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Removing Degeneracy May Require a Large Dimension Increase

Authors Ji{\v r}{\'\i} Matou{\v s}ek, Petr {\v S}kovro{\v n},

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                                 T HEORY OF C OMPUTING, Volume 3 (2007), pp. 159–177
                                             http://theoryofcomputing.org




Removing Degeneracy May Require a Large
          Dimension Increase∗
                                           Jiřı́ Matoušek                             Petr Škovroň
                                    Received: March 12, 2007; published: September 26, 2007.



        Abstract: Many geometric algorithms are formulated for input objects in general position;
        sometimes this is for convenience and simplicity, and sometimes it is essential for the al-
        gorithm to work at all. For arbitrary inputs this requires removing degeneracies, which has
        usually been solved by relatively complicated and computationally demanding perturbation
        methods.
            The result of this paper can be regarded as an indication that the problem of removing
        degeneracies has no simple “abstract” solution. We consider LP-type problems, a successful
        axiomatic framework for optimization problems capturing, e. g., linear programming and
        the smallest enclosing ball of a point set. For infinitely many integers D we construct a D-
        dimensional LP-type problem such that in order to remove degeneracies from it, we have
        to increase the dimension to at least (1 + ε)D, where ε > 0 is an absolute constant.
            The proof consists of showing that certain posets cannot be covered by pairwise disjoint
        copies of Boolean algebras under some restrictions on their placement. To this end, we
        prove that certain systems of linear inequalities are unsolvable, which seems to require
        surprisingly precise calculations.

   ∗ An extended abstract of this paper has appeared in Proceedings of European Conference on Combinatorics, Graph Theory

and Applications 2007 (Eurocomb), pp. 107–113.


ACM Classification: F.2.2
AMS Classification: 68U05, 06A07, 68R99
Key words and phrases: LP-type problem, degeneracy, general position, geometric computation, par-
tially ordered set


Authors retain copyright to their work and grant Theory of Computing unlimited rights
to publish the work electronically and in hard copy. Use of the work is permitted as
long as the author(s) and the journal are properly acknowledged. For the detailed
copyright statement, see http://theoryofcomputing.org/copyright.html.

c 2007 Jiřı́ Matoušek and Petr Škovroň                                                          DOI: 10.4086/toc.2007.v003a008
                                     J. M ATOU ŠEK AND P. Š KOVRO Ň

1    Introduction
Geometric computation and degeneracy. Many descriptions of algorithms in computational geom-
etry or in geometric optimization, as well as numerous proofs in discrete geometry, start with a sentence
similar to “Let us assume that the given points are in general position.” General position may mean
that no three among the points are collinear, or we may also require than no four are cocircular, etc.,
depending on the considered problem. Violations of general positions, such as three points on a line, are
referred to as degeneracies.
     Assuming the input to be nondegenerate (i. e., in general position) usually simplifies the description,
analysis, and implementation of a geometric algorithm significantly. For many algorithms, this assump-
tion can be avoided with some extra work and careful attention to detail (a case study, arguing in favor of
expending such extra work, is Burnikel et al. [4]). However, for some algorithms, the nondegeneracy as-
sumption is not only a convenient simplification, but rather an essential condition for correctness and/or
for running time analysis, which seems difficult to circumvent—we will mention an example below.
     General methods have been developed for removing degeneracies in geometric algorithms, based on
infinitesimal perturbations of the input (Edelsbrunner and Mücke [8], Yap [22], Emiris and Canny [9]).
Roughly speaking, the coordinates of each input object are changed by a suitable function of a real
parameter ε > 0, and the considered algorithm is executed with these new input objects, treating ε as
a formal quantity, smaller than any concrete nonzero real number occurring in the algorithm. These
approaches can actually be implemented, but they have several drawbacks: They slow down the compu-
tations significantly (typically by a large constant factor, but sometimes even much more), they increase
space requirements, and sometimes it may be difficult or impossible to reconstruct the correct result for
the original input from the result for the perturbed input—see [4] for a discussion.
     Removing degeneracies means “breaking ties” in some sense. Of course, the ties cannot be broken
arbitrarily, since geometric algorithms almost always depend on some kind of global consistency of the
input. Still, one might hope for some simpler, perhaps combinatorial, way of removing degeneracies.
     To illustrate what we have in mind, let us recall that the famous simplex method of linear program-
ming may also suffer from degeneracy—namely, for many pivoting rules the simplex method may get
into an infinite loop (to cycle) for certain highly degenerate inputs. (However, unlike degeneracy in
the geometric computations mentioned above, cycling of the simplex method is not an issue in prac-
tice.) There are two well-known pivot rules that provably avoid cycling: the lexicographic rule, which
is conceptually an infinitesimal perturbation, and Bland’s rule, which is a combinatorial rule working
solely with indices of variables and constraints, as opposed to geometric properties of the input. So
our question is, whether there is something like a general “Bland’s rule” that would allow one to avoid
degeneracies in (some interesting classes of) geometric algorithms.
     The present work can be regarded as an indication that a simple, general, and efficient combinatorial
method is unlikely to exist.


LP-type problems. We investigate the problem of removing degeneracies in a class of optimization
problems known as LP-type problems (or “generalized linear programming problems”). This axiomatic
framework, invented by Sharir and Welzl in 1992 [19], has become a well-established tool in the field
of geometric optimization; see [17, 1, 2, 3, 14, 5, 15, 11] for more applications and results on LP-type

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problems, as well as e. g. [21, 12, 10, 18] for the investigation of other, related frameworks.
     Once it is shown that a particular optimization problem is an LP-type problem and certain algo-
rithmic primitives are implemented for it, several efficient algorithms are immediately at disposal: the
Sharir–Welzl algorithm, two other randomized optimization algorithms due to Clarkson [7] (see [13, 6]
for a discussion of how it fits the LP-type framework), a deterministic version of it [6], and an algorithm
for computing the minimum solution that violates at most k of the given n constraints [16] (this is the
promised example of an algorithm where nondegeneracy appears crucial).
     An LP-type problem is given by a finite set H of constraints and a value w(G) ∈ R for every subset
G ⊆ H. Intuitively, w(G) is the minimum value of a solution that satisfies all constraints in G. As our
running example, we will use the problem of computing the smallest disk containing a given planar point
set. Here H is a finite point set in R2 and w(G) is the radius of the smallest circular disk that encloses
all points of G. The general definition is as follows:

Definition 1.1. An LP-type problem is a pair (H, w), where H is a finite set and w : 2H → R is a mapping
satisfying the following two conditions:1
      Monotonicity:        For all F ⊆ G ⊆ H we have w(F) ≤ w(G).
      Locality:            For all F ⊆ G ⊆ H and all h ∈ H,
                           if w(F) = w(G) = w(F ∪ {h}) then w(G ∪ {h}) = w(G).

    For the smallest enclosing disk problem, monotonicity is obvious, while verifying locality requires
the nontrivial but well known geometric result that the smallest enclosing disk is unique for every set.
    The most important parameter of an LP-type problem, essentially controlling the behavior of algo-
rithms dealing with the given problem, is the combinatorial dimension.

Definition 1.2. Let (H, w) be an LP-type problem and let G ⊆ H. A basis of G is any inclusion-minimal
subset B ⊆ G with w(B) = w(G). A set B ⊆ H is called a basis in (H, w) if it is a basis of some G ⊆ H.
The combinatorial dimension of (H, w) is the maximum cardinality of a basis.

    If (H, w) is a smallest enclosing disk problem, then the combinatorial dimension is at most 3 (since
for every point set G in the plane there is a subset B of at most 3 points of G such that G and B have
the same smallest enclosing disk). Similarly, a higher-dimensional version, the smallest enclosing ball
problem of a point set in Rd , has combinatorial dimension at most d + 1.

Degeneracy in LP-type problems. What should be considered a degeneracy in the smallest enclos-
ing disk problem? A reasonable answer is a subproblem with an “overdetermined” solution, which
means a set G whose minimum enclosing disk is determined by two distinct inclusion-minimal subsets
B1 , B2 ⊆ G. For example, B1 and B2 can be two different diametrical pairs determining the same disk.
Nondegeneracy for an arbitrary LP-type problem can be defined in a similar way [16].
   1 Actually, the usual definition of an LP-type problem is more general: the mapping w can also attain a special value −∞,

which is considered smaller than all real numbers, and for which the locality axiom is not required. Moreover, instead of R,
one can use an arbitrary linearly ordered set, but this brings nothing new, just sometimes a more convenient notation. We will
stick to the definition above since it is simpler, and it will be easy to check that the more general definition doesn’t change
anything in our result.


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                                                        d             c


                                                       a              b


       Figure 1: A degenerate LP-type problem where removing degeneracy increases dimension.


Definition 1.3. We call an LP-type problem (H, w) nondegenerate if w(B1 ) 6= w(B2 ) for any two distinct
bases B1 , B2 .

    Consequently, in a nondegenerate LP-type problem, every G ⊆ H has exactly one basis.2
    For removing degeneracies, we want to break the ties w(B1 ) = w(B2 ) by slightly modifying the
values of w, while retaining all strict inequalities among the original values:

Definition 1.4. An LP-type problem (H, w0 ) is a refinement of an LP-type problem (H, w) on the same
set of constraints if for all F, G ⊆ H with w(F) < w(G) we have w0 (F) < w0 (G).

    We thus formalize “removing degeneracies” of an LP-type problem (H, w) as the question of finding
a nondegenerate refinement of (H, w).
    At first sight it might seem that in order to produce a nondegenerate refinement, it should suffice to
impose some suitable linear order on every group of bases sharing the same value of w—perhaps one
could even take an arbitrary ordering.
    However, some thought reveals that things are not that simple. As was observed in [16], sometimes
we also have to create new bases, and even larger ones than those present in (H, w). Namely, consider
the smallest enclosing disk problem with H = {a, b, c, d} forming the vertices of a square (Figure 1).
The set H has two bases B1 = {a, c} and B2 = {b, d}, and the combinatorial dimension of the problem
is 2. We will refer to this particular 2-dimensional LP-type problem as the square example and denote
it by (Hsq , wsq ). It is easily checked (we will do so in Section 2) that any nondegenerate refinement has
dimension at least 3. Hence removing degeneracies necessarily increases the dimension by 2.
    In a preliminary report [20] containing some of the results of the present paper, an LP-type problem
was presented where removing degeneracy forces dimension increase by 2. Here we exhibit LP-type
problems where the required increase is arbitrarily large.

Theorem 1.5. There exists a positive constant ε > 0 such that for infinitely many values of D, there
is an LP-type problem of combinatorial dimension D, for which every nondegenerate refinement has
combinatorial dimension at least (1 + ε)D.

    The example of an LP-type problem as in the theorem is obtained by an “iterated join” of the square
example. We also show that an essentially equivalent example can be represented as a linear program in
the usual sense (a highly degenerate linear program).
   2 Another, seemingly weaker, notion of nondegeneracy naturally coming to mind is to require that every G ⊆ H has a unique

basis. However, any LP-type problem satisfies this latter condition can easily be converted into an LP-type problem of the same
dimension that is nondegenerate in the sense of Definition 1.3 [16]. So these definitions are essentially equivalent.


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    The result can also be understood as telling us that for degenerate LP-type problems, the combina-
torial dimension doesn’t convey a full “dimensionality” information about the problem. An alternative
dimension parameter might be the smallest possible dimension of a nondegenerate refinement; however,
this appears quite hard to determine.
    The main open question is, can the smallest possible dimension of a nondegenerate refinement be
bounded by some function of the dimension of the original degenerate LP-type problem? In particular,
does every 2-dimensional LP-type problem have a nondegenerate refinement of dimension bounded by
a universal constant? We suspect that it is not the case, but it seems that the methods of the present paper
are not sufficient to yield such a result. The structure of 2-dimensional LP-type problems, say, appears
both quite restricted and hard to describe, and at present we have no candidate for an LP-type problem
where removing degeneracies might require increasing the dimension by more than a small constant
factor.


2    Structure of nondegenerate LP-type problems
Let (H, w) be an LP-type problem. We consider the partially ordered set (poset) (2H , ⊆), a Boolean
algebra. For every x ∈ R, we define the set system Px = {G ⊆ H : w(G) = x}. The Px for all x ∈ R form
a partition of 2H . Monotonicity implies that Px has no “holes”: If F ⊂ M ⊂ G and x = w(F) = w(G),
then w(M) = x as well. The following lemma shows that for nondegenerate LP-type problems, each Px
is actually a copy of a Boolean algebra.

Lemma 2.1 (Cube lemma). Let (H, w) be a nondegenerate LP-type problem. For every x ∈ R with
Px 6= 0/ there exist two (uniquely determined) sets B,C ⊆ H such that Px = {F ⊆ H : B ⊆ F ⊆ C}. The
set B is the basis of all F ∈ Px .

    We call the set {F ⊆ H : B ⊆ F ⊆ C} a cube, we use the notation [B,C] for it, we call B the bottom
vertex and C the top vertex of the cube [B,C], and |C \ B| is the dimension of the cube.

Proof. We choose G ∈ Px arbitrarily, we let B be the basis of G, and we set
                                     n                             o
                               C = h ∈ H : w(B) = w(B ∪ {h}) .

We claim that this choice of B and C satisfies the desired conditions. First we prove that w(B) = w(C).
Letting C \ B = {c1 , . . . , cm }, we check by induction that w(B) = w(B ∪ {c1 , . . . , ci }), i = 0, 1, . . . , m.
Indeed, the induction step from i to i + 1 follows immediately from the locality axiom with F = B, G =
B ∪ {c1 , . . . , ci }, and h = ci+1 . Now when we have w(B) = w(C), monotonicity implies that [B,C] ⊆ Px .
    Now let us assume w(F) = w(B) for some F ⊆ H. Let B0 be a basis of F; we have w(B0 ) = w(F) =
w(B), and thus B = B0 by nondegeneracy. In particular, B ⊆ F. For every f ∈ F we have w(B) ≤
w(B ∪ { f }) ≤ w(F) = w(B), so w(B) = w(B ∪ { f }), and hence f ∈ C; thus F ⊆ C. Since F was an
arbitrary set in Px and we have obtained B ⊆ F ⊆ C, we conclude with Px ⊆ [B,C].
    The uniqueness of B and C follows from a simple observation that every cube has a unique top and
bottom vertex.

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                                                      abcd


                                     abc        acd          abd        bcd


                                           ac                      bd


                          Figure 2: The poset Pwsq (Hsq ) for the square example.


    To see how this lemma can be used, let us check the claim made in the introduction: every nonde-
generate refinement of the square example (Hsq , wsq ) has dimension at least 3. The poset Pwsq (Hsq ) of all
subsets of Hsq with the same smallest enclosing circle as that of Hsq consists of all subsets of {a, b, c, d}
containing {a, c} or {b, d}; see Figure 2.
    In any nondegenerate refinement, Pwsq (Hsq ) has to be expressed as a disjoint union of cubes, and if the
dimension of the refinement were 2, all of these cubes would have to have a 2-element set as the bottom
vertex. In order to cover {a, b, c, d}, we have to use a 2-dimensional cube, say [{a, c}, {a, b, c, d}]. To
cover the remaining sets {b, d}, {a, b, d}, and {b, c, d} by disjoint cubes, we must use at least one of the
0-dimensional (single-vertex) cubes [{a, b, d}, {a, b, d}] or [{b, c, d}, {b, c, d}] with a 3-element bottom
vertex. Therefore a combinatorial dimension of any nondegenerate refinement of (Hsq , wsq ) is at least 3.


3    The construction
We begin by defining a binary operation on LP-type problems.
Definition 3.1. Let (H1 , w1 ) and (H2 , w2 ) be LP-type problems, and assume H1 ∩ H2 = 0./ We define
a new LP-type problem, denoted by (H, w) = (H1 , w1 ) ∗ (H2 , w2 ) and called the join of (H1 , w1 ) and
(H2 , w2 ): H := H1 ∪ H2 and w(G) := w1 (G ∩ H1 ) + w2 (G ∩ H2 ) for all G ⊆ H.
Lemma 3.2. The join (H, w) = (H1 , w1 ) ∗ (H2 , w2 ) is indeed an LP-type problem, and dim(H, w) =
dim(H1 , w1 ) + dim(H2 , w2 ).
Proof. First we observe that if F ⊆ G and w(F) = w(G), then wi (F ∩ Hi ) = wi (G ∩ Hi ), i = 1, 2. Indeed,
since F ∩ Hi ⊆ G ∩ Hi , we have wi (F ∩ Hi ) ≤ wi (G ∩ Hi ), and to get equality of the sum, equality must
hold in both components.
    Now we verify the axioms for (H, w). Monotonicity is obvious, and for locality, let F ⊆ G ⊆ H and
h ∈ H satisfy w(F) = w(G) = w(F ∪ {h}). Supposing h ∈ H1 , we have w1 (F ∩ H1 ) = w1 (G ∩ H1 ) =
w1 ((F ∩ H1 ) ∪ {h}) by the observation above, and locality in (H1 , w1 ) yields w1 ((G ∩ H1 ) ∪ {h}) =
w1 (G ∩ H1 ). Then

        w(G ∪ {h}) = w1 ((G ∩ H1 ) ∪ {h}) + w2 (G ∩ H2 ) = w1 (G ∩ H1 ) + w2 (G ∩ H2 ) = w(G) ,

so (H, w) is indeed an LP-type problem.

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    Now we check dim(H, w) ≥ dim(H1 , w1 ) + dim(H2 , w2 ). Let Bi be a basis in (Hi , wi ) witnessing
dim(Hi , wi ). It suffices to check that B = B1 ∪ B2 is a basis in (H, w); that is, w(A) < w(B) for every
proper subset of B. Letting Ai = A ∩ Hi , we have Ai ⊆ Bi with at least one of the inclusions proper, say
A1 ⊂ B1 . Since B1 is a basis, we have w1 (A1 ) < w1 (B1 ) and w(A) < w(B) follows.
    For the opposite inequality dim(H, w) ≤ dim(H1 , w1 ) + dim(H2 , w2 ), we choose a basis B in (H, w)
with |B| = dim(H, w) and set Bi = B ∩ Hi . It suffices to check that Bi is a basis in (Hi , wi ). Let us consider
a proper subset A1 ⊂ B1 ; then

                       w1 (B1 ) + w2 (B2 ) = w(B1 ∪ B2 ) > w(A1 ∪ B2 ) = w1 (A1 ) + w2 (B2 ) ,

and we get w1 (A1 ) < w1 (B1 ) as needed. The lemma is proved.




The example. For the proof of Theorem 1.5 we define, for a natural number m, an LP-type problem
Lm as the m-fold join of the square example (Hsq , wsq ). More formally, we choose distinct elements
a1 , . . . , am , b1 , . . . , bm , c1 , . . . , cm , d1 , . . . , dm , we let Hi = {ai , bi , ci , di }, and we let wi : Hi → R be a
“copy” of the value function wsq from the square example, defined on Hi . We let

                                        Lm = (H, w) = (H1 , w1 ) ∗ · · · ∗ (Hm , wm )

(we note that the operation of join is clearly associative). We have |H| = 4m and by the above lemma,
Lm is an LP-type problem of combinatorial dimension D = 2m. It is easy to check that by taking a join
of m suitable nondegenerate refinements of the square example we obtain a nondegenerate refinement
of (H, w) of combinatorial dimension 3m.
    We want to bound from below the dimension of any nondegenerate refinement of Lm . Similar to
the warm-up argument for (Hsq , wsq ), any nondegenerate refinement L0 = (H, w0 ) of Lm of dimension
D0 yields a covering of the poset Pw(H) = {G ⊆ H : w(G) = w(H)} by disjoint cubes [B j ,C j ], where
each bottom vertex B j satisfies |B j | ≤ D0 . We will deal with this combinatorial problem in the next two
sections.


The case m = 2.        The 4-dimensional LP-type problem L2 is analyzed in [20], and it is shown that
every nondegenerate refinement has dimension at least 6. The corresponding poset Pw(H) is illustrated
in Figure 3. Interestingly, this Pw(H) does admit a cover by disjoint cubes with bottom vertices of
cardinality at most 5; see Figure 4. However, the covers corresponding to a nondegenerate refinement
have to satisfy an additional condition, called acyclicity, and a case analysis in [20] verifies that every
acyclic cover must have a bottom vertex of cardinality 6 or larger. Here we won’t define acyclicity; we
just remark that arbitrary covers by disjoint cubes correspond to nondegenerate violator spaces, which is
a generalization of LP-type problems investigated in [11]. One can thus say that L2 has a 5-dimensional
nondegenerate refinement in the realm of violator spaces, but not in the realm of LP-type problems. On
the other hand, the subsequent proof of Theorem 1.5 doesn’t use acyclicity in any way and thus it applies
equally well to violator spaces.

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                                                                                            8



                                                                                            7



                                                                                            6



                                                                                            5



                                                                                            4


Figure 3: The poset Pw(H) for m = 2. The numbers in the right indicate sizes of the corresponding sets.



                                                                                            8



                                                                                            7



                                                                                            6



                                                                                            5



                                                                                            4


Figure 4: A covering of Pw(H) by disjoint cubes with all bottom vertices of size at most 5; a 4-
dimensional cube is marked by circles around its vertices.



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4    Setting up a linear system
The basic strategy for the proof of Theorem 1.5 is simple. Let Lm = (H, w) be the example constructed
above and let us suppose that the poset P := Pw(H) ⊂ 2H can be covered by disjoint cubes [B j ,C j ]
with |B j | ≤ D0 . Since dim(H, w) = D = 2m and all bases of (H, w) have exactly this size, we have
2m ≤ |B j | ≤ |C j | ≤ |H| = 4m for all j. Let xd,k denote the number of cubes with |B j | = 2m + d and
|C j | = 2m + k, d ≤ ∆ := D0 − 2m, d ≤ k ≤ 2m. A cube [B j ,C j ] with |B j | = 2m + d and |C j | = 2m + k
contains sets of cardinality 2m + `, d ≤ ` ≤ k, and the number of sets of this cardinality in [B j ,C j ] equals
 k−d                                                                            a
                                                                                 
 `−d (this formula is actually valid for all ` if we adopt the convention that b = 0 for b < 0 or b > a).
If we let
                                  F(m, `) = |{G ∈ P : |G| = 2m + `}| ,
we get that the xk,d have to satisfy the following system of linear equations:
                           ∆      2m         
                                        k−d
                         ∑ ∑                   xd,k = F(m, `) ,          ` = 0, 1, . . . , 2m .                            (4.1)
                         d=0 k=max(d,`) ` − d

     We are going to prove that with ∆ = dεDe, where ε is a sufficiently small positive constant, this
system of equations for variables xk,d has no nonnegative real solution, provided that m is sufficiently
large.
     To see that an approach based on counting sets of individual cardinalities may help us to prove
nonexistence of the covering of P, note that already the proof in the end of Section 2 may be rephrased
in terms of counting. In the poset in Figure 2, the vector of numbers of sets of cardinality 2, 3, and
4 is (2, 4, 1). However, this vector cannot be obtained as a nonnegative linear combination of vectors
(1, 0, 0), (1, 1, 0), and (1, 2, 1), which give numbers of sets of the respective cardinalities in cubes with
the allowed cardinality of the bottom vertex.
     First we evaluate F(m, `).
Lemma 4.1. We have                                                     
                                                            m
                                 F(m, `) = ∑                             2m+`−3s ,
                                               s   s, ` − 2s, m − ` + s
                                                                                   n
                                                                                                = k1 !kn!2 !k3 ! is a multinomial
                                                                                            
with the sum being over all s with 0 ≤ 2s ≤ ` and s ≥ ` − m (here              k1 ,k2 ,k3
coefficient, k1 + k2 + k3 = n).
Proof. First we observe, reasoning as in the proof of Lemma 3.2, that a set B ⊆ H is a basis of H in
Lm if and only if each Bi = B ∩ Hi is a basis of Hi in (Hi , wi ). Hence the bases of H are the sets B with
B ∩ Hi = {ai , ci } or B ∩ Hi = {bi , di } for all i = 1, 2, . . . , m. A set G ⊆ H is in P iff it contains at least one
of these bases; i. e., if it contains at least one of the pairs {ai , ci }, {bi , di } for all i.
    For G ∈ P of cardinality 2m + ` let sr = |{i ∈ {1, 2, . . . , m} : |G ∩ Hi | = r}|, r = 2, 3, 4. We have
s2 + s3 + s4 = m and 2s2 + 3s3 + 4s4 = |G| = 2m + `. Calculation shows that s2 = m − ` + s4 and s3 =
` − 2s4 .
    For counting the number of
                               m
                                   possible ways of choosing G, we first fix s = s4 . Then s2 and s3 are fixed
as well, and there are s2 ,s3 ,s4 ways to choose the indices i contributing to each sr (in other words, to
choose which are the Hi where G takes 2, 3, or 4 elements, respectively). Knowing that |G ∩ Hi | = 2,

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there are two possibilities for G ∩ Hi , for |G ∩ Hi | = 3 we have 4 possibilities, and for |G ∩ Hi | there is just
one possibility. Therefore, once |G ∩ Hi | has been fixed for all i, there are 2s2 · 4s3 = 2m+`−3s4 possibilities
for G. Summation over s = s4 yields the statement of the lemma (the conditions on the range of s in the
summation correspond to the obvious restrictions s2 , s3 , s4 ≥ 0).


5    Unsolvability of the linear system
We recall that for finishing the proof of Theorem 1.5, it suffices to show that for ∆ := d2εme and m
sufficiently large, the linear system (4.1) has no nonnegative solution x = (xd,k )∆d=0 2m
                                                                                        k=d .
    Before starting with the formal proof, which is a sequence of somewhat frightening calculations,
we say a few words about how it was found. We started by testing the solvability for concrete values of
parameters via linear programming. We used the function LinearProgramming in Mathematica, which
uses arbitrary precision arithmetic and computes the solution exactly; this allowed us to deal with m up
to about 1000 (other LP solvers we tried failed for large instances because of insufficient accuracy). By
the Farkas lemma, the unsolvability is always witnessed by a linear combination of the equations that
has nonnegative coefficients on the left-hand side and negative right-hand side. By minimizing the sum
of absolute values of (suitably normalized) coefficients providing such a linear combination, we found
that the unsolvability was witnessed, in all examples we tried, by a linear combination of only 3 of the
equations. For simplifying the analytic approach, we then tried 3 consecutive equations, and found that
such combinations work as well, provided that the index of the middle equation is chosen in a suitable
range. These numerical results encouraged us to try finer and finer estimates, until we finally reached
the following proof.

Proof of the unsolvability of (4.1).         We set, somewhat arbitrarily, t = m/2, assuming m even (we
suspect that t = τm for any fixed τ ∈ (0, 1) would work, but we haven’t checked). We will show that for
sufficiently large m already the system of the three consecutive equations with ` = t − 1, t, and t + 1 has
no nonnegative solution. To this end, we find a linear combination of these three equations, with suitable
coefficients α, β , γ, such that the resulting equation has all coefficients on the left-hand side nonnegative,
while the right-hand side is strictly negative. We will assume that β is negative and we normalize the
coefficients so that β = −1 (we need not justify this assumption since we are free to choose α, β , γ as
we wish). Explicitly, to have the coefficient of the variable xd,k in the resulting equation nonnegative,
we need that the following system of (∆ + 1)(2m − ∆/2 + 1) inequalities is satisfied:
                                                    
                    k−d            k−d             k−d
              α               −            +γ               ≥ 0 , 0 ≤ d ≤ ∆ , d ≤ k ≤ 2m .                (5.1)
                 t −d −1           t −d          t −d +1

To have right-hand side strictly negative, we need the following inequality:

                                αF(m,t − 1) − F(m,t) + γF(m,t + 1) < 0 .                                     (5.2)

    Our basic intuition behind the proof is a “continuous” one: For m large and fixed, the left-hand side
of (5.2) is something like a “weighted second derivative” of F(m,t) according to t, while on the left-
hand side of (5.1) we have the same kind of the weighted second derivatives of the binomial coefficients.

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                 R EMOVING D EGENERACY M AY R EQUIRE A L ARGE D IMENSION I NCREASE

So our goal is to prove that the graph of F(m,t) “bends less” than the graph of any of the binomial
coefficients involved, and hence F cannot be built as a positive linear combination of the binomial
coefficients.
    However, this initial intuition is a quite rough one, since the choice of suitable α and γ turns out to
be surprisingly subtle. Namely, we need to choose α = α0 + α1 /t and γ = γ0 + γ1 /t, where
                               √                              √
                                 10 + 2                         10 − 2
                        α0 =             ≈ 1.29057 , γ0 =              ≈ 0.193713
                                   4                             6
are uniquely determined real constants and α1 , γ1 are constants in certain ranges. For concreteness we
set α1 = 1 and γ1 = 1/8.
    We get (5.1) from the following lemma:
Lemma 5.1. There is a positive constant ε > 0 such that, with the above choice of t, α, and γ, Equa-
tion (5.1) holds for all d ≤ ∆ = d2εme and for all k ≥ d, provided that m, and hence t, are sufficiently
large.
                                                                                   y
                                                                                      y       y
                                                                                                   
Proof. We use the substitution y = k − d and x = t − d. We thus want to show α x−1     − x + γ x+1   ≥ 0.
    For y < x − 1 all three terms are 0, and so we may assume y ≥ x − 1. We rewrite the left-hand side to
                        y!                                                              
                                     αx(x + 1) − (x + 1)(y − x + 1) + γ(y − x + 1)(y − x) .
              (x + 1)!(y − x + 1)!
Let us denote by f (α, γ, y, x) the expression in parentheses; we want to show that it is nonnegative.
    Let us choose constants α10 < α1 and γ10 < γ1 . Assuming ε in the lemma sufficiently small, we have d
sufficiently small compared to x, and hence α = α0 + α1 /(x + d) ≥ α0 + α10 /x and γ = γ0 + γ1 /(x + d) ≥
γ0 + γ10 /x.
    Since f is nondecreasing in α and in γ (for the relevant y and x), it suffices to check that
                                                    α10       γ0
                                        f (α0 +         , γ0 + 1 , y, x) ≥ 0 ,
                                                    x          x
and we will verify this for all sufficiently large real x and all real y. One of the properties of α0 and
γ0 needed here is α0 γ0 = 1/4. Things can be simplified a little by the substitution y = x(z + 1). Then
f (α0 + α10 /x, γ0 + γ10 /x, x(z + 1), x) is a polynomial in x and z. For x fixed it is a quadratic polynomial in
z, and the coefficient of z2 is γ0 x2 + γ10 x > 0 (this calculation and the subsequent ones were done using
Mathematica). Therefore, it has a unique minimum, which can be found by setting the first derivative
(according to z) to 0. This minimum occurs at
                                                     x2 + (1 − γ0 )x − γ10
                                     z0 = z0 (x) =                         .
                                                        2x(γ0 x + γ10 )
                                α0       γ0
Substituting this into f (α0 + x1 , γ0 + x1 , z(x + 1), x) yields a function of x of the form
                                     −γ0 − 2γ02 + 4α10 γ02 + γ10
                                                                 x + O(1) ,
                                               4γ02
the O(.) notation referring to x → ∞. Calculation shows that the coefficient of x is a positive real number
(for α10 and γ10 sufficiently close to α1 and γ1 , respectively). Hence f is indeed positive for the considered
values of the variables.

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                                             J. M ATOU ŠEK AND P. Š KOVRO Ň

Remark. It is easy to check that if α, γ are positive constants, then the inequality f (α, γ, x, y) ≥ 0
holds for all y and all sufficiently large x if and only if αγ > 14 . However, for such α and γ (5.2) fails.
We are thus forced to choose α and γ depending on x such that αγ → 14 as x → ∞.

     We now proceed to establish (5.2). We set
                                                                   
                                                         m
                                     Q(m,t, s) =                     2m+t−3s ,
                                                 s,t − 2s, m − t + s

so that F(m,t) = ∑s Q(m,t, s). First we look for the s maximizing Q(m,t, s). Let

                                                Q(m,t, s)      (t − 2s + 1)(t − 2s + 2)
                                r(m,t, s) =                  =
                                               Q(m,t, s − 1)        8s(m − t + s)

be the ratio of two consecutive terms. As a function of s it is decreasing, and so Q(m,t, s) is maximum
for the largest s with r(m,t, s) ≥ 1.
    We stick to our choice t = m/2. It is more convenient to use t as a parameter;  √ let us write r̃(t, s) =
r(2t,t, s) and Q̃(t, s) = Q(2t,t, s), and let us note that m −t = t. If we let σ = ( 10 − 3)/2 ≈ 0.0811388
be the positive root of the equation (1 − 2σ )2 = 8σ (1 + σ ) and s0 = bσtc, then

                                                     (t − 2s0 + 1)(t − 2s0 + 2)
                                    r̃(t, s0 ) =
                                                          8s0 (m − t + s0 )
                                                            (1 − 2σ + O(t −1 ))2
                                               =
                                                     8(σ + O(t −1 ))(1 + σ + O(t −1 ))
                                                     (1 − 2σ )2
                                               =                 + O(t −1 )
                                                     8σ (1 + σ )
                                               = 1 + O(t −1 ) .

     Next, we need an estimate on the rate of decrease of Q̃(t, s0 + a) as |a| increases.

                      4
Lemma 5.2. Let c0 = 1−2σ + σ1 + 1+σ
                                 1
                                    ≈ 18.0244. Suppose that a = o(t 2/3 ). Then

                                          Q̃(t, s0 + a)                   2
                                                        = (1 + o(1))e−c0 a /2t ,
                                            Q̃(t, s0 )

where o(.) refers to t → ∞ and the convergence is uniform in a.3

   3 That is, there exists a function ζ : (0, ∞) → [0, ∞) with ζ (t) → 0 as t → ∞ such that


                                                             Q̃(t, s0 + a)
                                            1 − ζ (t) ≤                   2    ≤ 1 + ζ (t)
                                                          Q̃(t, s0 )e−c0 a /2t

for all relevant a and t.


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                R EMOVING D EGENERACY M AY R EQUIRE A L ARGE D IMENSION I NCREASE

Proof. We will be summing over j = 1, 2, . . . , a in the proof. Let us write ξ = j/t; thus ξ = o(1). We
have, using 1 + x = ex + O(x2 ),

                                                     r̃(t, s0 + j)
                    r̃(t, s0 + j) = (1 + O(t −1 ))
                                                        r̃(t, s0 )
                                                                                     
                                                        1 − t−2s2 0j +1 1 − t−2s2 0j +2
                                    = (1 + O(t −1 ))                            
                                                              1 + sj0 1 + t+sj 0
                                                                      2
                                                                            2
                                                             1 −          ξ
                                    = (1 + O(t −1 ))               1−2σ
                                                       1 + σ1 ξ 1 + 1+σ      1
                                                                                
                                                                               ξ
                                                                       2
                                                                                     2
                                                                 e− 1−2σ ξ + O(ξ 2 )
                                    = (1 + O(t −1 ))  1                   1            
                                                        e σ ξ + O(ξ 2 ) e 1+σ ξ + O(ξ 2 )
                                                                                  2       1    1
                                    = (1 + O(t −1 ))(1 + O(ξ 2 ))e−( 1−2σ + σ + 1+σ )ξ
                                    = (1 + O(t −1 ) + O(ξ 2 ))e−c0 ξ .

Then, using ln(1 + x) = x + O(x2 ),
                                                   a
                              Q̃(t, s0 + a)
                         ln
                                Q̃(t, s0 )
                                              =   ∑ ln r̃(t, s0 + j)
                                                  j=1
                                                   a                                         3
                                                             c0 j            a               a
                                              =        ∑−     t
                                                                        +O
                                                                              t
                                                                                      +O
                                                                                               t2
                                                       j=1

                                                  c0 a2    a a3
                                                                                 
                                              = −       +O  +                         .
                                                   2t      t t2
The lemma follows.

   Next, we consider the expression D̃(t, s) = αQ(m,t − 1, s) − Q(m,t, s) + γQ(m,t + 1, s) with m = 2t,
α = α0 + α1 /t, and γ = γ0 + γ1 /t as above. The idea is to show that for s close to s0 we have D̃(t, s)
negative, while for s further from s0 it can be positive but it is sufficiently small compared to −D̃(t, s0 ).
Again, the calculation has to be done rather precisely in order to work.
Lemma 5.3. Let us suppose that a = o(t), and let s0 = bσtc be as above. Then
                                                              C  c1 2          
                              D̃(t, s0 + a) = Q̃(t, s0 + a)         a − 1 + o(1) ,
                                                              t t
where C is a certain positive constant whose value will not be important,
                                       √
                           c1 = (14584 10 + 46192)/5877 ≈ 15.70710522 ,

the o(.) notation refers to t → ∞, and the convergence is uniform in a.

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                                        J. M ATOU ŠEK AND P. Š KOVRO Ň

Proof. Similar to the proof of Lemma 5.1 we rewrite
                                                             1
                          D̃(t, s) = Q̃(t, s) ·                            g(α, γ,t, s) ,
                                                  2(t + 1 − 2s)(t + s + 1)

with g(α, γ,t, s) = α(t − 2s + 1)(t − 2s) − 2(t − 2s + 1)(t + s + 1) + 4γ(t + s)(t + s + 1). With the constant
σ as above, g(α0 + α1 /t, γ0 + γ1 /t,t, σt) becomes a polynomial in t, which is a√priori quadratic, √ but the
constants  α   and γ   are chosen so that the coefficient at t 2 , which equals 14 − 5 10 + (26 − 8 10)α +
       √ 0           0                                                                                     0
(11 − 2 10)γ0 , vanishes. (This, together with α0 γ0 = 41 , are the two conditions that uniquely determine
                                                                              √
α0 and γ0 .) The coefficient of the linear term equals −c2 = (191 − 62 10)/8 ≈ −0.632652 (and thus
g(α, γ,t, s) is indeed negative and of order t for s sufficiently near to σt).
    More quantitatively, expanding and simplifying gives

                  g(α0 + α1 /t, γ0 + γ1 /t,t, σt + b) = −c2t + c3 b2 + O(b2 /t + b + 1)
                 √
with c3 = (14 + 5 10)/3. For a = b + σt − s0 = b + σt − bσtc ≤ b + 1 we then obtain

                   g(α0 + α1 /t, γ0 + γ1 /t,t, s0 + a) = −c2t + c3 a2 + O(a2 /t + a + 1) .

Therefore, using a = o(t), we arrive at

                                                       −c2t + c3 a2 + O(a2 /t + a + 1)
                      D̃(t, s0 + a) = Q̃(t, s0 + a) ·
                                                          2(t + 1 − 2s)(t + s + 1)
                                                          2               
                                                       C c3 a
                                     = Q̃(t, s0 + a) ·          − 1 + o(1)
                                                       t    c2t

as required.


    We are ready to prove (5.2). For our choice of α, γ, and t we have

                        αF(m,t − 1) − F(m,t) + γF(m,t + 1) = ∑ D̃(t, s0 + a) .
                                                                          a

For concreteness let us set a0 = t 3/5 . We will show that

                                        ∑ D̃(t, s0 + a) ≤ − √t Q̃(t, s0 )
                                                                  δ
                                      |a|≤a0

for a constant δ > 0. Now for a > a0 we have

                  |D̃(t, s0 + a)| ≤ α Q̃(t − 1, s0 + a) + Q̃(t, s0 + a) + γ Q̃(t + 1, s0 + a) .

We have Q(t − 1, s0 + a) ≤ Q̃(t − 1, s0 + a0 ), which is smaller than Q̃(t − 1, s0 ) by a factor exponential
in t (see Lemma 5.2). A similar argument applies for t and t + 1 and for a < −a0 and thus the sum over
|a| > a0 is negligible.

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                   R EMOVING D EGENERACY M AY R EQUIRE A L ARGE D IMENSION I NCREASE

    Combining Lemmas 5.2 and 5.3, we get that for |a| ≤ a0 we have

                                                                                            c1 a2
                                                                                                             
                                               C                  2
                     D̃(t, s0 + a) = Q̃(t, s0 ) (1 + φt (a))e−c0 a /2t                            − 1 + ψt (a) ,
                                               t                                              t

where φt (a) and ψt (a) are some functions converging to 0 as t → ∞, uniformly in a.
   We will show that

                                                         c1 a2              √
                                                                      
                          ∑ (1 + φt (a))e −c0 a2 /2t
                                                      1−
                                                           t
                                                               − ψt (a) = Ω( t ) .                                         (5.3)
                        |a|≤a       0


Let us fix an arbitrarily small ν > 0 and let us assume that t has been chosen so large that |φt (a)| ≤ ν,
|ψt (a)| ≤ ν for all a. Then the left-hand side of (5.3) is bounded from below by

           ∑ e−c a /2t (1 − c1 a2 /t) − ∑ (1 + |φt (a)|)e−c a /2t |ψt (a)| − ∑ |φt (a)|e−c a /2t
                         2                                                          2                                  2
                     0                                                          0                                  0

          |a|≤a0                                   |a|≤a0                                             |a|≤a0


                                                               c1 a2
                                                                    
                                          ∑    −c0 a2 /2t
                                                                       − 3ν ∑ e−c0 a /2t .
                                                                                    2
                                    ≥         e            1 −
                                        |a|≤a
                                                                 t         |a|≤a
                                            0                                                0


By basic properties of Riemann integration and uniform continuity arguments it is routine to check that
both of these sums converge to the corresponding integrals as t → ∞. So it suffices to bound from below
                                                Z a0                          Z a0
                                                            2            c1                      2
                                    (1 − 3ν)           e−c0 a /2t da −                  a2 e−c0 a /2t da .
                                                 −a0                      t    −a0


Since a20 /t = t 1/5 → ∞ as t → ∞ and the integrands decrease exponentially in a2 /t, we make only a
negligible error by taking both integrals from −∞ to ∞. We have
                                          Z ∞                                             √
                                                      2
                                                e−c0 a /2t da = (1 − 3ν) 2πt/c0 ≈ 0.590419 t
                                                                        p
                             (1 − 3ν)
                                           −∞

while                               Z ∞
                               c1                  2           √   −3/2 √            √
                                          a2 e−c0 a /2t da = c1 2πc0     t ≈ 0.514513 t .
                                t    −∞

This finally proves (5.2).


6    A geometric representation by a linear program
It turns out that an LP-type problem L̂m = (H, ŵ), which is similar to Lm and which can also be used as
an example establishing Theorem 1.5, can be represented as a linear program. To see that our proof of
Theorem 1.5 works for L̂m as well, it will be enough to verify that its poset Pŵ(H) of maximum-weight
sets is isomorphic to Pw(H) of Lm , and this will follow from the discussion below.

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                                         J. M ATOU ŠEK AND P. Š KOVRO Ň

                                                                                   b   d

                                                  d
                                            b         z      a c
                                     c
                                 a




                                 xabcd


                                                       xab
                                                                                           x
                                                      0
                                            xbc                              xad
                                     xcd
                          y


             Figure 5: A linear program in R3 essentially representing the square example.


    We begin by setting up the following linear program with variables x, y, z (η > 0 is a very small
positive real number):
                                 minimize z + ηy + η 2 x subject to
                                     a:          x + 4y − 2z       ≤   1
                                     b:         3x + 8y + 2z       ≤   5
                                     c:         3x − 8y + 2z       ≤   −3
                                     d:         −x − 4y − 2z       ≤   −3
                                                       x, y, z     ≥   0 .
The corresponding LP-type problem (Hsq , ŵsq ) has the set Hsq = {a, b, c, d} of four constraints corre-
sponding to the four inequalities of the linear program. The value ŵsq (G) of any subset G ⊆ Hsq is the
minimum of the linear program where the constraints of Hsq \ G have been deleted (we stress that the
implicit nonnegativity constraints x, y, z ≥ 0 are always present, even for G = 0).  / In this way, ŵsq (G) is
well defined for every G.
     The linear program is illustrated in Figure 5. For better visualization, the picture shows the unit
cube [0, 1]3 , and intersections of the bounding planes of the constraints with the facets x = 0 and x = 1
of the cube. The minimum of the linear programs containing both the constraints a and c or both the
constraints b and d is attained at the point xabcd = (0, 1/2, 1/2); thus, ŵsq (Hsq ) = 1/2. It can be checked
that for every subset G of constraints containing neither {a, c} nor {b, d}, the minimum is attained at a
point with z = 0, and thus with ŵsq < 1/2 (the picture shows the minima for all G of cardinality 2). Thus
L̂ is a 2-dimensional LP-type problem with the poset Pŵsq (Hsq ) isomorphic to Pwsq (Hsq ) for the square
example.

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                  R EMOVING D EGENERACY M AY R EQUIRE A L ARGE D IMENSION I NCREASE

    Next, we observe that if (H, w) is an LP-type problem corresponding to a linear program with vari-
ables x1 , . . . , xn and with objective min ∑ ci xi , and (H 0 , w0 ) is an LP-type problem corresponding to a
linear program with variables x10 , . . . , xn0 and with objective min ∑ c0i xi0 , then the join (H, w) ∗ (H 0 , w0 ) cor-
responds to the linear program obtained by putting the constraints of both linear programs together and
with objective min(∑ ci xi + ∑ c0i xi0 ). Indeed, it suffices to check that the value function in (H, w) ∗ (H 0 , w0 )
coincides with the value function obtained from the combined linear program, and this is immediate. In
particular, the m-fold join L̂m of m disjoint copies of (Hsq , ŵsq ) corresponds to the following linear
program in 3m variables:

                               minimize ∑m                   2
                                           i=1 (zi + ηyi + η xi ) subject to
                                                               
                                     xi + 4yi − 2zi ≤ 1       
                                    3xi + 8yi + 2zi ≤ 5 
                                                               
                                                               
                                    3xi − 8yi + 2zi ≤ −3            i = 1, 2, . . . , m .
                                   −xi − 4yi − 2zi ≤ −3 
                                                               
                                                               
                                                               
                                            xi , yi , zi ≥ 0
                                                               

We could have presented the example for Theorem 1.5 in this form, but we find the abstract construction
of join more transparent.

Acknowledgment
We would like to thank the anonymous referees for careful reading and a number of valuable comments.


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AUTHORS
      Jiřı́ Matoušek [About the author]
      Department of Applied Mathematics and
      Institute of Theoretical Computer Science (ITI)
      Charles University, Malostranské nám. 25
      118 00 Praha 1, Czech Republic
      matousek kam mff cuni cz
      http://kam.mff.cuni.cz/~matousek/

      Petr Škovroň [About the author]
      Department of Applied Mathematics
      Charles University, Malostranské nám. 25
      118 00 Praha 1, Czech Republic
      xofon kam mff cuni cz
      http://kam.mff.cuni.cz/~xofon/


ABOUT THE AUTHORS
      J I Ř Í M ATOU ŠEK has been at Charles University in Prague steadily from 1981 on, although
              his role there has somehow progressed from student to professor. He has also often
              enjoyed the hospitality of nice people at many places, most extensively in Berlin, Ger-
              many and Zürich, Switzerland. He has written papers and textbooks in various areas
              of discrete mathematics and theoretical computer science, such as discrete and compu-
              tational geometry, low-distortion embeddings of metric spaces, geometric discrepancy,
              and topological combinatorics, often just out of curiosity. From his youth he has re-
              tained an aversion towards questionnaires and biographical sketches, although many
              other things he has already forgotten.

      P ETR Š KOVRO Ň is just finishing his Ph. D. in the Department of Applied Mathematics of
          Charles University, Prague under the supervision of Jiřı́ Matoušek. He is about to enter
          a computer industry career. In his free time he likes to play jazz on the flute.


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