'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'

def __new__(cls, corners, faces=[], pgroup=[]):

    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)
    obj._faces = FiniteSet(*faces)
    if (pgroup and (pgroup[0].size != len(corners))):
        raise ValueError('Permutation    size    unequal    to    number    of    corners.')
    obj._pgroup = PermutationGroup((pgroup or [Perm(range(len(corners)))]))
    return obj
