤ i ≤ m, hence
m
X
di ≥ 1 + 2(m − 1) = 2m − 1. (3.1)
i=1
But by Theorems 1.9 and 3.6, we have
m
X
di = 2n = 2(m − 1) = 2m − 2
i=1
which contradicts inequality (3.1). Conclude that T has at least two leaves.
Theorem 3.12. If T is a tree of order m and G is a graph with minimum degree
δ(G) ≥ m − 1, then T is isomorphic to a subgraph of G.
Proof. Use an inductive argument on the number of vertices. The result holds for m = 1
because K1 is a subgraph of every nontrivial graph. The result also holds for m = 2
since K2 is a subgraph of any graph with at least one edge.
Let m ≥ 3, let T1 be a tree of order m − 1, and let H be a graph with δ(H) ≥ m − 2.
Assume for induction that T1 is isomorphic to a subgraph of H. We need to show that
if T is a tree of order m and G is a graph with δ(G) ≥ m − 1, then T is isomorphic to a
subgraph of G. Towards that end, consider a leaf v of T and let u be a vertex of T such
that u is adjacent to v. Then T − v is a tree of order m − 1 and δ(G) ≥ m − 1 > m − 2.
Apply the inductive hypothesis to see that T − v is isomorphic to a subgraph T 0 of G.
Let u0 be the vertex of T 0 that corresponds to the vertex u of T under an isomorphism.
Since deg(u0 ) ≥ m − 1 and T 0 has m − 2 vertices distinct from u0 , it follows that u0 is
adjacent to some w ∈ V (G) such that w ∈ / V (T 0 ). Therefore T is isomorphic to the
graph obtained by adding the edge u0 w to T 0 .
Example 3.13. Consider a positive integer n. The Euler phi function ϕ(n) counts the
number of integers a, with 1 ≤ a ≤ n, such that gcd(a, n) = 1. The Euler phi sequence
of n is obtained by repeatedly iterating ϕ(n) with initial iteration value n. Continue
on iterating and stop when the output of ϕ(αk ) is 1, for some positive integer αk . The
number of terms generated by the iteration, including the initial iteration value n and
the final value of 1, is the length of ϕ(n).
(a) Let s0 = n, s1 , s2 , . . . , sk = 1 be the Euler phi sequence of n and produce a digraph G
of this sequence as follows. The vertex set of G is V = {s0 = n, s1 , s2 , . . . , sk = 1}
and the edge set of G is E = {si si+1 | 0 ≤ i < k}. Produce the digraphs of the Euler
phi sequences of 15, 22, 33, 35, 69, and 72. Construct the union of all such digraphs
and describe the resulting graph structure.
(b) For each n = 1, 2, . . . , 1000, compute the length of ϕ(n) and plot the pairs (n, ϕ(n))
on one set of axes.
Solution. The Euler phi sequence of 15 is
15, ϕ(15) = 8, ϕ(8) = 4, ϕ(4) = 2, ϕ(2) = 1.
116 Chapter 3. Trees and forests
The Euler phi sequences of 22, 33, 35, 69, and 72 can be similarly computed to obtain
their respective digraph representations. The union of all such digraphs is a directed tree
rooted at 1, as shown in Figure 3.11(a). Figure 3.11(b) shows a scatterplot of n versus
the length of ϕ(n).
1
2
12
4
10
8
length of ϕ(n)
8 10
6
15 20 24 22
4
33 44 35 72
2
0
0 200 400 600 800 1,000
69 n
(a) (b)
Figure 3.11: Union of digraphs of Euler phi sequences and scatterplot.
3.3 Minimum spanning trees
Suppose we want to design an electronic circuit connecting several components. If these
components represent the vertices of a graph and a wire connecting two components
represents an edge of the graph, then for economical reasons we will want to connect the
components together using the least amount of wire. The problem essentially amounts
to finding a minimum spanning tree in the graph containing these vertices.
But what is a spanning tree? We can characterize a spanning tree in several ways,
each leading to an algorithm for constructing a spanning tree. Let G be a connected
graph and let T be a subgraph of G. If T is a tree that contains all the vertices of
G, then T is called a spanning tree of G. We can think of T as a tree that is also an
edge-deletion subgraph of G. That is, we start with a connected graph G and delete an
edge from G such that the resulting edge-deletion subgraph T1 is still connected. If T1 is
a tree, then we have obtained a spanning tree of G. Otherwise, we delete an edge from
T1 to obtain an edge-deletion subgraph T2 that is still connected. If T2 is a tree, then
we are done. Otherwise, we repeat the above procedure until we obtain an edge-deletion
subgraph Tk of G such that Tk is connected, Tk is a tree, and it contains all vertices
of G. Each edge removal does not decrease the number of vertices and must also leave
3.3. Minimum spanning trees 117
the resulting edge-deletion subgraph connected. Thus eventually the above procedure
results in a spanning tree of G. Our discussion is summarized in Algorithm 3.1.
Algorithm 3.1: Randomized spanning tree construction.
Input: A connected graph G.
Output: A spanning tree of G.
1 T ←G
2 while T is not a tree do
3 e ← random edge of T
4 if T − e is connected then
5 T ←T −e
6 return T
Another characterization of a spanning tree T of a connected graph G is that T is a
maximal set of edges of G that contains no cycle. Kruskal’s algorithm (see section 3.3.1)
exploits this condition to construct a minimum spanning tree (MST). A minimum span-
ning tree is a spanning tree of a weighted graph having lowest total weight among all
possible spanning trees of the graph. A third characterization of a spanning tree is that
it is a minimal set of edges that connect all vertices, a characterization that results in
yet another algorithm called Prim’s algorithm (see section 3.3.2) for constructing mini-
mum spanning trees. The task of determining a minimum spanning tree in a connected
weighted graph is called the minimum spanning tree problem. As early as 1926, Otakar
Borůvka stated [36, 37] this problem and offered a solution now known as Borůvka’s
algorithm (see section 3.3.3). See [92, 147] for a history of the minimum spanning tree
problem.
3.3.1 Kruskal’s algorithm
In 1956, Joseph B. Kruskal published [132] a procedure for constructing a minimum span-
ning tree of a connected weighted graph G = (V, E). Now known as Kruskal’s algorithm,
with a suitable implementation the procedure runs in O |E| · log |E| time. Variants
of Kruskal’s algorithm include the algorithm by Prim [171] and that by Loberman and
Weinberger [145].
Kruskal’s algorithm belongs to the class of greedy algorithms. As will be explained
below, when constructing a minimum spanning tree Kruskal’s algorithm considers only
the edge having minimum weight among all available edges. Given a weighted nontrivial
graph G = (V, E) that is connected, let w : E → R be the weight function of G. The
first stage is creating a “skeleton” of the tree T that is initially set to be a graph without
edges, i.e. T = (V, ∅). The next stage involves sorting the edges of G by weights in
nondecreasing order. In other words, we label the edges of G as follows:
E = {e1 , e2 , . . . , en }
where n = |E| and w(e1 ) ≤ w(e2 ) ≤ · · · ≤ w(en ). Now consider each edge ei for
i = 1, 2, . . . , n. We add ei to the edge set of T provided that ei does not result in T
having a cycle. The only way adding ei = ui vi to T would create a cycle is if both ui and
vi were endpoints of edges (not necessarily distinct) in the same connected component
of T . As long as the acyclic condition holds with the addition of a new edge to T , we
118 Chapter 3. Trees and forests
add that new edge. Following the acyclic test, we also test that the (updated) graph
T is a tree of G. As G is a graph of order |V |, apply Theorem 3.10 to see that if T
has size |V | − 1, then it is a spanning tree of G. Algorithm 3.2 provides pseudocode of
our discussion of Kruskal’s algorithm. When the algorithm halts, it returns a minimum
spanning tree of G. The correctness of Algorithm 3.2 is proven in Theorem 3.14.
Algorithm 3.2: Kruskal’s algorithm.
Input: A connected weighted graph G = (V, E) with weight function w.
Output: A minimum spanning tree of G.
1 m ← |V |
2 T ←∅
3 sort E = {e1 , e2 , . . . , en } by weights so that w(e1 ) ≤ w(w2 ) ≤ · · · ≤ w(en )
4 for i ← 1, 2, . . . , n do
5 if ei ∈/ E(T ) and T ∪ {ei } is acyclic then
6 T ← T ∪ {ei }
7 if |T | = m − 1 then
8 return T
Theorem 3.14. Correctness of Algorithm 3.2. If G is a nontrivial connected
weighted graph, then Algorithm 3.2 outputs a minimum spanning tree of G.
Proof. Let G be a nontrivial connected graph of order m and having weight function w.
Let T be a subgraph of G produced by Kruskal’s algorithm 3.2. By construction, T is a
spanning tree of G with
E(T ) = {e1 , e2 , . . . , em−1 }
where w(e1 ) ≤ w(e2 ) ≤ · · · ≤ w(em−1 ) so that the total weight of T is
m−1
X
w(T ) = w(ei ).
i=1
Suppose for contradiction that T is not a minimum spanning tree of G. Among all the
minimum spanning trees of G, let H be a minimum spanning tree of G such that H has
the most number of edges in common with T . As T and H are distinct subgraphs of G,
then T has at least an edge not belonging to H. Let ei ∈ E(T ) be the first edge not in
H. Construct the graph G0 = H + ei obtained by adding the edge ei to H. Note that
G0 has exactly one cycle C. Since T is acyclic, there exists an edge e0 ∈ E(C) such that
e0 is not in T . Construct the graph T0 = G0 − e0 obtained by deleting the edge e0 from
G0 . Then T0 is a spanning tree of G with
w(T0 ) = w(H) + w(ei ) − w(e0 )
and w(H) ≤ w(T0 ) and hence w(e0 ) ≤ w(ei ). By Kruskal’s algorithm 3.2, ei is an edge of
minimum weight such that {e1 , e2 , . . . , ei−1 } ∪ {ei } is acyclic. Furthermore, the subgraph
{e1 , e2 , . . . , ei−1 , e0 } of H is acyclic. Thus we have w(ei ) = w(e0 ) and w(T0 ) = w(H) and
so T is a minimum spanning tree of G. By construction, T0 has more edges in common
with T than H has with T , in contradiction of our hypothesis.
3.3. Minimum spanning trees 119
def kruskal ( G ):
"""
Implements Kruskal ’s algorithm to compute a MST of a graph .
INPUT :
G - a connected edge - weighted graph or digraph
whose vertices are assumed to be 0 , 1 , .... , n -1.
OUTPUT :
T - a minimum weight spanning tree .
If G is not explicitly edge - weighted then the algorithm
assumes all edge weights are 1. The tree T returned is
a weighted graph , even if G is not .
EXAMPLES :
sage : A = matrix ([[0 ,1 ,2 ,3] ,[0 ,0 ,2 ,1] ,[0 ,0 ,0 ,3] ,[0 ,0 ,0 ,0]])
sage : G = DiGraph (A , format = " adjacency_matrix " , weighted = True )
sage : TE = kruskal ( G ); TE . edges ()
[(0 , 1 , 1) , (0 , 2 , 2) , (1 , 3 , 1)]
sage : G . edges ()
[(0 , 1 , 1) , (0 , 2 , 2) , (0 , 3 , 3) , (1 , 2 , 2) , (1 , 3 , 1) , (2 , 3 , 3)]
sage : G = graphs . PetersenGraph ()
sage : TE = kruskal ( G ); TE . edges ()
[(0 , 1 , 1) , (0 , 4 , 1) , (0 , 5 , 1) , (1 , 2 , 1) , (1 , 6 , 1) , (2 , 3 , 1) ,
(2 , 7 , 1) , (3 , 8 , 1) , (4 , 9 , 1)]
TODO :
Add ’’ verbose ’’ option to make steps more transparent .
( Useful for teachers and students .)
"""
T_vertices = G . vertices () # a list of the form range ( n )
T_edges = []
E = G . edges () # a list of triples
# start ugly hack
Er = [ list ( x ) for x in E ]
E0 = []
for x in Er :
x . reverse ()
E0 . append ( x )
E0 . sort ()
E = []
for x in E0 :
x . reverse ()
E . append ( tuple ( x ))
# end ugly hack to get E is sorted by weight
for x in E : # find edges of T
TV = flatten ( T_edges )
u = x [0]
v = x [1]
if not ( u in TV and v in TV ):
T_edges . append ([ u , v ])
# find adj mat of T
if G . weighted ():
AG = G . w e i g h t e d _ a dj a c e n c y _ m a t r i x ()
else :
AG = G . adjacency_matrix ()
GV = G . vertices ()
n = len ( GV )
AT = []
for i in GV :
rw = [0]* n
for j in GV :
if [i , j ] in T_edges :
rw [ j ] = AG [ i ][ j ]
AT . append ( rw )
AT = matrix ( AT )
return Graph ( AT , format = " adjacency_matrix " , weighted = True )
Here is an example. We start with the grid graph. This is implemented in Sage such
that the vertices are given by the coordinates of the grid the graph lies on, as opposed
to 0, 1, . . . , n − 1. Since the above implementation of Kruskal’s algorithm assumes that
the vertices are V = {0, 1, . . . , n − 1}, we first redefine the graph suitable for running
Kruskal’s algorithm on it.
sage : G = graphs . GridGraph ([4 ,4])
120 Chapter 3. Trees and forests
sage : A = G . adjacency_matrix ()
sage : G = Graph (A , format = " adjacency_matrix " , weighted = True )
sage : T = kruskal ( G ); T . edges ()
[(0 , 1 , 1) , (0 , 4 , 1) , (1 , 2 , 1) , (1 , 5 , 1) , (2 , 3 , 1) , (2 , 6 , 1) , (3 ,7 , 1) ,
(4 , 8 , 1) , (5 , 9 , 1) , (6 , 10 , 1) , (7 , 11 , 1) , (8 , 12 , 1) , (9 , 13 , 1) ,
(10 , 14 , 1) , (11 , 15 , 1)]
An illustration of this graph is given in Figure 3.12.
Figure 3.12: Kruskal’s algorithm for the 4 × 4 grid graph.
3.3.2 Prim’s algorithm
Like Kruskal’s algorithm, Prim’s algorithm uses a greedy approach to computing a min-
imum spanning tree of a connected weighted graph G = (V, E), where n = |V | and
m = |E|. The algorithm was developed in 1930 by Czech mathematician V. Jarnı́k [113]
and later independently by R. C. Prim [171] and E. W. Dijkstra [65]. However, Prim
was the first to present an implementation that runs in time O(n2 ). Using 2-heaps, the
runtime can be reduced [124] to O(m log n). With a Fibonacci heap implementation [86],
the runtime can be reduced even further to O(m + n log n).
Pseudocode of Prim’s algorithm is given in Algorithm 3.3. For each v ∈ V , cost[v]
denotes the minimum weight among all edges connecting v to a vertex in the tree T ,
and parent[v] denotes the parent of v in T . During the algorithm’s execution, vertices v
that are not in T are organized in the minimum-priority queue Q, prioritized according
to cost[v]. Lines 1 to 3 set each cost[v] to a number that is larger than any weight in
the graph G, usually written ∞. The parent of each vertex is set to NULL because we
have not yet started constructing the MST T . In lines 4 to 6, we choose an arbitrary
vertex r from V and mark that vertex as the root of T . The minimum-priority queue
is set to be all vertices from V . We set cost[r] to zero, making r the only vertex so far
with a cost that is < ∞. During the first execution of the while loop from lines 7 to 12,
r is the first vertex to be extracted from Q and processed. Line 8 extracts a vertex u
from Q based on the key cost, thus moving u to the vertex set of T . Line 9 considers all
vertices adjacent to u. In an undirected graph, these are simply the neighbors of u. (In a
digraph, one could try to replace adj(u) with the out-neighbors oadj(u). Unfortunately,
in the digraph case the Prim algorithm in general fails to find a minimum spanning tree
with the same orientation as the original digraph.) The while loop updates the cost and
parent fields of each vertex v adjacent to u that is not in T . If parent[v] 6= NULL, then
cost[v] < ∞ and cost[v] is the weight of an edge connecting v to some vertex already in T .
Lines 13 to 14 construct the edge set of the minimum spanning tree and return this edge
set. The proof of correctness of Algorithm 3.3 is similar to the proof of Theorem 3.14.
3.3. Minimum spanning trees 121
Figure 3.13 shows the minimum spanning tree rooted at vertex 1 as a result of running
Prim’s algorithm over a digraph; Figure 3.14 shows the corresponding tree rooted at
vertex 5 of an undirected graph.
Algorithm 3.3: Prim’s algorithm.
Input: A weighted connected graph G = (V, E) with weight function w.
Output: A minimum spanning tree T of G.
1 for each v ∈ V do
2 cost[v] ← ∞
3 parent[v] ← NULL
4 r ← arbitrary vertex of V
5 cost[r] ← 0
6 Q←V
7 while Q 6= ∅ do
8 u ← extractMin(Q)
9 for each v ∈ adj(u) do
10 if v ∈ Q and w(u, v) < cost[v] then
11 parent[v] ← u
12
cost[v] ← w(u, v)
13 T ← (v, parent[v]) | v ∈ V − {r}
14 return T
def prim ( G ):
"""
Implements Prim ’s algorithm to compute a MST of a graph .
INPUT :
G - a connected graph .
OUTPUT :
T - a minimum weight spanning tree .
REFERENCES :
http :// en . wikipedia . org / wiki / Prim ’ s_algorithm
"""
T_vertices = [0] # assumes G . vertices = range ( n )
T_edges = []
E = G . edges () # a list of triples
V = G . vertices ()
# start ugly hack to sort E
Er = [ list ( x ) for x in E ]
E0 = []
for x in Er :
x . reverse ()
E0 . append ( x )
E0 . sort ()
E = []
for x in E0 :
x . reverse ()
E . append ( tuple ( x ))
# end ugly hack to get E is sorted by weight
for x in E :
u = x [0]
v = x [1]
if u in T_vertices and not ( v in T_vertices ):
T_edges . append ([ u , v ])
T_vertices . append ( v )
# found T_vertices , T_edges
# find adj mat of T
if G . weighted ():
AG = G . w e i g h t e d _ a dj a c e n c y _ m a t r i x ()
else :
AG = G . adjacency_matrix ()
122 Chapter 3. Trees and forests
GV = G . vertices ()
n = len ( GV )
AT = []
for i in GV :
rw = [0]* n
for j in GV :
if [i , j ] in T_edges :
rw [ j ] = AG [ i ][ j ]
AT . append ( rw )
AT = matrix ( AT )
return Graph ( AT , format = " adjacency_matrix " , weighted = True )
sage : A = matrix ([[0 ,1 ,2 ,3] , [3 ,0 ,2 ,1] , [2 ,1 ,0 ,3] , [1 ,1 ,1 ,0]])
sage : G = DiGraph (A , format = " adjacency_matrix " , weighted = True )
sage : E = G . edges (); E
[(0 , 1 , 1) , (0 , 2 , 2) , (0 , 3 , 3) , (1 , 0 , 3) , (1 , 2 , 2) , (1 , 3 , 1) , (2 , 0 , 2) ,
(2 , 1 , 1) , (2 , 3 , 3) , (3 , 0 , 1) , (3 , 1 , 1) , (3 , 2 , 1)]
sage : prim ( G )
Multi - graph on 4 vertices
sage : prim ( G ). edges ()
[(0 , 1 , 1) , (0 , 2 , 2) , (1 , 3 , 1)]
1 1
1 2 1 2
3 1 3 1
2 2
0 1 1 2 0 1 1 2
2 2
3 1 3 1
1 3 1 3
3 3
(a) Original digraph. (b) 1st iteration of while loop.
1 1
1 2
3 1
2
0 1 1 2 0 1 2
2
3 1 1
1 3 1
3 3
(c) 2nd iteration of while loop. (d) Final MST.
Figure 3.13: Running Prim’s algorithm over a digraph.
sage : A = matrix ([[0 ,7 ,0 ,5 ,0 ,0 ,0] , [0 ,0 ,8 ,9 ,7 ,0 ,0] , [0 ,0 ,0 ,0 ,5 ,0 ,0] , \
3.3. Minimum spanning trees 123
... [0 ,0 ,0 ,0 ,15 ,6 ,0] , [0 ,0 ,0 ,0 ,0 ,8 ,9] , [0 ,0 ,0 ,0 ,0 ,0 ,11] , [0 ,0 ,0 ,0 ,0 ,0 ,0]])
sage : G = Graph (A , format = " adjacency_matrix " , weighted = True )
sage : E = G . edges (); E
[(0 , 1 , 7) , (0 , 3 , 5) , (1 , 2 , 8) , (1 , 3 , 9) , (1 , 4 , 7) , (2 , 4 , 5) ,
(3 , 4 , 15) , (3 , 5 , 6) , (4 , 5 , 8) , (4 , 6 , 9) , (5 , 6 , 11)]
sage : prim ( G ). edges ()
[(0 , 1 , 7) , (0 , 3 , 5) , (1 , 2 , 8) , (1 , 4 , 7) , (3 , 5 , 6) , (4 , 6 , 9)]
3.3.3 Borůvka’s algorithm
Borůvka’s algorithm [36, 37] is a procedure for finding a minimum spanning tree in a
weighted connected graph G = (V, E) for which all edge weights are distinct. It was first
published in 1926 by Otakar Borůvka but subsequently rediscovered by many others,
including Choquet [57] and Florek et al. [81]. If G has order n = |V | and size m = |E|,
it can be shown that Borůvka’s algorithm runs in time O(m log n).
Algorithm 3.4: Borůvka’s algorithm.
Input: A weighted connected graph G = (V, E) with weight function w. All the
edge weights of G are distinct.
Output: A minimum spanning tree T of G.
1 n ← |V |
2 T ← Kn
3 while |E(T )| < n − 1 do
4 for each component T 0 of T do
5 e0 ← edge of minimum weight that leaves T 0
6 E(T ) ← E(T ) ∪ e0
7 return T
Algorithm 3.4 provides pseudocode of Borůvka’s algorithm. Given a weighted con-
nected graph G = (V, E) all of whose edge weights are distinct, the initialization steps
in lines 1 and 2 construct a spanning forest T of G, i.e. the subgraph of G containing
all of the latter’s vertices and no edges. The initial forest has n components, each being
the trivial graph K1 . The while loop from lines 3 to 6 constructs a spanning tree of
G via a recursive procedure similar to Theorem 3.4. For each component T 0 of T , we
consider all the out-going edges of T 0 and choose an edge e0 that has minimum weight
among all such edges. This edge is then added to the edge set of T . In this way, two
distinct components, each of which is a tree, are joined together by a bridge. At the
end of the while loop, our final graph is a minimum spanning tree of G. Note that the
forest-merging steps in the for loop from lines 4 to 6 are amenable to parallelization,
hence the alternative name to Borůvka’s algorithm: the parallel forest-merging method.
Example 3.15. Figure 3.15 illustrates the gradual construction of a minimum spanning
tree for the undirected graph given in Figure 3.15(a). In this case, we require two
iterations of the while loop in Borůvka’s algorithm in order to obtain the final minimum
spanning tree in Figure 3.15(d).
def which_index (x , L ):
"""
L is a list of sublists ( or tuple of sets or list
of tuples , etc ).
124 Chapter 3. Trees and forests
6 6
11 11
5 9 5 9
8 8
6 4 6 4
15 5 15 5
3 7 2 3 7 2
9 8 9 8
5 1 5 1
7 7
0 0
(a) Original undirected graph. (b) 1st iteration of while loop.
6 6
11 11
5 9 5 9
8 8
6 4 6 4
15 5 15 5
3 7 2 3 7 2
9 8 9 8
5 1 5 1
7 7
0 0
(c) 2nd iteration of while loop. (d) 3rd iteration of while loop.
6 6
11
5 9 5 9
8
6 4 6 4
15 5 5
3 7 2 3 7 2
9 8
5 1 5 1
7 7
0 0
(e) 4th iteration of while loop. (f) Final MST.
Figure 3.14: Running Prim’s algorithm over an undirected graph.
3.3. Minimum spanning trees 125
6 6
8
7 10
3 4 2 3 2
9.5
2
4 5 3 5 4 5
6
9 11
0 1 1 0 1
(a) Original undirected graph. (b) 0th iteration of while loop.
6 6
8
7 10 7 10
3 4 2 3 4 2
2 2
4 5 4 5
0 1 1 0 1 1
(c) 1st iteration of while loop. (d) 2nd iteration of while loop.
Figure 3.15: Recursive construction of MST via Borůvka’s algorithm.
126 Chapter 3. Trees and forests
Returns the index of the first sublist which x belongs
to , or None if x is not in flatten ( L ).
The 0 - th element in
Lx = [ L . index ( S ) for S in L if x in S ]
almost works , but if the list is empty then Lx [0]
throws an exception .
EXAMPLES :
sage : L = [[1 ,2 ,3] ,[4 ,5] ,[6 ,7 ,8]]
sage : which_index (3 , L )
0
sage : which_index (4 , L )
1
sage : which_index (7 , L )
2
sage : which_index (9 , L )
sage : which_index (9 , L ) == None
True
"""
for S in L :
if x in S :
return L . index ( S )
return None
def boruvka ( G ):
"""
Implements Boruvka ’s algorithm to compute a MST of a graph .
INPUT :
G - a connected edge - weighted graph with distinct weights .
OUTPUT :
T - a minimum weight spanning tree .
REFERENCES :
http :// en . wikipedia . org / wiki / Boruvka ’ s_algorithm
"""
T_vertices = [] # assumes G . vertices = range ( n )
T_edges = []
T = Graph ()
E = G . edges () # a list of triples
V = G . vertices ()
# start ugly hack to sort E
Er = [ list ( x ) for x in E ]
E0 = []
for x in Er :
x . reverse ()
E0 . append ( x )
E0 . sort ()
E = []
for x in E0 :
x . reverse ()
E . append ( tuple ( x ))
# end ugly hack to get E is sorted by weight
for e in E :
# create about | V |/2 edges of T " cheaply "
TV = T . vertices ()
if not ( e [0] in TV ) or not ( e [1] in TV ):
T . add_edge ( e )
for e in E :
# connect the " cheapest " components to get T
C = T . c o n n e c t e d _ c o m p o n e n t s _ s u b g r a p h s ()
VC = [ S . vertices () for S in C ]
if not ( e in T . edges ()) and ( which_index ( e [0] , VC ) != which_index ( e [1] , VC )):
if T . is_connected ():
break
T . add_edge ( e )
return T
Some examples using Sage:
sage : A = matrix ([[0 ,1 ,2 ,3] , [4 ,0 ,5 ,6] , [7 ,8 ,0 ,9] , [10 ,11 ,12 ,0]])
sage : G = DiGraph (A , format = " adjacency_matrix " , weighted = True )
sage : boruvka ( G )
Multi - graph on 4 vertices
sage : boruvka ( G ). edges ()
[(0 , 1 , 1) , (0 , 2 , 2) , (0 , 3 , 3)]
3.4. Binary trees 127
sage : A = matrix ([[0 ,2 ,0 ,5 ,0 ,0 ,0] , [0 ,0 ,8 ,9 ,7 ,0 ,0] , [0 ,0 ,0 ,0 ,1 ,0 ,0] ,\
... [0 ,0 ,0 ,0 ,15 ,6 ,0] , [0 ,0 ,0 ,0 ,0 ,3 ,4] , [0 ,0 ,0 ,0 ,0 ,0 ,11] , [0 ,0 ,0 ,0 ,0 ,0 ,0]])
sage : G = Graph (A , format = " adjacency_matrix " , weighted = True )
sage : E = G . edges (); E
[(0 , 1 , 2) , (0 , 3 , 5) , (1 , 2 , 8) , (1 , 3 , 9) , (1 , 4 , 7) ,
(2 , 4 , 1) , (3 , 4 , 15) , (3 , 5 , 6) , (4 , 5 , 3) , (4 ,6 , 4) , (5 , 6 , 11)]
sage : boruvka ( G )
Multi - graph on 7 vertices
sage : boruvka ( G ). edges ()
[(0 , 1 , 2) , (0 , 3 , 5) , (2 , 4 , 1) , (3 , 5 , 6) , (4 , 5 , 3) , (4 , 6 , 4)]
sage : A = matrix ([[0 ,1 ,2 ,5] , [0 ,0 ,3 ,6] , [0 ,0 ,0 ,4] , [0 ,0 ,0 ,0]])
sage : G = Graph (A , format = " adjacency_matrix " , weighted = True )
sage : boruvka ( G ). edges ()
[(0 , 1 , 1) , (0 , 2 , 2) , (2 , 3 , 4)]
sage : A = matrix ([[0 ,1 ,5 ,0 ,4] , [0 ,0 ,0 ,0 ,3] , [0 ,0 ,0 ,2 ,0] , [0 ,0 ,0 ,0 ,0] , [0 ,0 ,0 ,0 ,0]])
sage : G = Graph (A , format = " adjacency_matrix " , weighted = True )
sage : boruvka ( G ). edges ()
[(0 , 1 , 1) , (0 , 2 , 5) , (1 , 4 , 3) , (2 , 3 , 2)]
3.4 Binary trees
A binary tree is a rooted tree with at most two children per parent. Each child is
designated as either a left-child or a right-child . Thus binary trees are also 2-ary trees.
Some examples of binary trees are illustrated in Figure 3.16. Given a vertex v in a
binary tree T of height h, the left subtree of v is comprised of the subtree that spans
the left-child of v and all of this child’s descendants. The notion of a right-subtree of a
binary tree is similarly defined. Each of the left and right subtrees of v is itself a binary
tree with height ≤ h − 1. If v is the root vertex, then each of its left and right subtrees
has height ≤ h − 1, and at least one of these subtrees has height equal to h − 1.
(a) (b) (c) (d)
Figure 3.16: Examples of binary trees.
Theorem 3.16. If T is a complete binary tree of height h, then T has 2h+1 − 1 vertices.
Proof. Argue by induction on h. The assertion of the theorem is trivially true in the
base case h = 0. Let k ≥ 0 and assume for induction that any complete binary tree of
height k has order 2k+1 − 1. Suppose T is a complete binary tree of height k + 1 and
denote the left and right subtrees of T by T1 and T2 , respectively. Each Ti (for i = 1, 2)
is a complete binary tree of height k and by our induction hypothesis Ti has 2k+1 − 1
vertices. Thus T has order
1 + (2k+1 − 1) + (2k+1 − 1) = 2k+2 − 1
as required.
128 Chapter 3. Trees and forests
Theorem 3.16 provides a useful upper bound on the order of a binary tree of a given
height. This upper bound is stated in the following corollary.
Corollary 3.17. A binary tree of height h has at most 2h+1 − 1 vertices.
We now count the number of possible binary trees on n vertices. Let bn be the number
of binary trees of order n. For n = 0, we set b0 = 1. The trivial graph is the only binary
tree with one vertex, hence b1 = 1. Suppose n > 1 and let T be a binary tree on n
vertices. Then the left subtree of T has order 0 ≤ i ≤ n − 1 and the right subtree has
n − 1 − i vertices. As there are bi possible left subtrees and bn−1−i possible right subtrees,
T has a total of bi bn−1−i different combinations of left and right subtrees. Summing from
i = 0 to i = n − 1 and we have
n−1
X
bn = bi bn−1−i . (3.2)
i=0
Expression (3.2) is known as the Catalan recursion and the number bn is the n-th Catalan
number, which we know from problem 1.15 can be expressed in the closed form
1 2n
bn = . (3.3)
n+1 n
Figures 3.17 to 3.19 enumerate all the different binary trees on 2, 3, and 4 vertices,
respectively.
(a) (b)
Figure 3.17: The b2 = 2 binary trees on 2 vertices.
(a) (b) (c) (d) (e)
Figure 3.18: The b3 = 5 binary trees on 3 vertices.
The first few values of (3.3) are
b0 = 1, b1 = 1, b2 = 2, b3 = 5, b4 = 14
which are rather small and of manageable size if we want to explicitly enumerate all
different binary trees with the above orders. However, from n = 4 onwards the value
of bn increases very fast. Instead of enumerating all the bn different binary trees of a
specified order n, a related problem is generating a random binary tree of order n. That
is, we consider the set B as a sample space of bn different binary trees on n vertices,
and choose a random element from B. Such a random element can be generated using
Algorithm 3.5. The list parent holds all vertices with less than two children, each vertex
can be considered as a candidate parent to which we can add a child. An element of
parent is a two-tuple (v, k) where the vertex v currently has k children.
3.4. Binary trees 129
(a) (b) (c) (d) (e)
(f) (g) (h) (i) (j)
(k) (l) (m) (n)
Figure 3.19: The b4 = 14 binary trees on 4 vertices.
Algorithm 3.5: Random binary tree.
Input: Positive integer n.
Output: A random binary tree on n vertices.
1 if n = 1 then
2 return K1
3 v←0
4 T ← null graph
5 add v to T
6 parent ← [(v, 0)]
7 for i ← 1, 2, . . . , n − 1 do
8 (v, k) ← remove random element from parent
9 if k < 1 then
10 add (v, k + 1) to parent
11 add edge (v, i) to T
12 add (i, 0) to parent
13 return T
130 Chapter 3. Trees and forests
3.4.1 Binary codes
What is a code?
A code is a rule for converting data in one format, or well-defined tangible representation,
into sequences of symbols in another format. The finite set of symbols used is called the
alphabet. We shall identify a code as a finite set of symbols which are the image of the
alphabet under this conversion rule. The elements of this set are referred to as codewords.
For example, using the ASCII code, the letters in the English alphabet get converted
into numbers in the set {0, 1, . . . , 255}. If these numbers are written in binary, then
each codeword of a letter has length 8, i.e. eight bits. In this way, we can reformat or
encode a “string” into a sequence of binary symbols, i.e. 0’s and 1’s. Encoding is the
conversion process one way. Decoding is the reverse process, converting these sequences
of code-symbols back into information in the original format.
Codes are used for:
Economy. Sometimes this is called entropy encoding since there is an entropy
function which describes how much information a channel (with a given error rate)
can carry and such codes are designed to maximize entropy as best as possible. In
this case, in addition to simply being given an alphabet A, one might be given a
weighted alphabet, i.e. an alphabet for which each symbol a ∈ A is associated with
a nonnegative number wa ≥ 0 (in practice, this number represents the probability
that the symbol a occurs in a typical word).
Reliability. Such codes are called error-correcting codes, since such codes are de-
signed to communicate information over a noisy channel in such a way that the
errors in transmission are likely to be correctable.
Security. Such codes are called cryptosystems. In this case, the inverse of the
coding function c : A → B ∗ is designed to be computationally infeasible. In other
words, the coding function c is designed to be a trapdoor function.
Other codes are merely simpler ways to communicate information (e.g. flag semaphores,
color codes, genetic codes, braille codes, musical scores, chess notation, football diagrams,
and so on) and have little or no mathematical structure. We shall not study them.
Basic definitions
If every word in the code has the same length, the code is called a block code. If a
code is not a block code, then it is called a variable-length code. A prefix-free code is a
code (typically one of variable-length) with the property that there is no valid codeword
in the code that is a prefix or start of any other codeword.1 This is the prefix-free
condition.
One example of a prefix-free code is the ASCII code. Another example is
00, 01, 100.
On the other hand, a non-example is the code
00, 01, 010, 100
1
In other words, a codeword s = s1 · · · sm is a prefix of a codeword t = t1 · · · tn if and only if m ≤ n
and s1 = t1 , . . . , sm = tm . Codes that are prefix-free are easier to decode than codes that are not
prefix-free.
3.4. Binary trees 131
since the second codeword is a prefix of the third one. Another non-example is Morse
code recalled in Table 3.1, where we use 0 for “·” (“dit”) and 1 for “−” (“dah”). For
example, consider the Morse code for aand the Morse code for w. These codewords
violate the prefix-free condition.
A 01 N 10
B 1000 O 111
C 1010 P 0110
D 100 Q 1101
E 0 R 010
F 0010 S 000
G 110 T 1
H 0000 U 001
I 00 V 0001
J 0111 W 011
K 101 X 1001
L 0100 Y 1011
M 11 Z 1100
Table 3.1: Morse code
Gray codes
We begin with some history.2 Frank Gray (1887–1969) wrote about the so-called Gray
codes in a 1951 paper published in the Bell System Technical Journal and then in 1953
patented a device (used for television sets) based on his paper. However, the idea of
a binary Gray code appeared earlier. In fact, it appeared in an earlier patent (one by
Stibitz in 1943). It was also used in the French engineer E. Baudot’s telegraph machine
of 1878 and in a French booklet by L. Gros on the solution published in 1872 to the
Chinese ring puzzle.
The term “Gray code” is ambiguous. It is actually a large family of sequences of
n-tuples. Let Zm = {0, 1, . . . , m − 1}. More precisely, an m-ary Gray code of length
n (called a binary Gray code when m = 2) is a sequence of all possible (i.e. N = mn )
n-tuples
g1 , g2 , . . . , gN
where
each gi ∈ Znm ,
gi and gi+1 differ by 1 in exactly one coordinate.
In other words, an m-ary Gray code of length n is a particular way to order the set of
all mn n-tuples whose coordinates are taken from Zm . From the transmission/commu-
nication perspective, this sequence has two advantages:
It is easy and fast to produce the sequence, since successive entries differ in only
one coordinate.
2
This history comes from an unpublished section 7.2.1.1 (“Generating all n-tuples”) in volume 4 of
Donald Knuth’s The Art of Computer Programming.
132 Chapter 3. Trees and forests
An error is relatively easy to detect, since we can compare an n-tuple with the
previous one. If they differ in more than one coordinate, we conclude that an error
was made.
Example 3.18. Here is a 3-ary Gray code of length 2:
[0, 0], [1, 0], [2, 0], [2, 1], [1, 1], [0, 1], [0, 2], [1, 2], [2, 2]
and the sequence
[0, 0, 0], [1, 0, 0], [1, 1, 0], [0, 1, 0], [0, 1, 1], [1, 1, 1], [1, 0, 1], [0, 0, 1]
is a binary Gray code of length 3.
Gray codes have applications to engineering, recreational mathematics (solving the
Tower of Hanoi puzzle, The Brain puzzle, the Chinese ring puzzle, etc.), and to math-
ematics (e.g. aspects of combinatorics, computational group theory, and the computa-
tional aspects of linear codes).
Binary Gray codes
Consider the so-called n-hypercube graph Qn , whose first few instances are illustrated
in Figure 1.28. This can be envisioned as the graph whose vertices are the vertices of a
cube in n-space
{(x1 , . . . , xn ) | 0 ≤ xi ≤ 1}
and whose edges are those line segments in Rn connecting two neighboring vertices, i.e.
two vertices that differ in exactly one coordinate. A binary Gray code of length n can
be regarded as a path on the hypercube graph Qn that visits each vertex of the cube
exactly once. In other words, a binary Gray code of length n may be identified with a
Hamiltonian path on the graph Qn . For example, Figure 3.20 illustrates a Hamiltonian
path on Q3 .
Figure 3.20: Viewing Γ3 as a Hamiltonian path on Q3 .
How do we efficiently compute a Gray code? Perhaps the simplest way to state the
idea of quickly constructing the reflected binary Gray code Γn of length n is as follows:
Γ0 = [ ],
Γn = [0, Γn−1 ], [1, Γrev
n−1 ]
3.4. Binary trees 133
where Γrev
m means the Gray code in reverse order. For instance, we have
Γ0 = [ ],
Γ1 = [0], [1] ,
Γ2 = [0, 0], [0, 1], [1, 1], [1, 0]
and so on. This is a nice procedure for creating the entire list at once, which gets very
long very fast. An implementation of the reflected Gray code using Python is given
below.
def graycode ( length , modulus ):
"""
Returns the n - tuple reflected Gray code mod m .
EXAMPLES :
sage : graycode (2 ,4)
[[0 , 0] ,
[1 , 0] ,
[2 , 0] ,
[3 , 0] ,
[3 , 1] ,
[2 , 1] ,
[1 , 1] ,
[0 , 1] ,
[0 , 2] ,
[1 , 2] ,
[2 , 2] ,
[3 , 2] ,
[3 , 3] ,
[2 , 3] ,
[1 , 3] ,
[0 , 3]]
"""
n , m = length , modulus
F = range ( m )
if n == 1:
return [[ i ] for i in F ]
L = graycode (n -1 , m )
M = []
for j in F :
M = M +[ ll +[ j ] for ll in L ]
k = len ( M )
Mr = [0]* m
for i in range (m -1):
i1 = i * int ( k / m ) # this requires Python 3.0 or Sage
i2 = ( i +1)* int ( k / m )
Mr [ i ] = M [ i1 : i2 ]
Mr [m -1] = M [( m -1)* int ( k / m ):]
for i in range ( m ):
if is_odd ( i ):
Mr [ i ]. reverse ()
M0 = []
for i in range ( m ):
M0 = M0 + Mr [ i ]
return M0
Consider the reflected binary code of length 8, i.e. Γ8 . This has 28 = 256 codewords.
Sage can easily create the list plot of the coordinates (x, y), where x is an integer j ∈ Z256
that indexes the codewords in Γ8 and the corresponding y is the j-th codeword in Γ8
converted to decimal. This will give us some idea of how the Gray code “looks” in some
sense. The plot is given in Figure 3.21.
What if we only want to compute the i-th Gray codeword in the Gray code of length
n? Can it be computed quickly without computing the entire list? At least in the case of
the reflected binary Gray code, there is a very simple way to do this. The k-th element in
134 Chapter 3. Trees and forests
200
100
0
0 50 100 150 200 250
Figure 3.21: Scatterplot of Γ8 .
the above-described reflected binary Gray code of length n is obtained by simply adding
the binary representation of k to the binary representation of the integer part of k/2.
An example using Sage is given below.
def int2binary (m , n ):
’’’
returns GF (2) vector of length n obtained
from the binary repr of m , padded by 0 ’ s
( on the left ) to length n .
EXAMPLES :
sage : for j in range (8):
....: print int2binary (j ,3)+ int2binary ( int ( j /2) ,3)
....:
(0 , 0 , 0)
(0 , 0 , 1)
(0 , 1 , 1)
(0 , 1 , 0)
(1 , 1 , 0)
(1 , 1 , 1)
(1 , 0 , 1)
(1 , 0 , 0)
’’’
s = bin ( m )
k = len ( s )
F = GF (2)
b = [ F (0)]* n
for i in range (2 , k ):
b [n - k + i ] = F ( int ( s [ i ]))
return vector ( b )
def graycodeword (m , n ):
’’’
returns the k - th codeword in the reflected binary Gray code
of length n .
EXAMPLES :
sage : graycodeword (3 ,3)
(0 , 1 , 0)
’’’
return map ( int , int2binary (m , n )+ int2binary ( int ( m /2) , n ))
3.5 Huffman codes
An alphabet A is a finite set whose elements are referred to as symbols. A word (or string
or message) over A is a finite sequence of symbols in A and the length of the word is
3.5. Huffman codes 135
the number of symbols it contains. A word is usually written by concatenating symbols
together, e.g. a1 a2 · · · ak (ai ∈ A) is a word of length k.
A commonly occurring alphabet in practice is the binary alphabet B = {0, 1}. A
word over the binary alphabet is a finite sequence of 0’s and 1’s. If A is an alphabet, let
A∗ denote the set of all words in A. The length of a word is denoted by vertical bars.
That is, if w = a1 · · · ak is a word over A, then define |w| : A∗ → Z by
|w| = |a1 · · · ak | = k.
Let A and B be two alphabets. A code for A using B is an injection c : A → B ∗ . By
abuse of notation, we often denote the code simply by the set
C = c(A) = {c(a) | a ∈ A}.
The elements of C are called codewords. If B is the binary alphabet, then C is called a
binary code.
3.5.1 Tree representation
Any binary code can be represented by a tree, as Example 3.19 shows.
Example 3.19. Let B` be the binary code of length ≤ `. Represent codewords of B`
using trees.
Solution. Here is how to represent the code B` consisting of all binary strings of length
≤ `. Start with the root node ε being the empty string. The two children of this node,
v0 and v1 , correspond to the two strings of length 1. Label v0 with a “0” and v1 with
a “1”. The two children of v0 , i.e. v00 and v01 , correspond to the strings of length 2
which start with a 0. Similarly, the two children of v1 , i.e. v10 and v11 , correspond to
the strings of length 2 that each starts with a 1. Continue creating child nodes until we
reach length `, at which point we stop. There are a total of 2`+1 − 1 nodes in this tree
and 2` of them are leaves (vertices of a tree with degree 1, i.e. childless nodes). Note
that the parent of any node is a prefix to that node. Label each node vs with the string
“s”, where s is a binary sequence of length ≤ `. See Figure 3.22 for an example when
` = 2.
ε
0 1
00 01 10 11
Figure 3.22: Tree representation of the binary code B2 .
In general, if C is a code contained in B` , then to create the tree for C, start with
the tree for B` . First, remove all nodes associated to a binary string for which it and
136 Chapter 3. Trees and forests
all of its descendants are not in C. Next, remove all labels which do not correspond to
codewords in C. The resulting labeled graph is the tree associated to the binary code C.
For visualizing the construction of Huffman codes later, it is important to see that
we can reverse this construction to start from such a binary tree and recover a binary
code from it. The codewords are determined by the following rules:
The root node gets the empty codeword.
Each left-ward branch gets a 0 appended to the end of its parent. Each right-ward
branch gets a 1 appended to the end.
3.5.2 Uniquely decodable codes
If c : A → B∗ is a code, then we can extend c to A∗ by concatenation:
c(a1 a2 · · · ak ) = c(a1 )c(a2 ) · · · c(ak ).
If the extension c : A∗ → T ∗ is also an injection, then c is called uniquely decodable.
The property of unique decodability or decipherability informally means that any given
sequence of symbols has at most one interpretation as a sequence of codewords.
Example 3.20. Is the Morse code in Table 3.1 uniquely decodable? Why or why not?
Solution. Note that these Morse codewords all have lengths less than or equal to 4.
Other commonly occurring symbols used (the digits 0 through 9, punctuation symbols,
and some others) are also encodable in Morse code, but they use longer codewords.
Let A denote the English alphabet, B = {0, 1} the binary alphabet, and c : A → B∗
the Morse code. Since c(ET ) = 01 = c(A), it is clear that the Morse code is not uniquely
decodable.
In fact, prefix-free implies uniquely decodable.
Theorem 3.21. If a code c : A → B∗ is prefix-free, then it is uniquely decodable.
Proof. We use induction on the length of a message. We want to show that if x1 · · · xk
and y1 · · · y` are messages with c(x1 ) · · · c(xk ) = c(y1 ) · · · c(y` ), then x1 · · · xk = y1 · · · y` .
This in turn implies k = ` and xi = yi for all i.
The case of length 1 follows from the fact that c : A → B ∗ is injective (by the
definition of code).
Suppose that the statement of the theorem holds for all codes of length < m. We
must show that the length m case is true. Suppose c(x1 ) · · · c(xk ) = c(y1 ) · · · c(y` ), where
m = max(k, `). These strings are equal, so the substring c(x1 ) of the left-hand side
and the substring c(y1 ) of the right-hand side are either equal or one is contained in the
other. If, say, c(x1 ) is properly contained in c(y1 ), then c is not prefix-free. Likewise if
c(y1 ) is properly contained in c(x1 ). Therefore, c(x1 ) = c(y1 ), which implies x1 = y1 .
Now remove this codeword from both sides, so c(x2 ) · · · c(xk ) = c(y2 ) · · · c(y` ). By the
induction hypothesis, x2 · · · xk = y2 · · · y` . These facts together imply k = ` and xi = yi
for all i.
3.5. Huffman codes 137
P
Consider now a weighted alphabet (A, p), where p : A → [0, 1] satisfies a∈A p(a) =
1, and a code c : A → B∗ . In other words, p is a probability distribution on A. Think of
p(a) as the probability that the symbol a arises in a typical message. The average word
length L(c) is3 X
L(c) = p(a) · |c(a)|
a∈A
where | · | is the length of a codeword. Given a weighted alphabet (A, p) as above, a code
c : A → B∗ is called optimal if there is no such code with a smaller average word length.
Optimal codes satisfy the following amazing property. For a proof, which is very easy
and highly recommended for anyone who is curious to see more, refer to section 3.6 of
Biggs [29].
Lemma 3.22. Suppose c : A → B∗ is a binary optimal prefix-free code and let ` =
maxa∈A |c(a)| denote the maximum length of a codeword. The following statements
hold.
1. If |c(a0 )| > |c(a)|, then p(a0 ) ≤ p(a).
2. The subset of codewords of length `, i.e.
C` = {c ∈ c(A) | ` = |c(a)|}
contains two codewords of the form b0 and b1 for some b ∈ B∗ .
3.5.3 Huffman coding
The Huffman code construction is based on the second property in Lemma 3.22. Using
this property, in 1952 David Huffman [111] presented an optimal prefix-free binary code,
which has since been named Huffman code.
Here is the recursive/inductive construction of a Huffman code. We shall regard the
binary Huffman code as a tree, as described above. Suppose that the weighted alphabet
(A, p) has n symbols. We assume inductively that there is an optimal prefix-free binary
code for any weighted alphabet (A0 , p0 ) having < n symbols.
Huffman’s rule 1 Let a, a0 ∈ A be symbols with the smallest weights. Construct a new
weighted alphabet with a, a0 replaced by the single symbol a∗ = aa0 and having
weight p(a∗ ) = p(a) + p(a0 ). All other symbols and weights remain unchanged.
Huffman’s rule 2 For the code (A0 , p0 ) above, if a∗ is encoded as the binary string s,
then the encoded binary string for a is s0 and the encoded binary string for a0 is
s1.
The above two rules tell us how to inductively build the tree representation for the
Huffman code of (A, p) up from its leaves (associated to the low weight symbols).
Find two different symbols of lowest weight, a and a0 . If two such symbols do not
exist, stop. Replace the weighted alphabet with the new weighted alphabet as in
Huffman’s rule 1.
3
In probability terminology, this is the expected value E(X) of the random variable X, which assigns
to a randomly selected symbol in A the length of the associated codeword in c.
138 Chapter 3. Trees and forests
Add two nodes (labeled with a and a0 , respectively) to the tree, with parent a∗ (see
Huffman’s rule 1).
If there are no remaining symbols in A, label the parent a∗ with the empty set and
stop. Otherwise, go to the first step.
These ideas are captured in Algorithm 3.6, which outlines steps to construct a binary
tree corresponding to the Huffman code of an alphabet. Line 2 initializes a minimum-
priority queue Q with the symbols in the alphabet A. Line 3 creates an empty binary
tree that will be used to represent the Huffman code corresponding to A. The for loop
from lines 4 to 10 repeatedly extracts from Q two elements a and b of minimum weights.
We then create a new vertex z for the tree T and also let a and b be vertices of T . The
weight W [z] of z is the sum of the weights of a and b. We let z be the parent of a and b,
and insert the new edges za and zb into T . The newly created vertex z is now inserted
into Q with priority W [z]. After n − 1 rounds of the for loop, the priority queue has only
one element in it, namely the root r of the binary tree T . We extract r from Q (line 11)
and return it together with T (line 12).
Algorithm 3.6: Binary tree representation of Huffman codes.
Input: An alphabet A of n symbols. A weight list W of size n such that W [i] is
the weight of ai ∈ A.
Output: A binary tree T representing the Huffman code of A and the root r of T .
1 n ← |A|
2 Q←A /* minimum priority queue */
3 T ← empty tree
4 for i ← 1, 2, . . . , n − 1 do
5 a ← extractMin(Q)
6 b ← extractMin(Q)
7 z ← node with left child a and right child b
8 add the edges za and zb to T
9 W [z] ← W [a] + W [b]
10 insert z into priority queue Q
11 r ← extractMin(Q)
12 return (T, r)
The runtime analysis of Algorithm 3.6 depends on the implementation of the priority
queue Q. Suppose Q is a simple unsorted list. The initialization on line 2 requires O(n)
time. The for loop from line 4 to 10 is executed exactly n − 1 times. Searching Q to
determine the element of minimum weight requires time at most O(n). Determining two
elements of minimum weights requires time O(2n). The for loop requires time O(2n2 ),
which is also the time requirement for the algorithm. An efficient implementation of
the priority queue Q, e.g. as a binary minimum heap, can lower the running time of
Algorithm 3.6 down to O(n log2 (n)).
Algorithm 3.6 represents the Huffman code of an alphabet as a binary tree T rooted
at r. For an illustration of the process of constructing a Huffman tree, see Figure 3.23.
To determine the actual encoding of each symbol in the alphabet, we feed T and r to
Algorithm 3.7 to obtain the encoding of each symbol. Starting from the root r whose
designated label is the empty string ε, the algorithm traverses the vertices of T in a
3.5. Huffman codes 139
13 2 2 19 4 5 2 7 3
13 2 1 19 4 5 2 7 1 3 1 1
(a) (b)
13 4 19 4 5 2 7 3 13 4 19 4 5 5 7
2 2 2 2 2 3
1 1 1 1
(c) (d)
13 8 19 5 5 7 13 8 19 10 7
4 4 2 3 4 4 5 5
2 2 2 2 2 3
1 1 1 1
(e) (f)
13 15 19 10 23 15 19
7 8 5 5 10 13 7 8
4 4 2 3 5 5 4 4
2 2 2 3 2 2
1 1 1 1
(g) (h)
57
23 34 23 34
10 13 15 19 10 13 15 19
5 5 7 8 5 5 7 8
2 3 4 4 2 3 4 4
2 2 2 2
1 1 1 1
(i) (j)
Figure 3.23: Constructing a Huffman tree.
140 Chapter 3. Trees and forests
breadth-first search fashion. If v is an internal vertex with label e, the label of its left-
child is the concatenation e0 and for the right-child of v we assign the label e1. If v
happens to be a leaf vertex, we take its label to be its Huffman encoding. Any Huffman
encoding assigned to a symbol of an alphabet is not unique. Either of the two children of
an internal vertex can be designated as the left- (respectively, right-) child. The runtime
of Algorithm 3.7 is O(|V |), where V is the vertex set of T .
Algorithm 3.7: Huffman encoding of an alphabet.
Input: A binary tree T representing the Huffman code of an alphabet A. The
root r of T .
Output: A list H representing a Huffman code of A, where H[ai ] corresponds to
a Huffman encoding of ai ∈ A.
1 H ← [] /* list of Huffman encodings */
2 Q ← [r] /* queue of vertices */
3 while length(Q) > 0 do
4 root ← dequeue(Q)
5 if root is a leaf then
6 H[root] ← label of root
7 else
8 a ← left child of root
9 b ← right child of root
10 enqueue(Q, a)
11 enqueue(Q, b)
12 label of a ← label of root + 0
13 label of b ← label of root + 1
14 return H
Example 3.23. Consider the alphabet A = {a, b, c, d, e, f } with corresponding weights
w(a) = 19, w(b) = 2, w(c) = 40, w(d) = 25, w(e) = 31, and w(f ) = 3. Construct a
binary tree representation of the Huffman code of A and determine the encoding of each
symbol of A.
Solution. Use Algorithm 3.6 to construct a binary tree representation of the weighted
alphabet A. The resulting binary tree T is shown in Figure 3.24(a), where ai : wi is an
abbreviation for “vertex ai has weight wi ”. The binary tree is rooted at k. To encode
each alphabetic symbol, input T and k into Algorithm 3.7 to get the encodings shown
in Figure 3.24(b).
3.6 Tree traversals
In computer science, tree traversal refers to the process of examining each vertex in a tree
data structure. Starting at the root of an ordered tree T , we can traverse the vertices of
T in one of various ways.
A level-order traversal of an ordered tree T examines the vertices in increasing order
of depth, with vertices of equal depth being examined according to their prescribed
order. One way to think about level-order traversal is to consider vertices of T having
3.6. Tree traversals 141
k : 120 ε
i : 49 j : 71 0 1
h : 24 d : 25 e : 31 c : 40 00 01 10 11
g:5 a : 19 000 001
b:2 f :3 0000 0001
(a) (b)
Figure 3.24: Binary tree representation of an alphabet and its Huffman encodings.
the same depth as being ordered from left to right in decreasing order of importance.
If [v1 , v2 , . . . , vn ] lists the vertices from left to right at depth k, a decreasing order of
importance can be realized by assigning each vertex a numeric label using a labelling
function L : V (T ) → R such that L(v1 ) < L(v2 ) < · · · < L(vn ). In this way, a vertex
with a lower numeric label is examined prior to a vertex with a higher numeric label. A
level-order traversal of T , whose vertices of equal depth are prioritized according to L,
is an examination of the vertices of T from top to bottom, left to right. As an example,
the level-order traversal of the tree in Figure 3.25 is
42, 4, 15, 2, 3, 5, 7, 10, 11, 12, 13, 14.
Our discussion is formalized in Algorithm 3.8, whose general structure mimics that of
breadth-first search. For this reason, level-order traversal is also known as breadth-first
traversal. Each vertex is enqueued and dequeued exactly once. The while loop is executed
n times, hence we have a runtime of O(n). Another name for level-order traversal is top-
down traversal because we first visit the root node and then work our way down the
tree, increasing the depth as we move downward.
Pre-order traversal is a traversal of an ordered tree using a general strategy similar
to depth-first search. For this reason, pre-order traversal is also referred to as depth-first
traversal. Parents are visited prior to their respective children and siblings are visited
according to their prescribed order. The pseudocode for pre-order traversal is presented
in Algorithm 3.9. Note the close resemblance to Algorithm 3.8; the only significant
change is to use a stack instead of a queue. Each vertex is pushed and popped exactly
once, so the while loop is executed n times, resulting in a runtime of O(n). Using
Algorithm 3.9, a pre-order traversal of the tree in Figure 3.25 is
42, 4, 2, 3, 10, 11, 14, 5, 12, 13, 15, 7.
Whereas pre-order traversal lists a vertex v the first time we visit it, post-order
traversal lists v the last time we visit it. In other words, children are visited prior to their
respective parents, with siblings being visited in their prescribed order. The prefix “pre”
in “pre-order traversal” means “before”, i.e. visit parents before visiting children. On the
142 Chapter 3. Trees and forests
42
4 15
2 3 5 7
10 11 12 13
14
Figure 3.25: Traversing a tree.
Algorithm 3.8: Level-order traversal.
Input: An ordered tree T on n > 0 vertices.
Output: A list of the vertices of T in level-order.
1 L ← []
2 Q ← empty queue
3 r ← root of T
4 enqueue(Q, r)
5 while length(Q) > 0 do
6 v ← dequeue(Q)
7 append(L, v)
8 [u1 , u2 , . . . , uk ] ← ordering of children of v
9 for i ← 1, 2, . . . , k do
10 enqueue(Q, ui )
11 return L
Algorithm 3.9: Pre-order traversal.
Input: An ordered tree T on n > 0 vertices.
Output: A list of the vertices of T in pre-order.
1 L ← []
2 S ← empty stack
3 r ← root of T
4 push(S, r)
5 while length(S) > 0 do
6 v ← pop(S)
7 append(L, v)
8 [u1 , u2 , . . . , uk ] ← ordering of children of v
9 for i ← k, k − 1, . . . , 1 do
10 push(S, ui )
11 return L
3.6. Tree traversals 143
other hand, the prefix “post” in “post-order traversal” means “after”, i.e. visit parents
after having visited their children. The pseudocode for post-order traversal is presented
in Algorithm 3.10, whose general structure bears close resemblance to Algorithm 3.9.
The while loop of the former is executed n times because each vertex is pushed and
popped exactly once, resulting in a runtime of O(n). The post-order traversal of the tree
in Figure 3.25 is
2, 10, 14, 11, 3, 12, 13, 5, 4, 7, 15, 42.
Algorithm 3.10: Post-order traversal.
Input: An ordered tree T on n > 0 vertices.
Output: A list of the vertices of T in post-order.
1 L ← []
2 S ← empty stack
3 r ← root of T
4 push(S, r)
5 while length(S) > 0 do
6 if top(S) is unmarked then
7 mark top(S)
8 [u1 , u2 , . . . , uk ] ← ordering of children of top(S)
9 for i ← k, k − 1, . . . , 1 do
10 push(S, ui )
11 else
12 v ← pop(S)
13 append(L, v)
14 return L
Instead of traversing a tree T from top to bottom as is the case with level-order
traversal, we can reverse the direction of our traversal by traversing a tree from bottom
to top. Called bottom-up traversal , we first visit all the leaves of T and consider the
subtree T1 obtained by vertex deletion of those leaves. We then recursively perform
bottom-up traversal of T1 by visiting all of its leaves and obtain the subtree T2 resulting
from vertex deletion of those leaves of T1 . Apply bottom-up traversal to T2 and its vertex
deletion subtrees until we have visited all vertices, including the root vertex. The result
is a procedure for bottom-up traversal as presented in Algorithm 3.11. In lines 3 to 5,
we initialize the list C to contain the number of children of vertex i. This takes O(m)
time, where m = |E(T )|. Lines 6 to 14 extract all the leaves of T and add them to the
queue Q. From lines 15 to 23, we repeatedly apply bottom-up traversal to subtrees of
T . As each vertex is enqueued and dequeued exactly once, the two loops together run
in time O(n) and therefore Algorithm 3.11 has a runtime of O(n + m). As an example,
a bottom-up traversal of the tree in Figure 3.25 is
2, 7, 10, 12, 13, 14, 15, 5, 11, 3, 4, 42.
Yet another common tree traversal technique is called in-order traversal . However, in-
order traversal is only applicable to binary trees, whereas the other traversal techniques
we considered above can be applied to any tree with at least one vertex. Given a binary
tree T having at least one vertex, in-order traversal first visits the root of T and consider
144 Chapter 3. Trees and forests
Algorithm 3.11: Bottom-up traversal.
Input: An ordered tree T on n > 0 vertices.
Output: A list of the vertices of T in bottom-up order.
1 Q ← empty queue
2 r ← root of T
3 C ← [0, 0, . . . , 0] /* n copies of 0 */
4 for each edge (u, v) ∈ E(T ) do
5 C[u] ← C[u] + 1
6 R ← empty queue
7 enqueue(R, r)
8 while length(R) > 0 do
9 v ← dequeue(R)
10 for each w ∈ children(v) do
11 if C[w] = 0 then
12 enqueue(Q, w)
13 else
14 enqueue(R, w)
15 L ← []
16 while length(Q) > 0 do
17 v ← dequeue(Q)
18 append(L, v)
19 if v 6= r then
20 C[parent(v)] ← C[parent(v)] − 1
21 if C[parent(v)] = 0 then
22 u ← parent(v)
23 enqueue(Q, u)
24 return L
Algorithm 3.12: In-order traversal.
Input: A binary tree T on n > 0 vertices.
Output: A list of the vertices of T in in-order.
1 L ← []
2 S ← empty stack
3 v ← root of T
4 while True do
5 if v 6= NULL then
6 push(S, v)
7 v ← left-child of v
8 else
9 if length(S) = 0 then
10 exit the loop
11 v ← pop(S)
12 append(L, v)
13 v ← right-child of v
14 return L
3.7. Problems 145
its left- and right-children. We then recursively apply in-order traversal to the left and
right subtrees of the root vertex. Notice the symmetry in our description of in-order
traversal: start at the root, then traverse the left and right subtrees in in-order. For this
reason, in-order traversal is sometimes referred to as symmetric traversal. Our discussion
is summarized in Algorithm 3.12. In the latter algorithm, if a vertex does not have a
left-child, then the operation of finding its left-child returns NULL. The same holds when
the vertex does not have a right-child. Since each vertex is pushed and popped exactly
once, it follows that in-order traversal runs in time O(n). Using Algorithm 3.12, an
in-order traversal of the tree in Figure 3.24(b) is
0000, 000, 0001, 00, 001, 0, 01, ε, 10, 1, 11.
3.7 Problems
When solving problems, dig at the roots instead of just hacking at the leaves.
— Anthony J. D’Angelo, The College Blue Book
3.1. Construct all nonisomorphic trees of order 7.
3.2. Let G be a weighted connected graph and let T be a subgraph of G. Then T is a
maximum spanning tree of G provided that the following conditions are satisfied:
(a) T is a spanning tree of G.
(b) The total weight of T is maximum among all spanning trees of G.
Modify Kruskal’s, Prim’s, and Borůvka’s algorithms to return a maximum spanning
tree of G.
3.3. Describe and present pseudocode of an algorithm to construct all spanning trees
of a connected graph. What is the worst-case runtime of your algorithm? How
many of the constructed spanning trees are nonisomorphic to each other? Repeat
the exercise for minimum and maximum spanning trees.
3.4. Consider an undirected, connected simple graph G = (V, E) of order n and size m
and having an integer weight function w : E → Z given by w(e) > 0 for all e ∈ E.
Suppose that G has N minimum spanning trees. Yamada et al. [210] provide
an O(N m ln n) algorithm to construct all the N minimum spanning trees of G.
Describe and provide pseudocode of the Yamada-Kataoka-Watanabe algorithm.
Provide runtime analysis and prove the correctness of this algorithm.
3.5. The solution of Example 3.3 relied on the following result: Let T = (V, E) be a tree
rooted at v0 and suppose v0 has exactly two children. If maxv∈V deg(v) = 3 and
v0 is the only vertex with degree 2, then T is a binary tree. Prove this statement.
Give examples of graphs that are binary trees but do not satisfy the conditions
of the result. Under which conditions would the above test return an incorrect
answer?
3.6. What is the worst-case runtime of Algorithm 3.1?
3.7. Figure 3.5 shows two nonisomorphic spanning trees of the 4 × 4 grid graph.
146 Chapter 3. Trees and forests
(a) For each n = 1, 2, . . . , 7, construct all nonisomorphic spanning trees of the
n × n grid graph.
(b) Explain and provide pseudocode of an algorithm for constructing all spanning
trees of the n × n grid graph, where n > 0.
(c) In general, if n is a positive integer, how many nonisomorphic spanning trees
are there in the n × n grid graph?
(d) Describe and provide pseudocode of an algorithm to generate a random span-
ning tree of the n × n grid graph. What is the worst-case runtime of your
algorithm?
3.8. Theorem 3.4 shows how to recursively construct a new tree from a given collection
of trees, hence it can be considered as a recursive definition of trees. To prove the-
orems based upon recursive definitions, we use a proof technique called structural
induction. Let S(C) be a statement about the collection of structures C, each of
which is defined by a recursive definition. In the base case, prove S(C) for the
basis structure(s) C. For the inductive case, let X be a structure formed using
the recursive definition from the structures Y1 , Y2 , . . . , Yk . Assume for induction
that the statements S(Y1 ), S(Y2 ), . . . , S(Yk ) hold and use the inductive hypotheses
S(Yi ) to prove S(X). Hence conclude that S(X) is true for all X. Apply structural
induction to show that any graph constructed using Theorem 3.4 is indeed a tree.
3.9. In Kruskal’s Algorithm 3.2, line 5 requires that the addition of a new edge to T
does not result in T having a cycle. A tree by definition has no cycles. Suppose
line 5 is changed to:
if ei ∈
/ E(T ) and T ∪ {ei } is a tree then
With this change, explain why Algorithm 3.2 would return a minimum spanning
tree or why the algorithm would fail to do so.
3.10. This problem is concerned with improving the runtime of Kruskal’s Algorithm 3.2.
Explain how to use a priority queue to obviate the need for sorting the edges by
weight. Investigate the union-find data structure. Explain how to use union-find
to ensure that the addition of each edge results in an acyclic graph.
3.11. Figure 3.26 shows a weighted version of the Chvátal graph, which has 12 ver-
tices and 24 edges. Use this graph as input to Kruskal’s, Prim’s, and Borůvka’s
algorithms and compare the resulting minimum spanning trees.
3.12. Algorithm 3.1 presents a randomized procedure to construct a spanning tree of a
given connected graph via repeated edge deletion.
(a) Describe and present pseudocode of a randomized algorithm to grow a span-
ning tree via edge addition.
(b) Would Algorithm 3.1 still work if the input graph G has self-loops or multiple
edges? Explain why or why not. If not, modify Algorithm 3.1 to handle the
case where G has self-loops and multiple edges.
(c) Repeat the previous exercise for Kruskal’s, Prim’s, and Borůvka’s algorithms.
3.7. Problems 147
0
11.4 40.7
17.1 14.4
5.6 35.4
5
1 4
14.5 15
6 9
11.8 8.5 48 3.7
11 10
0.2 9.1
27.1 17
6.9 43.2 10.2 42.7
7 22.1 8
22 36.6
2 3
44.2
Figure 3.26: Weighted Chvátal graph.
Algorithm 3.13: Random spanning tree of Kn .
Input: A positive integer n representing the order of Kn , with vertex set
V = {0, 1, . . . , n − 1}.
Output: A random spanning tree of Kn .
1 if n = 1 then
2 return K1
3 P ← random permutation of V
4 T ← null tree
5 for i ← 1, 2, . . . , n − 1 do
6 j ← random element from {0, 1, . . . , i − 1}
7 add edge (P [j], P [i]) to T
8 return T
148 Chapter 3. Trees and forests
3.13. Algorithm 3.13 constructs a random spanning tree of the complete graph Kn on
n > 0 vertices. Its runtime is dependent on efficient algorithms for obtaining a
random permutation of a set of objects, and choosing a random element from a
given set.
(a) Describe and analyze the runtime of a procedure to construct a random per-
mutation of a set of nonnegative integers.
(b) Describe an algorithm for randomly choosing an element of a set of nonnega-
tive integers. Analyze the runtime of this algorithm.
(c) Taking into consideration the previous two algorithms, what is the runtime of
Algorithm 3.13?
3.14. We want to generate a random undirected, connected simple graph on n vertices
and having m edges. Start by generating a random spanning tree T of Kn . Then
add random edges to T until the requirements are satisfied.
(a) Present pseudocode to realize the above procedure. What is the worst-case
runtime of your algorithm?
(b) Modify your algorithm to handle the case where m < n − 1. Why must
m ≥ n − 1?
(c) Modify your algorithm to handle the case where each edge has a weight within
the closed interval [α, β].
3.15. Enumerate all the different binary trees on 5 vertices.
3.16. Algorithm 3.5 generates a random binary tree on n > 0 vertices. Modify this
algorithm so that it generates a random k-ary tree of order n > 0, where k ≥ 3.
3.17. Show by giving an example that the Morse code is not prefix-free.
3.18. Consider the alphabet A = {a, b, c} with corresponding probabilities (or weights)
p(a) = 0.5, p(b) = 0.3, and p(c) = 0.2. Generate two different Huffman codes for
A and illustrate the tree representations of those codes.
3.19. Find the Huffman code for the letters of the English alphabet weighted by the
frequency of common American usage.4
3.20. Let G = (V1 , E2 ) be a graph and T = (V2 , E2 ) a spanning tree of G. Show that
there is a one-to-one correspondence between fundamental cycles in G and edges
not in T .
3.21. Let G = (V, E) be the 3 × 3 grid graph and T1 = (V1 , E1 ), T2 = (V2 , E2 ) be
spanning trees of G in Example 3.1. Find a fundamental cycle in G for T1 that is
not a fundamental cycle in G for T2 .
3.22. Usually there exist many spanning trees of a graph. Classify those graphs for
which there is only one spanning tree. In other words, find necessary and sufficient
conditions for a graph G such that if T is a spanning tree of G then T is unique.
4
You can find this on the Internet or in the literature. Part of this exercise is finding this frequency
distribution yourself.
3.7. Problems 149
3.23. Convert the function graycodeword into a pure Python function.
3.24. Example 3.13 verifies that for any positive integer n > 1, repeated iteration of
the Euler phi function ϕ(n) eventually produces 1. Show that this is the case or
provide an explanation why it is in general false.
3.25. The Collatz conjecture [134] asserts that for any integer n > 0, repeated iteration
of the function ( 3n+1
2
, if n is odd,
T (n) =
n
2
, if n is even
eventually produces the value 1. For example, repeated iteration of T (n) starting
from n = 22 results in the sequence
22, 11, 17, 26, 13, 20, 10, 5, 8, 4, 2, 1. (3.4)
One way to think about the Collatz conjecture is to consider the digraph G
produced by considering (ai , T (ai )) as a directed edge of G. Then the Collatz
conjecture can be rephrased to say that there is some integer k > 0 such that
(ak , T (ak )) = (2, 1) is a directed edge of G. The graph obtained in this man-
ner is called the Collatz graph of T (n). Given a collection of positive integers
α1 , α2 , . . . , αk , let Gαi be the Collatz graph of the function T (αi ) with initial iter-
ation value αi . Then the union of the Gαi is the directed tree
[
Gαi
i
rooted at 1, called the Collatz tree of (α1 , α2 , . . . , αk ). Figure 3.27 shows such a
tree for the collection of initial iteration values 1024, 336, 340, 320, 106, 104, and
96. See Lagarias [135, 136] for a comprehensive survey of the Collatz conjecture.
(a) The Collatz sequence of a positive integer n > 1 is the integer sequence pro-
duced by repeated iteration of T (n) with initial iteration value n. For example,
the Collatz sequence of n = 22 is the sequence (3.4). Write a Sage function
to produce the Collatz sequence of an integer n > 1.
(b) The Collatz length of n > 1 is the number of terms in the Collatz sequence of
n, inclusive of the starting iteration value and the final integer 1. For instance,
the Collatz length of 22 is 12, that of 106 is 11, and that of 51 is 18. Write
a Sage function to compute the Collatz length of a positive integer n > 1. If
n > 1 is a vertex in a Collatz tree, verify that the Collatz length of n is the
distance d(n, 1).
(c) Describe the Collatz graph produced by the function T (n) with initial iteration
value n = 1.
(d) Fix a positive integer n > 1 and let Li be the Collatz length of the integer
1 ≤ i ≤ n. Plot the pairs (i, Li ) on one set of axes.
3.26. The following result was first published in Wiener [208]. Let T = (V, E) be a tree
of order n > 0. For each edge e ∈ E, let n1 (e) and n2 (e) = n − n1 (e) be the orders
150 Chapter 3. Trees and forests
1
2
4
8
16 5
32 3 10
21 64 6 20
42 128 12 13 40
84 85 256 24 26 80
168 170 512 48 52 53 160
336 340 1024 96 104 106 320
Figure 3.27: The union of Collatz graphs is a tree.
3.7. Problems 151
of the two components of the edge-deletion subgraph T − e. Show that the Wiener
number of T is X
W (T ) = n1 (e) · n2 (e).
e∈E
3.27. The following result [156] was independently discovered in the late 1980s by Merris
and McKay, and is known as the Merris-McKay theorem. Let T be a tree of order
n and let L be its Laplacian matrix having eigenvalues λ1 , λ2 , . . . , λn . Show that
the Wiener number of T is
n−1
X 1
W (T ) = n .
i=1
λ i
3.28. For each of the algorithms below: (i) justify whether or not it can be applied
to multigraphs or multidigraphs; (ii) if not, modify the algorithm so that it is
applicable to multigraphs or multidigraphs.
(a) Randomized spanning tree construction Algorithm 3.1.
(b) Kruskal’s Algorithm 3.2.
(c) Prim’s Algorithm 3.3.
(d) Borůvka’s Algorithm 3.4.
3.29. Section 3.6 provides iterative algorithms for the following tree traversal techniques:
(a) Level-order traversal: Algorithm 3.8.
(b) Pre-order traversal: Algorithm 3.9.
(c) Post-order traversal: Algorithm 3.10.
(d) Bottom-up traversal: Algorithm 3.11.
(e) In-order traversal: Algorithm 3.12.
Rewrite each of the above as recursive algorithms.
3.30. In cryptography, the Merkle signature scheme [154] was introduced in 1987 as an
alternative to traditional digital signature schemes such as the Digital Signature
Algorithm or RSA. Buchmann et al. [46] and Szydlo [187] provide efficient algo-
rithms for speeding up the Merkle signature scheme. Investigate this scheme and
how it uses binary trees to generate digital signatures.
3.31. Consider the finite alphabet A = {a1 , a2 , . . . , ar }. If C is a subset of A∗ , then we say
that C is an r-ary code and call r the radix of the code. McMillan’s theorem [151],
first published in 1956, relates codeword lengths to unique decipherability. In
particular, let C = {c1 , c2 , . . . , cn } be an r-ary code where each ci has length `i . If
C is uniquely decipherable, McMillan’s theorem states that the codeword lengths
`i must satisfy Kraft’s inequality
Xn
1
`i
≤ 1.
i=1
r
Prove McMillan’s theorem.
152 Chapter 3. Trees and forests
3.32. A code C = {c1 , c2 , . . . , cn } is said to be instantaneous if each codeword ci can be
interpreted as soon as it is received. For example, given the the code {01, 010} and
the string 01010, upon receiving the first 0 we are unable to decide whether that
element belong to 01 or 010. However, the code {1, 01} is instantaneous because
given the string 1101 and the first 1, we can interpret the latter as the codeword
1. Prove that a code is instantaneous if and only if it is prefix-free.
3.33. Kraft’s inequality and the accompanying Kraft’s theorem were first published [131]
in 1949 in the Master’s thesis of Leon Gordon Kraft. Kraft’s theorem relates the
inequality to instantaneous codes. Let C = {c1 , c2 , . . . , cn } be an r-ary code where
each codeword ci has length `i . Kraft’s theorem states that C is an instantaneous
code if and only if the codeword lengths satisfy
Xn
1
`i
≤ 1.
i=1
r
Prove Kraft’s theorem.
3.34. Let T be a nontrivial tree and P
let ni count the number of vertices of T that have
degree i. Show that T has 2 + ∞ i=3 (i − 2)ni leaves.
3.35. If a forest F has k trees totalling n vertices altogether, how many edges does F
contain?
3.36. The Lucas number Ln , named after Édouard Lucas, has the following recursive
definition:
2, if n = 0,
Ln = 1, if n = 1,
Ln−1 + Ln−2 , if n > 1.
√
(a) If ϕ = (1 + 5)/2 is the golden ratio, show that
Ln = ϕn + (−ϕ)−n .
(b) Let τ (Wn ) be the number of spanning trees of the wheel graph. Benjamin
and Yerger [23] provide a combinatorial proof that τ (Wn ) = L2n − 2. Present
the Benjamin-Yerger combinatorial proof.
Chapter 4
Tree data structures
— Randall Munroe, xkcd, http://xkcd.com/835/
In Chapters 2 and 3, we discussed various algorithms that rely on priority queues as
one of their fundamental data structures. Such algorithms include Dijkstra’s algorithm,
Prim’s algorithm, and the algorithm for constructing Huffman trees. The runtime of
any algorithm that uses priority queues crucially depends on an efficient implementation
of the priority queue data structure. This chapter discusses the general priority queue
data structure and various efficient implementations based on trees. Section 4.1 provides
some theoretical underpinning of priority queues and considers a simple implementation
of priority queues as sorted lists. Section 4.2 discusses how to use binary trees to realize
an efficient implementation of priority queues called a binary heap. Although very useful
in practice, binary heaps do not lend themselves to being merged in an efficient manner,
a setback rectified in section 4.3 by a priority queue implementation called binomial
heaps. As a further application of binary trees, section 4.4 discusses binary search trees
as a general data structure for managing data in a sorted order.
153
154 Chapter 4. Tree data structures
4.1 Priority queues
A priority queue is essentially a queue data structure with various accompanying rules
regarding how to access and manage elements of the queue. Recall from section 2.2.1
that an ordinary queue Q has the following basic accompanying functions for accessing
and managing its elements:
dequeue(Q) — Remove the front of Q.
enqueue(Q, e) — Append the element e to the end of Q.
If Q is now a priority queue, each element is associated with a key or priority p ∈ X
from a totally ordered set X. A binary relation denoted by an infix operator, say “≤”,
is defined on all elements of X such that the following properties hold for all a, b, c ∈ X:
Totality: We have a ≤ b or b ≤ a.
Antisymmetry: If a ≤ b and b ≤ a, then a = b.
Transitivity: If a ≤ b and b ≤ c, then a ≤ c.
If the above three properties hold for the relation “≤”, then we say that “≤” is a total
order on X and that X is a totally ordered set. In all, if the key of each element of
Q belongs to the same totally ordered set X, we use the total order defined on X to
compare the keys of the queue elements. For example, the set Z of integers is totally
ordered by the “less than or equal to” relation. If the key of each e ∈ Q is an element
of Z, we use the latter relation to compare the keys of elements of Q. In the case of an
ordinary queue, the key of each queue element is its position index.
To extract from a priority queue Q an element of lowest priority, we need to define
the notion of smallest priority or key. Let pi be the priority or key assigned to element
ei of Q. Then pmin is the lowest key if pmin ≤ p for any element key p. The element with
corresponding key pmin is the minimum priority element. Based upon the notion of key
comparison, we define two operations on a priority queue:
insert(Q, e, p) — Insert into Q the element e with key p.
extractMin(Q) — Extract from Q an element having the smallest priority.
An immediate application of priority queues is sorting a finite sequence of items.
Suppose L is a finite list of n > 0 items on which a total order is defined. Let Q be
an empty priority queue. In the first phase of the priority queue sorting algorithm,
we extract each element e ∈ L from L and insert e into Q with key e itself. In other
words, each element e is its own key. This first phase of the sorting algorithm requires
n element extractions from L and n element insertions into Q. The second phase of
the algorithm involves extracting elements from Q via the extractMin operation. Queue
elements are extracted via extractMin and inserted back into L in the order in which
they are extracted from Q. Algorithm 4.1 presents pseudocode of our discussion. The
runtime of Algorithm 4.1 depends on how the priority queue Q is implemented.
4.2. Binary heaps 155
Algorithm 4.1: Sorting a sequence via priority queue.
Input: A finite list L of n > 0 elements on which a total order is defined.
Output: The same list L sorted by the total order relation defined on its
elements.
1 Q ← []
2 for i ← 1, 2, . . . , n do
3 e ← dequeue(L)
4 insert(Q, e, e)
5 for i ← 1, 2, . . . , n do
6 e ← extractMin(Q)
7 enqueue(L, e)
4.1.1 Sequence implementation
A simple way to implement a priority queue is to maintain a sorted sequence. Let
e0 , e1 , . . . , en be a sequence of n + 1 elements with corresponding keys κ0 , κ1 , . . . , κn and
suppose that the κi all belong to the same totally ordered set X having total order ≤.
Using the total order, we assume that the κi are sorted as
κ0 ≤ κ1 ≤ · · · ≤ κn
and ei ≤ ej if and only if κi ≤ κj . Then we consider the queue Q = [e0 , e1 , . . . , en ] as a
priority queue in which the head is always the minimum element and the tail is always
the maximum element. Extracting the minimum element is simply a dequeue operation
that can be accomplished in constant time O(1). However, inserting a new element into
Q takes linear time.
Let e be an element with corresponding key κ ∈ X. Inserting e into Q requires that
we maintain elements of Q sorted according to the total order ≤. If Q is empty, we
simply enqueue e into Q. Suppose now that Q is a nonempty priority queue. If κ ≤ κ0 ,
then e becomes the new head of Q. If κn ≤ κ, then e becomes the new tail of Q. Inserting
a new head or tail into Q each requires constant time O(1). However, if κ1 ≤ κ ≤ κn−1
then we need to traverse Q starting from e1 , searching for a position at which to insert e.
Let ei be the queue element at position i within Q. If κ ≤ κi then we insert e into Q at
position i, thus moving ei to position i + 1. Otherwise we next consider ei+1 and repeat
the above comparison process. By hypothesis, κ1 ≤ κ ≤ κn−1 and therefore inserting e
into Q takes a worst-case runtime of O(n).
4.2 Binary heaps
A sequence implementation of priority queues has the advantage of being simple to
understand. Inserting an element into a sequence-based priority queue requires linear
time, which can quickly become infeasible for queues containing hundreds of thousands
or even millions of elements. Can we do any better? Rather than using a sorted sequence,
we can use a binary tree to realize an implementation of priority queues that is much more
efficient than a sequence-based implementation. In particular, we use a data structure
called a binary heap, which allows for element insertion in logarithmic time.
In [209], Williams introduced the heapsort algorithm and described how to implement
a priority queue using a binary heap. A basic idea is to consider queue elements as
156 Chapter 4. Tree data structures
internal vertices in a binary tree T , with external vertices or leaves being “place-holders”.
The tree T satisfies two further properties:
1. A relational property specifying the relative ordering and placement of queue ele-
ments.
2. A structural property that specifies the structure of T .
The relational property of T can be expressed as follows:
Definition 4.1. Heap-order property. Let T be a binary tree and let v be a vertex of
T other than the root. If p is the parent of v and these vertices have corresponding keys
κp and κv , respectively, then κp ≤ κv .
The heap-order property is defined in terms of the total order used to compare the
keys of the internal vertices. Taking the total order to be the ordinary “less than or
equal to” relation, it follows from the heap-order property that the root of T is always
the vertex with a minimum key. Similarly, if the total order is the usual “greater than
or equal to” relation, then the root of T is always the vertex with a maximum key. In
general, if ≤ is a total order defined on the keys of T and u and v are vertices of T , we
say that u is less than or equal to v if and only if u ≤ v. Furthermore, u is said to be
a minimum vertex of T if and only if u ≤ v for all vertices of T . From our discussion
above, the root is always a minimum vertex of T and is said to be “at the top of the
heap”, from which we derive the name “heap” for this data structure.
Another consequence of the heap-order property becomes apparent when we trace
out a path from the root of T to any internal vertex. Let r be the root of T and let v be
any internal vertex of T . If r, v0 , v1 , . . . , vn , v is an r-v path with corresponding keys
κr , κv0 , κv1 , . . . , κvn , κv
then we have
κr ≤ κv0 ≤ κv1 ≤ · · · ≤ κvn ≤ κv .
In other words, the keys encountered on the path from r to v are arranged in nonde-
creasing order.
The structural property of T is used to enforce that T be of as small a height as
possible. Before stating the structural property, we first define the level of a binary tree.
Recall that the depth of a vertex in T is its distance from the root. Level i of a binary
tree T refers to all vertices of T that have the same depth i. We are now ready to state
the heap-structure property.
Definition 4.2. Heap-structure property. Let T be a binary tree with height h.
Then T satisfies the heap-structure property if T is nearly a complete binary tree. That
is, level 0 ≤ i ≤ h − 1 has 2i vertices, whereas level h has ≤ 2h vertices. The vertices at
level h are filled from left to right.
If a binary tree T satisfies both the heap-order and heap-structure properties, then
T is referred to as a binary heap. By insisting that T satisfy the heap-order property,
we are able to determine the minimum vertex of T in constant time O(1). Requiring
that T also satisfy the heap-structure property allows us to determine the last vertex
of T . The last vertex of T is identified as the right-most internal vertex of T having
the greatest depth. Figure 4.1 illustrates various examples of binary heaps. The heap-
structure property together with Theorem 3.16 result in the following corollary on the
height of a binary heap.
4.2. Binary heaps 157
0
2 3
4 6 8 10
(a)
0
2 3
6 4 8 10
17 13 19 24 23
(b)
1
3 2
6 5 8 10
13 17
(c)
Figure 4.1: Examples of binary heaps with integer keys.
158 Chapter 4. Tree data structures
Corollary 4.3. A binary heap T with n internal vertices has height
h = lg(n + 1) .
Proof. Level h − 1 has at least one internal vertex. Apply Theorem 3.16 to see that T
has at least
2h−2+1 − 1 + 1 = 2h−1
internal vertices. On the other hand, level h − 1 has at most 2h−1 internal vertices.
Another application of Theorem 3.16 shows that T has at most
2h−1+1 − 1 = 2h − 1
internal vertices. Thus n is bounded by
2h−1 ≤ n ≤ 2h − 1.
Taking logarithms of each side in the latter bound results in
lg(n + 1) ≤ h ≤ lg n + 1
and the corollary follows.
0 2 3 6 4 8 10 17 13 19 24 23
0 2 3 4 6 8 10
(a) (b)
1 3 2 6 5 8 10 13 17
(c)
Figure 4.2: Sequence representations of various binary heaps.
4.2.1 Sequence representation
Any binary heap can be represented as a binary tree. Each vertex in the tree must know
about its parent and its two children. However, a more common approach is to represent
a binary heap as a sequence such as a list, array, or vector. Let T be a binary heap
consisting of n internal vertices and let L be a list of n elements. The root vertex is
represented as the list element L[0]. For each index i, the children of L[i] are L[2i + 1]
and L[2i + 2] and the parent of L[i] is
i−1
L .
2
4.2. Binary heaps 159
With a sequence representation of a binary heap, each vertex needs not know about
its parent and children. Such information can be obtained via simple arithmetic on
sequence indices. For example, the binary heaps in Figure 4.1 can be represented as the
corresponding lists in Figure 4.2. Note that it is not necessary to store the leaves of T
in the sequence representation.
4.2.2 Insertion and sift-up
We now consider the problem of inserting a vertex v into a binary heap T . If T is empty,
inserting a vertex simply involves the creation of a new internal vertex. We let that
new internal vertex be v and let its two children be leaves. The resulting binary heap
augmented with v has exactly one internal vertex and satisfies both the heap-order and
heap-structure properties, as shown in Figure 4.3. In other words, any binary heap with
one internal vertex trivially satisfies the heap-order property.
v
(a) (b)
Figure 4.3: Inserting a vertex into an empty binary heap.
Let T now be a nonempty binary heap, i.e. T has at least one internal vertex, and
suppose we want to insert into T an internal vertex v. We must identify the correct leaf
of T at which to insert v. If the n internal vertices of T are r = v0 , v1 , . . . , vn−1 , then by
the sequence representation of T we can identify the last internal vertex vn−1 in constant
time. The correct leaf at which to insert v is the sequence element immediately following
vn−1 , i.e. the element at position n in the sequence representation of T . We replace with
v the leaf at position n in the sequence so that v now becomes the last vertex of T .
The binary heap T augmented with the new last vertex v satisfies the heap-structure
property, but may violate the heap-order property. To ensure that T satisfies the heap-
order property, we perform an operation on T called sift-up that involves possibly moving
v up through various levels of T . Let κv be the key of v and let κp(v) be the key of
v’s parent. If the relation κp(v) ≤ κv holds, then T satisfies the heap-order property.
Otherwise we swap v with its parent, effectively moving v up one level to be at the
position previously occupied by its parent. The parent of v is moved down one level
and now occupies the position where v was previously. With v in its new position, we
perform the same key comparison process with v’s new parent. The key comparison and
swapping continue until the heap-order property holds for T . In the worst case, v would
become the new root of T after undergoing a number of swaps that is proportional to the
height of T . Therefore, inserting a new internal vertex into T can be achieved in time
O(lg n). Figure 4.4 illustrates the insertion of a new internal vertex into a nonempty
binary heap and the resulting sift-up operation to maintain the heap-order property.
Algorithm 4.2 presents pseudocode of our discussion for inserting a new internal vertex
into a nonempty binary heap. The pseudocode is adapted from Howard [110], which
provides a C implementation of binary heaps.
160 Chapter 4. Tree data structures
1 1
2 3 2 3
6 4 8 10 6 4 8 10
17 13 19 24 23 17 13 19 24 23 0
(a) (b)
1 1
2 3 2 3
6 4 8 10 6 4 0 10
17 13 19 24 23 0 17 13 19 24 23 8
(c) (d)
1 1
2 3 2 0
6 4 0 10 6 4 3 10
17 13 19 24 23 8 17 13 19 24 23 8
(e) (f)
1 0
2 0 2 1
6 4 3 10 6 4 3 10
17 13 19 24 23 8 17 13 19 24 23 8
(g) (h)
Figure 4.4: Insert and sift-up in a binary heap.
4.2. Binary heaps 161
Algorithm 4.2: Inserting a new internal vertex into a binary heap.
Input: A nonempty binary heap T , in sequence representation, having n internal
vertices. An element v that is to be inserted as a new internal vertex of T .
Output: The binary heap T augmented with the new internal vertex v.
1 i←n
2 while i > 0 do
3 p ← b(i − 1)/2c
4 if κT [p] ≤ κv then
5 exit the loop
6 else
7 T [i] ← T [p]
8 i←p
9 T [i] ← v
10 return T
4.2.3 Deletion and sift-down
The process for deleting the minimum vertex of a binary heap bears some resemblance
to that of inserting a new internal vertex into the heap. Having removed the minimum
vertex, we must then ensure that the resulting binary heap satisfies the heap-order
property. Let T be a binary heap. By the heap-order property, the root of T has a
key that is minimum among all keys of internal vertices in T . If the root r of T is the
only internal vertex of T , i.e. T is the trivial binary heap, we simply remove r and T now
becomes the empty binary heap or the trivial tree, for which the heap-order property
vacuously holds. Figure 4.5 illustrates the case of removing the root of a binary heap
having one internal vertex.
r
(a) (b)
Figure 4.5: Deleting the root of a trivial binary heap.
We now turn to the case where T has n > 1 internal vertices. Let r be the root
of T and let v be the last internal vertex of T . Deleting r would disconnect T . So we
instead replace the key and information at r with the key and other relevant information
pertaining to v. The root r now has the key of the last internal vertex, and v becomes
a leaf.
At this point, T satisfies the heap-structure property but may violate the heap-order
property. To restore the heap-order property, we perform an operation on T called sift-
down that may possibly move r down through various levels of T . Let c(r) be the child of
r with key that is minimum among all the children of r, and let κr and κc(r) be the keys of
r and c(r), respectively. If κr ≤ κc(r) , then the heap-order property is satisfied. Otherwise
we swap r with c(r), moving r down one level to the position previously occupied by
c(r). Furthermore, c(r) is moved up one level to the position previously occupied by r.
With r in its new position, we perform the same key comparison process with a child of
162 Chapter 4. Tree data structures
r that has minimum key among all of r’s children. The key comparison and swapping
continue until the heap-order property holds for T . In the worst case, r would percolate
all the way down to the level that is immediately above the last level after undergoing a
number of swaps that is proportional to the height of T . Therefore, deleting the minimum
vertex of T can be achieved in time O(lg n). Figure 4.6 illustrates the deletion of the
minimum vertex of a binary heap with at least two internal vertices and the resulting
sift-down process that percolates vertices down through various levels of the heap in order
to maintain the heap-order property. Algorithm 4.3 summarizes our discussion of the
process for extracting the minimum vertex of T while also ensuring that T satisfies the
heap-order property. The pseudocode is adapted from the C implementation of binary
heaps in Howard [110]. With some minor changes, Algorithm 4.3 can be used to change
the key of the root vertex and maintain the heap-order property for the resulting binary
tree.
Algorithm 4.3: Extract the minimum vertex of a binary heap.
Input: A binary heap T , given in sequence representation, having n > 1 internal
vertices.
Output: Extract the minimum vertex of T . With one vertex removed, T must
satisfy the heap-order property.
1 root ← T [0]
2 n←n−1
3 v ← T [n]
4 i←0
5 j←0
6 while True do
7 left ← 2i + 1
8 right ← 2i + 2
9 if left < n and κT [left] ≤ κv then
10 if right < n and κT [right] ≤ κT [left] then
11 j ← right
12 else
13 j ← left
14 else if right < n and κT [right] ≤ κv then
15 j ← right
16 else
17 T [i] ← v
18 exit the loop
19 T [i] ← T [j]
20 i←j
21 return root
4.2.4 Constructing a binary heap
Given a collection of n vertices v0 , v1 , . . . , vn−1 with corresponding keys κ0 , κ1 , . . . , κn−1 ,
we want to construct a binary heap containing exactly those vertices. A basic approach
is to start with a trivial tree and build up a binary heap via successive insertions. As each
4.2. Binary heaps 163
1 23
2 3 2 3
6 4 8 10 6 4 8 10
17 13 19 24 23 17 13 19 24
(a) (b)
23 2
2 3 23 3
6 4 8 10 6 4 8 10
17 13 19 24 17 13 19 24
(c) (d)
2 2
23 3 4 3
6 4 8 10 6 23 8 10
17 13 19 24 17 13 19 24
(e) (f)
2 2
4 3 4 3
6 23 8 10 6 19 8 10
17 13 19 24 17 13 23 24
(g) (h)
Figure 4.6: Delete and sift-down in a binary heap.
164 Chapter 4. Tree data structures
insertion requires O(lg n) time, the method of binary heap construction via successive
insertion of each of the n vertices requires O(n · lg n) time. It turns out we could do a
bit better and achieve the same result in linear time.
Algorithm 4.4: Heapify a binary tree.
Input: A binary tree T , given in sequence representation, having n > 1 internal
vertices.
Output: The binary tree T heapified so that it satisfies the heap-order property.
1 for i ← bn/2c − 1, . . . , 0 do
2 v ← T [i]
3 j←0
4 while True do
5 left ← 2i + 1
6 right ← 2i + 2
7 if left < n and κT [left] ≤ κv then
8 if right < n and κT [right] ≤ κT [left] then
9 j ← right
10 else
11 j ← left
12 else if right < n and κT [right] ≤ κv then
13 j ← right
14 else
15 T [i] ← v
16 exit the while loop
17 T [i] ← T [j]
18 i←j
19 return T
A better approach starts by letting v0 , v1 , . . . , vn−1 be the internal vertices of a binary
tree T . The tree T need not satisfy the heap-order property, but it must satisfy the heap-
structure property. Suppose T is given in sequence representation so that we have the
correspondence vi = T [i] and the last internal vertex of T has index n − 1. The parent
of T [n − 1] has index
n−1
j= .
2
Any vertex of T with sequence index beyond n − 1 is a leaf. In other words, if an internal
vertex has index > j, then the children of that vertex are leaves and have indices ≥ n.
Thus any internal vertex with index ≥ bn/2c has leaves for its children. Conclude that
internal vertices with indices
jnk jnk jnk
, + 1, + 2, . . . , n − 1 (4.1)
2 2 2
have only leaves for their children.
Our next task is to ensure that the heap-order property holds for T . If v is an
internal vertex with index in (4.1), then the subtree rooted at v is trivially a binary
heap. Consider the indices from bn/2c − 1 all the way down to 0 and let i be such an
index, i.e. let 0 ≤ i ≤ bn/2c − 1. We heapify the subtree of T rooted at T [i], effectively
4.3. Binomial heaps 165
performing a sift-down on this subtree. Once we have heapified all subtrees rooted at
T [i] for 0 ≤ i ≤ bn/2c − 1, the resulting tree T is a binary heap. Our discussion is
summarized in Algorithm 4.4.
Earlier in this section, we claimed that Algorithm 4.4 can be used to construct a
binary heap in worst-case linear time. To prove this, let T be a binary tree satisfying the
heap-structure property and having n internal vertices. By Corollary 4.3, T has height
h = dlg(n + 1)e. We perform a sift-down for at most 2i vertices of depth i, where each
sift-down for a subtree rooted at a vertex of depth i takes O(h − i) time. Then the total
time for Algorithm 4.4 is
! !
X X 2−i
O 2i (h − i) = O 2h
0≤i<h 0≤i<h
2h−i
!
X k
= O 2h
k>0
2k
= O 2h+1
= O(n)
P
where we used the closed form k>0 k/2k = 2 for a geometric series and Theorem 3.16.
4.3 Binomial heaps
We are given two binary heaps T1 and T2 and we want to merge them into a single heap.
We could start by choosing to insert each element of T2 into T1 , successively extracting
the minimum element from T2 and insert that minimum element into T1 . If T1 and T2
have m and n elements, respectively, we would perform n extractions from T2 totalling
!
X
O lg k
0<k≤n
time and inserting all of the extracted elements from T2 into T1 requires a total runtime
of !
X
O lg k . (4.2)
n≤k<n+m
We approximate the addition of the two sums by
Z n+m k=n+m
k ln k − k
lg k dk = +C
0 ln 2 k=0
for some constant C. The above method of successive extraction and insertion therefore
has a total runtime of
(n + m) ln(n + m) − n − m
O
ln 2
for merging two binary heaps.
Alternatively, we could slightly improve the latter runtime for merging T1 and T2 by
successively extracting the last internal vertex of T2 . The whole process of extracting
166 Chapter 4. Tree data structures
all elements from T2 in this way takes O(n) time and inserting each of the extracted
elements into T1 still requires the runtime in expression (4.2). We approximate the sum
in (4.2) by
Z k=n+m k=n+m
k ln k − k
lg k dk = +C
k=n ln 2 k=n
for some constant C. Therefore the improved extraction and insertion method requires
(n + m) ln(n + m) − n ln n − m
O −n
ln 2
time in order to merge T1 and T2 .
Can we improve on the latter runtime for merging two binary heaps? It turns out we
can by using a type of mergeable heap called binomial heap that supports merging two
heaps in logarithmic time.
4.3.1 Binomial trees
A binomial heap can be considered as a collection of binomial trees. The binomial tree
of order k is denoted Bk and defined recursively as follows:
1. The binomial tree of order 0 is the trivial tree.
2. The binomial tree of order k > 0 is a rooted tree, where from left to right the
children of the root of Bk are roots of Bk−1 , Bk−2 , . . . , B0 .
Various examples of binomial trees are shown in Figure 4.7. The binomial tree Bk can
also be defined as follows. Let T1 and T2 be two copies of Bk−1 with root vertices r1
and r2 , respectively. Then Bk is obtained by letting, say, r1 be the left-most child of r2 .
Lemma 4.4 lists various basic properties of binomial trees. Property (3) of Lemma 4.4
uses the binomial coefficient, from whence Bk derives its name.
Lemma 4.4. Basic properties of binomial trees. Let Bk be a binomial tree of
order k ≥ 0. Then the following properties hold:
1. The order of Bk is 2k .
2. The height of Bk is k.
k
3. For 0 ≤ i ≤ k, we have i
vertices at depth i.
4. The root of Bk is the only vertex with maximum degree ∆(Bk ) = k. If the children
of the root are numbered k − 1, k − 2, . . . , 0 from left to right, then child i is the
root of the subtree Bi .
Proof. We use induction on k. The base case for each of the above properties is B0 ,
which trivially holds.
(1) By our inductive hypothesis, Bk−1 has order 2k−1 . Since Bk is comprised of two
copies of Bk−1 , conclude that Bk has order
2k−1 + 2k−1 = 2k .
4.3. Binomial heaps 167
(a) B0 (b) B1 (c) B2 (d) B3
(e) B4
(f) B5
Figure 4.7: Binomial trees Bk for k = 0, 1, 2, 3, 4, 5.
168 Chapter 4. Tree data structures
(2) The binomial tree Bk is comprised of two copies of Bk−1 , the root of one copy
being the left-most child of the root of the other copy. Then the height of Bk is one
greater than the height of Bk−1 . By our inductive hypothesis, Bk−1 has height k − 1 and
therefore Bk has height (k − 1) + 1 = k.
(3) Denote by D(k, i) the number of vertices of depth i in Bk . As Bk is comprised
of two copies of Bk−1 , a vertex at depth i in Bk−1 appears once in Bk at depth i and a
second time at depth i + 1. By our inductive hypothesis,
D(k, i) = D(k − 1, i) + D(k − 1, i − 1)
k−1 k−1
= +
i i−1
k
=
i
where we used Pascal’s formula which states that
n+1 n n
= +
r r−1 r
for any positive integers n and r with r ≤ n.
(4) This property follows from the definition of Bk .
Corollary 4.5. If a binomial tree has order n ≥ 0, then the degree of any vertex i is
bounded by deg(i) ≤ lg n.
Proof. Apply properties (1) and (4) of Lemma 4.4.
4.3.2 Binomial heaps
In 1978, Jean Vuillemin [199] introduced binomial heaps as a data structure for im-
plementing priority queues. Mark R. Brown [44, 45] subsequently extended Vuillemin’s
work, providing detailed analysis of binomial heaps and introducing an efficient imple-
mentation.
A binomial heap H can be considered as a collection of binomial trees. Each vertex
in H has a corresponding key and all vertex keys of H belong to a totally ordered set
having total order ≤. The heap also satisfies the following binomial heap properties:
Heap-order property. Let Bk be a binomial tree in H. If v is a vertex of Bk
other than the root and p is the parent of v and having corresponding keys κv and
κp , respectively, then κp ≤ κv .
Root-degree property. For any integer k ≥ 0, H contains at most one binomial
tree whose root has degree k.
If H is comprised of the binomial trees Bk0 , Bk1 , . . . , Bkn for nonnegative integers ki ,
we can consider H as a forest made up of the trees Bki . We can also represent H as a tree
in the following way. List the binomial trees of H as Bk0 , Bk1 , . . . , Bkn in nondecreasing
order of root degrees, i.e. the root of Bki has order less than or equal to the root of Bkj
if and only if ki ≤ kj . The root of H is the root of Bk0 and the root of each Bki has
for its child the root of Bki+1 . Both the forest and tree representations are illustrated in
Figure 4.8 for the binomial heap comprised of the binomial trees B0 , B1 , B3 .
4.3. Binomial heaps 169
(a) Binomial heap as a forest. (b) Binomial heap as a tree.
Figure 4.8: Forest and tree representations of a binomial heap.
The heap-order property for binomial heaps is analogous to the heap-order property
for binary heaps. In the case of binomial heaps, the heap-order property implies that
the root of a binomial tree has a key that is minimum among all vertices in that tree.
However, the similarity more or less ends there. In a tree representation of a binomial
heap, the root of the heap may not necessarily have the minimum key among all vertices
of the heap.
The root-degree property can be used to derive an upper bound on the number of
binomial trees in a binomial heap. If H is a binomial heap with n vertices, then H has
at most 1 + blg nc binomial trees. To prove this result, note that (see Theorem 2.1 and
Corollary 2.1.1 in [174, pp.40–42]) n can be uniquely written in binary representation as
the polynomial
n = ak 2k + ak−1 2k−1 + · · · + a1 21 + a0 20 .
Pblg nc i
The binary representation of n requires 1 + blg nc bits, hence n = i=0 ai 2 . Apply
property (1) of Lemma 4.4 to see that the binomial tree Bi is in H if and only if the i-th
bit is bi = 1. Conclude that H has at most 1 + blg nc binomial trees.
4.3.3 Construction and management
Let H be a binomial heap comprised of the binomial trees Bk0 , Bk1 , . . . , Bkn where the
root of Bki has order less than or equal to the root of Bkj if and only if ki ≤ kj .
Denote by rki the root of the binomial tree Bki . If v is a vertex of H, denote by
child[v] the left-most child of v and by sibling[v] we mean the sibling immediately to
the right of v. Furthermore, let parent[v] be the parent of v and let degree[v] denote
the degree of v. If v has no children, we set child[v] = NULL. If v is one of the roots
rki , we set parent[v] = NULL. And if v is the right-most child of its parent, then we set
sibling[v] = NULL.
The roots rk0 , rk1 , . . . , rkn can be organized as a linked list, called a root list, with
two functions for accessing the next root and the previous root. The root immediately
following rki is denoted next[rki ] = sibling[v] = rki+1 and the root immediately before rki
is written prev[rki ] = rki−1 . For rk0 and rkn , we set next[rkn ] = sibling[v] = NULL and
prev[rk0 ] = NULL. We also define the function head[H] that simply returns rk0 whenever
H has at least one element, and head[H] = NULL otherwise.
170 Chapter 4. Tree data structures
Minimum vertex
To find the minimum vertex, we find the minimum among rk0 , rk1 , . . . , rkm because by
definition the root rki is the minimum vertex of the binomial tree Bki . If H has n vertices,
we need to check at most 1 + blg nc vertices to find the minimum vertex of H. Therefore
determining the minimum vertex of H takes O(lg n) time. Algorithm 4.5 summarizes
our discussion.
Algorithm 4.5: Determine the minimum vertex of a binomial heap.
Input: A binomial heap H of order n > 0.
Output: The minimum vertex of H.
1 u ← NULL
2 v ← head[H]
3 min ← ∞
4 while v 6= NULL do
5 if κv < min then
6 min ← κv
7 u←v
8 v ← sibling[v]
9 return u
Merging heaps
Recall that Bk is constructed by linking the root of one copy of Bk−1 with the root of
another copy of Bk−1 . When merging two binomial heaps whose roots have the same
degree, we need to repeatedly link the respective roots. The root linking procedure runs
in constant time O(1) and is rather straightforward, as presented in Algorithm 4.6.
Algorithm 4.6: Linking the roots of binomial heaps.
Input: Two copies of Bk−1 , one rooted at u and the other at v.
Output: The respective roots of two copies of Bk−1 linked, with one root
becoming the parent of the other.
1 parent[u] ← v
2 sibling[u] ← child[v]
3 child[v] ← u
4 degree[v] ← degree[v] + 1
Besides linking the roots of two copies of Bk−1 , we also need to merge the root lists
of two binomial heaps H1 and H2 . The resulting merged list is sorted in nondecreasing
order of degree. Let L1 be the root list of H1 and let L2 be the root list of H2 . First
we create an empty list L. As the lists Li are already sorted in nondecreasing order of
vertex degree, we use merge sort to merge the Li into a single sorted list. The whole
procedure for merging the Li takes linear time O(n), where n = |L1 | + |L2 | − 1. Refer to
Algorithm 4.7 for pseudocode of the procedure just described.
Having clarified the root linking and root lists merging procedures, we are now ready
to describe a procedure for merging two nonempty binomial heaps H1 and H2 into a
4.3. Binomial heaps 171
Algorithm 4.7: Merging two root lists.
Input: Two root lists L1 and L2 , each containing the roots of binomial trees in
the binomial heaps H1 and H2 , respectively. Each root list Li is sorted in
increasing order of vertex degree.
Output: A single list L that merges the root lists Li and sorted in nondecreasing
order of degree.
1 i←1
2 j←1
3 L ← []
4 n ← |L1 | + |L2 | − 1
5 append(L1 , ∞)
6 append(L2 , ∞)
7 for k ← 0, 1, . . . , n do
8 if deg(L1 [i]) ≤ deg(L2 [j]) then
9 append(L, L1 [i])
10 i←i+1
11 else
12 append(L, L2 [j])
13 j ←j+1
14 return L
single binomial heap H. Initially there are at most two copies of B0 , one from each of
the Hi . If two copies of B0 are present, we let the root of one be the parent of the other
as per Algorithm 4.6, producing B1 as a result. From thereon, we generally have at most
three copies of Bk for some integer k > 0: one from H1 , one from H2 , and the third from
a previous merge of two copies of Bk−1 . In the presence of two or more copies of Bk , we
merge two copies as per Algorithm 4.6 to produce Bk+1 . If Hi has ni vertices, then Hi
has at most 1 + blg ni c binomial trees, from which it is clear that merging H1 and H2
requires
max(1 + blg n1 c, 1 + blg n2 c)
steps. Letting N = max(n1 , n2 ), we see that merging H1 and H2 takes logarithmic time
O(lg N ). The operation of merging two binomial heaps is presented in pseudocode as
Algorithm 4.8, which is adapted from Cormen et al. [61, p.463] and the C implementation
of binomial queues in [110]. A word of warning is order here. Algorithm 4.8 is destructive
in the sense that it modifies the input heaps Hi in-place without making copies of those
heaps.
Vertex insertion
Let v be a vertex with corresponding key κv and let H1 be a binomial heap of n vertices.
The single vertex v can be considered as a binomial heap H2 comprised of exactly the
binomial tree B0 . Then inserting v into H1 is equivalent to merging the heaps Hi and
can be accomplished in O(lg n) time. Refer to Algorithm 4.9 for pseudocode of this
straightforward procedure.
172 Chapter 4. Tree data structures
Algorithm 4.8: Merging two binomial heaps.
Input: Two binomial heaps H1 and H2 .
Output: A binomial heap H that results from merging the Hi .
1 H ← empty binomial heap
2 head[H] ← merge sort the root lists of H1 and H2
3 if head[H] = NULL then
4 return H
5 prevv ← NULL
6 v ← head[H]
7 nextv ← sibling[v]
8 while nextv 6= NULL do
9 if degree[v] 6= degree[nextv] or (sibling[nextv] 6= NULL and
degree[sibling[nextv]] = degree[v]) then
10 prevv ← v
11 v ← nextv
12 else if κv ≤ κnextv then
13 sibling[v] ← sibling[nextv]
14 link the roots nextv and v as per Algorithm 4.6
15 else
16 if prevv = NULL then
17 head[H] ← nextv
18 else
19 sibling[prevv] ← nextv
20 link the roots v and nextv as per Algorithm 4.6
21 v ← nextv
22 nextv ← sibling[v]
23 return H
Algorithm 4.9: Insert a vertex into a binomial heap.
Input: A binomial heap H and a vertex v.
Output: The heap H with v inserted into it.
1 H1 ← empty binomial heap
2 head[H1 ] ← v
3 parent[v] ← NULL
4 child[v] ← NULL
5 sibling[v] ← NULL
6 degree[v] ← 0
7 H ← merge H and H1 as per Algorithm 4.8
4.4. Binary search trees 173
Delete minimum vertex
Extracting the minimum vertex from a binomial heap H consists of several phases.
Let H be comprised of the binomial trees Bk0 , Bk1 , . . . , Bkm with corresponding roots
rk0 , rk1 , . . . , rkm and let n be the number of vertices in H. In the first phase, from among
the rki we identify the root v with minimum key and remove v from H, an operation
that runs in O(lg n) time because we need to process at most 1 + blg nc roots. With the
binomial tree Bk rooted at v thus severed from H, we now have a forest consisting of the
heap without Bk (denote this heap by H1 ) and the binomial tree Bk . By construction,
v is the root of Bk and the children of v from left to right can be considered as roots
of binomial trees as well, say B`s , B`s−1 , . . . , B`0 where `s > `s−1 > · · · > `0 . Now
sever the root v from its children. The B`j together can be viewed as a binomial heap
H2 with, from left to right, binomial trees B`0 , B`1 , . . . , B`s . Finally the binomial heap
resulting from removing v can be obtained by merging H1 and H2 in O(lg n) time as per
Algorithm 4.8. In total we can extract the minimum vertex of H in O(lg n) time. Our
discussion is summarized in Algorithm 4.10 and an illustration of the extraction process
is presented in Figure 4.9.
Algorithm 4.10: Extract the minimum vertex from a binomial heap.
Input: A binomial heap H.
Output: The minimum vertex of H removed.
1 v ← extract minimum vertex from root list of H
2 H2 ← empty binomial heap
3 L ← list of v’s children reversed
4 head[H2 ] ← L[0]
5 H ← merge H and H2 as per Algorithm 4.8
6 return v
4.4 Binary search trees
A binary search tree (BST) is a rooted binary tree T = (V, E) having vertex weight
function κ : V → R. The weight of each vertex v is referred to as its key, denoted κv .
Each vertex v of T satisfies the following properties:
Left subtree property. The left subtree of v contains only vertices whose keys
are at most κv . That is, if u is a vertex in the left subtree of v, then κu ≤ κv .
Right subtree property. The right subtree of v contains only vertices whose
keys are at least κv . In other words, any vertex u in the right subtree of v satisfies
κv ≤ κu .
Recursion property. Both the left and right subtrees of v must also be binary
search trees.
The above are collectively called the binary search tree property. See Figure 4.10 for
an example of a binary search tree. Based on the binary search tree property, we can
use in-order traversal (see Algorithm 3.12) to obtain a listing of the vertices of a binary
search tree sorted in nondecreasing order of keys.
174 Chapter 4. Tree data structures
70 65 60 40 1
67 66 68 41 43 44 2 3 5 7
69 45 48 49 3 6 9 8 12 15
47 4 8 11 9
10
(a)
70 65 60 40 1
67 66 68 41 43 44 2 3 5 7
69 45 48 49 3 6 9 8 12 15
47 4 8 11 9
10
(b)
70 65 60 40 7 5 3 2
67 66 68 41 43 44 15 8 12 3 6 9
69 45 48 49 9 4 8 11
47 10
(c)
7 5 3 2
70 65 15 60 8 12 40 3 6 9
67 66 68 9 41 43 44 4 8 11
69 45 48 49 10
47
(d)
Figure 4.9: Extracting the minimum vertex from a binomial heap.
4.4. Binary search trees 175
10
5 15
4 7 13 20
3 5 5 8 11
Figure 4.10: A binary search tree.
4.4.1 Searching
Given a BST T and a key k, we want to locate a vertex (if one exists) in T whose key is k.
The search procedure for a BST is reminiscent of the binary search algorithm discussed
in problem 2.10. We begin by examining the root v0 of T . If κv0 = k, the search is
successful. However, if κv0 6= k then we have two cases to consider. In the first case, if
k < κv0 then we search the left subtree of v0 . The second case occurs when k > κv0 , in
which case we search the right subtree of v0 . Repeat the process until a vertex v in T
is found for which k = κv or the indicated subtree is empty. Whenever the target key
is different from the key of the vertex we are currently considering, we move down one
level of T . Thus if h is the height of T , it follows that searching T takes a worst-case
runtime of O(h). The above procedure is presented in pseudocode as Algorithm 4.11.
Note that if a vertex v does not have a left subtree, the operation of locating the root
of v’s left subtree should return NULL. A similar comment applies when v does not have
a right subtree. Furthermore, from the structure of Algorithm 4.11, if the input BST is
empty then NULL is returned. See Figure 4.11 for an illustration of locating vertices with
given keys in a BST.
Algorithm 4.11: Locate a key in a binary search tree.
Input: A binary search tree T and a target key k.
Output: A vertex in T with key k. If no such vertex exists, return NULL.
1 v ← root[T ]
2 while v 6= NULL and k 6= κv do
3 if k < κv then
4 v ← leftchild[v]
5 else
6 v ← rightchild[v]
7 return v
From the binary search tree property, deduce that a vertex of a BST T with minimum
key can be found by starting from the root of T and repeatedly traversing left subtrees.
When we have reached the left-most vertex v of T , querying for the left subtree of v
should return NULL. At this point, we conclude that v is a vertex with minimum key.
Each query for the left subtree moves us one level down T , resulting in a worst-case
runtime of O(h) with h being the height of T . See Algorithm 4.12 for pseudocode of the
procedure.
The procedure for finding a vertex with maximum key is analogous to that for finding
176 Chapter 4. Tree data structures
10 10
5 15 5 15
4 7 13 20 4 7 13 20
3 5 5 8 11 19 23 3 5 5 8 11 19 23
2 3 9 22 26 2 3 9 22 26
(a) Vertex with key 6: search fail. (b) Vertex with key 22: search success.
Figure 4.11: Finding vertices with given keys in a BST.
10 10
5 15 5 15
4 7 13 20 4 7 13 20
3 5 5 8 11 19 23 3 5 5 8 11 19 23
2 3 9 22 26 2 3 9 22 26
(a) Minimum vertex. (b) Maximum vertex.
Figure 4.12: Locating minimum and maximum vertices in a BST.
10 10
5 15 5 15
4 7 13 20 4 7 13 20
3 5 5 8 11 19 23 3 5 5 8 11 19 23
2 3 9 22 26 2 3 9 22 26
(a) Successor of 9. (b) Predecessor of 11.
Figure 4.13: Searching for successor and predecessor.
4.4. Binary search trees 177
one with minimum key. Starting from the root of T , we repeatedly traverse right subtrees
until we encounter the right-most vertex, which by the binary search tree property has
maximum key. This procedure has the same worst-case runtime of O(h). Figure 4.12
illustrates the process of locating the minimum and maximum vertices of a BST.
Algorithm 4.12: Finding a vertex with minimum key in a BST.
Input: A nonempty binary search tree T .
Output: A vertex of T with minimum key.
1 v ← root of T
2 while leftchild[v] 6= NULL do
3 v ← leftchild[v]
4 return v
Corresponding to the notions of left- and right-children, we can also define successors
and predecessors as follows. Suppose v is not a maximum vertex of a nonempty BST
T . The successor of v is a vertex in T distinct from v with the smallest key greater
than or equal to κv . Similarly, for a vertex v that is not a minimum vertex of T , the
predecessor of v is a vertex in T distinct from v with the greatest key less than or equal
to κv . The notions of successors and predecessors are concerned with relative key order,
not a vertex’s position within the hierarchical structure of a BST. For instance, from
Figure 4.10 we see that the successor of the vertex u with key 8 is the vertex v with key
10, i.e. the root, even though v is an ancestor of u. The predecessor of the vertex a with
key 4 is the vertex b with key 3, i.e. the minimum vertex, even though b is a descendant
of a.
We now describe a method to systematically locate the successor of a given vertex.
Let T be a nonempty BST and v ∈ V (T ) not a maximum vertex of T . If v has a right
subtree, then we find a minimum vertex of v’s right subtree. In case v does not have
a right subtree, we backtrack up one level to v’s parent u = parent(v). If v is the root
of the right subtree of u, we backtrack up one level again to u’s parent, making the
assignments v ← u and u ← parent(u). Otherwise we return v’s parent. Repeat the
above backtracking procedure until the required successor is found. Our discussion is
summarized in Algorithm 4.13. Each time we backtrack to a vertex’s parent, we move
up one level, hence the worst-case runtime of Algorithm 4.13 is O(h) with h being the
height of T . The procedure for finding predecessors is similar. Refer to Figure 4.13 for
an illustration of locating successors and predecessors.
4.4.2 Insertion
Inserting a vertex v into a BST T is rather straightforward. If T is empty, we let v be the
root of T . Otherwise T has at least one vertex. In that case, we need to locate a vertex
in T that can act as a parent and “adopt” v as a child. To find a candidate parent, let
u be the root of T . If κv < κu then we assign the root of the left subtree of u to u itself.
Otherwise we assign the root of the right subtree of u to u. We then carry on the above
key comparison process until the operation of locating the root of a left or right subtree
returns NULL. At this point, a candidate parent for v is the last non-NULL value of u. If
κv < κu then we let v be u’s left-child. Otherwise v is the right-child of u. After each key
comparison, we move down at most one level so that in the worst-case inserting a vertex
178 Chapter 4. Tree data structures
Algorithm 4.13: Finding successors in a binary search tree.
Input: A nonempty binary search tree T and a vertex v that is not a maximum
of T .
Output: The successor of v.
1 if rightchild[v] 6= NULL then
2 return minimum vertex of v’s right subtree as per Algorithm 4.12
3 u ← parent(v)
4 while u 6= NULL and v = rightchild[u] do
5 v←u
6 u ← parent(u)
7 return u
into T takes O(h) time, where h is the height of T . Algorithm 4.14 presents pseudocode
of our discussion and Figure 4.14 illustrates how to insert a vertex into a BST.
Algorithm 4.14: Inserting a vertex into a binary search tree.
Input: A binary search tree T and a vertex x to be inserted into T .
Output: The same BST T but augmeneted with x.
1 u ← NULL
2 v ← root of T
3 while v 6= NULL do
4 u←v
5 if κx < κv then
6 v ← leftchild[v]
7 else
8 v ← rightchild[v]
9 parent[x] ← u
10 if u = NULL then
11 root[T ] ← x
12 else
13 if κx < κu then
14 leftchild[u] ← x
15 else
16 rightchild[u] ← x
4.4.3 Deletion
Whereas insertion into a BST is straightforward, removing a vertex requires much more
work. Let T be a nonempty binary search tree and suppose we want to remove v ∈ V (T )
from T . Having located the position that v occupies within T , we need to consider three
separate cases: (1) v is a leaf; (2) v has one child; (3) v has two children.
1. If v is a leaf, we simply remove v from T and the procedure is complete. The
resulting tree without v satisfies the binary search tree property.
4.4. Binary search trees 179
10 10
5 15 5 15
4 7 13 20 4 7 13 20
3 5 5 8 11 19 23 3 5 5 8 11 19 23
2 3 9 22 26 2 3 9 12 22 26
(a) (b)
Figure 4.14: Inserting into a binary search tree.
Algorithm 4.15: Deleting a vertex from a binary search tree.
Input: A nonempty binary search tree T and a vertex x ∈ V (T ) to be removed
from T .
Output: The same BST T but without x.
1 u ← NULL
2 v ← NULL
3 if leftchild[x] 6= NULL or rightchild[x] 6= NULL then
4 v←x
5 else
6 v ← successor of x
7 if leftchild[v] 6= NULL then
8 u ← leftchild[v]
9 else
10 u ← rightchild[v]
11 if u 6= NULL then
12 parent[u] ← parent[v]
13 if parent[v] = NULL then
14 root[T ] ← u
15 else
16 if v = leftchild[parent[v]] then
17 leftchild[parent[v]] ← u
18 else
19 rightchild[parent[v]] ← u
20 if v 6= x then
21 κx ← κv
22 copy v’s auxilary data into x
180 Chapter 4. Tree data structures
2. Suppose v has the single child u. Removing v would disconnect T , a situation that
can be prevented by splicing out u and letting u occupy the position previously
held by v. The resulting tree with v removed as described satisfies the binary
search tree property.
3. Finally suppose v has two children and let s and p be the successor and predecessor
of v, respectively. It can be shown that s has no left-child and p has no right-child.
We can choose to either splice out s or p. Say we choose to splice out s. Then we
remove v and let s hold the position previously occupied by v. The resulting tree
with v thus removed satisfies the binary search tree property.
The above procedure is summarized in Algorithm 4.15, which is adapted from [61, p.262].
Figure 4.15 illustrates the various cases to be considered when removing a vertex from
a BST. Note that in Algorithm 4.15, the process of finding the successor dominates the
runtime of the entire algorithm. Other operations in the algorithm take at most constant
time. Therefore deleting a vertex from a binary search tree can be accomplished in worst-
case O(h) time, where h is the height of the BST under consideration.
4.5 AVL trees
To motivate the need for AVL trees, note the lack of a structural property for binary
search trees similar to the structural property for binary heaps. Unlike binary heaps,
a BST is not required to have as small a height as possible. As a consequence, any
given nonempty collection C = {v0 , v1 , . . . , vk } of weighted vertices can be represented by
various BSTs with different heights; see Figure 4.16. Some BST representations of C have
heights smaller than other BST representations of C. Those BST representations with
smaller heights can result in reduced time for basic operations such as search, insertion,
and deletion and out-perform BST representations having larger heights. To achieve
logarithmic or near-logarithmic time complexity for basic operations, it is desirable to
maintain a BST with as small a height as possible.
Adelson-Velskiı̆ and Landis [1] introduced in 1962 a criterion for constructing and
maintaining binary search trees having logarithmic heights. Recall that the height of a
tree is the maximum depth of the tree. Then the Adelson-Velskiı̆-Landis criterion can
be expressed as follows.
Definition 4.6. Height-balance property. Let T be a binary tree and suppose v is
an internal vertex of T . Let h` be the height of the left subtree of v and let hr be the
height of v’s right subtree. Then v is said to be height-balanced if |h` − hr | ≤ 1. For each
internal vertex u of T , if u is height-balanced then the whole tree T is height-balanced.
Binary trees having the height-balance property are called AVL trees. The structure
of such trees is such that given any internal vertex v of an AVL tree, the heights of the
left and right subtrees of v differ by at most 1. Complete binary trees are trivial examples
of AVL trees, as are nearly complete binary trees. A less trivial example of AVL trees
are what is known as Fibonacci trees, so named because the construction of Fibonacci
trees bears some resemblance to how Fibonacci numbers are produced. Fibonacci trees
can be constructed recursively in the following manner. The Fibonacci tree F0 of height
0 is the trivial tree. The Fibonacci tree F1 of height 1 is a binary tree whose left and
right subtrees are both F0 . For n > 1, the Fibonacci tree Fn of height n is a binary
4.5. AVL trees 181
10 10
5 15 5 15
4 7 13 20 4 7 13 20
3 5 5 8 11 19 23 3 5 5 8 11 19 23
2 3 9 22 26 2 3 22 26
(a) Target vertex 9 is a leaf. (b) Leaf deleted.
10 10
5 15 5 15
4 7 13 20 4 7 12 20
3 5 5 8 12 19 23 3 5 5 8 11 19 23
2 3 11 22 26 2 3 22 26
(c) Target vertex 13 has one child. (d) Vertex deleted.
10
5 15 10
4 7 13 20 5 16
3 5 5 8 12 18 23 4 7 13 20
2 3 11 17 19 22 26 3 5 5 8 12 18 23
16 2 3 11 17 19 22 26
(e) Target vertex 15 has two children. (f) Vertex deleted.
Figure 4.15: Deleting a vertex from a binary search tree.
182 Chapter 4. Tree data structures
20 4
15 5
13 7
13
10 10
10 7 15
7 13
5 15 5 10 20
5 15
4 7 13 20 4 4 20
(a) (b) (c) (d)
Figure 4.16: Different structural representations of a BST.
tree whose left and right subtrees are Fn−2 and Fn−1 , respectively. Refer to Figure 4.17
for examples of Fibonacci trees; Figure 4.18 shows F6 together with subtree heights for
vertex labels.
(a) F0 (b) F1 (c) F2 (d) F3 (e) F4
(f) F5
Figure 4.17: Fibonacci trees of heights n = 0, 1, 2, 3, 4, 5.
Theorem 4.7. Logarithmic height. The height h of an AVL tree with n internal
vertices is bounded by
lg(n + 1) ≤ h < 2 · lg n + 1.
Proof. Any binary tree of height h has at most 2i leaves. From the proof of Corollary 4.3,
we see that n is bounded by 2h−1 ≤ n ≤ 2h − 1 and in particular n + 1 ≤ 2h . Take the
logarithm of both sides to get h ≥ lg(n + 1).
4.5. AVL trees 183
6
4 5
2 3 3 4
0 1 1 2 1 2 2 3
0 0 0 0 0 1 0 0 0 1 0 1 1 2
0 0 0 0 0 0 0 0 0 1
0 0
Figure 4.18: Fibonacci tree F6 with subtree heights for vertex labels.
Now instead of deriving an upper bound for h, we find the minimum order of an AVL
tree and from there derive the required upper bound for h. Let T be an AVL tree of
minimum order. One subtree of T has height h−1. The other subtree has height h−1 or
h − 2. Our objective is to construct T to have as small a number of vertices as possible.
Without loss of generality, let the left and right subtrees of T have heights h − 2 and
h − 1, respectively. The Fibonacci tree Fh of height h fits the above requirements for T .
If N (h) denote the number of internal vertices of Fh , then N (h) = 1+N (h−1)+N (h−2)
is strictly increasing so
N (h) > N (h − 2) + N (h − 2) = 2 · N (h − 2). (4.3)
Repeated application of (4.3) shows that
N (h) > 2i · N (h − 2i) (4.4)
for any integer i such that h − 2i ≥ 1. Choose i so that h − 2i = 1 or h − 2i = 2, say the
former. Substitute i = (h − 1)/2 into (4.4) yields N (h) > 2(h−1)/2 . That is, n > 2(h−1)/2
and taking logarithm of both sides yields h < 2 · lg n + 1.
An immediate consequence of Theorem 4.7 is that any binary search tree implemented
as an AVL tree should have at most logarithmic height. Contrast this with a general BST
of order N1 , whose height can be as low as logarithmic in N1 or as high as linear in N1 .
Translating to search time, we see that searching a general BST using Algorithm 4.11
is in the worst case O(N1 ), which is no better than searching a sorted list. However,
if N2 is the order of an AVL tree endowed with the binary search tree property, then
searching the AVL tree using Algorithm 4.11 has worst-case O(lg N2 ) runtime. While the
worst-case runtime of searching a general BST can vary between O(lg N1 ) and O(N1 ),
that for an AVL tree with the binary search tree property is at most O(lg N2 ).
4.5.1 Insertion
The algorithm for insertion into a BST can be modified and extended to support insertion
into an AVL tree. Let T be an AVL tree having the binary search tree property, and v
184 Chapter 4. Tree data structures
a vertex to be inserted into T . In the trivial case, T is the null tree so inserting v into T
is equivalent to letting T be the trivial tree rooted at v. Consider now the case where T
has at least one vertex. Apply Algorithm 4.14 to insert v into T and call the resulting
augmented tree Tv . But our problem is not yet over; Tv may violate the height-balance
property. To complete the insertion procedure, we require a technique to restore, if
necessary, the height-balance property to Tv .
To see why the augmented tree Tv may not necessarily be height-balanced, let u be
the parent of v in Tv , where previously u was a vertex T (and possibly a leaf). In the
original AVL tree T , let Pu : r = u0 , u1 , . . . , uk = u be the path from the root r of T
to u with corresponding subtree heights H(ui ) = hi for i = 0, 1, . . . , k. An effect of the
insertion is to extend the path Pu to the longer path Pv : r = u0 , u1 , . . . , uk = u, v and
possibly increase subtree heights by one. One of two cases can occur with respect to Tv .
1. Height-balanced: Tv is height-balanced so no need to do anything further. A simple
way to detect this is to consider the subtree S rooted at u, the parent of v. If S
has two children, then no height adjustment need to take place for vertices in Pu ,
hence Tv is an AVL tree (see Figure 4.19). Otherwise we perform any necessary
height adjustment for vertices in Pu , starting from uk = u and working our way
up to the root r = u0 . After adjusting the height of ui , we test to see whether
ui (with its new height) is height-balanced. If each of the ui with their new heights
are height-balanced, then Tv is height-balanced.
2. Height-unbalanced: During the height adjustment phase, it may happen that some
uj with its new height is not height-balanced. Among all such height-unbalanced
vertices, let u` be the first height-unbalanced vertex detected during the process of
height adjustment starting from uk = u and going up towards r = u0 . We need to
rebalance the subtree rooted at u` . Then we continue on adjusting heights of the
remaining vertices in Pu , also performing height-rebalancing where necessary.
Case 1 is relatively straightforward, but it is case 2 that involves much intricate work.
4 4 4
2 3 2 3 2 3
0 1 1 2 0 1 1 2 0 1 1 2
0 0 0 1 0 0 0 0 1 0 0 0 0 1
0 0 0 0 0 0
(a) Insert a vertex. (b) Vertex inserted. (c) Vertex inserted.
Figure 4.19: Augmented tree is balanced after insertion; vertex labels are heights.
We now turn to the case where inserting a vertex v into a nonempty AVL tree T
results in an augmented tree Tv that is not height-balanced. A general idea for rebal-
ancing (and hence restoring the height-balance property to) Tv is to determine where in
Tv the height-balance property is first violated (the search phase), and then to locally
rebalance subtrees at and around the point of violation (the repair phase). A description
of the search phase follows. Let
Pv : r = u0 , u1 , . . . , uk = u, v
4.5. AVL trees 185
be the path from the root r of Tv (and hence of T ) to v. Traversing upward from v to r,
let z be the first height-unbalanced vertex. Among the children of z, let y be the child of
higher height and hence an ancestor of v. Similarly, among the children of y let x be the
child of higher height. In case a tie occurs, let x be the child of y that is also an ancestor
of v. As each vertex is an ancestor of itself, it is possible that x = v. Furthermore, x is
a grandchild of z because x is a child of y, which in turn is a child of z. The vertex z
is not height-balanced due to inserting v into the subtree rooted at y, hence the height
of y is 2 greater than its sibling (see Figure 4.20, where height-unbalanced vertices are
colored red). We have determined the location at which the height-balance property is
first violated.
5
4 4 2 4 z
2 3 3 z 3 0 1 1 3 y
0 1 1 2 0 2 y 1 2 0 0 0 2 x
0 0 0 1 1 x 0 0 1 1 0
0 0 0 v 0 0 0 v
(a) Insert a vertex. (b) Vertex inserted. (c) Vertex inserted.
Figure 4.20: Augmented tree is unbalanced after insertion; vertex labels are heights.
We now turn to the repair phase. The central question is: How are we to re-
store the height-balance property to the subtree rooted at z? By trinode restructur-
ing is meant the process whereby the height-balance property is restored; the prefix
“tri” refers to the three vertices x, y, z that are central to this process. A common
name for the trinode restructuring is rotation in view of the geometric interpretation
of the process. Figure 4.21 distinguishes four rotation possibilities, two of which are
symmetrical to the other two. The single left rotation in Figure 4.21(a) occurs when
height(x) = height(root(T0 )) + 1 and detailed in Algorithm 4.16. The single right rota-
tion in Figure 4.21(b) occurs when height(x) = height(root(T3 )) + 1; see Algorithm 4.17
for pseudocode. Figure 4.21(c) illustrates the case of a right-left double rotation and
occurs when height(root(T3 )) = height(root(T0 )); see Algorithm 4.18 for pseudocode to
handle the rotation. The fourth case is illustrated in Figure 4.21(d) and occurs when
height(root(T0 )) = height(root(T3 )); refer to Algorithm 4.19 for pseudocode to handle
this left-right double rotation. Each of the four algorithms mentioned above run in con-
stant time O(1) and preserves the in-order traversal ordering of all vertices in Tv . In all,
the insertion procedure is summarized in Algorithm 4.20. If h is the height T , locat-
ing and inserting the vertex v takes worst-case O(h) time, which is also the worst-case
runtime for the search-and-repair phase. Thus letting n be the number of vertices in T ,
insertion takes worst-case O(lg n) time.
4.5.2 Deletion
The process of removing a vertex from an AVL tree is similar to the insertion proce-
dure. However, instead of using the insertion algorithm for BST, we use the deletion
186 Chapter 4. Tree data structures
z
y y
x z x
T0
T1
T2 T3 T0 T1 T2 T3
(a) Left rotation of y over z.
z
y y
x x z
T3
T2
T0 T1 T0 T1 T2 T3
(b) Right rotation of y over z.
z
y x
x z y
T0
T3
T1 T2 T0 T1 T2 T3
(c) Double rotation: right rotation of x over y, then left rotation over z.
z
y x
x y z
T3
T0
T1 T2 T0 T1 T2 T3
(d) Double rotation: left rotation of x over y, then right rotation over z.
Figure 4.21: Rotations in the trinode restructuring process.
4.5. AVL trees 187
Algorithm 4.16: Single left rotation in the trinode restructure process.
Input: Three vertices x, y, z of an augmented AVL tree Tv , where z is the first
height-unbalanced vertex in the path from v up to the root of Tv . The left
subtree of z is denoted T0 and the left subtree of y is T1 . The left and
right subtrees of x are T2 and T3 , respectively.
Output: A single left rotation to height-balance the subtree rooted at z.
1 rightchild[parent[z]] ← y
2 parent[y] ← parent[z]
3 parent[z] ← y
4 leftchild[y] ← z
5 parent[root[T1 ]] ← z
6 rightchild[z] ← root[T1 ]
7 height[z] ← 1 + max(height[root[T0 ]], height[root[T1 ]])
8 height[x] ← 1 + max(height[root[T2 ]], height[root[T3 ]])
9 height[y] ← 1 + max(height[x], height[z])
Algorithm 4.17: Single right rotation in the trinode restructure process.
Input: Three vertices x, y, z of an augmented AVL tree Tv , where z is the first
height-unbalanced vertex in the path from v up to the root of Tv . The left
subtree of z is T3 and the right subtree of y is T2 . The left and right
subtrees of x are T0 and T1 , respectively.
Output: A single right rotation to height-balance the subtree rooted at z.
1 leftchild[parent[z]] ← y
2 parent[y] ← parent[z]
3 parent[z] ← y
4 rightchild[y] ← z
5 parent[root[T2 ]] ← z
6 leftchild[z] ← root[T2 ]
7 height[x] ← 1 + max(height[root[T0 ]], height[root[T1 ]])
8 height[z] ← 1 + max(height[root[T2 ]], height[root[T3 ]])
9 height[y] ← 1 + max(height[x], height[z])
188 Chapter 4. Tree data structures
Algorithm 4.18: Double rotation: right rotation followed by left rotation.
Input: Three vertices x, y, z of an augmented AVL tree Tv , where z is the first
height-unbalanced vertex in the path from v up to the root of Tv . The left
subtree of z is T0 and the right subtree of y is T3 . The roots of the left
and right subtrees of x are denoted T1 and T2 , respectively.
Output: A right-left double rotation to height-balance the subtree rooted at z.
1 rightchild[parent[z]] ← x
2 parent[x] ← parent[z]
3 parent[z] ← x
4 leftchild[x] ← z
5 rightchild[x] ← y
6 rightchild[z] ← root[T1 ]
7 parent[root[T1 ]] ← z
8 parent[y] ← x
9 leftchild[y] ← root[T2 ]
10 parent[root[T2 ]] ← y
11 height[z] ← 1 + max(height[root[T0 ]], height[root[T1 ]])
12 height[y] ← 1 + max(height[root[T2 ]], height[root[T3 ]])
13 height[x] ← 1 + max(height[y], height[z])
Algorithm 4.19: Double rotation: left rotation followed by right rotation.
Input: Three vertices x, y, z of an augmented AVL tree Tv , where z is the first
height-unbalanced vertex in the path from v up to the root of Tv . The left
subtree of y is T0 and the right subtree of z is T3 . The roots of the left
and right subtrees of x are denoted T1 and T2 , respectively.
Output: A left-right double rotation to height-balance the subtree rooted at z.
1 leftchild[parent[z]] ← x
2 parent[x] ← parent[z]
3 parent[z] ← x
4 rightchild[x] ← z
5 leftchild[z] ← root[T2 ]
6 parent[T2 ] ← z
7 leftchild[x] ← y
8 parent[y] ← x
9 rightchild[y] ← root[T1 ]
10 parent[root[T1 ]] ← y
11 height[z] ← 1 + max(height[root[T2 ]], height[root[T3 ]])
12 height[y] ← 1 + max(height[root[T0 ]], height[root[T1 ]])
13 height[x] ← 1 + max(height[y], height[z])
4.5. AVL trees 189
Algorithm 4.20: Insert a vertex into an AVL tree.
Input: An AVL tree T and a vertex v.
Output: The AVL tree T with v inserted into it.
1 insert v into T as per Algorithm 4.14
2 height[v] ← 0
3 u←v /* begin height adjustment */
4 x ← NULL
5 y ← NULL
6 z ← NULL
7 while parent[u] 6= NULL do
8 u ← parent[u]
9 if leftchild[u] 6= NULL and rightchild[u] 6= NULL then
10 h` ← height[leftchild[u]]
11 hr ← height[rightchild[u]]
12 height[u] ← 1 + max(h` , hr )
13 if |h` − hr | > 1 then
14 if height[rightchild[rightchild[u]]] = height[leftchild[u]] + 1 then
15 z←u
16 y ← rightchild[z]
17 x ← rightchild[y]
18 trinode restructuring as per Algorithm 4.16
19 continue with next iteration of loop
20 if height[leftchild[leftchild[u]]] = height[rightchild[u]] + 1 then
21 z←u
22 y ← leftchild[z]
23 x ← leftchild[y]
24 trinode restructuring as per Algorithm 4.17
25 continue with next iteration of loop
26 if height[rightchild[rightchild[u]]] = height[leftchild[u]] then
27 z←u
28 y ← rightchild[z]
29 x ← leftchild[y]
30 trinode restructuring as per Algorithm 4.18
31 continue with next iteration of loop
32 if height[leftchild[leftchild[u]]] = height[rightchild[u]] then
33 z←u
34 y ← leftchild[z]
35 x ← rightchild[y]
36 trinode restructuring as per Algorithm 4.19
37 continue with next iteration of loop
38 if leftchild[u] 6= NULL then
39 height[u] ← 1 + height[leftchild[u]]
40 continue with next iteration of loop
41 if rightchild[u] 6= NULL then
42 height[u] ← 1 + height[rightchild[u]]
43 continue with next iteration of loop
190 Chapter 4. Tree data structures
Algorithm 4.15 for BST to remove the target vertex from an AVL tree. The result-
ing tree may violate the height-balance property, which can be restored using trinode
restructuring.
Let T be an AVL tree having vertex v and suppose we want to remove v from T . In
the trivial case, T is the trivial tree whose sole vertex is v. Deleting v is simply removing
it from T so that T becomes the null tree. On the other hand, suppose T has at least
n > 1 vertices. Apply Algorithm 4.15 to remove v from T and call the resulting tree with
v removed Tv . It is possible that Tv does not satisfy the height-balance property. To
restore the height-balance property to Tv , let u be the parent of v in T prior to deleting
v from T . Having deleted v from T , let P : r = u0 , u1 , . . . , uk = u be the path from the
root r of Tv to u. Adjust the height of u and, traversing from u up to r, perform height
adjustment to each vertex in P and where necessary carry out trinode restructuring. The
resulting algorithm is very similar to Algorithm 4.20; see Algorithm 4.21 for pseudocode.
The deletion procedure via Algorithm 4.15 requires worst-case runtime O(lg n), where
n is the number of vertices in T , and the height-adjustment process runs in worst-case
O(lg n) time as well. Thus Algorithm 4.21 has worst-case runtime of O(lg n).
4.6 Problems
No problem is so formidable that you can’t walk away from it.
— Charles M. Schulz
4.1. Let Q be a priority queue of n > 1 elements, given in sequence representation.
From section 4.1.1, we know that inserting an element into Q takes O(n) time and
deleting an element from Q takes O(1) time.
(a) Suppose Q is an empty priority queue and let e0 , e1 , . . . , en be n + 1 elements
we want to insert into Q. What is the total runtime required to insert all the
ei into Q while also ensuring that the resulting queue is a priority queue?
(b) Let Q = [e0 , e1 , . . . , en ] be a priority queue of n + 1 elements. What is the
total time required to remove all the elements of Q?
4.2. Prove the correctness of Algorithms 4.2 and 4.3.
4.3. Describe a variant of Algorithm 4.3 for modifying the key of the root of a binary
heap, without extracting any vertex from the heap.
4.4. Section 4.2.2 describes how to insert an element into a binary heap T . The general
strategy is to choose the first leaf following the last internal vertex of T , replace
that leaf with the new element so that it becomes an internal vertex, and perform a
sift-up operation from there. If instead we choose any leaf of T and replace that leaf
with the new element, explain why we cannot do any better than Algorithm 4.2.
4.5. Section 4.2.3 shows how to extract the minimum vertex from a binary heap T .
Instead of replacing the root with the last internal vertex of T , we could replace
the root with any other vertex of T that is not a leaf and then proceed to maintain
the heap-structure and heap-order properties. Explain why the latter strategy is
not better than Algorithm 4.3.
4.6. Problems 191
Algorithm 4.21: Delete a vertex from an AVL tree.
Input: An AVL tree T and a vertex v ∈ V (T ).
Output: The AVL tree T with v removed from it.
1 u ← parent[v]
2 delete v from T as per Algorithm 4.15
3 adjust the height of u /* begin height adjustment */
4 x ← NULL
5 y ← NULL
6 z ← NULL
7 while parent[u] 6= NULL do
8 u ← parent[u]
9 if leftchild[u] 6= NULL and rightchild[u] 6= NULL then
10 h` ← height[leftchild[u]]
11 hr ← height[rightchild[u]]
12 height[u] ← 1 + max(h` , hr )
13 if |h` − hr | > 1 then
14 if height[rightchild[rightchild[u]]] = height[leftchild[u]] + 1 then
15 z←u
16 y ← rightchild[z]
17 x ← rightchild[y]
18 trinode restructuring as per Algorithm 4.16
19 continue with next iteration of loop
20 if height[leftchild[leftchild[u]]] = height[rightchild[u]] + 1 then
21 z←u
22 y ← leftchild[z]
23 x ← leftchild[y]
24 trinode restructuring as per Algorithm 4.17
25 continue with next iteration of loop
26 if height[rightchild[rightchild[u]]] = height[leftchild[u]] then
27 z←u
28 y ← rightchild[z]
29 x ← leftchild[y]
30 trinode restructuring as per Algorithm 4.18
31 continue with next iteration of loop
32 if height[leftchild[leftchild[u]]] = height[rightchild[u]] then
33 z←u
34 y ← leftchild[z]
35 x ← rightchild[y]
36 trinode restructuring as per Algorithm 4.19
37 continue with next iteration of loop
38 if leftchild[u] 6= NULL then
39 height[u] ← 1 + height[leftchild[u]]
40 continue with next iteration of loop
41 if rightchild[u] 6= NULL then
42 height[u] ← 1 + height[rightchild[u]]
43 continue with next iteration of loop
192 Chapter 4. Tree data structures
4.6. Let S be a sequence of n > 1 real numbers. How can we use algorithms described
in section 4.2 to sort S?
4.7. The binary heaps discussed in section 4.2 are properly called minimum binary
heaps because the root of the heap is always the minimum vertex. A correspond-
ing notion is that of maximum binary heaps, where the root is always the maximum
element. Describe algorithms analogous to those in section 4.2 for managing max-
imum binary heaps.
4.8. What is the total time required to extract all elements from a binary heap?
4.9. Numbers of the form nr are called binomial coefficients. They also count the
number of r-combinations from a set of n objects. Algorithm 4.22 presents pseu-
docode to generate all the r-combinations of a set of n distinct objects. What is the
worst-case runtime of Algorithm 4.22? Prove the correctness of Algorithm 4.22.
4.10. In contrast to enumerating all the r-combinations of a set of n objects, we may
only want to generate a random r-combination. Describe and present pseudocode
of a procedure to generate a random r-combination of {1, 2, . . . , n}.
4.11. A problem related to the r-combinations of the set S = {1, 2, . . . , n} is that of
generating the permutations of S. Algorithm 4.23 presents pseudocode to generate
all the permutations of S in increasing lexicographic order. Find the worst-case
runtime of this algorithm and prove its correctness.
4.12. Provide a description and pseudocode of an algorithm to generate a random per-
mutation of {1, 2, . . . , n}.
4.13. Takaoka [188] presents a general method for combinatorial generation that runs in
O(1) time. How can Takaoka’s method be applied to generating combinations and
permutations?
4.14. The proof of Lemma 4.4 relies on Pascal’s formula, which states that for any
positive integers n and r such that r ≤ n, the following identity holds:
n+1 n n
= + .
r r−1 r
Prove Pascal’s formula.
4.15. Let m, n, r be nonnegative integers such that r ≤ n. Prove the Vandermonde
convolution X r
m+n m n
= .
r k=0
k r−k
The latter equation, also known as Vandermonde’s identity, was already known
as early as 1303 in China by Chu Shi-Chieh. Alexandre-Théophile Vandermonde
independently discovered it and his result was published in 1772.
4.16. If m and n are nonnegative integers, prove that
X n
m+n+1 m+k
= .
n k=0
k
4.6. Problems 193
Algorithm 4.22: Generating all the r-combinations of {1, 2, . . . , n}.
Input: Two nonnegative integers n and r.
Output: A list L containing all the r-combinations of the set {1, 2, . . . , n} in
increasing lexicographic order.
1 L ← []
2 ci ← i for i = 1, 2, . . . , r
3 append(L, c1 c2 · · · cr )
4 for i ← 2, 3, . . . , nr do
5 m←r
6 max ← n
7 while cm = max do
8 m←m−1
9 max ← max − 1
10 cm ← cm + 1
11 cj ← cj−1 + 1 for j = m + 1, m + 2, . . . , r
12 append(L, c1 c2 · · · cr )
13 return L
Algorithm 4.23: Generating all the permutations of {1, 2, . . . , n}.
Input: A positive integer n.
Output: A list L containing all the permutations of {1, 2, . . . , n} in increasing
lexicographic order.
1 L ← []
2 ci ← i for i = 1, 2, . . . , n
3 append(L, c1 c2 · · · cn )
4 for i ← 2, 3, . . . , n! do
5 m←n−1
6 while cm > cm+1 do
7 m←m−1
8 k←n
9 while cm > ck do
10 k ←k−1
11 swap the values of cm and ck
12 p←m+1
13 q←n
14 while p < q do
15 swap the values of cp and cq
16 p←p+1
17 q ←q−1
18 append(L, c1 c2 · · · cn )
19 return L
194 Chapter 4. Tree data structures
4.17. Let n be a positive integer. How many distinct binomial heaps having n vertices
are there?
4.18. The algorithms described in section 4.3 are formally for minimum binomial heaps
because the vertex at the top of the heap is always the minimum vertex. Describe
analogous algorithms for maximum binomial heaps.
4.19. If H is a binomial heap, what is the total time required to extract all elements
from H?
4.20. Frederickson [85] describes an O(k) time algorithm for finding the k-th smallest
element in a binary heap. Provide a description and pseudocode of Frederickson’s
algorithm and prove its correctness.
4.21. Fibonacci heaps [86] allow for amortized O(1) time with respect to finding the
minimum element, inserting an element, and merging two Fibonacci heaps. Delet-
ing the minimum element takes amortized time O(lg n), where n is the number
of vertices in the heap. Describe and provide pseudocode of the above Fibonacci
heap operations and prove the correctness of the procedures.
4.22. Takaoka [189] introduces another type of heap called a 2-3 heap. Deleting the
minimum element takes amortized O(lg n) time with n being the number of vertices
in the 2-3 heap. Inserting an element into the heap takes amortized O(1) time.
Describe and provide pseudocode of the above 2-3 heap operations. Under which
conditions would 2-3 heaps be more efficient than Fibonacci heaps?
4.23. In 2000, Chazelle [54] introduced the soft heap, which can perform common heap
operations in amortized O(1) time. He then applied [53] the soft heap to realize a
very efficient implementation of an algorithm for finding minimum spanning trees.
In 2009, Kaplan and Zwick [123] provided a simple implementation and analysis of
Chazelle’s soft heap. Describe soft heaps and provide pseudocode of common heap
operations. Prove the correctness of the algorithms and provide runtime analyses.
Describe how to use soft heap to realize an efficient implementation of an algorithm
to produce minimum spanning trees.
4.24. Explain any differences between the binary heap-order property, the binomial heap-
order property, and the binary search tree property. Can in-order traversal be used
to list the vertices of a binary heap in sorted order? Explain why or why not.
4.25. Present pseudocode of an algorithm to find a vertex with maximum key in a binary
search tree.
4.26. Compare and contrast algorithms for locating minimum and maximum elements
in a list with their counterparts for a binary search tree.
4.27. Let T be a nonempty BST and suppose v ∈ V (T ) is not a minimum vertex of T .
If h is the height of T , describe and present pseudocode of an algorithm to find the
predecessor of v in worst-case time O(h).
4.28. Let L = [v0 , v1 , . . . , vn ] be the in-order listing of a BST T . Present an algorithm
to find the successor of v ∈ V (T ) in constant time O(1). How can we find the
predecessor of v in constant time as well?
4.6. Problems 195
4.29. Modify Algorithm 4.15 to extract a minimum vertex of a binary search tree. Now
do the same to extract a maximum vertex. How can Algorithm 4.15 be modified
to extract a vertex from a binary search tree?
4.30. Let v be a vertex of a BST and suppose v has two children. If s and p are the
successor and predecessor of v, respectively, show that s has no left-child and p has
no right-child.
4.31. Let L = [e0 , e1 , . . . , en ] be a list of n + 1 elements from a totally ordered set X with
total order ≤. How can binary search trees be used to sort L?
4.32. Describe and present pseudocode of a recursive algorithm for each of the following
operations on a BST.
(a) Find a vertex with a given key.
(b) Locate a minimum vertex.
(c) Locate a maximum vertex.
(d) Insert a vertex.
4.33. Are the algorithms presented in section 4.4 able to handle a BST having duplicate
keys? If not, modify the relevant algorithm(s) to account for the case where two
vertices in a BST have the same key.
4.34. The notion of vertex level for binary trees can be extended to general rooted trees
as follows. Let T be a rooted tree with n > 0 vertices and height h. Then level
0 ≤ i ≤ h of T consists of all those vertices in T that have the same depth i. If
each vertex at level i has i + m children for some fixed integer m > 0, what is the
number of vertices at each level of T ?
4.35. Compare the search, insertion, and deletion times of AVL trees and random binary
search trees. Provide empirical results of your comparative study.
4.36. Describe and present pseudocode of an algorithm to construct a Fibonacci tree of
height n for some integer n ≥ 0. Analyze the worst-case runtime of your algorithm.
4.37. The upper bound in Theorem 4.7 can be improved as follows. From the proof of
the theorem, we have the recurrence relation N (h) > N (h − 1) + N (h − 2).
(a) If h ≤ 2, show that there exists some c > 0 such that N (h) ≥ ch .
(b) Assume for induction that
N (h) > N (h − 1) + N (h − 2) ≥ ch−1 + ch−2
for some h > 2. If c > 0, show that c2 − c − 1 = 0 is a solution to the
recurrence relation ch−1 + ch−2 and that
√ !h
1+ 5
N (h) > .
2
196 Chapter 4. Tree data structures
(c) Use the previous two parts to show that
1
h< · lg n
lg ϕ
√
where ϕ = (1 + 5)/2 is the golden ratio and n counts the number of internal
vertices of an AVL tree of height h.
4.38. The Fibonacci sequence Fn is defined as follows. We have initial values F0 = 0
and F1 = 1. For n > 1, the n-th term in the sequence can be obtained via the
recurrence relation Fn = Fn−1 + Fn−2 . Show that
ϕn − (−1/ϕ)n
Fn = √ (4.5)
5
where ϕ is the golden ratio. The closed form solution (4.5) to the Fibonacci se-
quence is known as Binet’s formula, named after Jacques Philippe Marie Binet,
even through Abraham de Moivre knew about this formula long before Binet did.
Chapter 5
Distance and connectivity
— Spiked Math, http://spikedmath.com/382.html
5.1 Paths and distance
5.1.1 Distance and metrics
Consider an edge-weighted simple graph G = (V, E, i, h) without negative weight cycles.
Here E ⊆ V (2) , i : E → V (2) is an incidence function as in (1.2), which we regard
as the identity function, and h : E → V is an orientation function as in (1.3). Let
W : E → R be the weight function. (If G is not provided with a weight function on
the edges, we assume that each edge has unit weight.) If v1 , v2 ∈ V are two vertices
and P = (e1 , e2 , . . . , em ) is a v1 -v2 path (so v1 is incident to e1 and v2 is incident to em ),
define the weight of P to be the sum of the weights of the edges in P :
m
X
W (P ) = W (ei ).
i=1
The distance function d : V × V → R ∪ {∞} on G is defined by
d(v1 , v2 ) = ∞
if v1 and v2 lie in distinct connected components of G, and by
d(v1 , v2 ) = min W (P ) (5.1)
P
197
198 Chapter 5. Distance and connectivity
otherwise, where the minimum is taken over all paths P from v1 to v2 . By hypothesis, G
has no negative weight cycles so the minimum in (5.1) exists. It follows by definition of
the distance function that d(u, v) = ∞ if and only if there is no path between u and v.
How we interpret the distance function d depends on the meaning of the weight
function W . In practical applications, vertices can represent physical locations such as
cities, sea ports, or landmarks. An edge weight could be interpreted as the physical
distance in kilometers between two cities, the monetary cost of shipping goods from one
sea port to another, or the time required to travel from one landmark to another. Then
d(u, v) could mean the shortest route in kilometers between two cities, the lowest cost
incurred in transporting goods from one sea port to another, or the least time required
to travel from one landmark to another.
The distance function d is not in general a metric, i.e. the triangle inequality does
not in general hold for d. However, when the distance function is a metric then G is
called a metric graph. The theory of metric graphs, due to their close connection with
tropical curves, is an active area of research. For more information on metric graphs, see
Baker and Faber [12].
5.1.2 Radius and diameter
A new hospital is to be built in a large city. Construction has not yet started and a
number of urban planners are discussing the future location of the new hospital. What
is a possible location for the new hospital and how are we to determine this location?
This is an example of a class of problems known as facility location problems. Suppose
our objective in selecting a location for the hospital is to minimize the maximum response
time between the new hospital and the site of an emergency. To help with our decision
making, we could use the notion of the center of a graph.
The center of a graph G = (V, E) is defined in terms of the eccentricity of the graph
under consideration. The eccentricity : V → R is defined as follows. For any vertex
v, the eccentricity (v) is the greatest distance between v and any other vertex in G. In
symbols, the eccentricity is expressible as
(v) = max d(u, v).
u∈V
For example, in a tree T with root r the eccentricity of r is the height of T . In the graph
of Figure 5.1, the eccentricity of 2 is 5 and the shortest paths that yield (2) are
P1 : 2, 3, 4, 14, 15, 16
P2 : 2, 3, 4, 14, 15, 17.
The eccentricity of a vertex v can be thought of as an upper bound on the distance from
v to any other vertex in G. Furthermore, we have at least one vertex in G whose distance
from v is (v).
v 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
(v) 6 5 4 4 5 6 7 7 5 6 7 7 6 5 6 7 7
Table 5.1: Eccentricity distribution for the graph in Figure 5.1.
To motivate the notion of the radius of a graph, consider the distribution of eccentric-
ity among vertices of the graph G in Figure 5.1. The required eccentricity distribution
5.1. Paths and distance 199
16
11 12 15 17
10 13 14
1 2 3 4 5
6 9
7 8
Figure 5.1: Determine the eccentricity, center, radius, and diameter.
7
6
(v)
5
4
2 4 6 8 10 12 14 16
v
Figure 5.2: Eccentricity distribution of the graph in Figure 5.1. The horizontal axis
represents the vertex name, while the vertical axis is the corresponding eccentricity.
200 Chapter 5. Distance and connectivity
is shown in Table 5.1. Among the eccentricities in the latter table, the minimum eccen-
tricity is (3) = (4) = 4. An intuitive interpretation is that both of the vertices 3 and
4 have the shortest distance to any other vertices in G. We can invoke an analogy with
plane geometry as follows. If a circle has radius r, then the distance from the center
of the circle to any point within the circle is at most r. The minimum eccentricity in
graph theory plays a role similar to the radius of a circle. If an object is strategically
positioned—e.g. a vertex with minimum eccentricity or the center of a circle—then its
greatest distance to any other object is guaranteed to be minimum. With the above
analogy in mind, we define the radius of a graph G = (V, E), written rad(G), to be the
minimum eccentricity among the eccentricity distribution of G. In symbols,
rad(G) = min (v).
v∈V
The center of G, written C(G), is the set of vertices with minimum eccentricity. Thus
the graph in Figure 5.1 has radius 4 and center {3, 4}. As should be clear from the latter
example, the radius is a number whereas the center is a set. Refer to the beginning of
the section where we mentioned the problem of selecting a location for a new hospital.
We could use a graph to represent the geography of the city wherein the hospital is to
be situated and select a location that is in the center of the graph.
Consider now the maximum eccentricity of a graph. In (2.5) we defined the diameter
of a graph G = (V, E) by
diam(G) = max d(u, v).
u,v∈V
u6=v
The diameter of G can also be defined as the maximum eccentricity of any vertex in G:
diam(G) = max (v).
v∈V
In case G is disconnected, define its diameter to be diam(G) = ∞. To compute diam(G),
use the Floyd-Roy-Warshall algorithm (see section 2.6) to compute the shortest distance
between each pair of vertices. The maximum of these distances is the diameter. The set
of vertices of G with maximum eccentricity is called the periphery of G, written per(G).
The graph in Figure 5.1 has diameter 7 and periphery {7, 8, 11, 12, 16, 17}.
Theorem 5.1. Eccentricities of adjacent vertices. Let G = (V, E) be an undi-
rected, connected graph having nonnegative edge weights. If uv ∈ E and W is a weight
function for G, then |(u) − (v)| ≤ W (uv).
Proof. By definition, we have d(u, x) ≤ (u) and d(v, x) ≤ (v) for all x ∈ V . Let w ∈ V
such that d(u, w) = (u). Apply the triangle inequality to obtain
d(u, w) ≤ d(u, v) + d(v, w)
(u) ≤ W (uv) + d(v, w)
≤ W (uv) + (v)
from which we have (u) − (v) ≤ W (uv). Repeating the above argument with the role
of u and v interchanged yields (v) − (u) ≤ W (uv). Both (u) − (v) ≤ W (uv) and
(v) − (u) ≤ W (uv) together yields the inequality |(u) − (v)| ≤ W (uv) as required.
5.1. Paths and distance 201
5.1.3 Center of trees
Given a tree T of order ≥ 3, we want to derive a bound on the number of vertices that
comprise the center of T . A graph in general can have one, two, or more number of
vertices for its center. Indeed, for any integer n > 0 we can construct a graph whose
center has cardinality n. The cases for n = 1, 2, 3 are illustrated in Figure 5.3. But can
we do the same for trees? That is, given any positive integer n does there exist a tree
whose center has n vertices? It turns out that the center of a tree cannot have more
than two vertices, a result first discovered [119] by Camille Jordan in 1869.
(a) |C(G)| = 1 (b) |C(G)| = 2 (c) |C(G)| = 3
Figure 5.3: Constructing graphs with arbitrarily large centers.
Theorem 5.2. Jordan [119]. If a tree T has order ≥ 3, then the center of T is either
a single vertex or two adjacent vertices.
Proof. As all eccentric vertices of T are leaves (see problem 5.7), removing all the leaves
of T decreases the eccentricities of the remaining vertices by one. The tree comprised of
the surviving vertices has the same center as T . Continue pruning leaves as described
above and note that the tree comprised of the surviving vertices has the same center as
the previous tree. After a finite number of leaf pruning stages, we eventually end up
with a tree made up of either one vertex or two adjacent vertices. The vertex set of this
final tree is the center of T .
5.1.4 Distance matrix
In sections 1.3.4 and 2.3, the distance matrix D of a graph G was defined to be D = [dij ],
where dij = d(vi , vj ) and the vertices of G are indexed by V = {v0 , v1 , . . . , vk }. The
matrix D is square where we set dij = 0 for entries along the main diagonal. If there is
no path from vi to vj , then we set dij = ∞. If G is undirected, then D is symmetric and
is equal to its transpose, i.e. DT = D. To compute the distance matrix D, apply the
Floyd-Roy-Warshall algorithm to determine the distances between all pairs of vertices.
Refer to Figure 5.4 for examples of distance matrices of directed and undirected graphs.
In the remainder of this section, “graph” refers to an undirected graph unless otherwise
specified.
Instead of one distance matrix, we can define several distance matrices on G. Consider
an edge-weighted graph G = (V, E) without negative weight cycles and let
d : V × V → R ∪ {∞}
be a distance function of G. Let ∂ = diam(G) be the diameter of G and index the
vertices of G in some arbitrary but fixed manner, say V = {v0 , v1 , . . . , vn }. The sequence
202 Chapter 5. Distance and connectivity
3
5 2
0 1 2 ∞ 1 2
∞ 0 1 ∞ ∞ ∞
∞ 1 0 ∞ ∞ ∞
∞ 2 1 0 2 1
4 1
∞ ∞ ∞ ∞ 0 1
0
∞ ∞ ∞ ∞ 1 0
(a)
3
5 2
0 1 2 3 1 2
1 0 1 2 2 3
2 1 0 1 3 2
3 2 1 0 2 1
4 1
1 2 3 2 0 1
0
2 3 2 1 1 0
(b)
Figure 5.4: Distance matrices of directed and undirected graphs.
of distance matrices of G are a sequence of (n − 1) × (n − 1) matrices A1 , A2 , . . . , A∂
where (
1, if d(vi , vj ) = k,
(Ak )ij =
0, otherwise.
In particular, A1 is the usual adjacency matrix A. To compute the sequence of distance
matrices of G, use the Floyd-Roy-Warshall algorithm to compute the distance between
each pair of vertices and assign the resulting distance to the corresponding matrix Ai .
The distance matrix arises in several applications, including communication network
design [93] and network flow algorithms [65]. Thanks to Graham and Pollak [93], the
following unusual fact is known. If T is any tree then
det D(T ) = (−1)n−1 (n − 1)2n−2
where n denotes the order of T . In particular, the determinant of the distance matrix of
a tree is independent of the structure of the tree. This fact is proven in the paper [93],
but see also [72].
5.2 Vertex and edge connectivity
If G = (V, E) is a graph and U ⊆ V is a vertex set with the property that G − U
has more connected components than G, then we call U a vertex-cut. The term cut-
vertex or cut-point is used when the vertex-cut consists of exactly one vertex. For an
intuitive appreciation of vertex-cut, suppose G = (V, E) is a connected graph. Then
U ⊆ V is a vertex-cut if the vertex deletion subgraph G − U is disconnected. For
example, the cut-vertex of the graph in Figure 5.5 is the vertex 0. By κv (G) we mean
the vertex connectivity of a connected graph G, defined as the minimum number of
vertices whose removal would either disconnect G or reduce G to the trivial graph. The
5.2. Vertex and edge connectivity 203
vertex connectivity κv (G) is also written as κ(G). The vertex connectivity of the graph in
Figure 5.5 is κv (G) = 1 because we only need to remove vertex 0 in order to disconnect
the graph. The vertex connectivity of a connected graph G is thus the vertex-cut of
minimum cardinality. And G is said to be k-connected if κv (G) ≥ k. From the latter
definition, it immediately follows that if G has at least 3 vertices and is k-connected then
any vertex-cut of G has at least cardinality k. For instance, the graph in Figure 5.5 is
1-connected. In other words, G is k-connected if the graph remains connected even after
removing any k − 1 or fewer vertices from G.
0
1 2 3
Figure 5.5: A claw graph with 4 vertices.
Figure 5.6: The Petersen graph on 10 vertices.
Example 5.3. Here is a Sage example concerning κ(G) using the Petersen graph de-
picted in Figure 5.6. A linear programming Sage package, such as GLPK, must be
installed for the commands below to work.
sage : G = graphs . PetersenGraph ()
sage : len ( G . vertices ())
10
sage : G . vertex_connectivity ()
3.0
sage : G . delete_vertex (0)
sage : len ( G . vertices ())
9
sage : G . vertex_connectivity ()
2.0
The notions of edge-cut and cut-edge are similarly defined. Let G = (V, E) be a
graph and D ⊆ E an edge set such that the edge deletion subgraph G − D has more
components than G. Then D is called an edge-cut. An edge-cut D is said to be minimal
204 Chapter 5. Distance and connectivity
if no proper subset of D is an edge-cut. The term cut-edge or bridge is reserved for
the case where the set D is a singleton. Think of a cut-edge as an edge whose removal
from a connected graph would result in that graph being disconnected. Going back
to the case of the graph in Figure 5.5, each edge of the graph is a cut-edge. A graph
having no cut-edge is called bridgeless. An open question as of 2010 involving bridges
is the cycle double cover conjecture, due to Paul Seymour and G. Szekeres, which states
that every bridgeless graph admits a set of cycles that contains each edge exactly twice.
The edge connectivity of a connected graph G, written κe (G) and sometimes denoted
by λ(G), is the minimum number of edges whose removal would disconnect G. In other
words, κe (G) is the minimum cardinality among all edge-cuts of G. Furthermore, G is
said to be k-edge-connected if κe (G) ≥ k. A connected graph that is k-edge-connected
is guaranteed to be connected after removing ≤ k − 1 edges from it. When we have
removed k or more edges, then the graph would become disconnected. By convention, a
1-edge-connected graph is simply a connected graph. The graph in Figure 5.5 has edge
connectivity κe (G) = 1 and is 1-edge-connected.
Example 5.4. Here is a Sage example concerning λ(G) using the Petersen graph shown
in Figure 5.6. You must install an optional linear programming Sage package such as
GLPK for the commands below to work.
sage : G = graphs . PetersenGraph ()
sage : len ( G . vertices ())
10
sage : E = G . edges (); len ( E )
15
sage : G . edge_connectivity ()
3.0
sage : G . delete_edge ( E [0])
sage : len ( G . edges ())
14
sage : G . edge_connectivity ()
2.0
Vertex and edge connectivity are intimately related to the reliability and survivability
of computer networks. If a computer network G (which is a connected graph) is k-
connected, then it would remain connected despite the failure of at most k − 1 network
nodes. Similarly, G is k-edge-connected if the network remains connected after the failure
of at most k − 1 network links. In practical terms, a network with redundant nodes
and/or links can afford to endure the failure of a number of nodes and/or links and
still be connected, whereas a network with very few redundant nodes and/or links (e.g.
something close to a spanning tree) is more prone to be disconnected. A k-connected or
k-edge-connected network is more robust (i.e. can withstand) against node and/or link
failures than is a j-connected or j-edge-connected network, where j < k.
Proposition 5.5. If δ(G) is the minimum degree of an undirected connected graph G =
(V, E), then the edge connectivity of G satisfies λ(G) ≤ δ(G).
Proof. Choose a vertex v ∈ V whose degree is deg(v) = δ(G). Deleting the δ(G) edges
incident on v suffices to disconnect G as v is now an isolated vertex. It is possible that
G has an edge-cut whose cardinality is smaller than δ(G). Hence the result follows.
Let G = (V, E) be a graph and suppose X1 and X2 comprise a partition of V . A
partition-cut of G, denoted hX1 , X2 i, is the set of all edges of G with one endpoint in
X1 and the other endpoint in X2 . If G is a bipartite graph with bipartition X1 and X2 ,
then hX1 , X2 i is a partition-cut of G. It follows that a partition-cut is also an edge-cut.
5.2. Vertex and edge connectivity 205
Proposition 5.6. An undirected connected graph G is k-edge-connected if and only if
any partition-cut of G has at least k edges.
Proof. Assume that G is k-edge-connected. Then each edge-cut has at least k edges. As
a partition-cut is an edge-cut, then any partition-cut of G has at least k edges.
On the other hand, suppose each partition-cut has at least k edges. If D is a minimal
edge-cut of G and X1 and X2 are the vertex sets of the two components of G − D, then
D = hX1 , X2 i. To see this, note that D ⊆ hX1 , X2 i. If hX1 , X2 i − D 6= ∅ then choose
some e ∈ hX1 , X2 i such that e ∈/ D. The endpoints of e belong to the same component
of G − D, in contradiction of the definition of X1 and X2 . Thus any minimal edge-cut is
a partition-cut and conclude that any edge-cut has at least k edges.
Proposition 5.7. If G = (V, E) is an undirected connected graph with vertex connectivity
κ(G) and edge connectivity λ(G), then we have κ(G) ≤ λ(G).
Proof. Let S be an edge-cut of G with cardinality k = |S| = λ(G). Removing k suitably
chosen vertices of G suffice to delete the edges of S and hence disconnect G. It is also
possible to have a smaller vertex-cut elsewhere in G. Hence the inequality follows.
Taking together Propositions 5.5 and 5.7, we have Whitney’s inequality.
Theorem 5.8. Whitney’s inequality [207]. Let G be an undirected connected graph
with vertex connectivity κ(G), edge connectivity λ(G), and minimum degree δ(G). Then
we have the following inequality:
κ(G) ≤ λ(G) ≤ δ(G).
Proposition 5.9. Let G be an undirected connected graph that is k-connected for some
k ≥ 3. If e is an edge of G, then the edge-deletion subgraph G − e is (k − 1)-connected.
Proof. Let V = {v1 , v2 , . . . , vk−2 } be a set of k − 2 vertices in G − e. It suffice to show
the existence of a u-v walk in (G − e) − V for any distinct vertices u and v in (G − e) − V .
We need to consider two cases: (i) at least one of the endpoints of e is in V ; and (ii)
neither endpoints of e is in V .
(i) Assume that V has at least one endpoint of e. As G − V is 2-connected, any
distinct pair of vertices u and v in G − V is connected by a u-v path that excludes e.
Hence the u-v path is also in (G − e) − V .
(ii) Assume that neither endpoints of e is in V . If u and v are distinct vertices in
(G − e) − V , then either: (1) both u and v are endpoints of e; or (2) at least one of u
and v is an endpoint of e.
(1) Suppose u and v are both endpoints of e. As G is k-connected, then G has at
least k + 1 vertices so that the vertex set of G − {v1 , v2 , . . . , vk−2 , u, v} is nonempty.
Let w be a vertex of G − {v1 , v2 , . . . , vk−2 , u, v}. Then there is a u-w path in G −
{v1 , v2 , . . . , vk−2 , v} and a w-v path in G − {v1 , v2 , . . . , vk−2 , u}. Neither the u-w nor
the w-v paths contain e. The concatenation of these two paths is a u-v walk in
(G − e) − V .
(2) Now suppose at least one of u and v, say u, is an endpoint of e. Let w be the other
endpoint of e. As G is k-connected, then G − {v1 , v2 , . . . , vk−2 , w} is connected and
we can find a u-v path P in G − {v1 , v2 , . . . , vk−2 , w}. Furthermore P is a u-v path
in G − {v1 , v2 , . . . , vk−2 } that neither contain w nor e. Hence P is a u-v path in
(G − e) − V .
206 Chapter 5. Distance and connectivity
Conclude that G − e is (k − 1)-connected.
Repeated application of Proposition 5.9 results in the following corollary.
Corollary 5.10. Let G be an undirected connected graph that is k-connected for some
k ≥ 3. If E is any set of m edges of G, for m ≤ k − 1, then the edge-deletion subgraph
G − E is (k − m)-connected.
What does it mean for a communications network to be fault-tolerant? In 1932,
Hassler Whitney provided [207] a characterization of 2-connected graphs whereby he
showed that a graph G is 2-connected if and only if each pair of distinct vertices in G
has two different paths connecting those two vertices. A key to understanding Whitney’s
characterization of 2-connected graphs is the notion of internal vertex of a path. Given a
path P in a graph, a vertex along that path is said to be an internal vertex if it is neither
the initial nor the final vertex of P . In other words, a path P has an internal vertex if
and only if P has at least two edges. Building upon the notion of internal vertices, we
now discuss what it means for two paths to be internally disjoint. Let u and v be distinct
vertices in a graph G and suppose P1 and P2 are two paths from u to v. Then P1 and P2
are said to be internally disjoint if they do not share any common internal vertex. Two
u-v paths are internally disjoint in the sense that both u and v are the only vertices to
be found in common between those paths. The notion of internally disjoint paths can be
easily extended to a collection of u-v paths. Whitney’s characterization essentially says
that a graph is 2-connected if and only if any two u-v paths are internally disjoint.
Consider the notion of internally disjoint paths within the context of communications
network. As a first requirement for fault-tolerant communications network, we want the
network to remain connected despite the failure of any network node. By Whitney’s
characterization, this is possible if the original communications network is 2-connected.
That is, we say that a communications network is fault-tolerant provided that any pair
of distinct nodes is connected by two internally disjoint paths. The failure of any node
should at least guarantee that any two distinct nodes are still connected.
Theorem 5.11. Whitney’s characterization of 2-connected graphs [207]. Let
G be an undirected connected graph having at least 3 vertices. Then G is 2-connected if
and only if any two distinct vertices in G are connected by two internally disjoint paths.
Proof. (⇐=) For the case of necessity, argue by contraposition. That is, suppose G is not
2-connected. Let v be a cut-vertex of G, from which it follows that G−v is disconnected.
We can find two vertices w and x such that there is no w-x path in G − v. Therefore v
is an internal vertex of any w-x path in G.
(=⇒) For the case of sufficiency, let G be 2-connected and let u and v be any two
distinct vertices in G. Argue by induction on d(u, v) that G has at least two internally
disjoint u-v paths. For the base case, suppose u and v are connected by an edge e so that
d(u, v) = 1. Adapt the proof of Proposition 5.9 to see that G − e is connected. Hence we
can find a u-v path P in G − e such that P and e are two internally disjoint u-v paths
in G.
Assume for induction that G has two internally disjoint u-v paths where d(u, v) < k
for some k ≥ 2. Let w and x be two distinct vertices in G such that d(w, x) = k and
hence there is a w-x path in G of length k, i.e. we have a w-x path
W : w = w1 , w2 , . . . , wk−1 , wk = x.
5.3. Menger’s theorem 207
Note that d(w, wk−1 ) < k and apply the induction hypothesis to see that we have two
internally disjoint w-wk−1 paths in G; call these paths P and Q. As G is 2-connected,
we have a w-x path R in G − wk−1 and hence R is also a w-x path in G. Let z be the
vertex on R that immediately precedes x and assume without loss of generality that z
is on P . We claim that G has two internally disjoint w-x paths. One of these paths is
the concatenation of the subpath of P from w to z with the subpath of R from z to x.
If x is not on Q, then construct a second w-x path, internally disjoint from the first one,
as follows: concatenate the path Q with the edge wk−1 w. In case x is on Q, take the
subpath of Q from w to x as the required second path.
From Theorem 5.11, an undirected connected graph G is 2-connected if and only if any
two distinct vertices of G are connected by two internally disjoint paths. In particular,
let u and v be any two distinct vertices of G and let P and Q be two internally disjoint
u-v paths as guaranteed by Theorem 5.11. Starting from u, travel along the path P
to arrive at v. Then start from v and travel along the path Q to arrive at u. The
concatenation of the internally disjoint paths P and Q is hence a cycle passing through
u and v. We have proved the following corollary to Theorem 5.11.
Corollary 5.12. Let G be an undirected connected graph having at least 3 vertices. Then
G is 2-connected if and only if any two distinct vertices of G lie on a common cycle.
The following theorem provides further characterizations of 2-connected graphs, in
addition to Whitney’s characterization.
Theorem 5.13. Characterizations of 2-connected graphs. Let G = (V, E) be an
undirected connected graph having at least 3 vertices. Then the following are equivalent.
1. G is 2-connected.
2. If u, v ∈ V are distinct vertices of G, then u and v lie on a common cycle.
3. If v ∈ V and e ∈ E, then v and e lie on a common cycle.
4. If e1 , e2 ∈ E are distinct edges of G, then e1 and e2 lie on a common cycle.
5. If u, v ∈ V are distinct vertices and e ∈ E, then they lie on a common path.
6. If u, v, w ∈ V are distinct vertices, then they lie on a common path.
7. If u, v, w ∈ V are distinct vertices, then there is a path containing any two of these
vertices but excluding the third.
5.3 Menger’s theorem
Menger’s theorem has a number of different versions: an undirected, vertex-connectivity
version; a directed, vertex-connectivity version; an undirected, edge-connectivity version;
and a directed, edge-connectivity version. In this section, we will prove the undirected,
vertex-connectivity version. But first, let’s consider a few technical results that will be
of use for the purpose of this section.
Let u and v be distinct vertices in a connected graph G = (V, E) and let S ⊆ V .
Then S is said to be u-v separating if u and v lie in different components of the vertex
208 Chapter 5. Distance and connectivity
deletion subgraph G − S. The vertices u and v are positioned such that after removing
vertices in S from G and the corresponding edges, u and v are no longer connected nor
strongly connected to each other. It is clear by definition that u, v ∈ / S. We also say
that S separates u and v, or S is a vertex separating set. Similarly an edge set T ⊆ E is
u-v separating (or separates u and v) if u and v lie in different components of the edge
deletion subgraph G − T . But unlike the case of vertex separating sets, it is possible for
u and v to be endpoints of edges in T because the removal of edges does not result in
deleting the corresponding endpoints. The set T is also called an edge separating set. In
other words, S is a vertex cut and T is an edge cut. When it is clear from context, we
simply refer to a separating set. See Figure 5.7 for illustrations of separating sets.
(a) Original graph. (b) Vertex separated.
(c) Original graph. (d) Edge separated.
Figure 5.7: Vertex and edge separating sets. Blue-colored vertices are those we want
to separate. The red-colored vertices form a vertex separating set or vertex cut; the
red-colored edges constitute an edge separating set or edge cut.
Proposition 5.14. Consider two distinct, non-adjacent vertices u, v in a connected
graph G. If Puv is a collection of internally disjoint u-v paths in G and Suv is a u-
v separating set of vertices in G, then
|Puv | ≤ |Suv |. (5.2)
Proof. Each u-v path in Puv must include at least one vertex from Suv because Suv is
a vertex cut of G. Any two distinct paths in Puv cannot contain the same vertex from
Suv . Thus the number of internally disjoint u-v paths is at most |Suv |.
The bound (5.2) holds for any u-v separating set Suv of vertices in G. In particular,
we can choose Suv to be of minimum cardinality among all u-v separating sets of vertices
in G. Thus we have the following corollary. Menger’s Theorem 5.18 provides a much
stronger statement of Corollary 5.15, saying in effect that the two quantities max(|Puv |)
and min(|Suv |) are equal.
Corollary 5.15. Consider any two distinct, non-adjacent vertices u, v in a connected
graph G. Let max(|Puv |) be the maximum number of internally disjoint u-v paths in G
and denote by min(|Suv |) the minimum cardinality of a u-v separating set of vertices in
G. Then we have max(|Puv |) ≤ min(|Suv |).
5.3. Menger’s theorem 209
Corollary 5.16. Consider any two distinct, non-adjacent vertices u, v in a connected
graph G. Let Puv be a collection of internally disjoint u-v paths in G and let Suv be a u-v
separating set of vertices in G. If |Puv | = |Suv |, then Puv has maximum cardinality among
all collections of internally disjoint u-v paths in G and Suv has minimum cardinality
among all u-v separating sets of vertices in G.
Proof. Argue by contradiction. Let Quv be another collection of internally disjoint u-v
paths in G such that |Quv | ≥ |Puv |. Then |Puv | ≤ |Quv | ≤ |Suv | by Proposition 5.14.
We cannot have |Quv | > |Puv |, which would be contradictory to our hypothesis that
Puv = |Suv |. Thus |Quv | = |Puv |. Let Tuv be another u-v separating set of vertices in
G such that |Tuv | ≤ |Suv |. Then we have |Puv | ≤ |Tuv | ≤ |Suv | by Proposition 5.14.
We cannot have |Tuv | < |Suv | because we would then end up with |Puv | ≤ |Tuv | and
Puv = |Suv |, a contradiction. Therefore |Tuv | = |Suv |.
Lemma 5.17. Consider two distinct, non-adjacent vertices u, v in a connected graph G
and let k be the minimum number of vertices required to separate u and v. If G has a
u-v path of length 2, then G has k internally disjoint u-v paths.
Proof. Argue by induction on k. For the base case, assume k = 1. Hence G has a cut
vertex x such that u and v are disconnected in G − x. Any u-v path must contain x. In
particular, there can be only one internally disjoint u-v path.
Assume for induction that k ≥ 2. Let P : u, x, v be a path in G having length 2 and
suppose S is a smallest u-v separating set for G − x. Then S ∪ {x} is a u-v separating
set for G. By the minimality of k, we have |S| ≥ k − 1. By the induction hypothesis,
we have at least k − 1 internally disjoint u-v paths in G − x. As P is internally disjoint
from any of the latter paths, conclude that G has k internally disjoint u-v paths.
Theorem 5.18. Menger’s theorem. Let G be an undirected connected graph and let
u and v be distinct, non-adjacent vertices of G. Then the maximum number of internally
disjoint u-v paths in G equals the minimum number of vertices needed to separate u and
v.
Proof. Suppose that the maximum number of independent u-v paths in G is attained by
u-v paths P1 , . . . , Pk . To obtain a separating set W ⊂ V , we must at least remove one
point in each path Pi . This implies the minimum number of vertices needed to separate
u and v is at least k. Therefore, we have an upper bound:
#{indep. u − v paths} ≤ #{min. number of vertices needed to separate u and v}.
We show that equality holds. Let n denote the number of edges of G. The proof is by
induction on n. By hypothesis, n ≥ 2. If n = 2 the statement holds by inspection, since
in that case G is a line graph with 3 vertices V = {u, v, w} and 2 edges, E = {uw.wv}.
In that situation, there is only 1 u-v path (namely, uwv) and only one vertex separating
u and v (namely, w).
Suppose now n > 3 and assume the statement holds for each graph with < n edges.
Let
k = #{independent u − v paths}
and let
` = #{min. number of vertices needed to separate u and v},
210 Chapter 5. Distance and connectivity
so that k ≤ `. Let e ∈ E and let G/e be the contraction graph having edges E − {e}
and vertices the same as those of G, except that the endpoints of e have been identified.
Suppose that k < ` and G does not have ` independent u-v paths. The contraction
graph G/e does not have ` independent u-v paths either (where now, if e contains u or
v then we must appropriately redefine u or v, if needed). However, by the induction
hypothesis G/e does have the property that the maximum number of internally disjoint
u-v paths equals the minimum number of vertices needed to separate u and v. Therefore,
#{independent u − v paths in G/e}
< #{min. number of vertices needed to separate u and v in G}.
By induction,
#{independent u − v paths in G/e}
= #{min. number of vertices needed to separate u and v in G/e}.
Now, we claim we can pick e such that e does contain u or v and in such a way that
#{minimum number of vertices needed to separate u and v in G}
≥ #{minimum number of vertices needed to separate u and v in G/e}.
Proof: Indeed, since n > 3 any separating set realizing the minimum number of vertices
needed to separate u and v in G cannot contain both a vertex in G adjacent to u and a
vertex in G adjacent to v. Therefore, we may pick e accordingly. (Q.E.D. claim)
The result follows from the claim and the above inequalities.
The following statement is the undirected, edge-connectivity version of Menger’s the-
orem.
Theorem 5.19. Menger’s theorem (edge-connectivity form). Let G be an undi-
rected graph, and let s and t be vertices in G. Then, the maximum number of edge-
disjoint (s, t)-paths in G equals the minimum number of edges from E(G) whose deletion
separates s and t.
This is proven the same way as the previous version but using the generalized min-
cut/max-flow theorem (see Remark 9.16 above).
Theorem 5.20. Dirac’s theorem. Let G = (V, E) be an undirected k-connected graph
with |V | ≥ k + 1 vertices for k ≥ 3. If S ⊆ V is any set of k vertices, then G has a cycle
containing the vertices of S.
Proof.
5.4 Whitney’s Theorem
Theorem 5.21. Whitney’s theorem (vertex version). Suppose G = (V, E) is a
graph with |V | ≥ k + 1. The following are equivalent:
G is k-vertex-connected,
Any pair of distinct vertices v, w ∈ V are connected by at least k independent
paths.
5.5. Centrality of a vertex 211
Solution. ...
Theorem 5.22. Whitney’s theorem (edge version). Suppose G = (V, E) is a
graph with |V | ≥ k + 1. The following are equivalent:
the graph G is k-edge-connected,
any pair of vertices are connected by at least k edge-disjoint paths.
Solution. ...
Theorem 5.23. Whitney’s Theorem. Let G = (V, E) be a connected graph such that
|V | ≥ 3. Then G is 2-connected if and only if any pair u, v ∈ V has two internally
disjoint paths between them.
5.5 Centrality of a vertex
Louis, I think this is the beginning of a beautiful friendship.
— Rick from the 1942 film Casablanca
degree centrality
betweenness centrality; for efficient algorithms, see [39, 212]
closeness centrality
eigenvector centrality
The degree centrality of a graph G = (V, E) is the list parameterized by the vertex set
V of G whose v-th entry is the fraction of vertices connected to v ∈ V . The centrality of
a vertex within a graph determines the relative importance of that vertex to its graph.
Degree centrality measures the number of edges incident upon a vertex.
sage: G = graphs.RandomNewmanWattsStrogatz(6,2,1/2)
sage: G
Graph on 6 vertices
sage: D = G.degree_sequence()
sage: D
[5, 4, 3, 3, 3, 2]
sage: VG = G.vertices()
sage: VG
[0, 1, 2, 3, 4, 5]
sage: DC = [QQ(x)/len(VG) for x in D]
sage: DC
[5/6, 2/3, 1/2, 1/2, 1/2, 1/3]
212 Chapter 5. Distance and connectivity
Figure 5.8: A randomly choosen graph whose “central-most” vertex is 0, with degree
centrality equal to 5/6.
This graph is shown in Figure 5.8.
The closeness centrality is defined to be
1
.
average distance to all vertices
Closeness centrality is an inverse measure of centrality in that a larger value indicates a
less central vertex while a smaller value indicates a more central vertex.
Algorithm 5.1: Friendship graph.
Input: A positive integer n.
Output: The friendship graph Fn .
1 if n = 1 then
2 return C3
3 G ← null graph
4 N ← 2n + 1
5 for i ← 0, 1, . . . , N − 3 do
6 if i is odd then
7 add edges (i, i + 1) and (i, N − 1) to G
8 else
9 add edge (i, N − 1) to G
10 add edges (N − 2, 0) and (N − 2, N − 1) to G
11 return E
5.6. Network reliability 213
5.6 Network reliability
Whitney synthesis
Tutte’s synthesis of 3-connected graphs
Harary graphs
constructing an optimal k-connected n-vertex graph
5.7 The spectrum of a graph
We use the notes “The spectrum of a graph” by Andries Brouwer [43] as a basic reference.
Spectrum of a graph
Laplacian spectrum of a graph
Applications
Examples from Boolean functions
Let G = (V, E) be a (possibly directed) finite graph on n = |V | vertices. The
adjacency matrix of G is the n × n matrix A = A(G) = (av,w )v,w∈V with rows and
columns indexed by V and entries av,w denoting the number of edges from v to w.
The spectrum of G, spec(G), is by definition the spectrum of A, that is, its multi-set
of eigenvalues together with their multiplicities. The characteristic polynomial of G is
that of A, that is, the polynomial pA defined by pA (x) = det(A − xI).
5.7.1 The Laplacian spectrum
Recall from section 1.3.3 that, given a simple graph G with n vertices V = {v1 , . . . , vn },
its (vertex) Laplacian matrix L = (`i,j )n×n is defined as:
deg(vi ) if i = j
`i,j = −1 if i 6= j and vi is adjacent to vj
0 otherwise.
The Laplacian spectrum is by definition the spectrum of the vertex Laplacian of G, that
is, its multi-set of eigenvalues together with their multiplicities.
For a graph G and its Laplacian matrix L with eigenvalues λn ≤ λn−1 ≤ · · · ≤ λ1 :
For all i, λi ≥ 0 and λn = 0.
The number of times 0 appears as an eigenvalue in the Laplacian is the number of
connected components in the graph.
λn = 0 because every Laplacian matrix has an eigenvector of [1, 1, . . . , 1],
214 Chapter 5. Distance and connectivity
If we define a signed edge adjacency matrix M with element me,v for edge e ∈ E
(connecting vertex vi and vj , with i < j) and vertex v ∈ V given by
1, if v = vi
Mev = −1, if v = vj
0, otherwise
then the Laplacian matrix L satisifies L = M T M , where M T is the matrix trans-
pose of M .
These are left as an exercise.
5.7.2 Applications of the (ordinary) spectrum
The following is a basic fact about the largest eigenvalue of a graph.
Theorem 5.24. Each graph G has a real eigenvalue λ1 > 0 with nonnegative real cor-
responding eigenvector, and such that for each eigenvalue λ we have |λ| ≤ λ1 .
We shall mostly be interested in the case where G is undirected, without loops or
multiple edges. This means that A is symmetric, has zero diagonal (av,v = 0), and is a
0-1 matrix ( av,w ∈ {0, 1}).
A number λ is eigenvalue of A if and only if it is a zero of the polynomial pA .
Since A is real and symmetric, all its eigenvalues are real and A is diagonalizable. In
particular, for each eigenvalue, its algebraic multiplicity (that is, its multiplicity as a
root of the characteristic polynomial) coincides with its geometric multiplicity (that is,
the dimension of the corresponding eigenspace).
Theorem 5.25. Let G be a connected graph of diameter d. Then G has at least d + 1
distinct eigenvalues.
Proof. Let the distinct eigenvalues of the adjacency matrix A of G be λ1 , ..., λr . Then
(A − λ1 I)...(A − λr I) = 0, so that Ar is a linear combination of I, A, ..., Ar−1 . But if
the distance from the vertex v ∈ V to the vertex w ∈ V is r, then (Ai )v,w = 0 for
0 ≤ i ≤ r − 1 and (Ar )v,w > 0, contradiction. Hence d > r.
5.7.3 Examples from Boolean functions
Let f be a Boolean function on GF (2)n . The Cayley graph of f is defined to be the
graph
Γf = (GF (2)n , Ef ),
whose vertex set is GF (2)n and the set of edges is defined by
Ef = {(u, v) ∈ GF (2)n × GF (2)n | f (u + v) = 1}.
The adjacency matrix Af of this graph is the matrix whose entries are
Ai,j = f (b(i) + b(j)),
5.7. The spectrum of a graph 215
where b(k) is the binary representation of the integer k. Note Γf is a regular graph of
degree wt(f ), where wt denotes the Hamming weight of f when regarded as a vector of
values (of length 2n ).
Recall that, given a graph Γ and its adjacency matrix A, the spectrum Spec(Γ) is
the multi-set of eigenvalues of A.
The Walsh transform of a Boolean function f is an integer-valued function over
GF (2)n that can be defined as
X
Wf (u) = (−1)f (x)+hu,xi .
xinGF (2)n
A Boolean function f is bent if |Wf (a)| = 2n/2 (this only makes sense if n is even). This
property says, roughly speaking, that f is “as non-linear as possible.” The Hadamard
transform of a integer-valued function f is an integer-valued function over GF (2)n that
can be defined as
X
Hf (u) = f (x)(−1)hu,xi .
xinGF (2)n
It turns out that the spectrum of Γf is equal to the Hadamard transform of f when
regarded as a vector of (integer) 0, 1-values. (This nice fact seems to have first appeared
in [24], [25].)
Recall that a graph is regular of degree r (or r-regular) if every vertex has degree r.
We say that an r-regular graph Γ is a strongly regular graph with parameters (v, r, d, e)
(for nonnegative integers e, d) provided, for all vertices u, v the number of vertices
adjacent to both u, v is equal to
e, if u, v are adjacent,
d, if u, v are nonadjacent.
It turns out tht f is bent if and only if Γf is strongly regular and e = d (see [25], [8]).
The following Sage computations illustrate these and other theorems in [185], [24],
[25], [8].
First, consider the Boolean function f : GF (2)4 → GF (2) given by f (x0 , x1 , x2 ) =
x0 x1 + x2 x3 .
sage: V = GF(2)^4
sage: f = lambda x: x[0]*x[1]+x[2]*x[3]
sage: CartesianProduct(range(16), range(16))
Cartesian product of [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15],
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15]
sage: C = CartesianProduct(range(16), range(16))
sage: Vlist = V.list()
sage: E = [(x[0],x[1]) for x in C if f(Vlist[x[0]]+Vlist[x[1]])==1]
sage: len(E)
96
sage: E = Set([Set(s) for s in E])
sage: E = [tuple(s) for s in E]
sage: Gamma = Graph(E)
sage: Gamma
Graph on 16 vertices
216 Chapter 5. Distance and connectivity
sage: VG = Gamma.vertices()
sage: L1 = []
sage: L2 = []
sage: for v1 in VG:
....: for v2 in VG:
....: N1 = Gamma.neighbors(v1)
....: N2 = Gamma.neighbors(v2)
....: if v1 in N2:
....: L1 = L1+[len([x for x in N1 if x in N2])]
....: if not(v1 in N2) and v1!=v2:
....: L2 = L2+[len([x for x in N1 if x in N2])]
....:
....:
sage: L1; L2
[2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2,
2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2,
2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2,
2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2,
2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2]
[2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2,
2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2,
2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2,
2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2,
2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2,
2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2,
2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2,
2, 2, 2, 2]
This implies the graph is strongly regular with d = e = 2. Let us use Sage to
determine some properties of this graph.
sage: Gamma.spectrum()
[6, 2, 2, 2, 2, 2, 2, -2, -2, -2, -2, -2, -2, -2, -2, -2]
sage: [walsh_transform(f, a) for a in V]
[4, 4, 4, -4, 4, 4, 4, -4, 4, 4, 4, -4, -4, -4, -4, 4]
sage: Omega_f = [v for v in V if f(v)==1]
sage: len(Omega_f)
6
sage: Gamma.is_bipartite()
False
sage: Gamma.is_hamiltonian()
True
sage: Gamma.is_planar()
False
sage: Gamma.is_regular()
True
sage: Gamma.is_eulerian()
True
sage: Gamma.is_connected()
True
sage: Gamma.is_triangle_free()
5.7. The spectrum of a graph 217
False
sage: Gamma.diameter()
2
sage: Gamma.degree_sequence()
[6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6]
The picture of the graph is given in Figure 5.9.
Figure 5.9: A Cayley graph of a Boolean function.
sage: H = matrix(QQ, 16, 16, [(-1)^(Vlist[x[0]]).dot_product(Vlist[x[1]]) for x in C])
sage: H
[ 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1]
[ 1 -1 1 -1 1 -1 1 -1 1 -1 1 -1 1 -1 1 -1]
[ 1 1 -1 -1 1 1 -1 -1 1 1 -1 -1 1 1 -1 -1]
[ 1 -1 -1 1 1 -1 -1 1 1 -1 -1 1 1 -1 -1 1]
[ 1 1 1 1 -1 -1 -1 -1 1 1 1 1 -1 -1 -1 -1]
[ 1 -1 1 -1 -1 1 -1 1 1 -1 1 -1 -1 1 -1 1]
[ 1 1 -1 -1 -1 -1 1 1 1 1 -1 -1 -1 -1 1 1]
[ 1 -1 -1 1 -1 1 1 -1 1 -1 -1 1 -1 1 1 -1]
[ 1 1 1 1 1 1 1 1 -1 -1 -1 -1 -1 -1 -1 -1]
[ 1 -1 1 -1 1 -1 1 -1 -1 1 -1 1 -1 1 -1 1]
[ 1 1 -1 -1 1 1 -1 -1 -1 -1 1 1 -1 -1 1 1]
[ 1 -1 -1 1 1 -1 -1 1 -1 1 1 -1 -1 1 1 -1]
[ 1 1 1 1 -1 -1 -1 -1 -1 -1 -1 -1 1 1 1 1]
[ 1 -1 1 -1 -1 1 -1 1 -1 1 -1 1 1 -1 1 -1]
[ 1 1 -1 -1 -1 -1 1 1 -1 -1 1 1 1 1 -1 -1]
[ 1 -1 -1 1 -1 1 1 -1 -1 1 1 -1 1 -1 -1 1]
218 Chapter 5. Distance and connectivity
sage: flist = vector(QQ, [int(f(v)) for v in V])
sage: H*flist
(6, -2, -2, 2, -2, -2, -2, 2, -2, -2, -2, 2, 2, 2, 2, -2)
sage: A = matrix(QQ, 16, 16, [f(Vlist[x[0]]+Vlist[x[1]]) for x in C])
sage: A.eigenvalues()
[6, 2, 2, 2, 2, 2, 2, -2, -2, -2, -2, -2, -2, -2, -2, -2]
The Hadamard transform of the Boolean function does indeed determine the spec-
trum of its Cayley graph.
Here is another example: f : GF (2)3 → GF (2) given by f (x0 , x1 , x2 ) = x0 x1 + x2 .
sage: V = GF(2)^3
sage: f = lambda x: x[0]*x[1]+x[2]
sage: Omega_f = [v for v in V if f(v)==1]
sage: len(Omega_f)
4
sage: C = CartesianProduct(range(8), range(8))
sage: Vlist = V.list()
sage: E = [(x[0],x[1]) for x in C if f(Vlist[x[0]]+Vlist[x[1]])==1]
sage: E = Set([Set(s) for s in E])
sage: E = [tuple(s) for s in E]
sage: Gamma = Graph(E)
sage: Gamma
Graph on 8 vertices
sage:
sage: VG = Gamma.vertices()
sage: L1 = []
sage: L2 = []
sage: for v1 in VG:
....: for v2 in VG:
....: N1 = Gamma.neighbors(v1)
....: N2 = Gamma.neighbors(v2)
....: if v1 in N2:
....: L1 = L1+[len([x for x in N1 if x in N2])]
....: if not(v1 in N2) and v1!=v2:
....: L2 = L2+[len([x for x in N1 if x in N2])]
....:
sage: L1; L2
[2, 0, 2, 2, 2, 2, 0, 2, 2, 2, 0, 2, 2, 2, 2, 0, 0, 2, 2, 2,
2, 0, 2, 2, 2, 0, 2, 2, 2, 2, 0, 2]
[2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2]
This implies that the graph is not strongly regular, therefore f is not bent. (There
are other reasons why f cannot be bent as well.) Again, let us use Sage to determine
some properties of this graph.
sage: Gamma.spectrum()
[4, 2, 0, 0, 0, -2, -2, -2]
sage:
5.8. Expander graphs and Ramanujan graphs 219
sage: Gamma.is_bipartite()
False
sage: Gamma.is_hamiltonian()
True
sage: Gamma.is_planar()
False
sage: Gamma.is_regular()
True
sage: Gamma.is_eulerian()
True
sage: Gamma.is_connected()
True
sage: Gamma.is_triangle_free()
False
sage: Gamma.diameter()
2
sage: Gamma.degree_sequence()
[4, 4, 4, 4, 4, 4, 4, 4]
sage: H = matrix(QQ, 8, 8, [(-1)^(Vlist[x[0]]).dot_product(Vlist[x[1]]) for x in C])
sage: H
[ 1 1 1 1 1 1 1 1]
[ 1 -1 1 -1 1 -1 1 -1]
[ 1 1 -1 -1 1 1 -1 -1]
[ 1 -1 -1 1 1 -1 -1 1]
[ 1 1 1 1 -1 -1 -1 -1]
[ 1 -1 1 -1 -1 1 -1 1]
[ 1 1 -1 -1 -1 -1 1 1]
[ 1 -1 -1 1 -1 1 1 -1]
sage: flist = vector(QQ, [int(f(v)) for v in V])
sage: H*flist
(4, 0, 0, 0, -2, -2, -2, 2)
sage: Gamma.spectrum()
[4, 2, 0, 0, 0, -2, -2, -2]
sage: A = matrix(QQ, 8, 8, [f(Vlist[x[0]]+Vlist[x[1]]) for x in C])
sage: A.eigenvalues()
[4, 2, 0, 0, 0, -2, -2, -2]
Again, we see the Hadamard transform does indeed determine the graph spectrum.
The picture of the graph is given in Figure 5.10.
5.8 Expander graphs and Ramanujan graphs
In combinatorics, an expander graph is a sparse graph that has strong connectivity
properties. Expander graphs have many applications - for example, to cryptography,
and the theory of error-correcting codes.
The edge expansion h(G) of a graph G = (V, E) is defined as
|∂(S)|
h(G) = min ,
0<|S|≤
|V |
2
|S|
220 Chapter 5. Distance and connectivity
Figure 5.10: Another Cayley graph of a Boolean function.
where the minimum is over all nonempty sets S of at most |V |/2 vertices and ∂(S) is
the edge boundary of S, i.e., the set of edges with exactly one endpoint in S.
The vertex expansion (or vertex isoperimetric number) hout (G) of a graph G is defined
as
|∂out (S)|
hout (G) = min ,
|V |
0<|S|≤ 2 |S|
where ∂out (S) is the outer boundary of S, i.e., the set of vertices in V (G) \ S with at
least one neighbor in S.
sage: G = PSL(2, 5)
sage: X = G.cayley_graph()
sage: V = X.vertices()
sage: S = [V[1], V[3], V[7], V[10], V[13], V[ 14], V[23]]
sage: delS = X.edge_boundary(S)
sage: edge_expan_XS = len(delS)/len(S); RR(edge_expan_XS)
1.00000000000000
sage: S = [V[1], V[3], V[7], V[12], V[24], V[37]]
sage: delS = X.edge_boundary(S)
sage: edge_expan_XS = len(delS)/len(S); RR(edge_expan_XS)
1.50000000000000
sage: S = [V[2], V[8], V[13], V[27], V[32], V[44], V[57]]
sage: delS = X.edge_boundary(S)
sage: edge_expan_XS = len(delS)/len(S); RR(edge_expan_XS)
1.42857142857143
sage: S = [V[0], V[6], V[11], V[16], V[21], V[29], V[35], V[45],V[53]]
sage: delS = X.edge_boundary(S)
sage: edge_expan_XS = len(delS)/len(S); RR(edge_expan_XS)
1.77777777777778
sage: n = len(X.vertices())
sage: J = range(n)
5.8. Expander graphs and Ramanujan graphs 221
sage: J30 = Subsets(J, int(n/2))
sage: K = J30.random_element()
sage: K
{0, 2, 3, 4, 5, 6, 8, 9, 11, 13, 16, 18, 19, 21, 24, 25, 26, 28, 29,
30, 36, 37, 38, 40, 42, 45, 46, 49, 53, 57}
sage: S = [V[i] for i in K] # 30 vertices, randomly selected
sage: delS = [v for v in V if min([X.distance(a,v) for a in S]) == 1]
sage: RR(len(delS))/RR(len(S))
0.800000000000000
A family G = {G1 , G2 , . . .} of d-regular graphs is an edge expander family if there is a
constant c > 0 such that h(G) ≥ c for each G ∈ G. A vertex expander family is defined
similarly, using hout (G) instead.
5.8.1 Ramanujan graphs
Let G be a connected d-regular graph with n vertices, and let λ0 ≥ λ1 ≥ . . . ≥ λn−1
be the eigenvalues of the adjacency matrix of G. Because G is connected and d-regular,
its eigenvalues satisfy d = λ0 > λ1 ≥ . . . ≥ λn−1 ≥ −d. Whenever there exists λi with
|λi | < d, define
λ(G) = max |λi |.
|λi |<d
√
A d-regular graph G is a Ramanujan graph if λ(G) is defined and λ(G) ≤ 2 d − 1.
Let q be a prime power such that q ≡ 1 (mod 4). Note that this implies that the
finite field GF (q) contains a square root of −1.
Now let V = GF (q) and E = {{a, b} ∈ GF (q) × GF (q) | (a − b) ∈ GF (q)× )2 }. This
set is well defined since a − b = (−1) · (b − a), and since −1 is a square, it follows that
a−b is a square if and only if b − a is a square.
By definition G = (V, E) is the Paley graph of order q.
The following facts are known about Paley graphs.
√
q−1 −1± q
The eigenvalues of Paley graphs are 2
(with multiplicity 1) and 2
(both with
multiplicity q−1
2
).
It is known that a Paley graph is a Ramanujan graph.
It is known that the family of Paley graphs of prime order is a vertex expander
graph family.
If q = pr , where p is prime, then Aut(G) has order q(q − 1)/2.
Here is Sage code for the Paley graph1 :
def Paley(q):
K.<a> = GF(q)
return Graph([K, lambda i,j: i != j and (i-j).is_square()])
1
Thanks to Chris Godsil; see [90].
222 Chapter 5. Distance and connectivity
Below is an example.
sage: X = Paley(13)
sage: X.vertices()
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12]
sage: X.is_vertex_transitive()
True
sage: X.degree_sequence()
[6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6]
sage: X.spectrum()
[6, 1.302775637731995?, 1.302775637731995?, 1.302775637731995?,
1.302775637731995?, 1.302775637731995?, 1.302775637731995?,
-2.302775637731995?, -2.302775637731995?, -2.302775637731995?,
-2.302775637731995?, -2.302775637731995?, -2.302775637731995?]
sage: G = X.automorphism_group()
sage: G.cardinality()
78
sage: 13*12/2
78
5.9 Problems
When you don’t share your problems, you resent hearing the problems of other people.
— Chuck Palahniuk, Invisible Monsters, 1999
5.1. Let G = (V, E) be an undirected, unweighted simple graph. Show that V and the
distance function on G form a metric space if and only if G is connected.
5.2. Let u and v be two distinct vertices in the same connected component of G. If P
is a u-v path such that d(u, v) = (u), we say that P is an eccentricity path for u.
(a) If r is the root of a tree, show that the end-vertex of an eccentricity path for
r is a leaf.
(b) If v is a vertex of a tree distinct from the root r, show that any eccentricity
path for v must contain r or provide an example to the contrary.
(c) A vertex w is said to be an eccentric vertex of v if d(v, w) = (v). Intuitively,
an eccentric vertex of v can be considered as being as far away from v as
possible. If w is an eccentric vertex of v and vice versa, then v and w are said
to be mutually eccentric. See Buckley and Lau [48] for detailed discussions of
mutual eccentricity. If w is an eccentric vertex of v, explain why v is also an
eccentric vertex of w or show that this does not in general hold.
5.3. If u and v are vertices of a connected graph G such that d(u, v) = diam(G), show
that u and v are mutually eccentric.
5.4. If uv is an edge of a tree T and w is a vertex of T distinct from u and v, show that
|d(u, w) − d(w, v)| = W (uv) with W (uv) being the weight of uv.
5.9. Problems 223
5.5. If u and v are vertices of a tree T such that d(u, v) = diam(T ), show that u and v
are leaves.
5.6. Let v1 , v2 , . . . , vk be the leaves of a tree T . Show that per(T ) = {v1 , v2 , . . . , vk }.
5.7. Show that all the eccentric vertices of a tree are leaves.
5.8. If G is a connected graph, show that rad(G) ≤ diam(G) ≤ 2 · rad(G).
5.9. Let T be a tree of order ≥ 3. If the center of T has one vertex, show that diam(T ) =
2 · rad(T ). If the center of T has two vertices, show that diam(T ) = 2 · rad(T ) − 1.
5.10. Let G = (V, E) be a simple undirected, connected graph. Define the distance of a
vertex v ∈ V by X
d(v) = d(v, x)
x∈V
and define the distance of the graph G itself by
1X
d(G) = d(v).
2 v∈V
For any vertex v ∈ V , show that d(G) ≤ d(v) + d(G − v) with G − v being a vertex
deletion subgraph of G. This result appeared in Entringer et al. [74, p.284].
5.11. Determine the sequence of distance matrices for the graphs in Figure 5.4.
5.12. If G = (V, E) is an undirected connected graph and v ∈ V , prove the following
vertex connectivity inequality:
κ(G) − 1 ≤ κ(G − v) ≤ κ(G).
5.13. If G = (V, E) is an undirected connected graph and e ∈ E, prove the following
edge connectivity inequality:
λ(G) − 1 ≤ λ(G − e) ≤ λ(G).
5.14. Figure 5.11 depicts how common grape cultivars are related to one another; the
graph is adapted from Myles et al. [158]. The numeric code of each vertex can
be interpreted according to Table 5.2. Compute various distance and connectivity
measures for the graph in Figure 5.11.
5.15. Prove the characterizations of 2-connected graphs as stated in Theorem 5.13.
5.16. Let G = (V, E) be an undirected connected graph of order n and suppose that
deg(v) ≥ (n + k − 2)/2 for all v ∈ V and some fixed positive integer k. Show that
G is k-connected.
5.17. A vertex (or edge) separating set S of a connected graph G is minimum if S has
the smallest cardinality among all vertex (respectively edge) separating sets in G.
Similarly S is said to be maximum if it has the greatest cardinality among all
vertex (respectively edge) separating sets in G. For the graph in Figure 5.7(a),
determine the following:
224 Chapter 5. Distance and connectivity
17
5 3
39
4 38
42
49
40 36
11
26
18
51
12 37
8 7
47 30
2
24
48 43
33
41 45
27 19
14 32 15
52 9 44
13
20
23 10
6
46 25
29 28
34
31
53
0
16
35
1
50
21
22
Figure 5.11: Network of common grape cultivars.
5.9. Problems 225
code name code name code name
0 Alicante Bouschet 1 Aramon 2 Bequignol
3 Cabernet Franc 4 Cabernet Sauvignon 5 Carignan
6 Chardonnay 7 Chenin Blanc 8 Colombard
9 Donzillinho 10 Ehrenfelser 11 Fer Servadou
12 Flora 13 Gamay 14 Gelber Ortlieber
15 Grüner Veltliner 16 Kemer 17 Merlot
18 Meslier-Saint-Francois 19 Müller-Thurgau 20 Muscat Blanc
21 Muscat Hamburg 22 Muscat of Alexandria 23 Optima
24 Ortega 25 Osteiner 26 Péagudo
27 Perle 28 Perle de Csaba 29 Perlriesling
30 Petit Manseng 31 Petite Bouschet 32 Pinot Noir
33 Reichensteiner 34 Riesling 35 Rotberger
36 Roter Veltliner 37 Rotgipfler 38 Royalty
39 Ruby Cabernet 40 Sauvignon Blanc 41 Schönburger
42 Semillon 43 Siegerrebe 44 Sylvaner
45 Taminga 46 Teinturier du Cher 47 Tinta Madeira
48 Traminer 49 Trincadeiro 50 Trollinger
51 Trousseau 52 Verdelho 53 Wittberger
Table 5.2: Numeric code and actual name of common grape cultivars.
(a) A minimum vertex separating set.
(b) A minimum edge separating set.
(c) A maximum vertex separating set.
(d) A maximum edge separating set.
(e) The number of minimum vertex separating sets.
(f) The number of minimum edge separating sets.
Chapter 6
Optimal graph traversals
6.1 Eulerian graphs
Motivation: tracing out all the edges of a graph without lifting your pencil.
multigraphs and simple graphs
Eulerian tours
Eulerian trails
6.2 Hamiltonian graphs
— Randall Munroe, xkcd, http://xkcd.com/230/
Motivation: the eager tourist problem: visiting all major sites of a city in the least
time/distance.
Hamiltonian paths (or cycles)
Hamiltonian graphs
Theorem 6.1. Ore 1960. Let G be a simple graph with n ≥ 3 vertices. If deg(u) +
deg(v) ≥ n for each pair of non-adjacent vertices u, v ∈ V (G), then G is Hamiltonian.
Corollary 6.2. Dirac 1952. Let G be a simple graph with n ≥ 3 vertices. If deg(v) ≥
n/2 for all v ∈ V (G), then G is Hamiltonian.
226
6.3. The Chinese Postman Problem 227
6.3 The Chinese Postman Problem
See section 6.2 of Gross and Yellen [95].
de Bruijn sequences
de Bruijn digraphs
constructing a (2, n)-de Bruijn sequence
postman tours and optimal postman tours
constructing an optimal postman tour
6.4 The Traveling Salesman Problem
— Randall Munroe, xkcd, http://xkcd.com/399/
See section 6.4 of Gross and Yellen [95], and section 35.2 of Cormen et al. [61].
Gray codes and n-dimensional hypercubes
the Traveling Salesman Problem (TSP)
nearest neighbor heuristic for TSP
some other heuristics for solving TSP
Chapter 7
Planar graphs
A planar graph is a graph that can be drawn on a sheet of paper without any overlapping
between its edges.
It is a property of many “natural” graphs drawn on the earth’s surface, like for
instance the graph of roads, or the graph of internet fibers. It is also a necessary property
of graphs we want to build, like VLSI layouts.
Of course, the property of being planar does not prevent one from finding a drawing
with many overlapping between edges, as this property only asserts that there exists a
drawing (or embedding) of the graph avoiding it. Planarity can be characterized in many
different ways, one of the most satiating being Kuratowski’s theorem.
See chapter 9 of Gross and Yellen [95].
7.1 Planarity and Euler’s Formula
planarity, non-planarity, planar and plane graphs
crossing numbers
Theorem 7.1. The complete bipartite graph K3,n is non-planar for n ≥ 3.
Theorem 7.2. Euler’s Formula. Let G be a connected plane graph having n vertices,
e edges and f faces. Then n − e + f = 2.
7.2 Kuratowski’s Theorem
Kuratowski graphs
The objective of this section is to prove the following theorem.
Theorem 7.3. [133] Kuratowski’s Theorem. A graph is planar if and only if it
contains no subgraph homeomorphic to a subdivision of K5 or K3,3 .
Graph Minors : The reader may find interesting to notice that the previous re-
sult, first proved in 1930 as purely topological (there is no mention of graphs in Kura-
towski’s original paper), can be seen as a very special case of the Graph Minor Theorem
(Thm1.31).
228
7.2. Kuratowski’s Theorem 229
(a) Errera graph.
(b) Planar representation.
Figure 7.1: The Errera graph is planar.
230 Chapter 7. Planar graphs
It can easily be seen that if a graph G is planar, any of its subgraph is also planar.
Besides, planarity is still preserved under edge contraction. These two facts mean to-
gether that any minor of a planar graph is still planar graph, which makes of planarity
a minor-closed property. If we let P̄ denote the poset of all non-planar graph, ordered
with the minor partial order, we can now consider the set P̄min of its minimal elements
which, by the Graph Minor Theorem, is a finite set.
Actually, Kuratowski’s theorem asserts that P̄min = {K5 , K3,3 }.
7.3 Planarity algorithms
planarity testing for 2-connected graphs
planarity testing algorithm of Hopcroft and Tarjan [107]
planarity testing algorithm of Boyer and Myrvold [38]
Chapter 8
Graph coloring
— Spiked Math, http://spikedmath.com/210.html
See Jensen and Toft [114] for a survey of graph coloring problems.
See Dyer and Frieze [70] for an algorithm on randomly colouring random graphs.
8.1 Vertex coloring
Vertex coloring is a widespread center of interest in graph theory, which has many vari-
ants. Formally speaking, a coloring of the vertex set of a graph G is any function
f : V (G) 7→ {1, . . . , k} giving to each vertex a color among a set of cardinal k. Things
get much more difficult when we add to it the constraint under which a coloring becomes
a proper coloring : a coloring with k colors of a graph G is said to be proper if there are
no edges between any two vertices colored with the same color. This can be rephrased
in many different ways :
∀i ∈ {1, . . . , k}, G[f −1 (i)] is a stable set
∀u, v ∈ G, u 6= v, f (u) = f (v) ⇒ uv 6∈ E(G)
A proper coloring of G with k colors is a partition of V (G) into k independent sets
231
232 Chapter 8. Graph coloring
Chromatic numbers : quite clearly, it is always possible to find a proper coloring
of a graph G using one color per vertex. For this reason, the Coloring Problem is an
optimisation problem which amounts to finding the least number k of colors such that
G can be properly colored with k colors – is k-colorable. This integer, written χ(G), is
called the chromatic number of G.
Greedy coloring, and an easy upper bound : the simple fact that Graph Col-
oring is a NP-complete problem must not prevent one from trying to color it greedily.
One such method would be to iteratively pick, in a graph G, an uncolored vertex v, and
to color it with the smallest color available which is not yet used by one of its neighbors
(in order to keep it proper). Such a coloring will obviously stay proper until the whole
vertex set is colored, and never use more than ∆(G) + 1 different colors (where ∆(G)
is the maximal degree of G), as in the procedure no vertex will ever exclude more than
∆(G) colors.
Such an algorithm can be written in Sage in a few lines :
sage : g = graphs . RandomGNP (100 ,5/100)
sage : C = Set ( xrange (100))
sage : color = {}
sage : for u in g :
... interdits = Set ([ color [ v ] for v in g . neighbors ( u ) if color . has_key ( v )])
... color [ u ] = min (C - interdits )
Brook’s Theorem
heuristics for vertex coloring
8.2 Edge coloring
Edge coloring is the direct application of vertex coloring to the line graph of a graph G,
written L(G), which is the graph whose vertices are the edges of G, two vertices being
adjacent if and only if their corresponding edges share an endpoint. We write χ(L(G)) =
χ0 (G) the chromatic index of G. In this special case, however, the optimization problem
defined above, though still NP-Complete, is much better understood through Vizing’s
theorem.
Theorem 8.1 (Vizing). The edges of a graph G can be properly colored using at least
∆(G) colors and at most ∆(G) + 1
Notice that the lower bound can be easily proved : if a vertex v has a degree d(v),
then at least d(v) colors are required to color G as all the edges incident to v must
receive different colors. Besides, the upper bound of ∆(G) + 1 can not be deduced from
the greedy algorithm given in the previous section, as the maximal degree of L(G) is not
equal to ∆(G) but to max d(u) + d(v) − 2, which can reach 2∆(G) − 2 in regular graphs.
u∼v
algorithm for edge coloring by maximum matching
algorithm for sequential edge coloring
8.3 Applications of graph coloring
assignment problems
8.3. Applications of graph coloring 233
scheduling problems
matching problems
map coloring and the Four Color Problem
Chapter 9
Network flows
See Jungnickel [120], and chapter 12 of Gross and Yellen [95].
9.1 Flows and cuts
single source-single sink networks
feasible networks
maximum flow and minimum cut
Let G = (V, E, i, h) be an unweighted multidigraph, as in Definition 1.6.
If F is a field such as R or GF (q) or a ring such as Z, let
C 0 (G, F ) = {f : V → F }, C 1 (G, F ) = {f : E → F },
be the sets of F -valued functions defined on V and E, respectively. If F is a field then
these are F -inner product spaces with inner product
X
(f, g) = f (x)g(x), (X = V, resp. X = E), (9.1)
x∈X
and
dim C 0 (G, F ) = |V |, dim C 1 (G, F ) = |E|.
If you index the sets V and E in some arbitrary but fixed way and define, for 1 ≤ i ≤ |V |
and 1 ≤ j ≤ |E|,
1, v = vi , 1, e = ej ,
fi (v) = gj (e) =
0, otherwise, 0, otherwise,
then F = {fi } ⊂ C 0 (G, F ) is a basis of C 0 (G, F ) and G = {gj } ⊂ C 1 (G, F ) is a basis of
C 1 (G, F ).
We order the edges
E = {e1 , e2 , . . . , e|E| },
in some arbitrary but fixed way. A vector representation (or characteristic vector or
incidence vector) of a subgraph G0 = (V, E 0 ) of G = (V, E), E 0 ⊂ E, is a binary |E|-tuple
234
9.1. Flows and cuts 235
vec(G0 ) = (a1 , a2 , . . . , a|E| ) ∈ GF (2)|E| ,
where
0 1, if ei ∈ E 0 ,
ai = ai (E ) =
/ E 0.
0, if ei ∈
In particular, this defines a mapping
vec : {subgraphs of G = (V, E)} → GF (2)|E| .
For any non-trivial partition
V = V1 ∪ V2 , Vi 6= ∅, V1 ∩ V2 = ∅,
the set of all edges e = (v1 , v2 ) ∈ E, with vi ∈ Vi (i = 1, 2), is called a cocycle1 of G.
A cocycle with a minimal set of edges is a bond (or cut set) of G. An Euler subgraph is
either a cycle or a union of edge-disjoint cycles.
The set of cycles of G is denoted Z(G) and the set of cocycles is denoted Z ∗ (G).
The F -vector space spanned by the vector representations of all the cycles is called
the cycle space of G, denoted Z(G) = Z(G, F ). This is the kernel of the incidence matrix
of G (§14.2 in Godsil and Royle [91]). Define
1
D : CP (G, F ) → C 0 (G,
P F ),
(Df )(v) = h(e)=v f (e) − t(e)=v f (e).
With respect to these bases F and G, the matrix representing the linear transformation
D : C 1 (G, F ) → C 0 (G, F ) is the incidence matrix. An element of the kernel of D is
sometimes called a flow (see Biggs [26]) or circulation (see below). Therefore, this is
sometimes also referred to as the space of flows or the circulation space.
It may be regarded as a subspace of C 1 (G, F ) of dimension n(G). When F is a finite
field, sometimes2 the cycle space is called the cycle code of G.
Let F be a field such as R or GF (q), for some prime power q. Let G be a digraph.
Some define a circulation (or flow) on G = (V, E) to be a function
f : E → F,
satisfying3
P P
u∈V, (u,v)∈E f (u, v) = w∈V, (v,w)∈E f (v, w).
(Note: this is simply the condition that f belongs to the kernel of D.)
Suppose G has a subgraph H and f is a circulation of G such that f is a constant
function on H and 0 elsewhere. We call such a circulation a characteristic function of
H. For example, if G has a cycle C and if f is the characteristic function on C, then f
is a circulation.
The circulation space C is the F -vector space of circulation functions. The cycle space
“clearly” may be identified with a subspace of the circulation space, since the F -vector
1
Also called an edge cut subgraph or disconnecting set or seg or edge cutset.
2
Jungnickel and Vanstone in [121] call this the even graphical code of G.
3
Note: In addition, some authors add the condition f (e) ≥ 0 - see e.g., Chung [58].
236 Chapter 9. Network flows
space spanned by the characteristic functions of cycles may be identified with the cycle
space of G. In fact, these spaces are isomorphic. Under the inner product (9.1), i.e.,
X
(f, g) = f (e)g(e), (9.2)
e∈E
this vector space is an inner product space.
Example 9.1. This example is not needed but is presented for its independent inter-
est. Assume G = (V, E) is a strongly connected directed graph. Define the transition
probability matrix P for a digraph G by
dx , if (x, y) ∈ E,
P (x, y) =
0, otherwise,
where dx denotes the out-degree. The Perron-Frobenius Theorem states that there exists
a unique left eigenvector φ such that (when regarded as a function φ : V → R) φ(v) > 0,
P all v ∈ V and φP = ρφ, where ρ is the spectral radius of G. We scale φ so that
for
v∈V φ(v) = 1. (This vector is sometimes called the Perron vector.) Let Fφ (u, v) =
φ(v)P (u, v). Fact: Fφ is a circulation. For a proof, see F. Chung [58].
If the edges of E are indexed in some arbitrary but fixed way then a circulation
function restricted to a subgraph H of G may be identified with a vector representation
of H, as described above. Thefore, the circulation functions gives a coordinate-free
version of the cycle space.
The F -vector space spanned by the vector representations of all the segs is called the
cocycle space (or the cut space) of G, denoted Z ∗ (G) = Z ∗ (G, F ). This is the column
space of the transpose of the incidence matrix of G (§14.1 in Godsil and Royle [91]). It
may be regarded as a subspace of C 1 (G, F ) of dimension the rank of G, r(G). When F
is a finite field, sometimes the cocycle space is called the cocycle code of G.
Lemma 9.2. Under the inner product (9.1) on C 1 (G, F ), the cycle space is orthogonal
to the cocycle space.
Solution. One proof follows from Theorem 8.3.1 in Godsil and Royle [91].
Here is another proof. By Theorem 2.3 in Bondy and Murty [34, p.27], an edge of G is
an edge cut if and only if it is contained in no cycle. Therefore, the vector representation
of any cocycle is supported on a set of indices which is disjoint from the support of the
vector representation of any cycle. Therefore, there is a basis of the cycle space which is
orthogonal to a basis of the cocycle space.
Proposition 9.3. Let F = GF (2). The cycle code of a graph G = (V, E) is a linear
binary block code of length |E|, dimension equal to the nullity of the graph, n(G),
and minimum distance equal to the girth of G. If C ⊂ GF (2)|E| is the cycle code
associated to G and C ∗ is the cocycle code associated to G then C ∗ is the dual code of
C. In particular, the cocycle code of G is a linear binary block code of length |E|, and
dimension r(G) = |E| − n(G).
This follows from Hakimi-Bredeson [101] (see also Jungnickel-Vanstone [121]) in the
binary case4 .
4
It is likely true in the non-binary case as well, but no proof seems to be in the literature.
9.1. Flows and cuts 237
Solution. Let d denote the minimum distance of the code C. Let γ denote the girth of
G, i.e., the smallest cardinality of a cycle in G. If K is a cycle in G then the vector
vec(K) ∈ GF (2)|E| is an element of the cycle code C ⊂ GF (2)|E| . This implies d ≤ γ.
In the other direction, suppose K1 and K2 are cycles in G with associated support
vectors v1 = vec(K1 ), v2 = vec(K2 ). Assume that at least one of these cycles is a cycle
of minimum length, say K1 , so the weight of its corresponding support vector is equal to
the girth γ. The only way that wt(v1 + v2 ) < min{wt(v1 ), wt(v2 )} can occur is if K1 and
K2 have some edges in common. In this case, the vector v1 + v2 represents a subgraph
which is either a cycle or it is a union of disjoint cycles. In either case, by minimality of
K1 , these new cycles must be at least as long. Therefore, d ≥ γ, as desired.
Consider a spanning tree T of a graph G and its complementary subgraph T . For
each edge e of T the graph T ∪ e contains a unique cycle. The cycles which arise in this
way are called the fundamental cycles of G, denoted cyc(T, e).
Example 9.4. Consider the graph below, with edges labeled as indicated, together with
a spanning tree, depicted to its right, in Figure 9.4.
7 9
0 4
2 6 8 10
1 3 5
Figure 9.1: A graph and a spanning tree for it.
You can see from Figure 9.4 that:
by adding edge 2 to the tree, you get a cycle 0, 1, 2 with vector representation
g1 = (1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0),
by adding edge 6 to the tree, you get a cycle 4, 5, 6 with vector representation
g2 = (0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0),
by adding edge 10 to the tree, you get a cycle 8, 9, 10 with vector representation
g3 = (0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1).
The vectors {g1 , g2 , g3 } form a basis of the cycle space of G over GF (2).
The cocycle space of a graph G (also known as the bond space of G or the cut-set
space of G) is the F -vector space spanned by the characteristic functions of bonds.
Example 9.5. Consider the graph below, with edges labeled as indicated, together with
an example of a bond, depicted to its right, in Figure 9.5.
You can see from Figure 9.5 that:
238 Chapter 9. Network flows
7 9
0 4
2 6 8 10
1 3 5
Figure 9.2: A graph and a bond of it.
by removing edge 3 from the graph, you get a bond with vector representation
b1 = (0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0),
by removing edge 7 from the graph, you get a bond with vector representation
b2 = (0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0),
by removing edges 0, 1 from the graph, you get a bond with vector representation
b3 = (1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0),
by removing edges 1, 2 from the graph, you get a bond with vector representation
b4 = (0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0),
by removing edges 4, 5 from the graph, you get a bond with vector representation
b5 = (0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0),
by removing edges 4, 6 from the graph, you get a bond with vector representation
b6 = (0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0),
by removing edges 8, 9 from the graph, you get a bond with vector representation
b7 = (0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0),
by removing edges 9, 10 from the graph, you get a bond with vector representation
b8 = (0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1).
The vectors {b1 , b2 , b3 , b4 , b5 , b6 , b7 , b8 } form a basis of the cocycle space of G over GF (2).
Note that these vectors are orthogonal to the basis vectors of the cycle space in
Example 9.4. Note also that the xor sum of two cuts is not a cut.For example, if you xor
the bond 4, 5 with the bond 4, 6 then you get the subgraph foormed by the edges 5, 6
and that is not a disconnecting cut of G.
9.2. Chip firing games 239
9.2 Chip firing games
Chip firing games on graphs (which are just pure fun) relate to “abelian sandpile models”
from physics to “rotor-routing models” from theoretical computer scientists (designing
efficient computer multiprocessor circuits) to “self-organized criticality” (a subdiscipline
of dynamical systems) to “algebraic potential theory” on a graph [26] to cryptography
(via the Biggs cryptosystem). Moreover, it relates the concepts of the Laplacian of the
graph to the tree number to the circulation space of the graph to the incidence matrix,
as well as many other ideas. Some good references are [69], [167], [168], [105] and [27].
Basic set-up
A chip firing game always starts with a directed multigraph G having no loops. A con-
figuration is a vertex-weighting, i.e., a function s : V → R. The players are represented
by the vertices of G and the vertex-weights represent the number of chips each player
(represented by that vertex) has. The initial vertex-weighting is called the starting con-
figuration of G. Let vertex v have outgoing degree d+ (v). If the weight of vertex v is
≥ d+ (v) (so that player can afford to give away all their chips) then that vertex is called
active.
Here is some SAGE/Python code for determining the active vertices.
SAGE
def active_vertices(G, s):
"""
Returns the list of active vertices.
INPUT:
G - a graph
s - a configuration (implemented as a list
or a dictionary keyed on
the vertices of the graph)
EXAMPLES:
sage: A = matrix([[0,1,1,0,0],[1,0,1,0,0],[1,1,0,1,0],[0,0,0,0,1],[0,0,0,0,0]])
sage: G = Graph(A, format = "adjacency_matrix", weighted = True)
sage: s = {0: 3, 1: 1, 2: 0, 3: 1, 4: 1}
sage: active_vertices(G, s)
[0, 4]
"""
V = G.vertices()
degs = [G.degree(v) for v in V]
active = [v for v in V if degs[V.index(v)]<=s[v]]
return active
If v is active then when you fire v you must also change the configuration. The new
configuration s0 will satisfy s0 (v) = s(v) − d+ (v) and s0 (v 0 ) = s(v 0 ) + 1 for each neighbor
v 0 of v. In other words, v will give away one chip to each of its d+ (v) neighbors. If
x : V → {0, 1}|V | ⊂ R|V | is the representation vector (“characteristic function”) of a
vertex then this change can be expressed more compactly as
s0 = s − L ◦ x, (9.3)
where L is the vertex Laplacian. It turns out that the column sums of L are all 0, so
this operation does not change the total number of chips. We use the notation
v
s → s0 ,
240 Chapter 9. Network flows
to indicate that the configuration s0 is the result of firing vertex v in configuration s.
Example 9.6. Consider the graph
1
•
@
@
0• @•2
4• •3
Figure 9.3: A graph with 5 vertices.
This graph has incidence matrix
−1 −1 0 0 0
0 −1 −1 0 0
D= 1 0 −1 −1 0 ,
0 0 0 −1 −1
0 0 0 0 1
and Laplacian
2 −1 0 − 1 0 0
−1 2 −1 0 0
L=D· D=
t
−1 −1 3 −1 0 .
0 0 −1 2 −1
0 0 0 −1 1
Suppose the initial configuration is s = (3, 1, 0, 1, 1), i.e.,
player 0 has 3 dollars,
player 1 has 1 dollar,
player 2 has nothing,
player 3 has 1 dollar,
player 4 has 1 dollar.
Notice player 0 is active. If we fire 0 then we get the new configuration s0 = (1, 2, 1, 1, 1).
Indeed, if we compute s0 = s − Lx(0), we get:
3 2 −1 −1 0 0 1 3 2 1
1 −1 2 −1 0 0 0 1 −1 2
s0 =
0 −
−1 −1 3 −1 0
0 =
0 −
−1 =
1 .
1 0 0 −1 2 −1 0 1 0 1
1 0 0 0 −1 1 0 1 0 1
This can be written more concisely as
0
(3, 1, 0, 1, 1) → (1, 2, 1, 1, 1).
9.2. Chip firing games 241
We have the cycle
1 0 2
(1, 2, 1, 1, 1) → (2, 0, 2, 1, 1) → (0, 1, 3, 1, 1) → (1, 2, 0, 2, 1)
3 4
→ (1, 2, 1, 0, 2) → (1, 2, 1, 1, 1).
Chip-firing game variants
For simplicity, let G = (V, E) be an undirected graph with an indexed set of vertices
V = {v1 , . . . , vm } and an indexed set of vertices E = {e1 , . . . , en }.
One variant (the “sandpile model”) has a special vertex, called “the sink,” which has
special firing properties. In the sandpile variant, the sink is never fired. Another variant
(the “dollar game”) has a special vertex, called “the source,” which has special firing
properties. In the dollar game variant, the source is only fired when not other vertex is
active. We shall consider the dollar game variant here, following Biggs [28].
We select a distinguished vertex q ∈ V , called the “source5 ,” which has a special
property to be described below. For the dollar game, a configuration is a function
s : V → R for which
X
s(v) = 0,
v∈V
and s(v) ≥ 0 for all v ∈ V with v 6= q. A vertex v 6= q can be fired if and only
if deg(v) ≤ s(v) (i.e., it “has enough chips”). The equation (9.3) describes the new
configuration after firing a vertex.
Here is some SAGE/Python code for determining the configuration after firing an
active vertex.
SAGE
def fire(G, s, v0):
"""
Returns the configuration after firing the active vertex v.
INPUT:
G - a graph
s - a configuration (implemented as a list
or a dictionary keyed on
the vertices of the graph)
v - a vertex of the graph
EXAMPLES:
sage: A = matrix([[0,1,1,0,0],[1,0,1,0,0],[1,1,0,1,0],[0,0,0,0,1],[0,0,0,0,0]])
sage: G = Graph(A, format = "adjacency_matrix", weighted = True)
sage: s = {0: 3, 1: 1, 2: 0, 3: 1, 4: 1}
sage: fire(G, s, 0)
{0: 1, 1: 2, 2: 1, 3: 1, 4: 1}
"""
V = G.vertices()
j = V.index(v0)
s1 = copy(s)
if not(v0 in V):
raise ValueError, "the last argument must be a vertex of the graph."
if not(v0 in active_vertices(G, s)):
raise ValueError, "the last argument must be an active vertex of the graph."
degs = [G.degree(w) for w in V]
for w in V:
if w == v0:
s1[v0] = s[v0] - degs[j]
if w in G.neighbors(v0):
5
Biggs humorously calls q “the government.”
242 Chapter 9. Network flows
s1[w] = s[w]+1
return s1
We say s : V → R is a stable configuration if 0 ≤ s(v) < deg(v), for all v 6= q. The
source vertex q can only be fired when no other vertex can be fired, that is only in the
case when a stable configuration has been reached.
Here is some SAGE/Python code for determining the stable vertices.
SAGE
def stable_vertices(G, s, source = None):
"""
Returns the list of stable vertices.
INPUT:
G - a graph
s - a configuration (implemented as a list
or a dictionary keyed on
the vertices of the graph)
EXAMPLES:
sage: A = matrix([[0,1,1,0,0],[1,0,1,0,0],[1,1,0,1,0],[0,0,0,0,1],[0,0,0,0,0]])
sage: G = Graph(A, format = "adjacency_matrix", weighted = True)
sage: s = {0: 3, 1: 1, 2: 0, 3: 1, 4: 1}
sage: stable_vertices(G, s)
"""
V = G.vertices()
degs = [G.degree(v) for v in V]
if source==None:
stable = [v for v in V if degs[V.index(v)]>s[v]]
else:
stable = [v for v in V if degs[V.index(v)]>s[v] and v!=source]
return stable
Suppose we are in a configuration s1 . We say a sequence vertices S = (w1 , w2 , . . . , wk ),
wi ∈ V not necessarily distinct, is legal if,
w1 is active in configuration s1 ,
for each i with 1 ≤ i < k, si+1 is obtained from si by firing wi in configuration si ,
for each i with 1 ≤ i < k, wi+1 is active in the configuration si+1 defined in the
previous step,
the source vertex q occurs in S only if it immediately follows a stable configuration.
We call s1 or w1 the start of S. A configuration s is recurrent if there is a legal sequence
starting at s which leads back to s. A configuration is critical if it recurrent and stable.
Here is some SAGE/Python code for determining a stable vertex resulting from a
legal sequence of firings of a given configuration s. I think it returns the unique critical
configuration associated to s but have not proven this.
SAGE
def stabilize(G, s, source, legal_sequence = False):
"""
Returns the stable configuration of the graph originating from
the given configuration s. If legal_sequence = True then the
sequence of firings is also returned. By van den Heuvel [1],
the number of firings needed to compute a critical configuration
is < 3(S+2|E|)|V|ˆ2, where S is the sum of the positive
9.2. Chip firing games 243
weights in the configuration.
EXAMPLES:
sage: A = matrix([[0,1,1,0,0],[1,0,1,0,0],[1,1,0,1,0],[0,0,1,0,1],[0,0,0,1,0]])
sage: G = Graph(A, format="weighted_adjacency_matrix")
sage: s = {0: 3, 1: 1, 2: 0, 3: 1, 4: -5}
sage: stabilize(G, s, 4)
{0: 0, 1: 1, 2: 2, 3: 1, 4: -4}
REFERENCES:
[1] J. van den Heuvel, "Algorithmic aspects of a chip-firing
game," preprint.
"""
V = G.vertices()
E = G.edges()
fire_number = 3*len(V)ˆ2*(sum([s[v] for v in V if s[v]>0])+2*len(E))+len(V)
if legal_sequence:
seq = []
stab = []
ac = active_vertices(G,s)
for i in range(fire_number):
if len(ac)>0:
s = fire(G,s,ac[0])
if legal_sequence:
seq.append(ac[0])
else:
stab.append(s)
break
ac = active_vertices(G,s)
if len(stab)==0:
raise ValueError, "No stable configuration found."
if legal_sequence:
return stab[0], seq
else:
return stab[0]
The incidence matrix D and its transpose t D can be regarded as homomorphisms
D : C 1 (G, Z) → C 0 (G, Z) and t
D : C 0 (G, Z) → C 1 (G, Z).
We can also regard the Laplacian L = D · t D as a homomorphism C 0 (G, Z) → C 0 (G, Z).
Denote by σ : C 0 (G, Z) → Z the homomorphism defined by
X
σ(f ) = f (v).
v∈V
Denote by K(G) the set of critical configurations on a graph G.
Lemma 9.7. (Biggs [28]) The set K(G) of critical configurations on a connected graph
G is in bijective correspondence with the abelian group Ker(σ)/Im(Q).
If you accept this lemma (which we do not prove here) then you must believe that
there is a bijection f : K(G) → Ker(σ)/Im(Q). Now, a group operation • on K(G) an
be defined by
a • b = f −1 (f (a) + f (b)),
for all a, b ∈ Ker(σ)/Im(Q).
Example 9.8. Consider again the graph
This graph has incidence matrix
244 Chapter 9. Network flows
1
•
@
@
0• @•2
4◦ •3
Figure 9.4: A graph with 5 vertices.
−1 −1 0 0 0
0 −1 −1 0 0
D=
1 0 −1 −1 0 ,
0 0 0 −1 −1
0 0 0 0 1
and Laplacian
2 −1 0 − 1 0 0
−1 2 −1 0 0
L=D· D=
t
−1 −1 3 −1 0 .
0 0 −1 2 −1
0 0 0 −1 1
Suppose the initial configuration is s = (3, 1, 0, 1, −5), i.e.,
player 0 has 3 dollars,
player 1 has 1 dollar,
player 2 has nothing,
player 3 has 1 dollar,
player 4 is the source vertex q.
The legal sequence (0, 1, 0, 2, 1, 0, 3, 2, 1, 0) leads to the stable configuration (0, 1, 2, 1, −4).
If q is fired then the configuration (0, 1, 2, 2, −5) is achieved. This is recurrent since it is
contained in the cyclic legal sequence
3 2
(0, 1, 2, 2, −5) → (0, 1, 3, 0, −4) → (1, 2, 0, 1, −4)
1 0 q
→ (2, 0, 1, 1, −4) → (0, 1, 2, 1, −4) → (0, 1, 2, 2, −5).
In particular, the configuration (0, 1, 2, 1, −4) is also recurrent. Since it is both stable
and recurrent, it is critical.
The following result is of basic importance but I’m not sure who proved it first. It is
quoted in many of the papers on this topic in one form or another.
Theorem 9.9. (Biggs [27], Theorem 3.8) If s is an configuration and G is connected
then there is a unique critical configuration s0 which can be obtained by a sequence of
legal firings for starting at s.
9.3. Ford-Fulkerson theorem 245
The map defined by the above theorem is denoted
γ : C 0 (G, R) → K(G).
Another way to define multiplication • on on K(G) is
γ(s1 ) • γ(s2 ) = γ(s1 + s2 ),
where s1 + s2 is computed using addition on C 0 (G, R). According to Perkinson [167],
Theorem 2.16, the critical group satisfies the following isomorphism:
K(G) ∼
= Zm−1 /L,
where L is the integer lattice generated by the columns of the reduced Laplacian matrix6 .
If s is a configuration then we define
X
wt(s) = s(q)
v∈V,v6=q
to be the weight of the configuration. The level of the configuration is defined by
level(s) = wt(s) − |E| + deg(q).
Lemma 9.10. (Merino [153]) If s is a critical configuration then
0 ≤ level(s) ≤ |E| − |V | + 1.
This is proven in Theorem 3.4.5 in [153]. What is also proven in [153] is a statement
whcih computes the number of critical configurations of a given level in terms of the
Tutte polynomial of the associated graph.
9.3 Ford-Fulkerson theorem
The Ford-Fulkerson Theorem, or “Max-flow/Min-cut Theorem,” was proven by P. Elias,
A. Feinstein, and C.E. Shannon in 1956, and, independently, by L.R. Ford, Jr. and D.R.
Fulkerson in the same year. So it should be called the “Elias-Feinstein-Ford-Fulkerson-
Shannon Theorem,” to be precise about the authorship.
To explain the meaning of this theorem, we need to introduce some notation and
terminology.
Consider an edge-weighted simple digraph G = (V, E, i, h) without negative weight
cycles. Here E ⊂ V (2) , i is an incidence function as in (??), which we regard as the
identity function, and h is an orientation function as in (??). Let G be a network, with
two distinguished vertices, the “source” and the “sink.” Let s and t denote the source and
the sink of G, respectively. The capacity (or edge capacity) ) is a mapping c : E → R,
denoted by cuv or c(u, v), for (u, v) ∈ E and h(e) = u. If (u, v) ∈ E and h(e) = v
then we set, by convention, c(v, u) = −c(u, v). Thinking of a graph as a network of
pipes (representing the edges) transporting water with various junctions (representing
6
The reduced Laplacian matrix is obtained from the Laplacian matrix by removing the row and
column associated to the source vertex.
246 Chapter 9. Network flows
vertices), the capacity function represents the maximum amount of “flow” that can pass
through an edge.
A flow is a mapping f : E → R, denoted by fuv or f (u, v), subject to the following
two constraints:
f (u, v) ≤ c(u, v), for each (u, v) ∈ V (the “capacity constraint”),
P P
u∈V, (u,v)∈E f (u, v) = u∈V, (v,u)∈E f (v, u) , for each v ∈ V (conservation of flows).
An edge (u, v) ∈ E is f -saturated if f (u, v) = c(u, v). An edge (u, v) ∈ E is f -zero
if f (u, v) = 0. A path with available capacity is called an “augmenting path.” More
precisely, a directed path form s to t is f -augmenting, or f -unsaturated, if no forward
edge is f -saturated and no backward edge is f -zero.
The value of the flow is defined by
X X
|f | = f (s, v) − f (v, s),
v∈V v∈V
where s is the source. It represents the amount of flow passing from the source to the
sink. The maximum flow problem is to maximize |f |, that is, to route as much flow as
possible from s to t.
Example 9.11. Consider the digraph having adjacency matrix
0 1 1 0 0 0
−1 0 −1 1 0 1
−1 1 0 0 1 0
,
0 −1 0 0 0 1
0 0 −1 0 0 1
0 −1 0 −1 −1 0
depicted in Figure 9.5.
2 4
0
1 5
3
Figure 9.5: A digraph with 6 vertices.
Suppose that each edge has capacity 1. A maximum flow f is obtained by taking a
flow value of 1 along each edge of the path
p1 : (0, 1), (1, 5),
and a flow value of 1 along each edge of the path
9.3. Ford-Fulkerson theorem 247
p2 : (0, 2), (2, 4), (4, 5).
The maximum value of the flow in this case is |f | = 2.
This graph can be created in Sage using the commands
sage : B = matrix ([[0 ,1 ,1 ,0 ,0 ,0] ,[0 ,0 ,0 ,1 ,0 ,1] ,[0 ,1 ,0 ,0 ,1 ,0] ,[0 ,0 ,0 ,0 ,0 ,1] ,[0 ,0 ,0 ,0 ,0 ,1] ,[0 ,0 ,
sage : H = DiGraph (B , format = " adjacency_matrix " , weighted = True )
Type H.show(edgewlabels=True) if you want to see the graph with the capacities la-
beling the edges.
Given a capacitated digraph with capacity c and flow f , we define the residual digraph
Gf = (V, E) to be the digraph with capacity cf (u, v) = c(u, v) − f (u, v) and no flow. In
other words, Gf is the same graph but it has a different capacity cf and flow 0. This is
also called a residual network.
Define an s − t cut in our capacitated digraph G to be a partition C = (S, T ) of V
such that s ∈ S and t ∈ T . Recall the cut-set of C is the set
{(u, v) ∈ E | u ∈ S, v ∈ T }.
Lemma 9.12. Let G = (V, E) be a capacitated digraph with capacity c : E → R, and
let s and t denote the source and the sink of G, respectively. If C is an s − t cut and if
the edges in the cut-set of C are removed, then |f | = 0.
Exercise 9.13. Prove Lemma 9.12.
The capacity of an s − t cut C = (S, T ) is defined by
X
c(S, T ) = c(u, v).
(s,t)∈(S,T )
The minimum cut problem is to minimize the amount of capacity of an s − t cut.
The following theorem is due to P. Elias, A. Feinstein, L.R. Ford, Jr., D.R. Fulkerson,
C.E. Shannon.
Theorem 9.14. (max-flow min-cut theorem) The maximum value of an s-t flow is equal
to the minimum capacity of an s-t cut.
The intuitive explanation of this result is as follows.
Suppose that G = (V, E) is a graph where each edge has capacity 1. Let s ∈ V be the
source and t ∈ V be the sink. The maximum flow from s to t is the maximum number
of independent paths from s to t. Denote this maximum flow by m. Each s-t cut must
intersect each s-t path at least once. In fact, if S is a minimal s-t cut then for each edge
e in S there is an s-t path containing e. Therefore, |S| ≤ e.
On the other hand, since each edge has unit capacity, the maximum flow value can’t
exceed the number of edges separating s from t, so m ≤ |S|.
Remark 9.15. Although the notion of an independent path is important for the network-
theoretic proof of Menger’s theorem (which we view as a corollary to the Ford-Fulkerson
theorem on network flows on networks having capacity 1 on all edges), its significance
is less important for networks having arbitrary capacities. One must use caution in
generalizing the above intuitive argument to establish a rigorous proof of the general
version of the MFMC theorem.
248 Chapter 9. Network flows
Remark 9.16. This theorem can be generalized as follows. In addition to edge capacity,
suppose there is capacity at each vertex, that is, a mapping c : V → R, denoted by
v 7→ c(v), such that the flow f has to satisfy not only the capacity constraint and the
conservation of flows, but also the vertex capacity constraint
X
f (w, v) ≤ c(v),
w∈V
for each v ∈ V − {s, t}. Define an s − t cut to be the set of vertices and edges such
that for any path from s to t, the path contains a member of the cut. In this case, the
capacity of the cut is the sum the capacity of each edge and vertex in it. In this new
definition, the generalized max-flow min-cut theorem states that the maximum value of
an s − t flow is equal to the minimum capacity of an s − t cut..
The idea behind the Ford-Fulkerson algorithm is very simple: As long as there is a
path from the source to the sink, with available capacity on all edges in the path, we
send as much flow as we can alone along each of these paths. This is done inductively,
one path at a time.
Algorithm 9.1: Ford-Fulkerson algorithm.
Input: Graph G = (V, E) with flow capacity c, source s, and sink t.
Output: A flow f from s to t which is a maximum for all edges in E.
1 f (u, v) ← 0 for each edge uv ∈ E
2 while there is an s-t path p in Gf such that cf (e) > 0 for each edge e ∈ E do
3 find cf (p) = min{cf (u, v) | (u, v) ∈ p}
4 for each edge uv ∈ do
5 f (u, v) = f (u, v) + cf (p)
6 f (v, u) = f (v, u) − cf (p)
To prove the max-flow/min-cut theorem we will use the following lemma.
Lemma 9.17. Let G = (V, E) be a directed graph with edge capacity c : E → Z, a
source s ∈ V , and a sink t ∈ V . A flow f : E → Z is a maximum flow if and only if
there is no f -augmenting path in the graph.
In other words, a flow f in a capacitated network is a maximum flow if and only if
there is no f -augmenting path in the network.
Solution. One direction is easy. Suppose that the flow is a maximum. If there is an
f -augmenting path then the current flow can be increased using that path, so the flow
would not be a maximum. This contradiction proves the “only if” direction.
Now, suppose there is no f -augmenting path in the network. Let S be the set of
vertices v such that there is an f -unsaturated path from the source s to v. We know
s ∈ S and (by hypothesis) t ∈/ S. Thus there is a cut of the form (S, T ) in the network.
Let e = (v, w) be any edge in this cut, v ∈ S and w ∈ T . Since there is no f -unsaturated
path from s to w, e is f -saturated. Likewise, any edge in the cut (T, S) is f -zero.
Therefore, the current flow value is equal to the capacity of the cut (S, T ). Therefore,
the current flow is a maximum.
We can now prove the max-flow/min-cut theorem.
9.3. Ford-Fulkerson theorem 249
Solution. Let f be a maximum flow. If
S = {v ∈ V | there exists an f − saturated path from s to v},
then by the previous lemma, S 6= V . Since T = V − S is non-empty, there is a cut
C = (S, T ). Each edge of this cut C in the capacitated network G is f -saturated.
Here is some Python code7 which implements this. The class FlowNetwork is basically
a Sage Graph class with edge weights and an extra data structure representing the flow
on the graph.
class Edge :
def __init__ ( self ,U ,V , w ):
self . source = U
self . to = V
self . capacity = w
def __repr__ ( self ):
return str ( self . source ) + " ->" + str ( self . to ) + " : " + str ( self . capacity )
class FlowNetwork ( object ):
"""
This is a graph structure with edge capacities .
EXAMPLES :
g = FlowNetwork ()
map ( g . add_vertex , [ ’ s ’,’o ’,’p ’,’q ’,’r ’,’t ’])
g . add_edge ( ’ s ’,’o ’ ,3)
g . add_edge ( ’ s ’,’p ’ ,3)
g . add_edge ( ’ o ’,’p ’ ,2)
g . add_edge ( ’ o ’,’q ’ ,3)
g . add_edge ( ’ p ’,’r ’ ,2)
g . add_edge ( ’ r ’,’t ’ ,3)
g . add_edge ( ’ q ’,’r ’ ,4)
g . add_edge ( ’ q ’,’t ’ ,2)
print g . max_flow ( ’ s ’,’t ’)
"""
def __init__ ( self ):
self . adj , self . flow , = {} ,{}
def add_vertex ( self , vertex ):
self . adj [ vertex ] = []
def get_edges ( self , v ):
return self . adj [ v ]
def add_edge ( self , u ,v , w =0):
assert ( u != v )
edge = Edge (u ,v , w )
redge = Edge (v ,u ,0)
edge . redge = redge
redge . redge = edge
self . adj [ u ]. append ( edge )
self . adj [ v ]. append ( redge )
self . flow [ edge ] = self . flow [ redge ] = 0
def find_path ( self , source , sink , path ):
if source == sink :
return path
for edge in self . get_edges ( source ):
residual = edge . capacity - self . flow [ edge ]
if residual > 0 and not ( edge , residual ) in path :
result = self . find_path ( edge . to , sink , path + [ ( edge , residual ) ])
if result != None :
return result
def max_flow ( self , source , sink ):
path = self . find_path ( source , sink , [])
while path != None :
flow = min ( res for edge , res in path )
7
Please see http://en.wikipedia.org/wiki/Ford-Fulkersonwalgorithm.
250 Chapter 9. Network flows
for edge , res in path :
self . flow [ edge ] += flow
self . flow [ edge . redge ] -= flow
path = self . find_path ( source , sink , [])
return sum ( self . flow [ edge ] for edge in self . get_edges ( source ))
9.4 Edmonds and Karp’s algorithm
The objective of this section is to prove Edmond and Karp’s algorithm for the maximum
flow-minimum cut problem with polynomial complexity.
9.5 Goldberg and Tarjan’s algorithm
The objective of this section is to prove Goldberg and Tarjan’s algorithm for finding
maximal flows with polynomial complexity.
Chapter 10
Random graphs
A random graph can be thought of as being a member from a collection of graphs having
some common properties. Recall that Algorithm 3.5 allows for generating a random
binary tree having at least one vertex. Fix a positive integer n and let T be a collection
of all binary trees on n vertices. It can be infeasible to generate all members of T , so for
most purposes we are only interested in randomly generating a member of T . A binary
tree of order n generated in this manner is said to be a random graph.
This chapter is a digression into the world of random graphs and various models
for generating different types of random graphs. Unlike other chapters in this book, our
approach is rather informal and not as rigorous as in other chapters. We will discuss some
common models of random graphs and a number of their properties without being bogged
down in details of proofs. Along the way, we will demonstrate that random graphs can
be used to model diverse real-world networks such as social, biological, technological,
and information networks. The edited volume [159] provides some historical context
for the “new” science of networks. Bollobás [31] and Kolchin [130] provide standard
references on the theory of random graphs with rigorous proofs. For comprehensive
surveys of random graphs and networks that do not go into too much technical details,
see [14, 71, 203, 204]. On the other hand, surveys that cover diverse applications of
random graphs and networks and are geared toward the technical aspects of the subject
include [5, 30, 50, 62, 63, 67, 163].
10.1 Network statistics
Numerous real-world networks are large, having from thousands up to millions of vertices
and edges. Network statistics provide a way to describe properties of networks without
concerning ourselves with individual vertices and edges. A network statistic should
describe essential properties of the network under consideration, provide a means to
differentiate between different classes of networks, and be useful in network algorithms
and applications [41]. In this section, we discuss various common network statistics that
can be used to describe graphs underlying large networks.
10.1.1 Degree distribution
The degree distribution of a graph G = (V, E) quantifies the fraction of vertices in G
having a specific degree k. If v is any vertex of G, we denote this fraction by
p = Pr[deg(v) = k] (10.1)
251
252 Chapter 10. Random graphs
As indicated by the notation, we can think of (10.1) as the probability that a vertex v ∈ V
chosen uniformly at random has degree k. The degree distribution of G is consequently a
histogram of the degrees of vertices in G. Figure 10.1 illustrates the degree distribution
of the Zachary [215] karate club network. The degree distributions of many real-world
networks have the same general curve as depicted in Figure 10.1(b), i.e. a peak at low
degrees followed by a tail at higher degrees. See for example the degree distribution of
the neural network in Figure 10.2, that of a power grid network in Figure 10.3, and the
degree distribution of a scientific co-authorship network in Figure 10.4.
0.3
0.25
0.2
0.15
0.1
5 · 10−2
2 4 6 8 10 12 14 16
(a) Zachary karate club network. (b) Linear scaling.
10−0.6
10−0.8
10−1
10−1.2
10−1.4
100 100.2 100.4 100.6 100.8 101 101.2
(c) Log-log scaling.
Figure 10.1: The friendship network within a 34-person karate club. This is more com-
monly known as the Zachary [215] karate club network. The network is an undirected,
connected, unweighted graph having 34 vertices and 78 edges. The horizontal axis repre-
sents degree; the vertical axis represents the probability that a vertex from the network
has the corresponding degree.
10.1.2 Distance statistics
In chapter 5 we discussed various distance metrics such as radius, diameter, and eccen-
tricity. To that distance statistics collection we add the average or characteristic distance
d, defined as the arithmetic mean of all distances in a graph. Let G = (V, E) be a simple
graph with n = |V | and m = |E|, where G can be either directed or undirected. Then
10.1. Network statistics 253
·10−2
6
10−1.5
4
10−2
2
20 40 60 80 100 120 100 101 102
(a) Linear scaling. (b) Log-log scaling.
Figure 10.2: Degree distribution of the neural network of the Caenorhabditis elegans.
The network is a directed, not strongly connected, weighted graph with 297 vertices
and 2,359 edges. The horizontal axis represents degree; the vertical axis represents the
probability that a vertex from the network has the corresponding degree. The degree
distribution is derived from dataset by Watts and Strogatz [205] and White et al. [206].
0.3
10−1
0.2
10−2
0.1
10−3
5 10 15 100 100.2 100.4 100.6 100.8 101 101.2
(a) Linear scaling. (b) Log-log scaling.
Figure 10.3: Degree distribution of the Western States Power Grid of the United States.
The network is an undirected, connected, unweighted graph with 4,941 vertices and 6,594
edges. The horizontal axis represents degree; the vertical axis represents the probability
that a vertex from the network has the corresponding degree. The degree distribution is
derived from dataset by Watts and Strogatz [205].
254 Chapter 10. Random graphs
10−1
0.12
0.1
10−2
8 · 10−2
6 · 10−2 10−3
4 · 10−2
2 · 10−2 10−4
0
0 50 100 150 200 250 100 101 102
(a) Linear scaling. (b) Log-log scaling.
Figure 10.4: Degree distribution of the network of co-authorships between scientists
posting preprints on the condensed matter eprint archive at http://arxiv.org/archive/
cond-mat. The network is a weighted, disconnected, undirected graph having 40,421
vertices and 175,693 edges. The horizontal axis represents degree; the vertical axis
represents the probability that a vertex from the co-authorship network has the corre-
sponding degree. The degree distribution is derived from the 2005 update of the dataset
by Newman [161].
G has size at most n(n − 1) because for any distinct vertex pair u, v ∈ V we count the
edge from u to v and the edge from v to u. The characteristic distance of G is defined
by
1 X
d(G) = d(u, v)
n(n − 1) u6=v∈V
where the distance function d is given by
∞, if there is no path from u to v,
d(u, v) = 0, if u = v,
k, where k is the length of a shortest u-v path.
If G is strongly connected (respectively, connected for the undirected case) then our
distance function is of the form d : V × V → Z+ ∪ {0}, where the codomain is the
set of nonnegative integers. The case where G is not strongly connected (respectively,
disconnected for the undirected version) requires special care. One way is to compute
the characteristic distance for each component and then find the average of all such
characteristic distances. Call the resulting characteristic distance dc , where c means
component. Another way is to assign a large number as the distance of non-existing
shortest paths. If there is no u-v path, we let d(u, v) = n because n = |V | is larger than
the length of any shortest path between connected vertices. The resulting characteristic
distance is denoted db , where b means big number. Furthermore denote by dκ the number
of pairs (u, v) such that v is not reachable from u. For example, the Zachary [215] karate
club network has d = 2.4082 and dκ = 0; the C. elegans neural network [205, 206]
has db = 71.544533, dc = 3.991884, and dκ = 20, 268; the Western States Power Grid
network [205] has d = 18.989185 and dκ = 0; and the condensed matter co-authorship
network [161] has db = 7541.74656, dc = 5.499329, and dκ = 152, 328, 281.
10.2. Binomial random graph model 255
We can also define the concept of distance distribution similar to how the degree
distribution was defined in section 10.1.1. If ` is a positive integer with u and v being
connected vertices in a graph G = (V, E), denote by
p = Pr[d(u, v) = `] (10.2)
the fraction of ordered pairs of connected vertices in V × V having distance ` between
them. As is evident from the above notation, we can think of (10.2) as the probability
that a uniformly chosen connected pair (u, v) of vertices in G has distance `. The distance
distribution of G is hence a histogram of the distances between pairs of vertices in G.
Figure 10.5 illustrates distance distributions of various real-world networks.
0.3
0.4
0.2
0.3
0.2
0.1
0.1
0
1 2 3 4 5 2 4 6 8 10 12 14
(a) Zachary karate club network [215]. (b) C. elegans neural network [205, 206].
·10−2
0.3
5
4
0.2
3
2
0.1
1
0 0
10 20 30 40 2 4 6 8 10 12 14 16 18
(c) Power grid network [205]. (d) Condensed matter co-authorship net-
work [161].
Figure 10.5: Distance distributions for various real-world networks. The horizontal axis
represents distance and the vertical axis represents the probability that a uniformly
chosen pair of distinct vertices from the network has the corresponding distance between
them.
10.2 Binomial random graph model
In 1959, Gilbert [89] introduced a random graph model that now bears the name bino-
mial (or Bernoulli ) random graph model. First, we fix a positive integer n, a probability
256 Chapter 10. Random graphs
Algorithm 10.1: Generate a random graph in G(n, p).
Input: Positive integer n and a probability 0 < p < 1.
Output: A random graph from G(n, p).
1 G ← Kn
2 V ← {0, 1, . . . , n − 1}
3 E ← {2-combinations of V }
4 for each e ∈ E do
5 r ← draw uniformly at random from interval (0, 1)
6 if r < p then
7 add edge e to G
8 return G
p, and a vertex set V = {0, 1, . . . , n − 1}. By G(n, p) we mean a probability space over
the set of undirected simple graphs on n vertices. If G is any element of the probability
space G(n, p) and ij is any edge for distinct i, j ∈ V , then ij occurs as an edge of G
independently with probability p. In symbols, for any distinct pair i, j ∈ V we have
Pr[ij ∈ E(G)] = p
where all such events are mutually independent. Any graph G drawn uniformly at
random from G(n, p) is a subgraph of the complete graph Kn and it follows from (1.6)
that G has at most n2 edges. Then the probability that G has m edges is given by
n
pm (1 − p)( 2 )−m . (10.3)
Notice the resemblance of (10.3) to the binomial distribution. By G ∈ G(n, p) we mean
that G is a random graph of the space G(n, p) and having size distributed as (10.3).
To generate a random graph in G(n, p), start with G being a graph on n vertices but
G is Kn , the complement of the complete graph on n vertices.
no edges. That is, initially
Consider each of the n2 possible edges in some order and add it independently to G
with probability p. See Algorithm 10.1 for pseudocode of the procedure. The runtime
of Algorithm 10.1 depends on an efficient algorithm for generating all 2-combinations of
a set of n objects. We could adapt Algorithm 4.22 to our needs or search for a more
efficient algorithm; see problem 10.3 for discussion of an algorithm to generate a graph
in G(n, p) in quadratic time. Figure 10.6 illustrates some random graphs from G(25, p)
with p = i/6 for i = 0, 1, . . . , 5. See Figure 10.7 for results for graphs in G(2 · 104 , p).
The expected number of edges of any G ∈ G(n, p) is
n pn(n − 1)
α = E[|E|] = p · =
2 2
and the expected total degree is
n
β = E[# deg] = 2p · = pn(n − 1).
2
Then the expected degree of each edge is p(n − 1). From problem 1.7 we know that the
number of undirected simple graphs on n vertices is given by
2n(n−1)/2
10.2. Binomial random graph model 257
where (10.3) is the probability of any of these graphs being the output of the above
procedure. Let κ(n, m) be the number of graphs from G(n, p) that are connected and
have size m, and by Pr[Gκ ] is meant the probability that G ∈ G(n, p) is connected. Apply
expression (10.3) to see that
(n2 )
X n
Pr[Gκ ] = κ(n, i) · pi (1 − p)( 2 )−i
i=n−1
where n − 1 is the least number of edges of any undirected connected graph on n vertices,
i.e. the size of any spanning tree of a connected graph in G(n, p). Similarly define
Pr[κij ] to be the probability that two distinct vertices i, j of G ∈ G(n, p) are connected.
Gilbert [89] showed that as n → ∞, then we have
Pr[Gκ ] ∼ 1 − n(1 − p)n−1
and
Pr[κij ] ∼ 1 − 2(1 − p)n−1 .
Algorithm 10.2: Random oriented graph via G(n, p).
Input: Positive integer n and probability 0 < p < 1.
Output: A random oriented graph on n vertices.
1 G ← random graph in G(n, p) as per Algorithm 10.3
2 E ← edge set of G
3 G ← directed version of G
4 cutoff ← draw uniformly at random from interval (0, 1)
5 for each edge uv ∈ E do
6 r ← draw uniformly at random from interval (0, 1)
7 if r < cutoff then
8 remove uv from G
9 else
10 remove vu from G
11 return G
Example 10.1. Consider a digraph D = (V, E) without self-loops or multiple edges.
Then D is said to be oriented if for any distinct pair u, v ∈ V at most one of uv, vu is
an edge of D. Provide specific examples of oriented graphs.
Solution. If u, v ∈ V is any pair of distinct vertices of an oriented graph D = (V, E), we
have various possibilities:
1. uv ∈
/ E and vu ∈
/ E.
2. uv ∈ E and vu ∈
/ E.
3. uv ∈
/ E and vu ∈ E.
258 Chapter 10. Random graphs
(a) p = 0; α = 0, |E| = 0; β = 0, # deg = 0 (b) p = 1/6; α = 50, |E| = 44; β = 100,
# deg = 88
(c) p = 1/3; α = 100, |E| = 108; β = 200, (d) p = 1/2; α = 150, |E| = 156; β = 300,
# deg = 212 # deg = 312
(e) p = 2/3; α = 200, |E| = 185; β = 400, (f) p = 5/6; α = 250, |E| = 255; β = 500,
# deg = 370 # deg = 510
Figure 10.6: Binomial random graphs G(25, p) for various values of p.
10.2. Binomial random graph model 259
·108
4
α
β
3 α̂
β̂
2
1
0
0 0.2 0.4 0.6 0.8 1
Figure 10.7: Comparison of expected and experimental values of the number of edges
and total degree of random simple undirected graphs in G(n, p). The horizontal axis
represents probability points; the vertical axis represents the size and total degree (ex-
pected or experimental). Fix n = 20, 000 and consider r = 50 probability points chosen
as follows. Let pmin = 0.000001, pmax = 0.999999, and F = (pmax /pmin )1/(r−1) . For
i = 1, 2, . . . , r = 50 the i-th probability point pi is defined by pi = pmin F i−1 . Each
experiment consists in generating M = 500 random graphs from G(n, pi ). For each
Gi ∈ G(n, pi ), where i = 1, 2, . . . , 500, compute its actual size αi and actual total degree
βi . Then take the mean α̂ of the αi and the mean β̂ of the βi .
Let n > 0 be the number of vertices in D and let 0 < p < 1. Generate a random
oriented graph as follows. First we generate a binomial random graph G ∈ G(n, p) where
G is simple and undirected. Then we consider the digraph version of G and proceed to
randomly prune either uv or vu from G, for each distinct pair of vertices u, v. Refer to
Algorithm 10.2 for pseudocode of our discussion. A Sage implementation follows:
sage : G = graphs . RandomGNP (20 , 0.1)
sage : E = G . edges ( labels = False )
sage : G = G . to_directed ()
sage : cutoff = 0.5
sage : for u , v in E :
... r = random ()
... if r < cutoff :
... G . delete_edge (u , v )
... else :
... G . delete_edge (v , u )
which produced the random oriented graph in Figure 10.8.
Efficient generation of sparse G ∈ G(n, p)
The techniques discussed so far (Algorithms 10.1 and 10.9) for generating a random
graph from G(n, p) can be unsuitable when the number of vertices n is in the hundreds
of thousands or millions. In many applications of G(n, p) we are only interested in sparse
random graphs. A linear time algorithm to generate a random sparse graph from G(n, p)
is presented by Batagelj and Brandes [19].
The Batagelj-Brandes algorithm for generating a random sparse graph G ∈ G(n, p)
uses what is known as a geometric method to skip over certain edges. Fix a probability
0 < p < 1 that an edge will be in the resulting random sparse graph G. If e is an edge
260 Chapter 10. Random graphs
Figure 10.8: A random oriented graph generated using a graph in G(20, 0.1) and cutoff
probability 0.5.
of G, we can consider the events leading up to the choice of e as
e1 , e2 , . . . , ek
where in the i-th trial the event ei is a failure, for 1 ≤ i < k, but the event ek is the
first success after k − 1 successive failures. In probabilistic terms, we perform a series
of independent trials each having success probability p and stop when the first success
occurs. Letting X be the number of trials required until the first success occurs, then X
is a geometric random variable with parameter p and probability mass function
Pr[X = k] = p(1 − p)k−1 (10.4)
for integers k ≥ 1, where
∞
X
p(1 − p)k−1 = 1.
k=1
In other words, waiting times are geometrically distributed.
Suppose we want to generate a random number from a geometric distribution, i.e. we
want to simulate X such that
Pr[X = k] = p(1 − p)k−1 , k = 1, 2, 3, . . .
Note that
X̀
Pr[X = k] = 1 − Pr[X > ` − 1] = 1 − (1 − p)`−1 .
k=1
In other words, we can simulate a geometric random variable by generating r uniformly
at random from the interval (0, 1) and set X to that value of k for which
1 − (1 − p)k−1 < r < 1 − (1 − p)k
10.2. Binomial random graph model 261
or equivalently for which
(1 − p)k < 1 − r < (1 − p)k−1
where 1 − r and r are both uniformly distributed. Thus we can define X by
X = min{k | (1 − p)k < 1 − r}
ln(1 − r)
= min k k >
ln(1 − p)
ln(1 − r)
=1+ .
ln(1 − p)
That is, we can choose k to be
ln(1 − r)
k =1+
ln(1 − p)
which is used as a basis of Algorithm 10.3. In the latter algorithm, note that the vertex
set is V = {0, 1, . . . , n − 1} and candidate edges are generated in lexicographic order.
The Batagelj-Brandes Algorithm 10.3 has worst-case runtime O(n + m), where n and m
are the order and size, respectively, of the resulting graph.
Algorithm 10.3: Linear generation of a random sparse graph in G(n, p).
Input: Positive integer n and a probability 0 < p < 1.
Output: A random sparse graph from G(n, p).
1 G ← Kn
2 u←1
3 v ← −1
4 while u < n do
5 r ← draw uniformly at random from interval (0, 1)
6 v ← v + 1 + bln(1 − r)/ ln(1 − p)c
7 while v ≥ u and u < n do
8 v ←v−u
9 u←u+1
10 if u < n then
11 add edge uv to G
12 return G
Degree distribution
Consider a random graph G ∈ G(n, p) and let v be a vertex of G. With probability p, the
vertex v is incident with each of the remaining n − 1 vertices in G. Then the probability
that v has degree k is given by the binomial distribution
n−1 k
Pr[deg(v) = k] = p (1 − p)n−1−k (10.5)
k
262 Chapter 10. Random graphs
and the expected degree of v is E[deg(v)] = p(n − 1). Setting z = p(n − 1), we can
express (10.5) as
k n−1
n−1 z z
Pr[deg(v) = k] = 1−
k n−1−z n−1
and thus
zk
Pr[deg(v) = k] → exp(−z)
k!
as n → ∞. In the limit of large n, the probability that vertex v has degree k approaches
the Poisson distribution. That is, as n gets larger and larger any random graph in G(n, p)
has a Poisson degree distribution.
10.3 Erdős-Rényi model
Let N be a fixed nonnegative integer. The Erdős-Rényi [76, 77] (or uniform) random
graph model, denoted G(n, N ), is a probability space over the set of undirected simple
graphs on n vertices and exactly N edges. Hence G(n, N ) can be considered as a collection
n
of ( 2 ) undirected simple graphs on exactly N edges, each such graph being selected with
N
equal probability. A note of caution is in order here. Numerous papers on random graphs
refer to G(n, p) as the Erdős-Rényi random graph model, where in fact this binomial
random graph model should be called the Gilbert model in honor of E. N. Gilbert who
introduced [89] it in 1959. Whenever a paper makes a reference to the Erdős-Rényi
model, one should question whether the paper is referring to G(n, p) or G(n, N ).
To generate a graph in G(n, N ), start with G being a graph on n vertices but no
n
edges. Then choose N of the possible 2 edges independently and uniformly at random
and let the chosen edges be the edge set of G. Each graph G ∈ G(n, N ) is associated
with a probability
n
1 2
N
of being the graph resulting from the above procedure. Furthermore each of the n2
edges has a probability
n
1
2
of being chosen. Algorithm 10.4 presents a straightforward translation of the above
procedure into pseudocode.
The runtime of Algorithm 10.4 is probabilistic and can be analyzed via the geometric
distribution. If i is the number of edges chosen so far, then the probability of choosing
a new edge in the next step is
n
2
−i
n .
2
We repeatedly choose an edge uniformly at random from the collection of all possible
edges, until we come across the first edge that is not already in the graph. The number
of trials required until the first new edge is chosen can be modeled using the geometric
distribution with probability mass function (10.4). Given a geometric random variable
10.3. Erdős-Rényi model 263
Algorithm 10.4: Generation of random graph in G(n, N ).
n
Input: Positive integer n and integer N with 0 ≤ N ≤ 2
.
Output: A random graph from G(n, N ).
1 G←K
nn o
2 E ← e0 , e1 , . . . , e(n)−1
2
3 for i ← 0, 1, . . . , N − 1 do
4 r ← draw uniformly at random from 0, 1, . . . , n2 − 1
5 while er is an edge of G do
6 r ← draw uniformly at random from 0, 1, . . . , n2 − 1
7 add edge er to G
8 return G
X, we have the expectation
∞
X 1
E[X] = n · p(1 − p)n−1 = .
n=1
p
Therefore the expected number of trials until a new edge be chosen is
n
2
n
2
−i
from which the expected total runtime is
N Z
X n N n
2 2
n ≈ n dx
i=1 2
− i 0 2
− x
n
n 2
= · ln n
+C
2 2
− N
for some constant C. The denominator in the latter fraction becomes zero when n2 = N ,
which can be prevented by adding one to the denominator. Then we have the expected
total runtime !
XN n n
n
n
2 ∈Θ · ln n 2
i=1 2
−i 2 2
−N +1
which is O(N ) when N ≤ n2 /2, and O(N ln N ) when N = n2 . In other words,
Algorithm 10.4 has expected nlinear
runtime when the number N of required edges satisfies
n
N ≤ 2 /2. But for N > 2 /2, we obtain expected linear runtime by generating the
n
complete graph Kn and randomly delete 2 − N edges from the latter graph. Our
discussion is summarized in Algorithm 10.5.
264 Chapter 10. Random graphs
Algorithm 10.5: Generation of random graph in G(n, N ) in expected linear time.
Input: Positive integer n and integer N with 0 ≤ N ≤ n2 .
Output: A random graph from G(n, N ).
n
1 if N ≤ /2 then
2
2 return result of Algorithm 10.4
3 G ← Kn
n
4 for i ← 1, 2, . . . , − N do
2
5 e ← draw uniformly at random from E(G)
6 remove edge e from G
7 return G
10.4 Small-world networks
Vicky: Hi, Janice.
Janice: Hi, Vicky.
Vicky: How are you?
Janice: Good.
Harry: You two know each other?
Janice: Yeah, I met Vicky at the mall today.
Harry: Well, what a small world! You know, I wonder who else I know knows someone I
know that I don’t know knows that person I know.
— from the TV series Third Rock from the Sun, season 5, episode 22, 2000.
Many real-world networks exhibit the small-world effect: that most pairs of distinct
vertices in the network are connected by relatively short path lengths. The small-world
effect was empirically demonstrated [155] in a famous 1960s experiment by Stanley Mil-
gram, who distributed a number of letters to a random selection of people. Recipients
were instructed to deliver the letters to the addressees on condition that letters must be
passed to people whom the recipients knew on a first-name basis. Milgram found that
on average six steps were required for a letter to reach its target recipient, a number now
immortalized in the phrase “six degrees of separation” [97]. Figure 10.9 plots results of
an experimental study of the small-world problem as reported in [193]. The small-world
effect has been studied and verified for many real-world networks including
social: collaboration network of actors in feature films [7,205], scientific publication
authorship [51, 96, 160, 161];
information: citation network [173], Roget’s Thesaurus [127], word co-occurrence [66,
80];
technological: internet [55, 79], power grid [205], train routes [179], software [162,
196];
biological: metabolic network [116], protein interactions [115], food web [112, 148],
neural network [205, 206].
10.4. Small-world networks 265
15
10
frequency 5
0
0 2 4 6 8 10
number of intermediaries
Figure 10.9: Frequency distribution of the number of intermediaries required for letters
to reach their intended addressees. The distribution has a mean of 5.3, interpreted as the
average number of intermediaries required for a letter to reach its intended destination.
The plot is derived from data reported in [193].
Watts and Strogatz [202, 203, 205] proposed a network model that produces graphs
exhibiting the small-world effect. We will use the notation “ ” to mean “much greater
than”. Let n and k be positive integers such that n k ln n 1 (in particular,
0 < k < n/2) with k being even. Consider a probability 0 < p < 1. Starting from
an undirected k-circulant graph G = (V, E) on n vertices, the Watts-Strogatz model
proceeds to rewire each edge with probability p. The rewiring procedure, also called
edge swapping, works as follows. Let V be uniformly distributed. For each v ∈ V , let
e ∈ E be an edge having v as an endpoint. Choose another u ∈ V different from v. With
probability p, delete the edge e and add the edge vu. The rewiring must produce a simple
graph with the same order and size as G. As p → 1, the graph G goes from k-circulant
to exhibiting properties of graphs drawn uniformly from G(n, p). Small-world networks
are intermediate between k-circulant and binomial random graphs (see Figure 10.10).
The Watts-Strogatz model is said to provide a procedure for interpolating between the
latter two types of graphs.
(a) p = 0, k-circulant (b) p = 0.3, small-world (c) p = 1, random
Figure 10.10: With increasing randomness, k-circulant graphs evolve to exhibit prop-
erties of random graphs in G(n, p). Small-world networks are intermediate between
k-circulant graphs and random graphs in G(n, p).
266 Chapter 10. Random graphs
The last paragraph contains an algorithm for rewiring edges of a graph. While the
algorithm is simple, in practice it potentially skips over a number of vertices to be
considered for rewiring. If G = (V, E) is a k-circulant graph on n vertices and p is the
rewiring probability, the candidate vertices to be rewired follow a geometric distribution
with parameter p. This geometric trick, essentially the same speed-up technique used by
the Batagelj-Brandes Algorithm 10.3, can be used to speed up the rewiring algorithm.
To elaborate, suppose G has vertex set V = {0, 1, . . . , n − 1}. If r is chosen uniformly at
random from the interval (0, 1), the index of the vertex to be rewired can be obtained
from
ln(1 − r)
1+ .
ln(1 − p)
The above geometric method is incorporated into Algorithm 10.6 to generate a Watts-
Strogatz network in worst-case runtime O(nk +m), where n and k are as per the input of
the algorithm and m is the size of the k-circulant graph on n vertices. Note that lines 7
to 12 are where we avoid self-loops and multiple edges.
Algorithm 10.6: Watts-Strogatz network model.
Input: Positive integer n denoting the number of vertices. Positive even integer k
for the degree of each vertex, where n k ln n 1. In particular, k
should satisfy 0 < k < n/2. Rewiring probability 0 < p ≤ 1.
Output: A Watts-Strogatz network on n vertices.
1 M ← nk /* sum of all vertex degrees = twice number of edges */
2 r ← draw uniformly at random from interval (0, 1)
3 v ← 1 + bln(1 − r)/ ln(1 − p)c
4 E ← contiguous edge list of k-circulant graph on n vertices
5 while v ≤ M do
6 u ← draw uniformly at random from [0, 1, . . . , n − 1]
7 if v − 1 is even then
8 while E[v] = u or (u, E[v]) ∈ E do
9 u ← draw uniformly at random from [0, 1, . . . , n − 1]
10 else
11 while E[v − 2] = u or (E[v − 2], u) ∈ E do
12 u ← draw uniformly at random from [0, 1, . . . , n − 1]
13 E[v − 1] ← u
14 r ← draw uniformly at random from interval (0, 1)
15 v ← v + 1 + bln(1 − r)/ ln(1 − p)c
16 G ← Kn
17 add edges in E to G
18 return G
Characteristic path length
Watts and Strogatz [205] analyzed the structure of networks generated by Algorithm 10.6
via two quantities: the characteristic path length ` and the clustering coefficient C. The
characteristic path length quantifies the average distance between any distinct pair of
vertices in a Watts-Strogatz network. The quantity `(G) is thus said to be a global
10.4. Small-world networks 267
property of G. Watts and Strogatz characterized as small-world those networks that
exhibit high clustering coefficients and low characteristic path lengths.
Let G = (V, E) be a Watts-Strogatz network as generated by Algorithm 10.6, where
the vertex set is V = {0, 1, . . . , n − 1}. For each pair of vertices i, j ∈ V , let dij be the
distance from i to j. If there is no path from i to j or i = j, set dij = 0. Thus
0, if there is no path from i to j,
dij = 0, if i = j,
k, where k is the length of a shortest path from i to j.
Since G is undirected, we have dij = dji . Consequently when computing the distance
between each distinct pair of vertices, we should avoid double counting by computing dij
for i < j. Then the characteristic path length of G is defined by
1 1X
`(G) = · dij
n(n − 1)/2 2 i6=j
X (10.6)
1
= dij
n(n − 1) i6=j
which is averaged over all possible pairs of distinct vertices, i.e. the number of edges in
the complete graph Kn .
It is inefficient to compute the characteristic path length via equation (10.6) because
we would effectively sum n(n − 1) distance values. As G is undirected, note that
1X X X
dij = dij = dij .
2 i6=j i<j i>j
The latter equation holds for the following reason. Let D = [dij ] be a matrix of distances
for G, where i is the row index, j is the column index, and dij is the distance from i to j.
The required sum of distances can be obtained by summing all entries above (or below)
the main diagonal of D. Therefore the characteristic path length can be expressed as
2 X
`(G) = dij
n(n − 1) i<j
2 X
= dij
n(n − 1) i>j
which requires summing n(n−1)
2
distance values.
Let G = (V, E) be a Watts-Strogatz network with n = |V |. Set k 0 = k/2, where k is
as per Algorithm 10.6. As the rewiring probability p → 0, the average path length tends
to
n n
`→ 0 = .
4k 2k
In the special case p = 0, we have
n(n + k − 2)
`= .
2k(n − 1)
ln n
However as p → 1, we have ` → ln k
.
268 Chapter 10. Random graphs
Clustering coefficient
The clustering coefficient of a simple graph G quantifies the “cliquishness” of vertices
in G = (V, E). This quantity is thus said to be a local property of G. Watts and
Strogatz [205] defined the clustering coefficient as follows. Suppose n = |V | > 0 and let ni
count the number of neighbors of vertex i ∈ V , a quantity that is equivalent to the degree
of i, i.e. deg(i) = ni . The complete graph Kni on the ni neighbors of i has ni (ni − 1)/2
edges. The neighbor graph Ni of i is a subgraph of G, consisting of all vertices (6= i) that
are adjacent to i and preserving the adjacency relation among those vertices as found in
the supergraph G. For example, given the graph in Figure 10.11(a) the neighbor graph
of vertex 10 is shown in Figure 10.11(b). The local clustering coefficient Ci of i is the
ratio
Ni
Ci =
ni (ni − 1)/2
where Ni counts the number of edges in Ni . In case i has degree deg(i) < 2, we set the
local clustering coefficient of i to be zero. Then the clustering coefficient of G is defined
by
1X 1X Ni
C(G) = Ci = .
n i∈V n i∈V ni (ni − 1)/2
3 2 3
4 1 1
5 10 0
6 9 6
7 8 7 8
(a) Graph on 11 vertices. (b) N10
Figure 10.11: The neighbor graph of a vertex.
Consider the case where we have a k-circulant graph G = (V, E) on n vertices and a
rewiring probability p = 0. That is, we do not rewire any edge of G. Each vertex of G has
degree k. Let k 0 = k/2. Then the k neighbors of each vertex in G has 3k 0 (k 0 − 1)/2 edges
between them, i.e. each neighbor graph Ni has size 3k 0 (k 0 − 1)/2. Then the clustering
coefficient of G is
3(k 0 − 1)
.
2(2k 0 − 1)
When the rewiring probability is p > 0, Barrat and Weigt [18] showed that the clustering
coefficient of any graph G0 in the Watts-Strogatz network model (see Algorithm 10.6)
can be approximated by
0 3(k 0 − 1)
C(G ) ≈ 0
(1 − p)3 .
2(2k − 1)
10.5. Scale-free networks 269
Degree distribution
For a Watts-Strogatz network without rewiring, each vertex has the same degree k. It
easily follows that for each vertex v, we have the degree distribution
(
1, if i = k,
Pr[deg(v) = i] =
0, otherwise.
A rewiring probability p > 0 introduces disorder in the network and broadens the
degree distribution, while the expected degree is k. A k-circulant graph on n vertices
has nk/2 edges. With the rewiring probability p > 0, a total of pnk/2 edges would
be rewired. However note that only one endpoint of an edge is rewired, thus after the
rewiring process the degree of any vertex v is deg(v) ≥ k/2. Therefore with k > 2, a
Watts-Strogatz network has no isolated vertices.
For p > 0, Barrat and Weigt [18] showed that the degree of a vertex v can be written
as deg(v) = k/2 + ni with ni ≥ 0, where ni can be divided into two parts α and β as
follows. First α ≤ k/2 edges are left intact after the rewiring process, the probability of
this occurring is 1 − p for each edge. Second β = ni − α edges have been rewired towards
i, each with probability 1/n. The probability distribution of α is
k/2
P1 (α) = (1 − p)α pk/2−α
α
and the probability distribution of β is
β pnk/2−β
pnk/2 1 1
P2 (β) = 1−
β n n
where
(pk/2)β
P2 (β) → exp(−pk/2)
β!
for large n. Combine the above two factors to obtain the degree distribution
min{κ−k/2, k/2}
X k/2 (pk/2)κ−k/2−i
Pr[deg(v) = κ] = (1 − p)i pk/2−i exp(−pk/2)
i=0
i (κ − k/2 − i)!
for κ ≥ k/2.
10.5 Scale-free networks
The networks covered so far—Gilbert G(n, p) model, Erdős-Rényi G(n, N ) model, Watts-
Strogatz small-world model—are static. Once a network is generated from any of these
models, the corresponding model does not specify any means for the network to evolve
over time. Barabási and Albert [15] proposed a network model based on two ingredients:
1. Growth: at each time step, a new vertex is added to the network and connected
to a pre-determined number of existing vertices.
2. Preferential attachment: the newly added vertex is connected to an existing vertex
in proportion to the latter’s existing degree.
270 Chapter 10. Random graphs
Preferential attachment also goes by the colloquial name of the “rich-get-richer” effect
due to the work of Herbert Simon [184]. In sociology, preferential attachment is known
as the Matthew effect due to the following verse from the Book of Matthew, chapter 25
verse 29, in the Bible: “For to every one that hath shall be given but from him that
hath not, that also which he seemeth to have shall be taken away.” Barabási and Albert
observed that many real-world networks exhibit statistical properties of their proposed
model. One particularly significant property is that of power-law scaling, hence the
Barabási-Albert model is also called a model of scale-free networks. Note that it is only
the degree distributions of scale-free networks that are scale-free. In their empirical study
of the World Wide Web (WWW) and other real-world networks, Barabási and Albert
noted that the probability that a web page increases in popularity is directly proportional
to the page’s current popularity. Thinking of a web page as a vertex and the degree of a
page as the number of other pages that the current page links to, the degree distribution
of the WWW follows a power law function. Power-law scaling has been confirmed for
many real-world networks:
actor collaboration network [15]
citation [64, 173, 178] and co-authorship networks [161]
human sexual contacts network [118, 143]
the Internet [55, 79, 197] and the WWW [6, 17, 42]
metabolic networks [115, 116]
telephone call graphs [3, 4]
Figure 10.12 illustrates the degree distributions of various real-world networks, plotted
on log-log scales. Corresponding distributions for various simulated Barabási-Albert
networks are illustrated in Figure 10.13.
But how do we generate a scale-free graph as per the description in Barabási and
Albert [15]? The original description of the Barabási-Albert model as contained in [15]
is rather ambiguous with respect to certain details. First, the whole process is supposed
to begin with a small number of vertices. But as the degree of each of these vertices
is zero, it is unclear how the network is to grow via preferential attachment from the
initial pool of vertices. Second, Barabási and Albert neglected to clearly specify how to
select the neighbors for the newly added vertex. The above ambiguities are resolved by
Bollobás et al. [33], who gave a precise statement of a random graph process that realizes
the Barabási-Albert model. Fix a sequence of vertices v1 , v2 , . . . and consider the case
where each newly added vertex is to be connected to m = 1 vertex already in a graph.
Inductively define a random graph process (Gt1 )t≥0 as follows, where Gt1 is a digraph on
{vi | 1 ≤ i ≤ t}. Start with the null graph G01 or the graph G11 with one vertex and one
self-loop. Denote by degG (v) the total (in and out) degree of vertex v in the graph G.
For t > 1 construct Gt1 from Gt−11 by adding the vertex vt and a directed edge from vt to
vi , where i is randomly chosen with probability
(
degGt−1
1
(vs )/(2t − 1), if 1 ≤ s ≤ t − 1,
Pr[i = s] =
1/(2t − 1), if s = t.
10.5. Scale-free networks 271
10−1
10−1
10−2
10−2
−3
10
10−3
10−4
10−4
10−5
10−5
10−6
100 101 102 100 101 102 103
(a) US patent citation network. (b) Google web graph.
10−1 10−2
10−2
10−3
10−3
10−4
10−4
10−5
10−5
10−6
100 101 102 103 104 100 101 102 103
(c) LiveJournal friendship network. (d) Actor collaboration network.
Figure 10.12: Degree distributions of various real-world networks on log-log scales. The
horizontal axis represents degree and the vertical axis is the corresponding probability of
a vertex having that degree. The US patent citation network [141] is a directed graph on
3, 774, 768 vertices and 16, 518, 948 edges. It covers all citations made by patents granted
between 1975 and 1999. The Google web graph [142] is a digraph having 875, 713 vertices
and 5, 105, 039 edges. This dataset was released in 2002 by Google as part of the Google
Programming Contest. The LiveJournal friendship network [11, 142] is a directed graph
on 4, 847, 571 vertices and 68, 993, 773 edges. The actor collaboration network [15], based
on the Internet Movie Database (IMDb) at http://www.imdb.com, is an undirected graph
on 383, 640 vertices and 16, 557, 920 edges. Two actors are connected to each other if
they have starred in the same movie. In all of the above degree distributions, self-loops
are not taken into account and, where a graph is directed, we only consider the in-degree
distribution.
272 Chapter 10. Random graphs
10−1 10−1
10−2
10−2
10−3
10−3
10−4
10−4
10−5
10−5 0 10−6 0
10 101 102 103 104 10 101 102 103 104 105
(a) n = 105 vertices (b) n = 106 vertices
10−1 10−1
10−2 10−2
10−3 10−3
10−4 10−4
10−5
10−5
10−6
10−6
10−7
10−7 0 1 2 3 4 5 6
10 10 10 10 10 10 10 100 101 102 103 104 105 106
(c) n = 107 vertices (d) n = 2 · 107 vertices
Figure 10.13: Degree distributions of simulated graphs in the classic Barabási-Albert
model. The horizontal axis represents degree; the vertical axis is the corresponding
probability of a vertex having a particular degree. Each generated graph is directed and
has minimum out-degree m = 5. The above degree distributions are only for in-degrees
and do not take into account self-loops.
10.5. Scale-free networks 273
The latter process generates a forest. For m > 1 the graph evolves as per the case
m = 1; i.e. we add m edges from vt one at a time. This process can result in self-
n
loops and multiple edges. We write Gm for the collection of all graphs on n vertices
n
and minimal degree m in the Barabási-Albert model, where a random graph from Gm is
n n
denoted Gm ∈ Gm .
Now consider the problem of translating the above procedure into pseudocode. Fix a
positive integer n > 1 for the number of vertices in the scale-free graph to be generated
via preferential attachment. Let m ≥ 1 be the number of vertices that each newly added
vertex is to be connected to; this is equivalent to the minimum degree that any new vertex
will end up possessing. At any time step, let M be the contiguous edge list of all edges
created thus far in the above random graph process. It is clear that the frequency (or
number of occurrences) of a vertex is equivalent to the vertex’s degree. We can thus use
M as a pool to sample in constant time from the degree-skewed distribution. Batagelj and
Brandes [19] used the latter observation to construct an algorithm for generating scale-
free networks via preferential attachment; pseudocode is presented in Algorithm 10.7.
Note that the algorithm has linear runtime O(n + m), where n is the order and m the
size of the graph generated by the algorithm.
Algorithm 10.7: Scale-free network via preferential attachment.
Input: Positive integer n > 1 and minimum degree d ≥ 1.
Output: Scale-free network on n vertices.
1 G ← Kn /* vertex set is V = {0, 1, . . . , n − 1} */
2 M ← list of length 2nd
3 for v ← 0, 1, . . . , n − 1 do
4 for i ← 0, 1, . . . , d − 1 do
5 M [2(vd + i)] ← v
6 r ← draw uniformly at random from {0, 1, . . . , 2(vd + i)}
7 M [2(vd + i) + 1] ← M [r]
8 add edge (M [2i], M [2i + 1]) to G for i ← 0, 1, . . . , nd − 1
9 return G
On the evidence of computer simulation and various real-world networks, it was
suggested [15,16] that Pr[deg(v) = k] ∼ k −γ with γ = 2.9 ± 0.1. Letting n be the number
of vertices, Bollobás et al. [33] obtained Pr[deg(v) = k] asymptotically for all k ≤ n1/15
and showed as a consequence that γ = 3. In the process of doing so, Bollobás et al.
proved various results concerning the expected degree. Denote by #nm (k) the number of
vertices of Gnm with in-degree k (and consequently with total degree m + k). For the case
m = 1, we have the expectation
1
E[degGt1 (vt )] = 1 +
2t − 1
and for s < t we have
2t
E[degGt1 (vs )] = E[degGt−1 (vs )].
2t − 1 1
Taking the above two equations together, for 1 ≤ s ≤ n we have
n
Y 2i 4n−s+1 n!2 (2s − 2)!
E[degGn1 (vs )] = = .
i=s
2i − 1 (2n)!(s − 1)!2
274 Chapter 10. Random graphs
Furthermore for 0 ≤ k ≤ n1/15 we have
2m(m + 1)n
E[#nm (k)] ∼
(k + m)(k + m + 1)(k + m + 2)
uniformly in k.
As regards the diameter, with n as per Algorithm 10.7, computer simulation by
Barabási, Albert, and Jeong [6, 17] and heuristic arguments by Newman et al. [164]
suggest that a graph generated by the Barabási-Albert model has diameter approximately
ln n. As noted by Bollobás and Riordan [32], the approximation diam(Gnm ) ≈ ln n holds
for the case m = 1, but for m ≥ 2 they showed that as n → ∞ then diam(Gnm ) →
ln / ln ln n.
10.6 Problems
Where should I start? Start from the statement of the problem. What can I do? Visualize
the problem as a whole as clearly and as vividly as you can.
— G. Polya, from page 33 of [170]
10.1. Algorithm 10.8 presents a procedure to construct a random graph that is simple
and undirected; the procedure is adapted from pages 4–7 of Lau [138]. Analyze the
time complexity of Algorithm 10.8. Compare and contrast your results with that
for Algorithm 10.5.
10.2. Modify Algorithm 10.8 to generate the following random graphs.
(a) Simple weighted, undirected graph.
(b) Simple digraph.
(c) Simple weighted digraph.
10.3. Algorithm 10.1 can be considered as a template for generating random graphs in
G(n, p). The procedure does not specify how to generate all the 2-combinations of
a set of n > 1 objects. Here we discuss how to construct all such 2-combinations
and derive a quadratic time algorithm for generating random graphs in G(n, p).
(a) Consider a vertex set V = {0, 1, . . . , n − 1} with at least two elements and let
E be the set of all 2-combinations of V , where each 2-combination is written
ij. Show that ij ∈ E if and only if i < j.
(b) From the previous exercise, we know that if 0 ≤ i < n − 1 then there are
n − (i + 1) pairs jk where either i = j or i = k. Show that
n−2
X n2 − n
(n − i − 1) =
i=0
2
and conclude that Algorithm 10.9 has worst-case runtime O((n2 − n)/2).
10.4. Modify the Batagelj-Brandes Algorithm 10.3 to generate the following types of
graphs.
(a) Directed simple graphs.
10.6. Problems 275
Algorithm 10.8: Random simple undirected graph.
Input: Positive integers n and m specifying the order and size, respectively, of
the output graph.
Output: A random simple undirected graph with n vertices and m edges. If m
exceeds the size of Kn , then Kn is returned.
1 if n = 1 then
2 return K1
3 max ← n(n − 1)/2
4 if m > max then
5 return Kn
6 G ← null graph
7 A ← n × n adjacency matrix with entries aij
8 aij ← False for 0 ≤ i, j < n
9 i←0
10 while i < m do
11 u ← draw uniformly at random from {0, 1, . . . , n − 1}
12 v ← draw uniformly at random from {0, 1, . . . , n − 1}
13 if u = v then
14 continue with next iteration of loop
15 if u > v then
16 swap values of u and v
17 if auv = False then
18 add edge uv to G
19 auv ← True
20 i←i+1
21 return G
Algorithm 10.9: Quadratic generation of a random graph in G(n, p).
Input: Positive integer n and a probability 0 < p < 1.
Output: A random graph from G(n, p).
1 G ← Kn
2 V ← {0, 1, . . . , n − 1}
3 for i ← 0, 1, . . . , n − 2 do
4 for j ← i + 1, i + 2, . . . , n − 1 do
5 r ← draw uniformly at random from interval (0, 1)
6 if r < p then
7 add edge ij to G
8 return G
276 Chapter 10. Random graphs
Algorithm 10.10: Briggs’ algorithm for random graph in G(n, N ).
Input: Positive integers n and N such that 1 ≤ N ≤ n2 .
Output: A random graph from G(n, N ).
n
1 max ←
2
2 if n = 1 or N = max then
3 return Kn
4 G ← Kn
5 u ← 0
6 v ← 1
7 t ← 0 /* number of candidates processed so far */
8 k ← 0 /* number of edges selected so far */
9 while True do
10 r ← draw uniformly at random from {0, 1, . . . , max − t}
11 if r < N − k then
12 add edge uv to G
13 k ←k+1
14 if k = N then
15 return G
16 t←t+1
17 v ←v+1
18 if v = n then
19 u←u+1
20 v ←u+1
10.6. Problems 277
(b) Directed acyclic graphs.
(c) Bipartite graphs.
10.5. Repeat the previous problem for Algorithm 10.5.
10.6. In 2006, Keith M. Briggs provided [40] an algorithm that generates a random
graph in G(n, N ), inspired by Knuth’s Algorithm S (Selection sampling technique)
as found on page 142 of Knuth [128]. Pseudocode of Briggs’ procedure is presented
in Algorithm 10.10. Provide runtime analysis of Algorithm 10.10 and compare your
results with those presented in section 10.3. Under which conditions would Briggs’
algorithm be more efficient than Algorithm 10.5?
10.7. Briggs’ Algorithm 10.10 follows the general template of an algorithm that samples
without replacement n items from a pool of N candidates. Here 0 < n ≤ N and
the size N of the candidate pool is known in advance. However there are situations
where the value of N is not known beforehand, and we wish to sample without
replacement n items from the candidate pool. What we know is that the candidate
pool has enough members to allow us to select n items. Vitter’s algorithm R [198],
called reservoir sampling, is suitable for the situation and runs in O(n(1+ln(N/n)))
expected time. Describe and provide pseudocode of Vitter’s algorithm, prove its
correctness, and provide runtime analysis.
10.8. Repeat Example 10.1 but using each of Algorithms 10.1 and 10.5.
10.9. Diego Garlaschelli introduced [88] in 2009 a weighted version of the G(n, p) model,
called the weighted random graph model. Denote by GW (n, p) the weighted random
graph model. Provide a description and pseudocode of a procedure to generate a
graph in GW (n, p) and analyze the runtime complexity of the algorithm. Describe
various statistical physics properties of GW (n, p).
10.10. Latora and Marchiori [137] extended the Watts-Strogatz model to take into ac-
count weighted edges. A crucial idea in the Latora-Marchiori model is the concept
of network efficiency. Describe the Latora-Marchiori model and provide pseudocode
of an algorithm to construct Latora-Marchiori networks. Explain the concepts
of local and global efficiencies and how these relate to clustering coefficient and
characteristic path length. Compare and contrast the Watts-Strogatz and Latora-
Marchiori models.
10.11. The following model for “growing” graphs is known as the CHKNS model [49],1
named for its original proponents. Start with the trivial graph G at time step
t = 1. For each subsequent time step t > 1, add a new vertex to G. Furthermore
choose two vertices uniformly at random and with probability δ join them by an
undirected edge. The newly added edge does not necessarily have the newly added
vertex as an endpoint. Denote by dk (t) the expected number of vertices with degree
k at time t. Assuming that no self-loops are allowed, show that
d0 (t)
d0 (t + 1) = d0 (t) + 1 − 2δ
t
1
Or the “chickens” model, depending on how you pronounce “CHKNS”.
278 Chapter 10. Random graphs
and
dk−1 (t) dk (t)
dk (t + 1) = dk (t) + 2δ − 2δ .
t t
As t → ∞, show that the probability that a vertex be chosen twice decreases as
t−2 . If v is a vertex chosen uniformly at random, show that
(2δ)k
Pr[deg(v) = k] =
(1 + 2δ)k+1
and conclude that the CHKNS model has an exponential degree distribution. The
size of a component counts the number of vertices in the component itself. Let
Nk (t) be the expected number of components of size k at time t. Show that
N1 (t)
N1 (t + 1) = N1 (t) + 1 − 2δ
t
and for k > 1 show that
k−1
!
X iNi (t) (k − i)Nk−i (t) kNk (t)
Nk (t + 1) = Nk (t) + δ · − 2δ .
i=1
t t t
10.12. Algorithm 10.7 can easily be modified to generate other types of scale-free net-
works. Based upon the latter algorithm, Batagelj and Brandes [19] presented
a procedure for generating bipartite scale-free networks; see Algorithm 10.11 for
pseudocode. Analyze the runtime efficiency of Algorithm 10.11. Fix positive inte-
ger values for n and d, say n = 10, 000 and d = 4. Use Algorithm 10.11 to generate
a bipartite graph with your chosen values for n and d. Plot the degree distribution
of the resulting graph using a log-log scale and confirm that the generated graph
is scale-free.
10.13. Find the degree and distance distributions, average path lengths, and clustering
coefficients of the following network datasets:
(a) actor collaboration [15]
(b) co-authorship of condensed matter preprints [161]
(c) Google web graph [142]
(d) LiveJournal friendship [11, 142]
(e) neural network of the C. elegans [205, 206]
(f) US patent citation [141]
(g) Western States Power Grid of the US [205]
(h) Zachary karate club [215]
10.14. Consider the plots of degree distributions in Figures 10.12 and 10.13. Note the
noise in the tail of each plot. To smooth the tail, we can use the cumulative degree
distribution ∞
X
c
P (k) = Pr[deg(v) = i].
i=k
Given a graph with scale-free degree distribution P (k) ∼ k −α and α > 1, the
cumulative degree distribution follows P c (k) ∼ k 1−α . Plot the cumulative degree
distribution of each network dataset in Problem 10.13.
10.6. Problems 279
Algorithm 10.11: Bipartite scale-free network via preferential attachment.
Input: Positive integer n > 1 and minimum degree d ≥ 1.
Output: Bipartite scale-free multigraph. Each partition has n vertices and each
vertex has minimum degree d.
1 G ← K2n /* vertex set is {0, 1, . . . , 2n − 1} */
2 M1 ← list of length 2nd
3 M2 ← list of length 2nd
4 for v = 0, 1, . . . , n − 1 do
5 for i = 0, 1, . . . , d − 1 do
6 M1 [2(vd + i)] ← v
7 M2 [2(vd + i)] ← n + v
8 r ← draw uniformly at random from {0, 1, . . . , 2(vd + i)}
9 if r is even then
10 M1 [2(vd + i) + 1] ← M2 [r]
11 else
12 M1 [2(vd + i) + 1] ← M1 [r]
13 r ← draw uniformly at random from {0, 1, . . . , 2(vd + i)}
14 if r is even then
15 M2 [2(vd + i) + 1] ← M1 [r]
16 else
17 M2 [2(vd + i) + 1] ← M2 [r]
18 add edges (M1 [2i], M1 [2i + 1]) and (M2 [2i], M2 [2i + 1]) to G for i = 0, 1, . . . , nd − 1
19 return G
Chapter 11
Graph problems and their LP
formulations
This document is meant as an explanation of several graph theoretical functions defined
in Sage’s Graph Library (http://www.sagemath.org/), which use Linear Programming
to solve optimization of existence problems.
11.1 Maximum average degree
The average degree of a graph G is defined as ad(G) = 2|E(G)|
|V (G)|
. The maximum average
degree of G is meant to represent its densest part, and is formally defined as :
mad(G) = max ad(H)
H⊆G
Even though such a formulation does not show it, this quantity can be computed in
polynomial time through Linear Programming. Indeed, we can think of this as a simple
flow problem defined on a bipartite graph. Let D be a directed graph whose vertex set
we first define as the disjoint union of E(G) and V (G). We add in D an edge between
(e, v) ∈ E(G) × V (G) if and only if v is one of e’s endpoints. Each edge will then have a
flow of 2 (through the addition in D of a source and the necessary edges) to distribute
among its two endpoints. We then write in our linear program the constraint that each
vertex can absorb a flow of at most z (add to D the necessary sink and the edges with
capacity z).
Clearly, if H ⊆ G is the densest subgraph in G, its |E(H)| edges will send a flow
of 2|E(H)| to their |V (H)| vertices, such a flow being feasible only if z ≥ 2|E(H)|
|V (H)|
. An
elementary application of the max-flow/min-cut theorem, or of Hall’s bipartite matching
theorem shows that such a value for z is also sufficient. This LP can thus let us compute
the Maximum Average Degree of the graph.
Sage method : Graph.maximum_average_degree()
LP Formulation :
Minimize : z
Such that :
280
11.2. Traveling Salesman Problem 281
– a vertex can absorb at most z
X
∀v ∈ V (G), xe,v ≤ z
e∈E(G)
e∼v
– each edge sends a flow of 2
∀e = uv ∈ E(G), xe,u + xe,u = 2
xe,v real positive variable
Here is the corresponding Sage code:
sage : g = graphs . PetersenGraph ()
sage : p = M i x e d I n t e g e rL i n e a r P r o g r a m ( maximization = False )
sage : x = p . new_variable ( dim = 2 )
sage : p . set_objective ( p [ ’z ’ ])
sage : for v in g :
... p . add_constraint ( sum ([ x [ u ][ v ] for u in g . neighbors ( v ) ]) <= p [ ’z ’] )
sage : for u , v in g . edges ( labels = False ):
... p . add_constraint ( x [ u ][ v ] + x [ v ][ u ] == 2 )
sage : p . solve ()
3.0
REMARK : In many if not all the other LP formulations, this Linear Program
is used as a constraint. In those problems, we are always at some point looking for a
subgraph H of G such that H does not contain any cycle. The edges of G are in this
case variables, whose value can be equal to 0 or 1 depending on whether they belong
to such a graph H. Based on the observation that the Maximum Average Degree of a
tree on n vertices is exactly its average degree (= 2 − 2/n < 1), and that any cycles
in a graph ensures its average degree is larger than 2, we can then set the constraint
2
that z ≤ 2 − |V (G)| . This is a handy way to write in LP the constraint that “the set of
edges belonging to H is acyclic”. For this to work, though, we need to ensure that the
variables corresponding to our edges are binary variables.
11.2 Traveling Salesman Problem
Given a graph G whose edges are weighted by a function w : E(G) → R, a solution to
the T SP is a Hamiltonian (spanning) cycle whose weight (the sum of the weight of its
edges) is minimal. It is easy to define both the objective and the constraint that each
vertex must have exactly two neighbors, but this could produce solutions such that the
set of edges define the disjoint union of several cycles. One way to formulate this linear
program is hence to add the constraint that, given an arbitrary vertex v, the set S of
edges in the solution must contain no cycle in G − v, which amounts to checking that
the set of edges in S no adjacent to v is of maximal average degree strictly less than 2,
using the remark from section ??.
We will then, in this case, define variables representing the edges included in the
solution, along with variables representing the weight that each of these edges will send
to their endpoints.
LP Formulation :
282 Chapter 11. Graph problems and their LP formulations
Minimize X
w(e)be
e∈E(G)
Such that :
– Each vertex is of degree 2
X
∀v ∈ V (G), be = 2
e∈E(G)
e∼v
– No cycle disjoint from a special vertex v ∗
* Each edge sends a flow of 2 if it is taken
∀e = uv ∈ E(G − v ∗ ), xe,u + xe,v = 2be
* Vertices receive strictly less than 2
X 2
∀v ∈ V (G − v ∗ ), xe,v ≤ 2 −
e∈E(G)
|V (G)|
e∼v
Variables
– xe,v real positive variable (flow sent by the edge)
– be binary (is the edge in the solution ?)
Sage method : Graph.traveling_salesman_problem()
Here is the corresponding Sage corresponding to a simpler case – looking for an
Hamiltonian cycle in a graph:
sage : g = graphs . GridGraph ([4 ,4])
sage : p = M i x e d I n t e g e rL i n e a r P r o g r a m ( maximization = False )
sage : f = p . new_variable ()
sage : r = p . new_variable ()
sage : eps = 1/(2* Integer ( g . order ()))
sage : x = g . vertex_iterator (). next ()
sage : # reorders the edge as they can appear in the two different ways
sage : R = lambda x , y : (x , y ) if x < y else (y , x )
sage : # All the vertices have degree 2
sage : for v in g :
... p . add_constraint ( sum ([ f [ R (u , v )] for u in g . neighbors ( v )]) == 2)
sage : # r is greater than f
sage : for u , v in g . edges ( labels = None ):
... p . add_constraint ( r [( u , v )] + r [( v , u )] - f [ R (u , v )] >= 0)
sage : # no cycle which does not contain x
sage : for v in g :
... if v != x :
... p . add_constraint ( sum ([ r [( u , v )] for u in g . neighbors ( v )]) <= 1 - eps )
sage : p . set_objective ( None )
sage : p . set_binary ( f )
sage : p . solve () # optional - GLPK , CBC , CPLEX
0.0
11.3. Edge-disjoint spanning trees 283
sage : # We can now build the solution
sage : # found as a graph
sage : f = p . get_values ( f ) # optional - GLPK , CBC , CPLEX
sage : tsp = Graph () # optional - GLPK , CBC , CPLEX
sage : for e in g . edges ( labels = False ): # optional - GLPK , CBC , CPLEX
... if f [ R ( e [0] , e [1])] == 1: # optional - GLPK , CBC , CPLEX
... tsp . add_edge ( e ) # optional - GLPK , CBC , CPLEX
sage : tsp . is_regular ( k =2) and tsp . is_connected () # optional - GLPK , CBC , CPLEX
True
sage : tsp . order () == g . order () # optional - GLPK , CBC , CPLEX
True
11.3 Edge-disjoint spanning trees
This problem is polynomial by a result from Edmonds. Obviously, nothing ensures the
following formulation is a polynomial algorithm as it contains many integer variables,
but it is still a short practical way to solve it.
This problem amounts to finding, given a graph G and an integer k, edge-disjoint
spanning trees T1 , . . . , Tk which are subgraphs of G. In this case, we will chose to define
a spanning tree as an acyclic set of |V (G)| − 1 edges.
Sage method : Graph.edge_disjoint_spanning_trees()
LP Formulation :
Maximize : nothing
Such that :
– An edge belongs to at most one set
X
∀e ∈ E(G), be,k ≤ 1
i∈[1,...,k]
– Each set contains |V (G)| − 1 edges
X
∀i ∈ [1, . . . , k], be,k = |V (G)| − 1
e∈E(G)
– No cycles
* In each set, each edge sends a flow of 2 if it is taken
∀i ∈ [1, . . . , k], ∀e = uv ∈ E(G), xe,k,u + xe,k,u = 2be,k
* Vertices receive strictly less than 2
X 2
∀i ∈ [1, . . . , k], ∀v ∈ V (G), xe,k,v ≤ 2 −
e∈E(G)
|V (G)|
e∼v
Variables
– be,k binary (is edge e in set k ?)
– xe,k,u positive real (flow sent by edge e to vertex u in set k)
284 Chapter 11. Graph problems and their LP formulations
Here is the corresponding Sage code:
sage : g = graphs . RandomGNP (40 ,.6)
sage : p = M i x e d I n t e g e rL i n e a r P r o g r a m ()
sage : colors = range (2)
sage : # Sort an edge
sage : S = lambda (x , y ) : (x , y ) if x < y else (y , x )
sage : edges = p . new_variable ( dim = 2)
sage : r_edges = p . new_variable ( dim = 2)
sage : # An edge belongs to at most one tree
sage : for e in g . edges ( labels = False ):
... p . add_constraint ( sum ([ edges [ j ][ S ( e )] for j in colors ]) , max = 1)
sage : for j in colors :
... # each color class has g . order () -1 edges
... p . add_constraint (
... sum ([ edges [ j ][ S ( e )] for e in g . edges ( labels = None )])
... >= g . order () -1)
... # Each vertex is in the tree
... for v in g . vertices ():
... p . add_constraint (
... sum ([ edges [ j ][ S ( e )] for e in g . edges_incident (v , labels = None )])
... >= 1)
... # r_edges is larger than edges
... for u , v in g . edges ( labels = None ):
... p . add_constraint (
... r_edges [ j ][( u , v )] + r_edges [ j ][( v , u )]
... == edges [ j ][ S (( u , v ))] )
sage : # no cycles
sage : epsilon = (3* Integer ( g . order ()))**( -1)
sage : for j in colors :
... for v in g :
... p . add_constraint (
... sum ([ r_edges [ j ][( u , v )] for u in g . neighbors ( v )])
... <= 1 - epsilon )
sage : p . set_binary ( edges )
sage : p . set_objective ( None )
sage : p . solve () # optional - GLPK , CBC , CPLEX
0.0
sage : # We can now build the solution
sage : # found as a list of trees
sage : edges = p . get_values ( edges ) # optional - GLPK , CBC , CPLEX
sage : trees = [ Graph () for c in colors ] # optional - GLPK , CBC , CPLEX
sage : for e in g . edges ( labels = False ): # optional - GLPK , CBC , CPLEX
... for c in colors : # optional - GLPK , CBC , CPLEX
... if round ( edges [ c ][ S ( e )]) == 1: # optional - GLPK , CBC , CPLEX
... trees [ c ]. add_edge ( e ) # optional - GLPK , CBC , CPLEX
sage : all ([ trees [ j ]. is_tree () for j in colors ]) # optional - GLPK , CBC , CPLEX
True
11.4 Steiner tree
See Trietsch [194] for a relationship between Steiner trees and Euler’s problem of polygon
division. Finding a spanning tree in a Graph G can be done in linear time, whereas
computing a Steiner Tree is NP-hard. The goal is in this case, given a graph, a weight
function w : E(G) → R and a set S of vertices, to find the tree of minimum cost
connecting them all together. Equivalently, we will be looking for an acyclic subgraph
Hof G containing |V (H)| vertices and |E(H)| = |V (H)| − 1 edges, which contains each
vertex from S
11.4. Steiner tree 285
LP Formulation :
Minimize : X
w(e)be
e∈E(G)
Such that :
– Each vertex from S is in the tree
X
∀v ∈ S, be ≥ 1
e∈E(G)
e∼v
– c is equal to 1 when a vertex v is in the tree
∀v ∈ V (G), ∀e ∈ E(G), e ∼ v, be ≤ cv
– The tree contains |V (H)| vertices and |E(H)| = |V (H)| − 1 edges
X X
cv − be = 1
v∈G e∈E(G)
– No Cycles
* Each edge sends a flow of 2 if it is taken
∀e = uv ∈ E(G), xe,u + xe,u = 2be,k
* Vertices receive strictly less than 2
X 2
∀v ∈ V (G), xe,v ≤ 2 −
e∈E(G)
|V (G)|
e∼v
Variables :
– be binary (is e in the tree ?)
– cv binary (does the tree contain v ?)
– xe,v real positive variable (flow sent by the edge)
Sage method : Graph.steiner_tree()
Here is the corresponding Sage code:
sage : g = graphs . GridGraph ([10 ,10])
sage : vertices = [(0 ,2) ,(5 ,3)]
sage : from sage . numerical . mip import M i x e d I n t e ge r L i n e a r P r o g r a m
sage : p = M i x e d I n t e g er L i n e a r P r o g r a m ( maximization = False )
sage : # Reorder an edge
sage : R = lambda (x , y ) : (x , y ) if x < y else (y , x )
sage : # edges used in the Steiner Tree
sage : edges = p . new_variable ()
sage : # relaxed edges to test for acyclicity
sage : r_edges = p . new_variable ()
286 Chapter 11. Graph problems and their LP formulations
sage : # Whether a vertex is in the Steiner Tree
sage : vertex = p . new_variable ()
sage : # Which vertices are in the tree ?
sage : for v in g :
... for e in g . edges_incident (v , labels = False ):
... p . add_constraint ( vertex [ v ] - edges [ R ( e )] , min = 0)
sage : # We must have the given vertices in our tree
sage : for v in vertices :
... p . add_constraint (
... sum ([ edges [ R ( e )] for e in g . edges_incident (v , labels = False )]
... == 1)
sage : # The number of edges is equal to the number of vertices in our tree minus 1
sage : p . add_constraint (
... sum ([ vertex [ v ] for v in g ])
... - sum ([ edges [ R ( e )] for e in g . edges ( labels = None )])
... == 1)
sage : # There are no cycles in our graph
sage : for u , v in g . edges ( labels = False ):
... p . add_constraint (
... r_edges [( u , v )]+ r_edges [( v , u )] - edges [ R (( u , v ))]
... <= 0 )
sage : eps = 1/(5* Integer ( g . order ()))
sage : for v in g :
... p . add_constraint ( sum ([ r_edges [( u , v )] for u in g . neighbors ( v )]) , max = 1 - eps )
sage : p . set_objective ( sum ([ edges [ R ( e )] for e in g . edges ( labels = False )]))
sage : p . set_binary ( edges )
sage : p . solve () # optional - GLPK , CBC , CPLEX
6.0
sage : # We can now build the solution
sage : # found as a tree
sage : edges = p . get_values ( edges ) # optional - GLPK , CBC , CPLEX
sage : st = Graph () # optional - GLPK , CBC , CPLEX
sage : st . add_edges (
... [ e for e in g . edges ( labels = False )
... if edges [ R ( e )] == 1]) # optional - GLPK , CBC , CPLEX
sage : st . is_tree () # optional - GLPK , CBC , CPLEX
True
sage : all ([ v in st for v in vertices ]) # optional - GLPK , CBC , CPLEX
True
11.5 Linear arboricity
The linear arboricity of a graph G is the least number k such that the edges of G can
be partitioned into k classes, each of them being a forest of paths (the disjoints union
of paths – trees of maximal degree 2). The corresponding LP is very similar to the one
giving edge-disjoint spanning trees
LP Formulation :
Maximize : nothing
Such that :
– An edge belongs to exactly one set
X
∀e ∈ E(G), be,k = 1
i∈[1,...,k]
11.5. Linear arboricity 287
– Each class has maximal degree 2
X
∀v ∈ V (G), ∀i ∈ [1, . . . , k], be,k ≤ 2
e∈E(G)
e∼v
– No cycles
* In each set, each edge sends a flow of 2 if it is taken
∀i ∈ [1, . . . , k], ∀e = uv ∈ E(G), xe,k,u + xe,k,v = 2be,k
* Vertices receive strictly less than 2
X 2
∀i ∈ [1, . . . , k], ∀v ∈ V (G), xe,k,v ≤ 2 −
e∈E(G)
|V (G)|
e∼v
Variables
– be,k binary (is edge e in set k ?)
– xe,k,u positive real (flow sent by edge e to vertex u in set k)
Sage method : sage.graphs.graph_coloring.linear_arboricity()
Here is the corresponding Sage code :
sage : g = graphs . GridGraph ([4 ,4])
sage : k = 2
sage : p = M i x e d I n t e g e rL i n e a r P r o g r a m ()
sage : # c is a boolean value such that c [ i ][( u , v )] = 1
sage : # if and only if (u , v ) is colored with i
sage : c = p . new_variable ( dim =2)
sage : # relaxed value
sage : r = p . new_variable ( dim =2)
sage : E = lambda x , y : (x , y ) if x < y else (y , x )
sage : MAD = 1 -1/( Integer ( g . order ())*2)
sage : # Partition of the edges
sage : for u , v in g . edges ( labels = None ):
... p . add_constraint ( sum ([ c [ i ][ E (u , v )] for i in range ( k )]) , max =1 , min =1)
sage : for i in range ( k ):
... # r greater than c
... for u , v in g . edges ( labels = None ):
... p . add_constraint ( r [ i ][( u , v )] + r [ i ][( v , u )] - c [ i ][ E (u , v )] , max =0 , min =0)
... # Maximum degree 2
... for u in g . vertices ():
... p . add_constraint ( sum ([ c [ i ][ E (u , v )] for v in g . neighbors ( u )]) , max = 2)
... # no cycles
... p . add_constraint ( sum ([ r [ i ][( u , v )] for v in g . neighbors ( u )]) , max = MAD )
sage : p . set_objective ( None )
sage : p . set_binary ( c )
sage : c = p . get_values ( c )
sage : gg = g . copy ()
sage : gg . delete_edges ( g . edges ())
sage : answer = [ gg . copy () for i in range ( k )]
sage : add = lambda (u , v ) , i : answer [ i ]. add_edge (( u , v ))
sage : for i in range ( k ):
... for u , v in g . edges ( labels = None ):
... if c [ i ][ E (u , v )] == 1:
... add (( u , v ) , i )
288 Chapter 11. Graph problems and their LP formulations
11.6 H-minor
For more information on minor theory, please see
http://en.wikipedia.org/wiki/Minorw%28graphwtheory%29
It is a wonderful subject, and I do not want to begin talking about it when I know I
couldn’t freely fill pages with remarks :-)
For our purposes, we will just say that finding a minor H in a graph G, consists in :
1. Associating to each vertex h ∈ H a set Sh of representants in H, different vertices
h having disjoints representative sets
2. Ensuring that each of these sets is connected (can be contracted)
3. If there is an edge between h1 and h2 in H, there must be an edge between the
corresponding representative sets
Here is how we will address these constraints :
1. Easy
2. For any h, we can find a spanning tree in Sh (an acyclic set of |Sh | − 1 edges)
3. This one is very costly.
To each directed edge g1 g2 (I consider g1 g2 and g2 g1 as different) and every edge
h1 h2 is associated a binary variable which can be equal to one only if g1 represents
h1 and g2 represents g2 . We then sum all these variables to be sure there is at least
one edge from one set to the other.
11.6. H-minor 289
LP Formulation :
Maximize : nothing
Such that :
– A vertex g ∈ V (G) can represent at most one vertex h ∈ V (H)
X
∀g ∈ V (G), rsh,g ≤ 1
h∈V (H)
– An edge e can only belong to the tree of h if both its endpoints represent h
∀e = g1 g2 ∈ E(G), te,h ≤ rsh,g1 and te,h ≤ rsh,g2
– In each representative set, the number of vertices is one more than the number
of edges in the corresponding tree
X X
∀h, rsh,g − te,h = 1
g∈V (G) e∈E(G)
– No cycles in the union of the spanning trees
* Each edge sends a flow of 2 if it is taken
X
∀e = uv ∈ E(G), xe,u + xe,v = 2 te,h
h∈V (H)
* Vertices receive strictly less than 2
X 2
∀v ∈ V (G), xe,k,v ≤ 2 −
e∈E(G)
|V (G)|
e∼v
– arc(g1 ,g2 ),(h1 ,h2 ) can only be equal to 1 if g1 g2 is leaving the representative set
of h1 to enter the one of h2 . (note that this constraints has to be written both
for g1 , g2 , and then for g2 , g1 )
∀g1 , g2 ∈ V (G), g1 6= g2 , ∀h1 h2 ∈ E(H)
arc(g1 ,g2 ),(h1 ,h2 ) ≤ rsh1 ,g1 and arc(g1 ,g2 ),(h1 ,h2 ) ≤ rsh2 ,g2
– We have the necessary edges between the representative sets
∀h1 h2 ∈ E(H)
X
arc(g1 ,g2 ),(h1 ,h2 ) ≥ 1
∀g1 ,g2 ∈V (G),g1 6=g2
Variables
– rsh,g binary (does g represent h ? rs = “representative set”)
– te,h binary (does e belong to the spanning tree of the set representing h ?)
– xe,v real positive (flow sent from edge e to vertex v)
290 Chapter 11. Graph problems and their LP formulations
– arc(g1 ,g2 ),(h1 ,h2 ) binary (is edge g1 g2 leaving the representative set of h1 to enter
the one of h2 ?)
Here is the corresponding Sage code:
sage : g = graphs . PetersenGraph ()
sage : H = graphs . CompleteGraph (4)
sage : p = M i x e d I n t e g e rL i n e a r P r o g r a m ()
sage : # sorts an edge
sage : S = lambda (x , y ) : (x , y ) if x < y else (y , x )
sage : # rs = Representative set of a vertex
sage : # for h in H , v in G is such that rs [ h ][ v ] == 1 if and only if v
sage : # is a representant of h in g
sage : rs = p . new_variable ( dim =2)
sage : for v in g :
... p . add_constraint ( sum ([ rs [ h ][ v ] for h in H ]) , max = 1)
sage : # We ensure that the set of representatives of a
sage : # vertex h contains a tree , and thus is connected
sage : # edges represents the edges of the tree
sage : edges = p . new_variable ( dim = 2)
sage : # there can be a edge for h between two vertices
sage : # only if those vertices represent h
sage : for u , v in g . edges ( labels = None ):
... for h in H :
... p . add_constraint ( edges [ h ][ S (( u , v ))] - rs [ h ][ u ] , max = 0 )
... p . add_constraint ( edges [ h ][ S (( u , v ))] - rs [ h ][ v ] , max = 0 )
sage : # The number of edges of the tree in h is exactly the cardinal
sage : # of its representative set minus 1
sage : for h in H :
... p . add_constraint (
... sum ([ edges [ h ][ S ( e )] for e in g . edges ( labels = None )])
... - sum ([ rs [ h ][ v ] for v in g ])
... ==1 )
sage : # a tree has no cycle
sage : epsilon = 1/(5* Integer ( g . order ()))
sage : r_edges = p . new_variable ( dim =2)
sage : for h in H :
... for u , v in g . edges ( labels = None ):
... p . add_constraint (
... r_edges [ h ][( u , v )] + r_edges [ h ][( v , u )] >= edges [ h ][ S (( u , v ))])
... for v in g :
... p . add_constraint (
... sum ([ r_edges [ h ][( u , v )] for u in g . neighbors ( v )]) <= 1 - epsilon )
sage : # Once the representative sets are described , we must ensure
sage : # there are arcs corresponding to those of H between them
sage : h_edges = p . new_variable ( dim =2)
sage : for h1 , h2 in H . edges ( labels = None ):
... for v1 , v2 in g . edges ( labels = None ):
... p . add_constraint ( h_edges [( h1 , h2 )][ S (( v1 , v2 ))] - rs [ h2 ][ v2 ] , max = 0)
... p . add_constraint ( h_edges [( h1 , h2 )][ S (( v1 , v2 ))] - rs [ h1 ][ v1 ] , max = 0)
... p . add_constraint ( h_edges [( h2 , h1 )][ S (( v1 , v2 ))] - rs [ h1 ][ v2 ] , max = 0)
... p . add_constraint ( h_edges [( h2 , h1 )][ S (( v1 , v2 ))] - rs [ h2 ][ v1 ] , max = 0)
sage : p . set_binary ( rs )
sage : p . set_binary ( edges )
sage : p . set_objective ( None )
sage : p . solve () # optional - GLPK , CBC , CPLEX
0.0
sage : # We can now build the solution found as a
sage : # dictionary associating to each vertex of H
11.6. H-minor 291
sage : # the corresponding set of vertices in G
sage : rs = p . get_values ( rs )
sage : from sage . sets . set import Set
sage : rs_dict = {}
sage : for h in H :
... rs_dict [ h ] = [ v for v in g if rs [ h ][ v ]==1]
Appendix A
Asymptotic growth
Name Standard notation Intuitive notation Meaning
theta f (n) = Θ(g(n)) f (n) ∈ Θ(g(n)) f (n) ≈ c · g(n)
big oh f (n) = O(g(n)) f (n) ≤ O(g(n)) f (n) ≤ c · g(n)
omega f (n) = Ω(g(n)) f (n) ≥ Ω(g(n)) f (n) ≥ c · g(n)
little oh f (n) = o(g(n)) f (n) o(g(n)) f (n) g(n)
little omega f (n) = ω(g(n)) f (n) ω(g(n)) f (n) g(n)
tilde f (n) = Θ̃(g(n)) f (n) ∈ Θ̃(g(n)) f (n) ≈ logΘ(1) g(n)
Table A.1: Meaning of asymptotic notations.
Class lim f (n)/g(n) = Equivalent definition
n→∞
f (n) = Θ(g(n)) a constant f (n) = O(g(n)) and f (n) = Ω(g(n))
f (n) = o(g(n)) zero f (n) = O(g(n)) but f (n) 6= Ω(g(n))
f (n) = ω(g(n)) ∞ f (n) 6= O(g(n)) but f (n) = Ω(g(n))
Table A.2: Asymptotic behavior in the limit of large n.
292
Appendix B
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Index
A(G), 18 rad(G), 200
C 0 (G, F ), 234 , 35
C 1 (G, F ), 234 td, 91
Cn , 16 ε, 135, 138
En , 49 ϕ(n), 115
Gc , 34 f -augmenting, 246
Kn , 15 f -saturated, 246
Km,n , 17 f -unsaturated, 246
Ln , 35 f -zero, 246
Pn , 16, 35 k-connected, 203
Qn , 35 k-edge-connected, 204
Wn , 29 n-queens problem, 101
∆, 29 n-space, 132
∆(G), 10 graph6, 54, 56, 58, 59
adj, 4 sparse6, 56, 58
L, 22, 42 Lukaszewicz, J., 123
L(G), 8
Ni , 268 edge expander family, 221
∼
=, 22
deg, 5, 8 active vertex, 239
deg+ , 7 acyclic, 105, 117, 146
deg− , 7 Adelson-Velskiı̆, G. M., 180
δ(G), 10 adjacency matrix, 18
depth(v), 106 reduced, 20
diam(G), 200 signed edge, 214
dir, 92 algorithm
, 198 greedy, 76, 77, 117, 120
, 264 optimization, 111
height(T ), 106 random, 50, 51, 104, 146–148, 256, 257,
iadj, 5 261, 263, 264, 266, 273, 275, 276,
id, 5 279
κ(G), 203 recursive, 151
κe (G), 204 alphabet, 40, 41, 130, 134
κv (G), 202 binary, 135, 136
λ(G), 204 English, 130, 148
lg, 158 weighted, 130, 137
oadj, 5 Altito, Noelie, 85
od, 5 arcs, 3
ω, 14 Argentina, 94, 95
G, 34 ASCII, 56, 58, 130
per(G), 200 augmenting path, 246
310
Index 311
Australia, 94, 95 Jacques Philippe Marie, 196
Australian National University, 56 binomial
automata theory, 39, 85 coefficient, 166, 192
AVL tree, 180, 196 distribution, 256
height-balance property, 180 random graph, 258
tree, 166
backtrack, 66 binomial heap, 153, 165, 166, 168, 194
algorithm, 102 maximum, 194
Baker, Matthew, 198 minimum, 194
balanced bracket problem, 98, 99 order property, 168, 169, 194
Bangkok, 94, 95 properties, 168
Barabási-Albert model, 270 root-degree property, 168, 169
Batagelj, Vladimir, 259, 278 biology, 197
Batagelj-Brandes algorithm, 259, 261 bipartite graph, 16, 17, 50, 51, 278, 279
Baudot, E., 131 complete, 17
Beijing, 94, 95 bit, 58, 130, 135
Bell, Jordan, 102 least significant, 58
Bellman, Richard E., 77 most significant, 58
Bellman-Ford algorithm, 73, 77–80, 85, 86, parity, 58
93 bit vector, 56, 58, 59
Benjamin, Arthur T., 152 length, 58
bent, 215 bond, 32, 111, 235
Berlin, 94, 95 Borůvka
Bernoulli family, 107 algorithm, 117, 123, 125, 145, 146, 151
BFS, 60–64, 66, 70 Otakar, 117, 123
big-endian, 58, 59 bowtie graph, 13
Biggs, Norman, 137 braille, 130
binary heap, 153, 155, 180 branch cut, 108, 110
heap-structure property, 180 Brandes, Ulrik, 259, 278
maximum, 192 Brasilia, 94, 95
minimum, 192 Brazil, 94, 95
order property, 156, 194 breadth-first search, 60–64, 69, 70, 72–74,
sift-down, 161 89, 92, 98, 110, 140, 141
sift-up, 159 tree, 60, 63
structure property, 156 bridge, 32, 106, 112, 123, 204
binary search, 89, 91, 175 bridgeless, 204
binary search tree, 153, 173, 175, 180, 182, Briggs
183 algorithm, 277
left subtree property, 173 Keith M., 277
property, 173, 183, 194 BST, 173
recursion property, 173 bubble sort, 96, 97
right subtree property, 173 Buenos Aires, 94, 95
binary tree, 108, 110, 127–129, 148, 153 butterfly graph, 13
complete, 127, 180
nearly complete, 156, 180 Caenorhabditis elegans, 253
random, 129, 148 Canada, 94, 95
Binet canonical label, 24, 25
formula, 196 Cantor-Schröder-Bernstein theorem, 51
312 Index
capacity, 245 economy, 130
cut, 247 error-correcting, 20, 130
card, 66 linear, 132
cardinality, 8 optimal, 137
Carroll, Lewis, 2 prefix, 130
Cartesian product, 35, 36 prefix-free, 130, 136, 148
Catalan radix, 151
number, 49, 128 reliability, 130
recursion, 128 security, 130
Cayley graph of Boolean function, 214 tree representation, 135
characteristic polynomial, 213 uniquely decodable, 136
Chazelle, Bernard, 194 variable-length, 130
check matrix, 20 codeword, 130, 135
chemistry, 82, 197 length, 137
chess, 64, 101, 130 coding function, 130
chessboard, 64 Cohen, Danny, 58
knight, 64 Collatz
knight piece, 64 conjecture, 149
knight’s tour, 64–66, 101 graph, 149, 150
queen, 101, 102 length, 149
child sequence, 149
left, 127, 140, 145 tree, 149, 150
right, 127, 140, 145 color code, 130
China, 94, 95 coloring
Chinese ring puzzle, 131, 132 edge, 38
chip firing game, 239 vertex, 38, 39
chip-firing game, 239 combinatorial generation, 192
chip-firing games, 239 combinatorial graphs, 2
CHKNS model, 277 combinatorics, 132
Choquet, G., 123 communications network, 206
Chu Shi-Chieh, 192 complement, 34
Chvátal graph, 146, 147 complete graph, 15, 16, 147, 148, 256, 261,
circuit, 13 263, 266, 267, 275, 276
board, 60 component, 14, 28, 112
electronic, 116 connected, 117
circulation, 235 computer science, 39, 140
circulation space, 235, 236 condensed matter, 254
classification tree, 106, 107, 111 configuration, 239, 241
claw graph, 203 critical, 242
closed form, 128 level, 245
closeness centrality, 212 recurrent, 242
cocycle code, 236 stable, 242
cocycle space, 236 starting, 239
cocyle, 235 weight, 245
code, 130, 135 connected graph, 14, 111
r-ary, 151 connectivity, 98
binary, 130, 135, 136 cost, 71
block, 130 Coward, Noel, 11
Index 313
critical configurations, 243 disconnecting set, 32
critical group, 243, 245 distance, 53, 63, 64, 71, 72, 74, 78, 80, 106
cryptosystem, 130 characteristic, 254
cut function, 71, 72, 197, 198, 222
set, 32, 113 matrix, 22, 72
cut set, 235 minimum, 74
cut space, 236 total, 91
cut-edge, 203, 204 distance distribution, 255
cut-point, 202 distribution
cut-vertex, 202, 203 binomial, 261
cycle, 13, 72, 73, 87, 105, 106, 112, 114 geometric, 260, 262
fundamental, 114, 148 Poisson, 262
negative, 72, 73, 78, 80, 85–87, 104 uniform, 261
cycle code, 235 divide and conquer, 104
cycle double cover conjecture, 204 dollar game, 241
cycle graph, 16, 51 Dryden, John, 60
cycle space, 235 dynamic programming, 80
D’Angelo, Anthony J., 145 eccentricity, 198, 199
Dörrie, Heinrich, 49 mutual, 222
data structure, 53, 153 path, 222
de Moivre, Abraham, 196 vertex, 222
de Montmort, Pierre Rémond, 100 edge, 3
decode, 130 boundary, 220
degree, 5, 8 capacity, 245
matrix, 22 contraction, 33
maximum, 10, 110 cut, 32, 111, 208
minimum, 10, 115 deletion, 32
sequence, 24, 115 deletion subgraph, 31
weighted, 8 directed, 4
degree centrality, 211 endpoint, 117
degree distribution, 251–254, 261, 271, 272 expansion, 220
depth-first search, 60, 64, 66–70, 73, 89, 92, head, 6
98, 110, 141 incident, 3
tree, 66, 69 multigraph, 6
de Moivre, Abraham, 100 multiple, 3
DFA, 40, 41 tagging game, 111
DFS, 64, 66–70 tail, 6
diameter, 63, 64 weight, 6
Digital Signature Algorithm, 151 edge cut subgraph, 235
digraph, 5, 106 edge cutset, 235
weighted, 71 edge-cut, 203
Dijkstra Edmonds, Jack, 111
algorithm, 14, 74–78, 85, 86, 93, 153 eigenvalue, 151
E. W., 74, 120 element
Dirac’s theorem, 210 random, 128
directedness, 92 Elkies, Noam D., 66
disconnected graph, 14 encode, 130
314 Index
endianness, 58 Franklin graph, 23
England, 94, 95, 102 Frederickson, Greg N., 194
entropy FreeBSD, 53
encoding, 130 frequency distribution, 265
function, 130 friendship graph, 212
Erdős, Paul, 25 FRW, 78, 80
error rate, 130 function plot, 2
Euclidean algorithm, 88 fundamental cycles, 237
Euler
Leonhard, 1, 10, 48, 49, 115 Gallai, Tibor, 25
phi function, 115, 149 Garlaschelli, Diego, 277
phi sequence, 115, 116 genetic code, 130
polygon division problem, 48, 49 Germany, 94, 95
subgraph, 13 Gilbert, E. N., 262
Euler subgraph, 235 girth, 13
Eulerian trail, 1 Goldbach, Christian, 48
expander graph, 221 Goldberg, R., 78
golden ratio, 152, 196
Faber, Xander, 198
Graham, Ronald L., 202
family tree, 14, 15, 106, 107
graph, 3
fault-tolerant, 206
applications, 37
Fermat’s little theorem, 51
connected, 14, 70, 71
Fibonacci
number, 180 dense, 55, 80
sequence, 196 directed, 5
tree, 180, 182–184, 195 disconnected, 14
FIFO, 61, 66 intersection, 29
filesystem, 106 join, 29
hierarchy, 106 line, 8
finite automaton, 39, 40, 51 nonisomorphic, 49
deterministic, 40, 41 simple, 8
nondeterministic, 41, 42 sparse, 18, 55, 80, 85, 86
first in, first out, 61 traversal, 60
flag semaphore, 130 trivial, 115, 123
Florek, K., 123 undirected, 3
Florentine families, 38 union, 28, 29
flow, 235, 246 unweighted, 3
value, 246 weighted, 6, 71, 117, 120
flow chart, 53 graph isomorphism, 22, 25
flow space, 235 graph minor, 37
Floyd, Robert, 80 graphical sequence, 25, 27
Floyd-Roy-Warshall algorithm, 78, 80–82, Gray code, 131, 132
84, 85, 93, 201, 202 m-ary, 131
football, 130 binary, 131, 132
forbidden minor, 37 reflected, 132–134
Ford, Lester Randolph, Jr., 77 Gray, Frank, 131
forest, 105, 106 Gribkovskaia, Irina, 1
Foulds, L. R., 38 grid, 36
Index 315
graph, 103, 104, 106, 108, 119, 120, 146, unoriented, 21
148 indegree, 5
Gros, L., 131 unweighted, 7
group theory India, 94, 95
computational, 132 induction, 112, 115, 127, 136, 137, 146
Gulliver’s Travels, 58 structural, 146
infix notation, 99
Hadamard transform, 215 information channel, 130
Hakimi, S. L., 26 insertion sort, 97
Halskau Sr., Øyvind, 1 Internet, 270
Hamming distance, 35 topology, 271
Hampton Court Palace, 102 interpolation search, 91
handshaking lemma, 10 invariant, 24, 27
Havel, Václav, 26 isomorphism, 115
Havel-Hakimi isoperimetric number, 220
test, 27
theorem, 26 Japan, 94, 95
heap Jarnı́k, V., 120
2-heap, 120 Johnson
k-ary, 77 algorithm, 73, 85, 86, 93
binary, 77 Donald B., 85
binary minimum, 138 join, 114
Fibonacci, 77, 85, 120 Jordan, Camille, 201
heapsort, 155
hierarchical structure, 14, 105, 106 Königsberg, 1
Hoare, C. A. R., 98 graph, 2, 6
Hopcroft, John E., 64 seven bridges puzzle, 1, 10
Hopkins, Brian, 1 Kaliningrad, 1
Horner Kaplan, Haim, 194
method, 88 Kataoka, Seiji, 145
W. G., 88 Kinoshita, Harunobu, 104
house graph, 3 Kleene
Huffman algorithm, 85
David, 137 Stephen, 85
tree, 153 Klein, Felix, 2
Huffman code, 136–139, 141, 148 Kneser graph, 54–56
binary, 137 Knuth
encoding, 140 Algorithm S, 277
tree construction, 137 Donald E., 91, 96, 97, 131, 277
tree representation, 138, 139, 141 Kraft
Humpty Dumpty, 2 inequality, 151, 152
hypercube graph, 35, 36, 132 Leon Gordon, 152
theorem, 152
in-neighbor, 5 Kruskal
incidence algorithm, 117–120, 145, 146, 151
function, 6 Joseph B., 117
matrix, 21
incidence matrix ladder graph, 35
oriented, 21 Lagarias, Jeffrey C., 149
316 Index
Landis, E. M., 180 Matthew effect, 270
language, 42 max-flow min-cut theorem, 247
regular, 42 generalized, 248
Laplacian, 243 maximum flow problem, 246
reduced, 245 maze, 60, 64, 102
Laplacian matrix, 22, 151, 213 McKay, Brendan D., 56, 66
Laporte, Gilbert, 1 McMillan
last in, first out, 66 Brockway, 151
Latora, V., 277 theorem, 151
Latora-Marchiori model, 277 Menezes, Alfred J., 96
lattice, 36 Menger’s theorem, 207–209
Lee, C. Y., 60 merge sort, 170
legal firing sequence, 242 Merkle, Ralph C., 151
Lehman, A., 111 Merris-McKay theorem, 151
Lehmer, D. H., 50 mesh, 36, 37
level message, 134
binary tree, 156 metabolic network, 270
tree, 195 metric, 72, 198
LIFO, 66 function, 71
Lima, 94, 95 metric graph, 198
line graph, 8 metric space, 72
linear search, 89 finite, 72
Linux, 106 Milgram, Stanley, 264
list, 54, 61, 63, 66, 77, 138 minimum cut problem, 247
adjacency, 54, 55, 63 minimum spanning tree problem, 117
contiguous edge, 56, 266 molecular graph, 37, 38, 82
edge, 56 Montmort-Moivre strategy, 101
element, 54 Moore, Edward F., 60, 77
empty, 54 Morse code, 131, 136, 148
length, 54 Moscow, 94, 95
little-endian, 58 MST, 117
Loberman, H., 117 multi-undirected graph, 5
Loebbing, Martin, 66 multidigraph, 5, 40
London, 94, 95 multigraph, 5
Lucas adjacency, 7
M. Édouard, 64, 152 in-neighbor, 7
number, 152 out-neighbor, 7
Munroe, Randall, 53, 64, 74, 77, 105, 153,
Madrid, 94, 95 226, 227
Marchiori, M., 277 musical score, 130
marriage ties, 38
matrix, 18 neighbor graph, 268
adjacency, 19, 54, 55, 59 network, 39, 245
bi-adjacency, 20 biological, 253, 264
distance, 202 citation, 271
main diagonal, 59 collaboration, 271
transpose, 48 communication, 111
upper triangle, 59 flow, 39
Index 317
information, 264 permutation
social, 254, 264, 271 equivalent, 24
technological, 253, 264 random, 148
Zachary karate club, 252 Peru, 94, 95
New Delhi, 94, 95 Petersen
NFA, 41, 42 graph, 38, 39, 69, 70, 203, 204
node, 3 Julius, 69
noisy channel, 130 planar graph, 38, 49
null graph, 4, 270 plane, 103
Nuutila, Esko, 85 Pollak, O., 202
postfix notation, 99
operations research, 39 power grid, 253
order, 3 preferential attachment, 269, 270, 273
organism, 107, 111 prefix-free condition, 130
orientation, 6, 21 Pregel River, 1
probability, 92 Pretoria, 94, 95
oriented graph, 257 Prim
random, 257, 260 algorithm, 117, 120–122, 124, 145, 146,
Ottawa, 94, 95 151, 153
out-neighbor, 5, 61, 69, 73, 74 R. C., 117, 120
outdegree, 5 priority queue, 153, 154
unweighted, 7, 8 probability, 137
outer boundary, 220 expectation, 137
overfull graph, 48 sample space, 128
Oxley, James, 111 space, 256
Paley graph, 221 pseudorandom number, 50, 260
parallel forest-merging, 123 Python, 18
parallelization, 123 queue, 61, 63, 66, 70, 74, 141
partition, 50 dequeue, 61, 63, 141, 143
Pascal end, 61
formula, 168, 192 enqueue, 61, 63, 141, 143
path, 12, 13, 105, 106 front, 61
closed, 13 length, 61
distance, 71 minimum-priority, 85, 120, 138
even, 13 priority, 138
geodesic, 14 rear, 61
graph, 35 start, 61
Hamiltonian, 132 quicksort, 98
internally disjoint, 206
length, 71, 72, 106 Ramanujan graph, 221
odd, 13 random graph, 251
shortest, 53, 71–74, 76, 78, 80, 85 Bernoulli, 255
tree, 113, 114 binomial, 255, 265
weighted, 85 Erdős-Rényi, 262
path graph, 16 uniform, 262
pendant, 8, 114 weighted, 277
perfect square, 107 random variable
Perkal, J., 123 geometric, 260
318 Index
Rasmussen, Rune, 111 multigraphs, 8, 9
recurrence relation, 195, 196 switching game, 111, 112
recursion, 80, 111, 112, 123, 125, 137, 143, shellsort, 97
145, 146 Shimbel, A., 77
reduced Laplacian, 245 Shirali, Satish, 72
regular expression, 42 shortest path, 14
regular graph, 10, 51 Simon, Herbert, 270
k-circulant, 51, 52, 265, 266 simple graph, 8, 148, 265
r-regular, 10, 52 random, 256, 261, 263, 264, 275, 276
relative complement, 34 single-source shortest path, 74, 77
remainder, 58 six degrees of separation, 264
Renaissance, 38 size, 3
reservoir sampling, 277 component, 278
residual digraph, 247 tree, 112–114
residual network, 247 small-world, 51, 64, 266
reverse Polish notation, 99 algorithm, 266
rich-get-richer effect, 270 characteristic path length, 266
river crossing problem, 98 clustering coefficient, 266, 268
Robertson, Neil, 37 effect, 264
Robertson-Seymour theorem, 37 experimental results, 265
Roget’s Thesaurus, 264 network, 266
root directory, 106 social network, 197
root list, 169 social network analysis, 38
rotor-routing model, 239 South Africa, 94, 95
Roy, Bernard, 80 Spain, 94, 95
RSA, 151 spanning forest, 123
Runde, Volker, 72 spanning subgraph, 15
Russia, 1, 94, 95 spanning tree, 38, 106, 108, 111, 116–118,
146, 148, 152
sandpile model, 241 maximum, 145
saturated edge, 246 minimum, 116–118, 120–125, 146
scale-free network, 273, 278, 279 randomized construction, 146, 147, 151
scatterplot, 2, 116, 134 sparse graph, 259
Schulz, Charles M., 190 spectrum, 213
scientific collaboration, 254 Laplacian, 213
Sedgewick, Robert, 91, 96, 97, 102 stack, 66, 69, 70, 141
seg, 235 length, 66
selection sort, 96, 97 pop, 66, 69, 141, 143, 145
self-complementary graph, 34 push, 66, 69, 141, 143, 145
self-loop, 4 Stanley, Richard P., 66
separating set, 32, 208 star graph, 17
set, 54 state, 40, 41
n-set, 2 accepting, 40, 41
totally ordered, 154 diagram, 40, 41
Seymour final, 40, 41, 98
Paul, 37, 204 initial, 40, 41, 98
Shannon Steinhaus, H., 123
Claude E., 8, 111 Stevens, Brett, 102
Index 319
Stinson, Douglas R., 96 post-order, 141, 143, 151
string, 40, 130, 134 pre-order, 141, 142, 151
accepted, 42 treasure map, 1
empty, 138 tree, 14, 60, 105, 106, 114
Strogatz, Steven H., 264 2-ary, 127
strongly regular graph, 215 n-ary, 106
subgraph, 11, 15 binary, 106, 136, 138, 143
edge-deletion, 116, 117, 151 complete, 106, 111
subtree, 111, 143 depth, 106
left, 127, 145 directed, 106
right, 127, 145 expression, 106, 107
supergraph, 15 height, 106
Swift, Jonathan, 58 nonisomorphic, 106, 108, 109
Sydney, 94, 95 ordered, 106, 140
symbol, 134 recursive definition, 111, 112, 146
symbolic computation, 98 rooted, 14, 60, 69, 106, 110, 127
symmetric difference, 29, 30 subtree, 112
Szekeres, G., 204 traversal, 140, 142
triangle inequality, 72, 87, 198
Takaoka, Tadao, 192, 194 Tripathi, Amitabha, 25
Tanner graph, 20 trivial graph, 15, 106, 277
Tarjan, Robert Endre, 64 tuple, 2
Tarry, Gaston, 64
union
telegraph, 131
digraph, 115, 116
Thailand, 94, 95
union-find, 146
The Brain puzzle, 132
Unix, 106
Thoreau, Henry David, 8
unweighted degree, 7
threshold
USA, 94, 95, 253, 271
probability, 92
Through the Looking Glass, 2 value of flow, 246
Tokyo, 94, 95 van Oorschot, Paul C., 96
topology, 72 Vandermonde
total order, 154 Alexandre-Théophile, 192
Tower of Hanoi puzzle, 132 convolution, 192
trail, 12, 13 Vanstone, Scott A., 96
closed, 13 Vasudeva, Harkrishan L., 72
transition vending machine, 39–41
function, 40, 41 vertex, 3
table, 40 adjacent, 3
transition probability matrix, 236 child, 106, 110
transitive closure, 84, 85 cut, 32, 208
trapdoor function, 130 degree, 10
Trappe, Wade, 96 deletion, 31, 143
traveling salesman problem, 38, 53 deletion subgraph, 29, 202
traversal endpoint, 105, 106
bottom-up, 143, 144, 151 expansion, 220
in-order, 143–145, 151, 173 head, 4
level-order, 140–143, 151 internal, 106, 206
320 Index
isolated, 8, 54, 111 wheel graph, 29, 38, 152
isoperimetric number, 220 Whitney
leaf, 105, 106, 135, 143 Hassler, 206
multigraph, 6 inequality, 205
parent, 106 theorem, 210
root, 14, 106, 108, 110, 135 Wiener
set, 3 Harold, 82, 149
source, 74, 87 number, 82, 151
tail, 4 Williams, J. W. J., 155
union, 29 Wilson, Robin, 1
vertex connectivity, 202 wine, 212, 217, 220, 223–225
vertex-cut, 202 word, 40, 134
Vijay, Sujith, 25 World Wide Web, 270
Vitter
Yamada, Takeo, 104, 145
algorithm, 277
Yerger, Carl R., 152
Jeffrey Scott, 277
Vuillemin, Jean, 168 Zachary, Wayne W., 252
zero padding, 58
Wagner Zubrzycki, S., 123
conjecture, 37 Zwick, Uri, 194
Klaus, 37
walk, 11, 13
closed, 13
length, 11, 12
trivial, 11
Walsh transform, 215
Walther, Hansjoachim, 38
Warshall, Stephen, 80, 85
Washington DC, 94, 95
Washington, Lawrence C., 96
Watanabe, Kohtaro, 145
Watts, Duncan J., 264
Watts-Strogatz model, 265, 266, 277
Wegener, Ingo, 66
weight, 71, 72, 80, 117
correcting, 72, 73
function, 72, 86, 117
graph, 6
minimum, 76, 117, 120, 123
multigraph, 6
negative, 73, 77, 78
nonnegative, 71–74, 76, 85
path, 197
positive, 72
reweight, 85, 86
setting, 72, 73
unit, 71, 72
Weinberger, A., 117