from __future__ import print_function, division
from sympy.core import Basic, Tuple
from sympy.sets import FiniteSet
from sympy.core.compatibility import as_int, range
from sympy.combinatorics import Permutation as Perm
from sympy.combinatorics.perm_groups import PermutationGroup
from sympy.utilities.iterables import (minlex, unflatten, flatten)
rmul = Perm.rmul
[docs]class Polyhedron(Basic):
"""
Represents the polyhedral symmetry group (PSG).
The PSG is one of the symmetry groups of the Platonic solids.
There are three polyhedral groups: the tetrahedral group
of order 12, the octahedral group of order 24, and the
icosahedral group of order 60.
All doctests have been given in the docstring of the
constructor of the object.
References
==========
http://mathworld.wolfram.com/PolyhedralGroup.html
"""
_edges = None
def __new__(cls, corners, faces=[], pgroup=[]):
"""
The constructor of the Polyhedron group object.
It takes up to three parameters: the corners, faces, and
allowed transformations.
The corners/vertices are entered as a list of arbitrary
expressions that are used to identify each vertex.
The faces are entered as a list of tuples of indices; a tuple
of indices identifies the vertices which define the face. They
should be entered in a cw or ccw order; they will be standardized
by reversal and rotation to be give the lowest lexical ordering.
If no faces are given then no edges will be computed.
>>> from sympy.combinatorics.polyhedron import Polyhedron
>>> Polyhedron(list('abc'), [(1, 2, 0)]).faces
{(0, 1, 2)}
>>> Polyhedron(list('abc'), [(1, 0, 2)]).faces
{(0, 1, 2)}
The allowed transformations are entered as allowable permutations
of the vertices for the polyhedron. Instance of Permutations
(as with faces) should refer to the supplied vertices by index.
These permutation are stored as a PermutationGroup.
Examples
========
>>> from sympy.combinatorics.permutations import Permutation
>>> Permutation.print_cyclic = False
>>> from sympy.abc import w, x, y, z
Here we construct the Polyhedron object for a tetrahedron.
>>> corners = [w, x, y, z]
>>> faces = [(0, 1, 2), (0, 2, 3), (0, 3, 1), (1, 2, 3)]
Next, allowed transformations of the polyhedron must be given. This
is given as permutations of vertices.
Although the vertices of a tetrahedron can be numbered in 24 (4!)
different ways, there are only 12 different orientations for a
physical tetrahedron. The following permutations, applied once or
twice, will generate all 12 of the orientations. (The identity
permutation, Permutation(range(4)), is not included since it does
not change the orientation of the vertices.)
>>> pgroup = [Permutation([[0, 1, 2], [3]]), \
Permutation([[0, 1, 3], [2]]), \
Permutation([[0, 2, 3], [1]]), \
Permutation([[1, 2, 3], [0]]), \
Permutation([[0, 1], [2, 3]]), \
Permutation([[0, 2], [1, 3]]), \
Permutation([[0, 3], [1, 2]])]
The Polyhedron is now constructed and demonstrated:
>>> tetra = Polyhedron(corners, faces, pgroup)
>>> tetra.size
4
>>> tetra.edges
{(0, 1), (0, 2), (0, 3), (1, 2), (1, 3), (2, 3)}
>>> tetra.corners
(w, x, y, z)
It can be rotated with an arbitrary permutation of vertices, e.g.
the following permutation is not in the pgroup:
>>> tetra.rotate(Permutation([0, 1, 3, 2]))
>>> tetra.corners
(w, x, z, y)
An allowed permutation of the vertices can be constructed by
repeatedly applying permutations from the pgroup to the vertices.
Here is a demonstration that applying p and p**2 for every p in
pgroup generates all the orientations of a tetrahedron and no others:
>>> all = ( (w, x, y, z), \
(x, y, w, z), \
(y, w, x, z), \
(w, z, x, y), \
(z, w, y, x), \
(w, y, z, x), \
(y, z, w, x), \
(x, z, y, w), \
(z, y, x, w), \
(y, x, z, w), \
(x, w, z, y), \
(z, x, w, y) )
>>> got = []
>>> for p in (pgroup + [p**2 for p in pgroup]):
... h = Polyhedron(corners)
... h.rotate(p)
... got.append(h.corners)
...
>>> set(got) == set(all)
True
The make_perm method of a PermutationGroup will randomly pick
permutations, multiply them together, and return the permutation that
can be applied to the polyhedron to give the orientation produced
by those individual permutations.
Here, 3 permutations are used:
>>> tetra.pgroup.make_perm(3) # doctest: +SKIP
Permutation([0, 3, 1, 2])
To select the permutations that should be used, supply a list
of indices to the permutations in pgroup in the order they should
be applied:
>>> use = [0, 0, 2]
>>> p002 = tetra.pgroup.make_perm(3, use)
>>> p002
Permutation([1, 0, 3, 2])
Apply them one at a time:
>>> tetra.reset()
>>> for i in use:
... tetra.rotate(pgroup[i])
...
>>> tetra.vertices
(x, w, z, y)
>>> sequentially = tetra.vertices
Apply the composite permutation:
>>> tetra.reset()
>>> tetra.rotate(p002)
>>> tetra.corners
(x, w, z, y)
>>> tetra.corners in all and tetra.corners == sequentially
True
Notes
=====
Defining permutation groups
---------------------------
It is not necessary to enter any permutations, nor is necessary to
enter a complete set of transformations. In fact, for a polyhedron,
all configurations can be constructed from just two permutations.
For example, the orientations of a tetrahedron can be generated from
an axis passing through a vertex and face and another axis passing
through a different vertex or from an axis passing through the
midpoints of two edges opposite of each other.
For simplicity of presentation, consider a square --
not a cube -- with vertices 1, 2, 3, and 4:
1-----2 We could think of axes of rotation being:
| | 1) through the face
| | 2) from midpoint 1-2 to 3-4 or 1-3 to 2-4
3-----4 3) lines 1-4 or 2-3
To determine how to write the permutations, imagine 4 cameras,
one at each corner, labeled A-D:
A B A B
1-----2 1-----3 vertex index:
| | | | 1 0
| | | | 2 1
3-----4 2-----4 3 2
C D C D 4 3
original after rotation
along 1-4
A diagonal and a face axis will be chosen for the "permutation group"
from which any orientation can be constructed.
>>> pgroup = []
Imagine a clockwise rotation when viewing 1-4 from camera A. The new
orientation is (in camera-order): 1, 3, 2, 4 so the permutation is
given using the *indices* of the vertices as:
>>> pgroup.append(Permutation((0, 2, 1, 3)))
Now imagine rotating clockwise when looking down an axis entering the
center of the square as viewed. The new camera-order would be
3, 1, 4, 2 so the permutation is (using indices):
>>> pgroup.append(Permutation((2, 0, 3, 1)))
The square can now be constructed:
** use real-world labels for the vertices, entering them in
camera order
** for the faces we use zero-based indices of the vertices
in *edge-order* as the face is traversed; neither the
direction nor the starting point matter -- the faces are
only used to define edges (if so desired).
>>> square = Polyhedron((1, 2, 3, 4), [(0, 1, 3, 2)], pgroup)
To rotate the square with a single permutation we can do:
>>> square.rotate(square.pgroup[0])
>>> square.corners
(1, 3, 2, 4)
To use more than one permutation (or to use one permutation more
than once) it is more convenient to use the make_perm method:
>>> p011 = square.pgroup.make_perm([0, 1, 1]) # diag flip + 2 rotations
>>> square.reset() # return to initial orientation
>>> square.rotate(p011)
>>> square.corners
(4, 2, 3, 1)
Thinking outside the box
------------------------
Although the Polyhedron object has a direct physical meaning, it
actually has broader application. In the most general sense it is
just a decorated PermutationGroup, allowing one to connect the
permutations to something physical. For example, a Rubik's cube is
not a proper polyhedron, but the Polyhedron class can be used to
represent it in a way that helps to visualize the Rubik's cube.
>>> from sympy.utilities.iterables import flatten, unflatten
>>> from sympy import symbols
>>> from sympy.combinatorics import RubikGroup
>>> facelets = flatten([symbols(s+'1:5') for s in 'UFRBLD'])
>>> def show():
... pairs = unflatten(r2.corners, 2)
... print(pairs[::2])
... print(pairs[1::2])
...
>>> r2 = Polyhedron(facelets, pgroup=RubikGroup(2))
>>> show()
[(U1, U2), (F1, F2), (R1, R2), (B1, B2), (L1, L2), (D1, D2)]
[(U3, U4), (F3, F4), (R3, R4), (B3, B4), (L3, L4), (D3, D4)]
>>> r2.rotate(0) # cw rotation of F
>>> show()
[(U1, U2), (F3, F1), (U3, R2), (B1, B2), (L1, D1), (R3, R1)]
[(L4, L2), (F4, F2), (U4, R4), (B3, B4), (L3, D2), (D3, D4)]
Predefined Polyhedra
====================
For convenience, the vertices and faces are defined for the following
standard solids along with a permutation group for transformations.
When the polyhedron is oriented as indicated below, the vertices in
a given horizontal plane are numbered in ccw direction, starting from
the vertex that will give the lowest indices in a given face. (In the
net of the vertices, indices preceded by "-" indicate replication of
the lhs index in the net.)
tetrahedron, tetrahedron_faces
------------------------------
4 vertices (vertex up) net:
0 0-0
1 2 3-1
4 faces:
(0, 1, 2) (0, 2, 3) (0, 3, 1) (1, 2, 3)
cube, cube_faces
----------------
8 vertices (face up) net:
0 1 2 3-0
4 5 6 7-4
6 faces:
(0, 1, 2, 3)
(0, 1, 5, 4) (1, 2, 6, 5) (2, 3, 7, 6) (0, 3, 7, 4)
(4, 5, 6, 7)
octahedron, octahedron_faces
----------------------------
6 vertices (vertex up) net:
0 0 0-0
1 2 3 4-1
5 5 5-5
8 faces:
(0, 1, 2) (0, 2, 3) (0, 3, 4) (0, 1, 4)
(1, 2, 5) (2, 3, 5) (3, 4, 5) (1, 4, 5)
dodecahedron, dodecahedron_faces
--------------------------------
20 vertices (vertex up) net:
0 1 2 3 4 -0
5 6 7 8 9 -5
14 10 11 12 13-14
15 16 17 18 19-15
12 faces:
(0, 1, 2, 3, 4) (0, 1, 6, 10, 5) (1, 2, 7, 11, 6)
(2, 3, 8, 12, 7) (3, 4, 9, 13, 8) (0, 4, 9, 14, 5)
(5, 10, 16, 15, 14) (6, 10, 16, 17, 11) (7, 11, 17, 18, 12)
(8, 12, 18, 19, 13) (9, 13, 19, 15, 14)(15, 16, 17, 18, 19)
icosahedron, icosahedron_faces
------------------------------
12 vertices (face up) net:
0 0 0 0 -0
1 2 3 4 5 -1
6 7 8 9 10 -6
11 11 11 11 -11
20 faces:
(0, 1, 2) (0, 2, 3) (0, 3, 4)
(0, 4, 5) (0, 1, 5) (1, 2, 6)
(2, 3, 7) (3, 4, 8) (4, 5, 9)
(1, 5, 10) (2, 6, 7) (3, 7, 8)
(4, 8, 9) (5, 9, 10) (1, 6, 10)
(6, 7, 11) (7, 8, 11) (8, 9, 11)
(9, 10, 11) (6, 10, 11)
>>> from sympy.combinatorics.polyhedron import cube
>>> cube.edges
{(0, 1), (0, 3), (0, 4), '...', (4, 7), (5, 6), (6, 7)}
If you want to use letters or other names for the corners you
can still use the pre-calculated faces:
>>> corners = list('abcdefgh')
>>> Polyhedron(corners, cube.faces).corners
(a, b, c, d, e, f, g, h)
References
==========
[1] www.ocf.berkeley.edu/~wwu/articles/platonicsolids.pdf
"""
faces = [minlex(f, directed=False, is_set=True) for f in faces]
corners, faces, pgroup = args = \
[Tuple(*a) for a in (corners, faces, pgroup)]
obj = Basic.__new__(cls, *args)
obj._corners = tuple(corners) # in order given
obj._faces = FiniteSet(*faces)
if pgroup and pgroup[0].size != len(corners):
raise ValueError("Permutation size unequal to number of corners.")
# use the identity permutation if none are given
obj._pgroup = PermutationGroup((
pgroup or [Perm(range(len(corners)))] ))
return obj
@property
def corners(self):
"""
Get the corners of the Polyhedron.
The method ``vertices`` is an alias for ``corners``.
Examples
========
>>> from sympy.combinatorics import Polyhedron
>>> from sympy.abc import a, b, c, d
>>> p = Polyhedron(list('abcd'))
>>> p.corners == p.vertices == (a, b, c, d)
True
See Also
========
array_form, cyclic_form
"""
return self._corners
vertices = corners
@property
def array_form(self):
"""Return the indices of the corners.
The indices are given relative to the original position of corners.
Examples
========
>>> from sympy.combinatorics import Permutation, Cycle
>>> from sympy.combinatorics.polyhedron import tetrahedron
>>> tetrahedron.array_form
[0, 1, 2, 3]
>>> tetrahedron.rotate(0)
>>> tetrahedron.array_form
[0, 2, 3, 1]
>>> tetrahedron.pgroup[0].array_form
[0, 2, 3, 1]
See Also
========
corners, cyclic_form
"""
corners = list(self.args[0])
return [corners.index(c) for c in self.corners]
@property
def cyclic_form(self):
"""Return the indices of the corners in cyclic notation.
The indices are given relative to the original position of corners.
See Also
========
corners, array_form
"""
return Perm._af_new(self.array_form).cyclic_form
@property
def size(self):
"""
Get the number of corners of the Polyhedron.
"""
return len(self._corners)
@property
def faces(self):
"""
Get the faces of the Polyhedron.
"""
return self._faces
@property
def pgroup(self):
"""
Get the permutations of the Polyhedron.
"""
return self._pgroup
@property
def edges(self):
"""
Given the faces of the polyhedra we can get the edges.
Examples
========
>>> from sympy.combinatorics import Polyhedron
>>> from sympy.abc import a, b, c
>>> corners = (a, b, c)
>>> faces = [(0, 1, 2)]
>>> Polyhedron(corners, faces).edges
{(0, 1), (0, 2), (1, 2)}
"""
if self._edges is None:
output = set()
for face in self.faces:
for i in range(len(face)):
edge = tuple(sorted([face[i], face[i - 1]]))
output.add(edge)
self._edges = FiniteSet(*output)
return self._edges
[docs] def rotate(self, perm):
"""
Apply a permutation to the polyhedron *in place*. The permutation
may be given as a Permutation instance or an integer indicating
which permutation from pgroup of the Polyhedron should be
applied.
This is an operation that is analogous to rotation about
an axis by a fixed increment.
Notes
=====
When a Permutation is applied, no check is done to see if that
is a valid permutation for the Polyhedron. For example, a cube
could be given a permutation which effectively swaps only 2
vertices. A valid permutation (that rotates the object in a
physical way) will be obtained if one only uses
permutations from the ``pgroup`` of the Polyhedron. On the other
hand, allowing arbitrary rotations (applications of permutations)
gives a way to follow named elements rather than indices since
Polyhedron allows vertices to be named while Permutation works
only with indices.
Examples
========
>>> from sympy.combinatorics import Polyhedron, Permutation
>>> from sympy.combinatorics.polyhedron import cube
>>> cube.corners
(0, 1, 2, 3, 4, 5, 6, 7)
>>> cube.rotate(0)
>>> cube.corners
(1, 2, 3, 0, 5, 6, 7, 4)
A non-physical "rotation" that is not prohibited by this method:
>>> cube.reset()
>>> cube.rotate(Permutation([[1, 2]], size=8))
>>> cube.corners
(0, 2, 1, 3, 4, 5, 6, 7)
Polyhedron can be used to follow elements of set that are
identified by letters instead of integers:
>>> shadow = h5 = Polyhedron(list('abcde'))
>>> p = Permutation([3, 0, 1, 2, 4])
>>> h5.rotate(p)
>>> h5.corners
(d, a, b, c, e)
>>> _ == shadow.corners
True
>>> copy = h5.copy()
>>> h5.rotate(p)
>>> h5.corners == copy.corners
False
"""
if not isinstance(perm, Perm):
perm = self.pgroup[perm]
# and we know it's valid
else:
if perm.size != self.size:
raise ValueError('Polyhedron and Permutation sizes differ.')
a = perm.array_form
corners = [self.corners[a[i]] for i in range(len(self.corners))]
self._corners = tuple(corners)
[docs] def reset(self):
"""Return corners to their original positions.
Examples
========
>>> from sympy.combinatorics.polyhedron import tetrahedron as T
>>> T.corners
(0, 1, 2, 3)
>>> T.rotate(0)
>>> T.corners
(0, 2, 3, 1)
>>> T.reset()
>>> T.corners
(0, 1, 2, 3)
"""
self._corners = self.args[0]
def _pgroup_calcs():
"""Return the permutation groups for each of the polyhedra and the face
definitions: tetrahedron, cube, octahedron, dodecahedron, icosahedron,
tetrahedron_faces, cube_faces, octahedron_faces, dodecahedron_faces,
icosahedron_faces
(This author didn't find and didn't know of a better way to do it though
there likely is such a way.)
Although only 2 permutations are needed for a polyhedron in order to
generate all the possible orientations, a group of permutations is
provided instead. A set of permutations is called a "group" if::
a*b = c (for any pair of permutations in the group, a and b, their
product, c, is in the group)
a*(b*c) = (a*b)*c (for any 3 permutations in the group associativity holds)
there is an identity permutation, I, such that I*a = a*I for all elements
in the group
a*b = I (the inverse of each permutation is also in the group)
None of the polyhedron groups defined follow these definitions of a group.
Instead, they are selected to contain those permutations whose powers
alone will construct all orientations of the polyhedron, i.e. for
permutations ``a``, ``b``, etc... in the group, ``a, a**2, ..., a**o_a``,
``b, b**2, ..., b**o_b``, etc... (where ``o_i`` is the order of
permutation ``i``) generate all permutations of the polyhedron instead of
mixed products like ``a*b``, ``a*b**2``, etc....
Note that for a polyhedron with n vertices, the valid permutations of the
vertices exclude those that do not maintain its faces. e.g. the
permutation BCDE of a square's four corners, ABCD, is a valid
permutation while CBDE is not (because this would twist the square).
Examples
========
The is_group checks for: closure, the presence of the Identity permutation,
and the presence of the inverse for each of the elements in the group. This
confirms that none of the polyhedra are true groups:
>>> from sympy.combinatorics.polyhedron import (
... tetrahedron, cube, octahedron, dodecahedron, icosahedron)
...
>>> polyhedra = (tetrahedron, cube, octahedron, dodecahedron, icosahedron)
>>> [h.pgroup.is_group for h in polyhedra]
...
[True, True, True, True, True]
Although tests in polyhedron's test suite check that powers of the
permutations in the groups generate all permutations of the vertices
of the polyhedron, here we also demonstrate the powers of the given
permutations create a complete group for the tetrahedron:
>>> from sympy.combinatorics import Permutation, PermutationGroup
>>> for h in polyhedra[:1]:
... G = h.pgroup
... perms = set()
... for g in G:
... for e in range(g.order()):
... p = tuple((g**e).array_form)
... perms.add(p)
...
... perms = [Permutation(p) for p in perms]
... assert PermutationGroup(perms).is_group
In addition to doing the above, the tests in the suite confirm that the
faces are all present after the application of each permutation.
References
==========
http://dogschool.tripod.com/trianglegroup.html
"""
def _pgroup_of_double(polyh, ordered_faces, pgroup):
n = len(ordered_faces[0])
# the vertices of the double which sits inside a give polyhedron
# can be found by tracking the faces of the outer polyhedron.
# A map between face and the vertex of the double is made so that
# after rotation the position of the vertices can be located
fmap = dict(zip(ordered_faces,
range(len(ordered_faces))))
flat_faces = flatten(ordered_faces)
new_pgroup = []
for i, p in enumerate(pgroup):
h = polyh.copy()
h.rotate(p)
c = h.corners
# reorder corners in the order they should appear when
# enumerating the faces
reorder = unflatten([c[j] for j in flat_faces], n)
# make them canonical
reorder = [tuple(map(as_int,
minlex(f, directed=False, is_set=True)))
for f in reorder]
# map face to vertex: the resulting list of vertices are the
# permutation that we seek for the double
new_pgroup.append(Perm([fmap[f] for f in reorder]))
return new_pgroup
tetrahedron_faces = [
(0, 1, 2), (0, 2, 3), (0, 3, 1), # upper 3
(1, 2, 3), # bottom
]
# cw from top
#
_t_pgroup = [
Perm([[1, 2, 3], [0]]), # cw from top
Perm([[0, 1, 2], [3]]), # cw from front face
Perm([[0, 3, 2], [1]]), # cw from back right face
Perm([[0, 3, 1], [2]]), # cw from back left face
Perm([[0, 1], [2, 3]]), # through front left edge
Perm([[0, 2], [1, 3]]), # through front right edge
Perm([[0, 3], [1, 2]]), # through back edge
]
tetrahedron = Polyhedron(
range(4),
tetrahedron_faces,
_t_pgroup)
cube_faces = [
(0, 1, 2, 3), # upper
(0, 1, 5, 4), (1, 2, 6, 5), (2, 3, 7, 6), (0, 3, 7, 4), # middle 4
(4, 5, 6, 7), # lower
]
# U, D, F, B, L, R = up, down, front, back, left, right
_c_pgroup = [Perm(p) for p in
[
[1, 2, 3, 0, 5, 6, 7, 4], # cw from top, U
[4, 0, 3, 7, 5, 1, 2, 6], # cw from F face
[4, 5, 1, 0, 7, 6, 2, 3], # cw from R face
[1, 0, 4, 5, 2, 3, 7, 6], # cw through UF edge
[6, 2, 1, 5, 7, 3, 0, 4], # cw through UR edge
[6, 7, 3, 2, 5, 4, 0, 1], # cw through UB edge
[3, 7, 4, 0, 2, 6, 5, 1], # cw through UL edge
[4, 7, 6, 5, 0, 3, 2, 1], # cw through FL edge
[6, 5, 4, 7, 2, 1, 0, 3], # cw through FR edge
[0, 3, 7, 4, 1, 2, 6, 5], # cw through UFL vertex
[5, 1, 0, 4, 6, 2, 3, 7], # cw through UFR vertex
[5, 6, 2, 1, 4, 7, 3, 0], # cw through UBR vertex
[7, 4, 0, 3, 6, 5, 1, 2], # cw through UBL
]]
cube = Polyhedron(
range(8),
cube_faces,
_c_pgroup)
octahedron_faces = [
(0, 1, 2), (0, 2, 3), (0, 3, 4), (0, 1, 4), # top 4
(1, 2, 5), (2, 3, 5), (3, 4, 5), (1, 4, 5), # bottom 4
]
octahedron = Polyhedron(
range(6),
octahedron_faces,
_pgroup_of_double(cube, cube_faces, _c_pgroup))
dodecahedron_faces = [
(0, 1, 2, 3, 4), # top
(0, 1, 6, 10, 5), (1, 2, 7, 11, 6), (2, 3, 8, 12, 7), # upper 5
(3, 4, 9, 13, 8), (0, 4, 9, 14, 5),
(5, 10, 16, 15, 14), (6, 10, 16, 17, 11), (7, 11, 17, 18,
12), # lower 5
(8, 12, 18, 19, 13), (9, 13, 19, 15, 14),
(15, 16, 17, 18, 19) # bottom
]
def _string_to_perm(s):
rv = [Perm(range(20))]
p = None
for si in s:
if si not in '01':
count = int(si) - 1
else:
count = 1
if si == '0':
p = _f0
elif si == '1':
p = _f1
rv.extend([p]*count)
return Perm.rmul(*rv)
# top face cw
_f0 = Perm([
1, 2, 3, 4, 0, 6, 7, 8, 9, 5, 11,
12, 13, 14, 10, 16, 17, 18, 19, 15])
# front face cw
_f1 = Perm([
5, 0, 4, 9, 14, 10, 1, 3, 13, 15,
6, 2, 8, 19, 16, 17, 11, 7, 12, 18])
# the strings below, like 0104 are shorthand for F0*F1*F0**4 and are
# the remaining 4 face rotations, 15 edge permutations, and the
# 10 vertex rotations.
_dodeca_pgroup = [_f0, _f1] + [_string_to_perm(s) for s in '''
0104 140 014 0410
010 1403 03104 04103 102
120 1304 01303 021302 03130
0412041 041204103 04120410 041204104 041204102
10 01 1402 0140 04102 0412 1204 1302 0130 03120'''.strip().split()]
dodecahedron = Polyhedron(
range(20),
dodecahedron_faces,
_dodeca_pgroup)
icosahedron_faces = [
[0, 1, 2], [0, 2, 3], [0, 3, 4], [0, 4, 5], [0, 1, 5],
[1, 6, 7], [1, 2, 7], [2, 7, 8], [2, 3, 8], [3, 8, 9],
[3, 4, 9], [4, 9, 10 ], [4, 5, 10], [5, 6, 10], [1, 5, 6],
[6, 7, 11], [7, 8, 11], [8, 9, 11], [9, 10, 11], [6, 10, 11]]
icosahedron = Polyhedron(
range(12),
icosahedron_faces,
_pgroup_of_double(
dodecahedron, dodecahedron_faces, _dodeca_pgroup))
return (tetrahedron, cube, octahedron, dodecahedron, icosahedron,
tetrahedron_faces, cube_faces, octahedron_faces,
dodecahedron_faces, icosahedron_faces)
(tetrahedron, cube, octahedron, dodecahedron, icosahedron,
tetrahedron_faces, cube_faces, octahedron_faces,
dodecahedron_faces, icosahedron_faces) = _pgroup_calcs()
