Hide keyboard shortcuts

Hot-keys on this page

r m x p   toggle line displays

j k   next/prev highlighted chunk

0   (zero) top of page

1   (one) first highlighted chunk

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

22

23

24

25

26

27

28

29

30

31

32

33

34

35

36

37

38

39

40

41

42

43

44

45

46

47

48

49

50

51

52

53

54

55

56

57

58

59

60

61

62

63

64

65

66

67

68

69

70

71

72

73

74

75

76

77

78

79

80

81

82

83

84

85

86

87

88

89

90

91

92

93

94

95

96

97

98

99

100

101

102

103

104

105

106

107

108

109

110

111

112

113

114

115

116

117

118

119

120

121

122

123

124

125

126

127

128

129

130

131

132

133

134

135

136

137

138

139

140

141

142

143

144

145

146

147

148

149

150

151

152

153

154

155

156

157

158

159

160

161

162

163

164

165

166

167

168

169

170

171

172

173

174

175

176

177

178

179

180

181

182

183

184

185

186

187

188

189

190

191

192

193

194

195

196

197

198

199

200

201

202

203

204

205

206

207

208

209

210

211

212

213

214

215

216

217

218

219

220

221

222

223

224

225

226

227

228

229

230

231

232

233

234

235

236

237

238

239

240

241

242

243

244

245

246

247

248

249

250

251

252

253

254

255

256

257

258

259

260

261

262

263

264

265

266

267

268

269

270

271

272

273

274

275

276

277

278

279

280

281

282

283

284

285

286

287

288

289

290

291

292

293

294

295

296

297

298

299

300

301

302

303

304

305

306

307

308

309

310

311

312

313

314

315

316

317

318

319

320

321

322

323

324

325

326

327

328

329

330

331

332

333

334

335

336

337

338

339

340

341

342

343

344

345

346

347

348

349

350

351

352

353

354

355

356

357

358

359

360

361

362

363

364

365

366

367

368

369

370

371

372

373

374

375

376

377

378

379

380

381

382

383

384

385

386

387

388

389

390

391

392

393

394

395

396

397

398

399

400

401

402

403

404

405

406

407

408

409

410

411

412

413

414

415

416

417

418

419

420

421

422

423

424

425

426

427

428

429

430

431

432

433

434

435

436

437

438

439

440

441

442

443

444

445

446

447

448

449

450

451

452

453

454

455

456

457

458

459

460

461

462

463

464

465

466

467

468

469

470

471

472

473

474

475

476

477

478

479

480

481

482

483

484

485

486

487

488

489

490

491

492

493

494

495

496

497

498

499

500

501

502

503

504

505

506

507

508

509

510

511

512

513

514

515

516

517

518

519

520

521

522

523

524

525

526

527

528

529

530

531

532

533

534

535

536

537

538

539

540

541

542

543

544

545

546

547

548

549

550

551

552

553

554

555

556

557

558

559

560

561

562

563

564

565

566

567

568

569

570

571

572

573

574

575

576

577

578

579

580

581

582

583

584

585

586

587

588

589

590

591

592

593

594

595

596

597

598

599

600

601

602

603

604

605

606

607

608

609

610

611

612

613

614

615

616

617

618

619

620

621

622

623

624

625

626

627

628

629

630

631

632

633

634

635

636

637

638

639

640

641

642

643

644

645

646

647

648

649

650

651

652

653

654

655

656

657

658

659

660

661

662

663

664

665

666

667

668

669

670

671

672

673

674

675

676

677

678

679

680

681

682

683

684

685

686

687

688

689

690

691

692

693

694

695

696

697

698

699

700

701

702

703

704

705

706

707

708

709

710

711

712

713

714

715

716

717

718

719

720

721

722

723

724

725

726

727

728

729

730

731

732

733

734

735

736

737

738

739

740

741

742

743

744

745

746

747

748

749

750

751

752

753

754

755

756

757

758

759

760

761

762

763

764

765

766

767

768

769

770

771

772

773

774

775

776

777

778

779

780

781

782

783

784

785

786

787

788

789

790

791

792

793

794

795

796

797

798

799

800

801

802

803

804

805

806

807

808

809

810

811

812

813

814

815

816

817

818

819

820

821

822

823

824

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 

 

 

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 

 

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) 

 

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()