Fix zillions of MSVC compiler warnings. Sigh.
[sgt/puzzles] / cube.c
1 /*
2 * cube.c: Cube game.
3 */
4
5 #include <stdio.h>
6 #include <stdlib.h>
7 #include <string.h>
8 #include <assert.h>
9 #include <math.h>
10
11 #include "puzzles.h"
12
13 #define MAXVERTICES 20
14 #define MAXFACES 20
15 #define MAXORDER 4
16 struct solid {
17 int nvertices;
18 float vertices[MAXVERTICES * 3]; /* 3*npoints coordinates */
19 int order;
20 int nfaces;
21 int faces[MAXFACES * MAXORDER]; /* order*nfaces point indices */
22 float normals[MAXFACES * 3]; /* 3*npoints vector components */
23 float shear; /* isometric shear for nice drawing */
24 float border; /* border required around arena */
25 };
26
27 static const struct solid tetrahedron = {
28 4,
29 {
30 0.0F, -0.57735026919F, -0.20412414523F,
31 -0.5F, 0.28867513459F, -0.20412414523F,
32 0.0F, -0.0F, 0.6123724357F,
33 0.5F, 0.28867513459F, -0.20412414523F,
34 },
35 3, 4,
36 {
37 0,2,1, 3,1,2, 2,0,3, 1,3,0
38 },
39 {
40 -0.816496580928F, -0.471404520791F, 0.333333333334F,
41 0.0F, 0.942809041583F, 0.333333333333F,
42 0.816496580928F, -0.471404520791F, 0.333333333334F,
43 0.0F, 0.0F, -1.0F,
44 },
45 0.0F, 0.3F
46 };
47
48 static const struct solid cube = {
49 8,
50 {
51 -0.5F,-0.5F,-0.5F, -0.5F,-0.5F,+0.5F,
52 -0.5F,+0.5F,-0.5F, -0.5F,+0.5F,+0.5F,
53 +0.5F,-0.5F,-0.5F, +0.5F,-0.5F,+0.5F,
54 +0.5F,+0.5F,-0.5F, +0.5F,+0.5F,+0.5F,
55 },
56 4, 6,
57 {
58 0,1,3,2, 1,5,7,3, 5,4,6,7, 4,0,2,6, 0,4,5,1, 3,7,6,2
59 },
60 {
61 -1.0F,0.0F,0.0F, 0.0F,0.0F,+1.0F,
62 +1.0F,0.0F,0.0F, 0.0F,0.0F,-1.0F,
63 0.0F,-1.0F,0.0F, 0.0F,+1.0F,0.0F
64 },
65 0.3F, 0.5F
66 };
67
68 static const struct solid octahedron = {
69 6,
70 {
71 -0.5F, -0.28867513459472505F, 0.4082482904638664F,
72 0.5F, 0.28867513459472505F, -0.4082482904638664F,
73 -0.5F, 0.28867513459472505F, -0.4082482904638664F,
74 0.5F, -0.28867513459472505F, 0.4082482904638664F,
75 0.0F, -0.57735026918945009F, -0.4082482904638664F,
76 0.0F, 0.57735026918945009F, 0.4082482904638664F,
77 },
78 3, 8,
79 {
80 4,0,2, 0,5,2, 0,4,3, 5,0,3, 1,4,2, 5,1,2, 4,1,3, 1,5,3
81 },
82 {
83 -0.816496580928F, -0.471404520791F, -0.333333333334F,
84 -0.816496580928F, 0.471404520791F, 0.333333333334F,
85 0.0F, -0.942809041583F, 0.333333333333F,
86 0.0F, 0.0F, 1.0F,
87 0.0F, 0.0F, -1.0F,
88 0.0F, 0.942809041583F, -0.333333333333F,
89 0.816496580928F, -0.471404520791F, -0.333333333334F,
90 0.816496580928F, 0.471404520791F, 0.333333333334F,
91 },
92 0.0F, 0.5F
93 };
94
95 static const struct solid icosahedron = {
96 12,
97 {
98 0.0F, 0.57735026919F, 0.75576131408F,
99 0.0F, -0.93417235896F, 0.17841104489F,
100 0.0F, 0.93417235896F, -0.17841104489F,
101 0.0F, -0.57735026919F, -0.75576131408F,
102 -0.5F, -0.28867513459F, 0.75576131408F,
103 -0.5F, 0.28867513459F, -0.75576131408F,
104 0.5F, -0.28867513459F, 0.75576131408F,
105 0.5F, 0.28867513459F, -0.75576131408F,
106 -0.80901699437F, 0.46708617948F, 0.17841104489F,
107 0.80901699437F, 0.46708617948F, 0.17841104489F,
108 -0.80901699437F, -0.46708617948F, -0.17841104489F,
109 0.80901699437F, -0.46708617948F, -0.17841104489F,
110 },
111 3, 20,
112 {
113 8,0,2, 0,9,2, 1,10,3, 11,1,3, 0,4,6,
114 4,1,6, 5,2,7, 3,5,7, 4,8,10, 8,5,10,
115 9,6,11, 7,9,11, 0,8,4, 9,0,6, 10,1,4,
116 1,11,6, 8,2,5, 2,9,7, 3,10,5, 11,3,7,
117 },
118 {
119 -0.356822089773F, 0.87267799625F, 0.333333333333F,
120 0.356822089773F, 0.87267799625F, 0.333333333333F,
121 -0.356822089773F, -0.87267799625F, -0.333333333333F,
122 0.356822089773F, -0.87267799625F, -0.333333333333F,
123 -0.0F, 0.0F, 1.0F,
124 0.0F, -0.666666666667F, 0.745355992501F,
125 0.0F, 0.666666666667F, -0.745355992501F,
126 0.0F, 0.0F, -1.0F,
127 -0.934172358963F, -0.12732200375F, 0.333333333333F,
128 -0.934172358963F, 0.12732200375F, -0.333333333333F,
129 0.934172358963F, -0.12732200375F, 0.333333333333F,
130 0.934172358963F, 0.12732200375F, -0.333333333333F,
131 -0.57735026919F, 0.333333333334F, 0.745355992501F,
132 0.57735026919F, 0.333333333334F, 0.745355992501F,
133 -0.57735026919F, -0.745355992501F, 0.333333333334F,
134 0.57735026919F, -0.745355992501F, 0.333333333334F,
135 -0.57735026919F, 0.745355992501F, -0.333333333334F,
136 0.57735026919F, 0.745355992501F, -0.333333333334F,
137 -0.57735026919F, -0.333333333334F, -0.745355992501F,
138 0.57735026919F, -0.333333333334F, -0.745355992501F,
139 },
140 0.0F, 0.8F
141 };
142
143 enum {
144 TETRAHEDRON, CUBE, OCTAHEDRON, ICOSAHEDRON
145 };
146 static const struct solid *solids[] = {
147 &tetrahedron, &cube, &octahedron, &icosahedron
148 };
149
150 enum {
151 COL_BACKGROUND,
152 COL_BORDER,
153 COL_BLUE,
154 NCOLOURS
155 };
156
157 enum { LEFT, RIGHT, UP, DOWN, UP_LEFT, UP_RIGHT, DOWN_LEFT, DOWN_RIGHT };
158
159 #define GRID_SCALE 48.0F
160 #define ROLLTIME 0.1F
161
162 #define SQ(x) ( (x) * (x) )
163
164 #define MATMUL(ra,m,a) do { \
165 float rx, ry, rz, xx = (a)[0], yy = (a)[1], zz = (a)[2], *mat = (m); \
166 rx = mat[0] * xx + mat[3] * yy + mat[6] * zz; \
167 ry = mat[1] * xx + mat[4] * yy + mat[7] * zz; \
168 rz = mat[2] * xx + mat[5] * yy + mat[8] * zz; \
169 (ra)[0] = rx; (ra)[1] = ry; (ra)[2] = rz; \
170 } while (0)
171
172 #define APPROXEQ(x,y) ( SQ(x-y) < 0.1 )
173
174 struct grid_square {
175 float x, y;
176 int npoints;
177 float points[8]; /* maximum */
178 int directions[8]; /* bit masks showing point pairs */
179 int flip;
180 int blue;
181 int tetra_class;
182 };
183
184 struct game_params {
185 int solid;
186 /*
187 * Grid dimensions. For a square grid these are width and
188 * height respectively; otherwise the grid is a hexagon, with
189 * the top side and the two lower diagonals having length d1
190 * and the remaining three sides having length d2 (so that
191 * d1==d2 gives a regular hexagon, and d2==0 gives a triangle).
192 */
193 int d1, d2;
194 };
195
196 struct game_state {
197 struct game_params params;
198 const struct solid *solid;
199 int *facecolours;
200 struct grid_square *squares;
201 int nsquares;
202 int current; /* index of current grid square */
203 int sgkey[2]; /* key-point indices into grid sq */
204 int dgkey[2]; /* key-point indices into grid sq */
205 int spkey[2]; /* key-point indices into polyhedron */
206 int dpkey[2]; /* key-point indices into polyhedron */
207 int previous;
208 float angle;
209 int completed;
210 int movecount;
211 };
212
213 game_params *default_params(void)
214 {
215 game_params *ret = snew(game_params);
216
217 ret->solid = CUBE;
218 ret->d1 = 4;
219 ret->d2 = 4;
220
221 return ret;
222 }
223
224 int game_fetch_preset(int i, char **name, game_params **params)
225 {
226 game_params *ret = snew(game_params);
227 char *str;
228
229 switch (i) {
230 case 0:
231 str = "Cube";
232 ret->solid = CUBE;
233 ret->d1 = 4;
234 ret->d2 = 4;
235 break;
236 case 1:
237 str = "Tetrahedron";
238 ret->solid = TETRAHEDRON;
239 ret->d1 = 2;
240 ret->d2 = 1;
241 break;
242 case 2:
243 str = "Octahedron";
244 ret->solid = OCTAHEDRON;
245 ret->d1 = 2;
246 ret->d2 = 2;
247 break;
248 case 3:
249 str = "Icosahedron";
250 ret->solid = ICOSAHEDRON;
251 ret->d1 = 3;
252 ret->d2 = 3;
253 break;
254 default:
255 sfree(ret);
256 return FALSE;
257 }
258
259 *name = dupstr(str);
260 *params = ret;
261 return TRUE;
262 }
263
264 void free_params(game_params *params)
265 {
266 sfree(params);
267 }
268
269 game_params *dup_params(game_params *params)
270 {
271 game_params *ret = snew(game_params);
272 *ret = *params; /* structure copy */
273 return ret;
274 }
275
276 static void enum_grid_squares(game_params *params,
277 void (*callback)(void *, struct grid_square *),
278 void *ctx)
279 {
280 const struct solid *solid = solids[params->solid];
281
282 if (solid->order == 4) {
283 int x, y;
284
285 for (x = 0; x < params->d1; x++)
286 for (y = 0; y < params->d2; y++) {
287 struct grid_square sq;
288
289 sq.x = (float)x;
290 sq.y = (float)y;
291 sq.points[0] = x - 0.5F;
292 sq.points[1] = y - 0.5F;
293 sq.points[2] = x - 0.5F;
294 sq.points[3] = y + 0.5F;
295 sq.points[4] = x + 0.5F;
296 sq.points[5] = y + 0.5F;
297 sq.points[6] = x + 0.5F;
298 sq.points[7] = y - 0.5F;
299 sq.npoints = 4;
300
301 sq.directions[LEFT] = 0x03; /* 0,1 */
302 sq.directions[RIGHT] = 0x0C; /* 2,3 */
303 sq.directions[UP] = 0x09; /* 0,3 */
304 sq.directions[DOWN] = 0x06; /* 1,2 */
305 sq.directions[UP_LEFT] = 0; /* no diagonals in a square */
306 sq.directions[UP_RIGHT] = 0; /* no diagonals in a square */
307 sq.directions[DOWN_LEFT] = 0; /* no diagonals in a square */
308 sq.directions[DOWN_RIGHT] = 0; /* no diagonals in a square */
309
310 sq.flip = FALSE;
311
312 /*
313 * This is supremely irrelevant, but just to avoid
314 * having any uninitialised structure members...
315 */
316 sq.tetra_class = 0;
317
318 callback(ctx, &sq);
319 }
320 } else {
321 int row, rowlen, other, i, firstix = -1;
322 float theight = (float)(sqrt(3) / 2.0);
323
324 for (row = 0; row < params->d1 + params->d2; row++) {
325 if (row < params->d1) {
326 other = +1;
327 rowlen = row + params->d2;
328 } else {
329 other = -1;
330 rowlen = 2*params->d1 + params->d2 - row;
331 }
332
333 /*
334 * There are `rowlen' down-pointing triangles.
335 */
336 for (i = 0; i < rowlen; i++) {
337 struct grid_square sq;
338 int ix;
339 float x, y;
340
341 ix = (2 * i - (rowlen-1));
342 x = ix * 0.5F;
343 y = theight * row;
344 sq.x = x;
345 sq.y = y + theight / 3;
346 sq.points[0] = x - 0.5F;
347 sq.points[1] = y;
348 sq.points[2] = x;
349 sq.points[3] = y + theight;
350 sq.points[4] = x + 0.5F;
351 sq.points[5] = y;
352 sq.npoints = 3;
353
354 sq.directions[LEFT] = 0x03; /* 0,1 */
355 sq.directions[RIGHT] = 0x06; /* 1,2 */
356 sq.directions[UP] = 0x05; /* 0,2 */
357 sq.directions[DOWN] = 0; /* invalid move */
358
359 /*
360 * Down-pointing triangle: both the up diagonals go
361 * up, and the down ones go left and right.
362 */
363 sq.directions[UP_LEFT] = sq.directions[UP_RIGHT] =
364 sq.directions[UP];
365 sq.directions[DOWN_LEFT] = sq.directions[LEFT];
366 sq.directions[DOWN_RIGHT] = sq.directions[RIGHT];
367
368 sq.flip = TRUE;
369
370 if (firstix < 0)
371 firstix = ix & 3;
372 ix -= firstix;
373 sq.tetra_class = ((row+(ix&1)) & 2) ^ (ix & 3);
374
375 callback(ctx, &sq);
376 }
377
378 /*
379 * There are `rowlen+other' up-pointing triangles.
380 */
381 for (i = 0; i < rowlen+other; i++) {
382 struct grid_square sq;
383 int ix;
384 float x, y;
385
386 ix = (2 * i - (rowlen+other-1));
387 x = ix * 0.5F;
388 y = theight * row;
389 sq.x = x;
390 sq.y = y + 2*theight / 3;
391 sq.points[0] = x + 0.5F;
392 sq.points[1] = y + theight;
393 sq.points[2] = x;
394 sq.points[3] = y;
395 sq.points[4] = x - 0.5F;
396 sq.points[5] = y + theight;
397 sq.npoints = 3;
398
399 sq.directions[LEFT] = 0x06; /* 1,2 */
400 sq.directions[RIGHT] = 0x03; /* 0,1 */
401 sq.directions[DOWN] = 0x05; /* 0,2 */
402 sq.directions[UP] = 0; /* invalid move */
403
404 /*
405 * Up-pointing triangle: both the down diagonals go
406 * down, and the up ones go left and right.
407 */
408 sq.directions[DOWN_LEFT] = sq.directions[DOWN_RIGHT] =
409 sq.directions[DOWN];
410 sq.directions[UP_LEFT] = sq.directions[LEFT];
411 sq.directions[UP_RIGHT] = sq.directions[RIGHT];
412
413 sq.flip = FALSE;
414
415 if (firstix < 0)
416 firstix = ix;
417 ix -= firstix;
418 sq.tetra_class = ((row+(ix&1)) & 2) ^ (ix & 3);
419
420 callback(ctx, &sq);
421 }
422 }
423 }
424 }
425
426 static int grid_area(int d1, int d2, int order)
427 {
428 /*
429 * An NxM grid of squares has NM squares in it.
430 *
431 * A grid of triangles with dimensions A and B has a total of
432 * A^2 + B^2 + 4AB triangles in it. (You can divide it up into
433 * a side-A triangle containing A^2 subtriangles, a side-B
434 * triangle containing B^2, and two congruent parallelograms,
435 * each with side lengths A and B, each therefore containing AB
436 * two-triangle rhombuses.)
437 */
438 if (order == 4)
439 return d1 * d2;
440 else
441 return d1*d1 + d2*d2 + 4*d1*d2;
442 }
443
444 struct grid_data {
445 int *gridptrs[4];
446 int nsquares[4];
447 int nclasses;
448 int squareindex;
449 };
450
451 static void classify_grid_square_callback(void *ctx, struct grid_square *sq)
452 {
453 struct grid_data *data = (struct grid_data *)ctx;
454 int thisclass;
455
456 if (data->nclasses == 4)
457 thisclass = sq->tetra_class;
458 else if (data->nclasses == 2)
459 thisclass = sq->flip;
460 else
461 thisclass = 0;
462
463 data->gridptrs[thisclass][data->nsquares[thisclass]++] =
464 data->squareindex++;
465 }
466
467 char *new_game_seed(game_params *params)
468 {
469 struct grid_data data;
470 int i, j, k, m, area, facesperclass;
471 int *flags;
472 char *seed, *p;
473
474 /*
475 * Enumerate the grid squares, dividing them into equivalence
476 * classes as appropriate. (For the tetrahedron, there is one
477 * equivalence class for each face; for the octahedron there
478 * are two classes; for the other two solids there's only one.)
479 */
480
481 area = grid_area(params->d1, params->d2, solids[params->solid]->order);
482 if (params->solid == TETRAHEDRON)
483 data.nclasses = 4;
484 else if (params->solid == OCTAHEDRON)
485 data.nclasses = 2;
486 else
487 data.nclasses = 1;
488 data.gridptrs[0] = snewn(data.nclasses * area, int);
489 for (i = 0; i < data.nclasses; i++) {
490 data.gridptrs[i] = data.gridptrs[0] + i * area;
491 data.nsquares[i] = 0;
492 }
493 data.squareindex = 0;
494 enum_grid_squares(params, classify_grid_square_callback, &data);
495
496 facesperclass = solids[params->solid]->nfaces / data.nclasses;
497
498 for (i = 0; i < data.nclasses; i++)
499 assert(data.nsquares[i] >= facesperclass);
500 assert(data.squareindex == area);
501
502 /*
503 * So now we know how many faces to allocate in each class. Get
504 * on with it.
505 */
506 flags = snewn(area, int);
507 for (i = 0; i < area; i++)
508 flags[i] = FALSE;
509
510 for (i = 0; i < data.nclasses; i++) {
511 for (j = 0; j < facesperclass; j++) {
512 unsigned long divisor = RAND_MAX / data.nsquares[i];
513 unsigned long max = divisor * data.nsquares[i];
514 unsigned long n;
515
516 do {
517 n = rand();
518 } while (n >= max);
519
520 n /= divisor;
521
522 assert(!flags[data.gridptrs[i][n]]);
523 flags[data.gridptrs[i][n]] = TRUE;
524
525 /*
526 * Move everything else up the array. I ought to use a
527 * better data structure for this, but for such small
528 * numbers it hardly seems worth the effort.
529 */
530 while ((int)n < data.nsquares[i]-1) {
531 data.gridptrs[i][n] = data.gridptrs[i][n+1];
532 n++;
533 }
534 data.nsquares[i]--;
535 }
536 }
537
538 /*
539 * Now we know precisely which squares are blue. Encode this
540 * information in hex. While we're looping over this, collect
541 * the non-blue squares into a list in the now-unused gridptrs
542 * array.
543 */
544 seed = snewn(area / 4 + 40, char);
545 p = seed;
546 j = 0;
547 k = 8;
548 m = 0;
549 for (i = 0; i < area; i++) {
550 if (flags[i]) {
551 j |= k;
552 } else {
553 data.gridptrs[0][m++] = i;
554 }
555 k >>= 1;
556 if (!k) {
557 *p++ = "0123456789ABCDEF"[j];
558 k = 8;
559 j = 0;
560 }
561 }
562 if (k != 8)
563 *p++ = "0123456789ABCDEF"[j];
564
565 /*
566 * Choose a non-blue square for the polyhedron.
567 */
568 {
569 unsigned long divisor = RAND_MAX / m;
570 unsigned long max = divisor * m;
571 unsigned long n;
572
573 do {
574 n = rand();
575 } while (n >= max);
576
577 n /= divisor;
578
579 sprintf(p, ":%d", data.gridptrs[0][n]);
580 }
581
582 sfree(data.gridptrs[0]);
583 sfree(flags);
584
585 return seed;
586 }
587
588 static void add_grid_square_callback(void *ctx, struct grid_square *sq)
589 {
590 game_state *state = (game_state *)ctx;
591
592 state->squares[state->nsquares] = *sq; /* structure copy */
593 state->squares[state->nsquares].blue = FALSE;
594 state->nsquares++;
595 }
596
597 static int lowest_face(const struct solid *solid)
598 {
599 int i, j, best;
600 float zmin;
601
602 best = 0;
603 zmin = 0.0;
604 for (i = 0; i < solid->nfaces; i++) {
605 float z = 0;
606
607 for (j = 0; j < solid->order; j++) {
608 int f = solid->faces[i*solid->order + j];
609 z += solid->vertices[f*3+2];
610 }
611
612 if (i == 0 || zmin > z) {
613 zmin = z;
614 best = i;
615 }
616 }
617
618 return best;
619 }
620
621 static int align_poly(const struct solid *solid, struct grid_square *sq,
622 int *pkey)
623 {
624 float zmin;
625 int i, j;
626 int flip = (sq->flip ? -1 : +1);
627
628 /*
629 * First, find the lowest z-coordinate present in the solid.
630 */
631 zmin = 0.0;
632 for (i = 0; i < solid->nvertices; i++)
633 if (zmin > solid->vertices[i*3+2])
634 zmin = solid->vertices[i*3+2];
635
636 /*
637 * Now go round the grid square. For each point in the grid
638 * square, we're looking for a point of the polyhedron with the
639 * same x- and y-coordinates (relative to the square's centre),
640 * and z-coordinate equal to zmin (near enough).
641 */
642 for (j = 0; j < sq->npoints; j++) {
643 int matches, index;
644
645 matches = 0;
646 index = -1;
647
648 for (i = 0; i < solid->nvertices; i++) {
649 float dist = 0;
650
651 dist += SQ(solid->vertices[i*3+0] * flip - sq->points[j*2+0] + sq->x);
652 dist += SQ(solid->vertices[i*3+1] * flip - sq->points[j*2+1] + sq->y);
653 dist += SQ(solid->vertices[i*3+2] - zmin);
654
655 if (dist < 0.1) {
656 matches++;
657 index = i;
658 }
659 }
660
661 if (matches != 1 || index < 0)
662 return FALSE;
663 pkey[j] = index;
664 }
665
666 return TRUE;
667 }
668
669 static void flip_poly(struct solid *solid, int flip)
670 {
671 int i;
672
673 if (flip) {
674 for (i = 0; i < solid->nvertices; i++) {
675 solid->vertices[i*3+0] *= -1;
676 solid->vertices[i*3+1] *= -1;
677 }
678 for (i = 0; i < solid->nfaces; i++) {
679 solid->normals[i*3+0] *= -1;
680 solid->normals[i*3+1] *= -1;
681 }
682 }
683 }
684
685 static struct solid *transform_poly(const struct solid *solid, int flip,
686 int key0, int key1, float angle)
687 {
688 struct solid *ret = snew(struct solid);
689 float vx, vy, ax, ay;
690 float vmatrix[9], amatrix[9], vmatrix2[9];
691 int i;
692
693 *ret = *solid; /* structure copy */
694
695 flip_poly(ret, flip);
696
697 /*
698 * Now rotate the polyhedron through the given angle. We must
699 * rotate about the Z-axis to bring the two vertices key0 and
700 * key1 into horizontal alignment, then rotate about the
701 * X-axis, then rotate back again.
702 */
703 vx = ret->vertices[key1*3+0] - ret->vertices[key0*3+0];
704 vy = ret->vertices[key1*3+1] - ret->vertices[key0*3+1];
705 assert(APPROXEQ(vx*vx + vy*vy, 1.0));
706
707 vmatrix[0] = vx; vmatrix[3] = vy; vmatrix[6] = 0;
708 vmatrix[1] = -vy; vmatrix[4] = vx; vmatrix[7] = 0;
709 vmatrix[2] = 0; vmatrix[5] = 0; vmatrix[8] = 1;
710
711 ax = (float)cos(angle);
712 ay = (float)sin(angle);
713
714 amatrix[0] = 1; amatrix[3] = 0; amatrix[6] = 0;
715 amatrix[1] = 0; amatrix[4] = ax; amatrix[7] = ay;
716 amatrix[2] = 0; amatrix[5] = -ay; amatrix[8] = ax;
717
718 memcpy(vmatrix2, vmatrix, sizeof(vmatrix));
719 vmatrix2[1] = vy;
720 vmatrix2[3] = -vy;
721
722 for (i = 0; i < ret->nvertices; i++) {
723 MATMUL(ret->vertices + 3*i, vmatrix, ret->vertices + 3*i);
724 MATMUL(ret->vertices + 3*i, amatrix, ret->vertices + 3*i);
725 MATMUL(ret->vertices + 3*i, vmatrix2, ret->vertices + 3*i);
726 }
727 for (i = 0; i < ret->nfaces; i++) {
728 MATMUL(ret->normals + 3*i, vmatrix, ret->normals + 3*i);
729 MATMUL(ret->normals + 3*i, amatrix, ret->normals + 3*i);
730 MATMUL(ret->normals + 3*i, vmatrix2, ret->normals + 3*i);
731 }
732
733 return ret;
734 }
735
736 game_state *new_game(game_params *params, char *seed)
737 {
738 game_state *state = snew(game_state);
739 int area;
740
741 state->params = *params; /* structure copy */
742 state->solid = solids[params->solid];
743
744 area = grid_area(params->d1, params->d2, state->solid->order);
745 state->squares = snewn(area, struct grid_square);
746 state->nsquares = 0;
747 enum_grid_squares(params, add_grid_square_callback, state);
748 assert(state->nsquares == area);
749
750 state->facecolours = snewn(state->solid->nfaces, int);
751 memset(state->facecolours, 0, state->solid->nfaces * sizeof(int));
752
753 /*
754 * Set up the blue squares and polyhedron position according to
755 * the game seed.
756 */
757 {
758 char *p = seed;
759 int i, j, v;
760
761 j = 8;
762 v = 0;
763 for (i = 0; i < state->nsquares; i++) {
764 if (j == 8) {
765 v = *p++;
766 if (v >= '0' && v <= '9')
767 v -= '0';
768 else if (v >= 'A' && v <= 'F')
769 v -= 'A' - 10;
770 else if (v >= 'a' && v <= 'f')
771 v -= 'a' - 10;
772 else
773 break;
774 }
775 if (v & j)
776 state->squares[i].blue = TRUE;
777 j >>= 1;
778 if (j == 0)
779 j = 8;
780 }
781
782 if (*p == ':')
783 p++;
784
785 state->current = atoi(p);
786 if (state->current < 0 || state->current >= state->nsquares)
787 state->current = 0; /* got to do _something_ */
788 }
789
790 /*
791 * Align the polyhedron with its grid square and determine
792 * initial key points.
793 */
794 {
795 int pkey[4];
796 int ret;
797
798 ret = align_poly(state->solid, &state->squares[state->current], pkey);
799 assert(ret);
800
801 state->dpkey[0] = state->spkey[0] = pkey[0];
802 state->dpkey[1] = state->spkey[0] = pkey[1];
803 state->dgkey[0] = state->sgkey[0] = 0;
804 state->dgkey[1] = state->sgkey[0] = 1;
805 }
806
807 state->previous = state->current;
808 state->angle = 0.0;
809 state->completed = FALSE;
810 state->movecount = 0;
811
812 return state;
813 }
814
815 game_state *dup_game(game_state *state)
816 {
817 game_state *ret = snew(game_state);
818
819 ret->params = state->params; /* structure copy */
820 ret->solid = state->solid;
821 ret->facecolours = snewn(ret->solid->nfaces, int);
822 memcpy(ret->facecolours, state->facecolours,
823 ret->solid->nfaces * sizeof(int));
824 ret->nsquares = state->nsquares;
825 ret->squares = snewn(ret->nsquares, struct grid_square);
826 memcpy(ret->squares, state->squares,
827 ret->nsquares * sizeof(struct grid_square));
828 ret->dpkey[0] = state->dpkey[0];
829 ret->dpkey[1] = state->dpkey[1];
830 ret->dgkey[0] = state->dgkey[0];
831 ret->dgkey[1] = state->dgkey[1];
832 ret->spkey[0] = state->spkey[0];
833 ret->spkey[1] = state->spkey[1];
834 ret->sgkey[0] = state->sgkey[0];
835 ret->sgkey[1] = state->sgkey[1];
836 ret->previous = state->previous;
837 ret->angle = state->angle;
838 ret->completed = state->completed;
839 ret->movecount = state->movecount;
840
841 return ret;
842 }
843
844 void free_game(game_state *state)
845 {
846 sfree(state);
847 }
848
849 game_state *make_move(game_state *from, int x, int y, int button)
850 {
851 int direction;
852 int pkey[2], skey[2], dkey[2];
853 float points[4];
854 game_state *ret;
855 float angle;
856 int i, j, dest, mask;
857 struct solid *poly;
858
859 /*
860 * All moves are made with the cursor keys.
861 */
862 if (button == CURSOR_UP)
863 direction = UP;
864 else if (button == CURSOR_DOWN)
865 direction = DOWN;
866 else if (button == CURSOR_LEFT)
867 direction = LEFT;
868 else if (button == CURSOR_RIGHT)
869 direction = RIGHT;
870 else if (button == CURSOR_UP_LEFT)
871 direction = UP_LEFT;
872 else if (button == CURSOR_DOWN_LEFT)
873 direction = DOWN_LEFT;
874 else if (button == CURSOR_UP_RIGHT)
875 direction = UP_RIGHT;
876 else if (button == CURSOR_DOWN_RIGHT)
877 direction = DOWN_RIGHT;
878 else
879 return NULL;
880
881 /*
882 * Find the two points in the current grid square which
883 * correspond to this move.
884 */
885 mask = from->squares[from->current].directions[direction];
886 if (mask == 0)
887 return NULL;
888 for (i = j = 0; i < from->squares[from->current].npoints; i++)
889 if (mask & (1 << i)) {
890 points[j*2] = from->squares[from->current].points[i*2];
891 points[j*2+1] = from->squares[from->current].points[i*2+1];
892 skey[j] = i;
893 j++;
894 }
895 assert(j == 2);
896
897 /*
898 * Now find the other grid square which shares those points.
899 * This is our move destination.
900 */
901 dest = -1;
902 for (i = 0; i < from->nsquares; i++)
903 if (i != from->current) {
904 int match = 0;
905 float dist;
906
907 for (j = 0; j < from->squares[i].npoints; j++) {
908 dist = (SQ(from->squares[i].points[j*2] - points[0]) +
909 SQ(from->squares[i].points[j*2+1] - points[1]));
910 if (dist < 0.1)
911 dkey[match++] = j;
912 dist = (SQ(from->squares[i].points[j*2] - points[2]) +
913 SQ(from->squares[i].points[j*2+1] - points[3]));
914 if (dist < 0.1)
915 dkey[match++] = j;
916 }
917
918 if (match == 2) {
919 dest = i;
920 break;
921 }
922 }
923
924 if (dest < 0)
925 return NULL;
926
927 ret = dup_game(from);
928 ret->current = i;
929
930 /*
931 * So we know what grid square we're aiming for, and we also
932 * know the two key points (as indices in both the source and
933 * destination grid squares) which are invariant between source
934 * and destination.
935 *
936 * Next we must roll the polyhedron on to that square. So we
937 * find the indices of the key points within the polyhedron's
938 * vertex array, then use those in a call to transform_poly,
939 * and align the result on the new grid square.
940 */
941 {
942 int all_pkey[4];
943 align_poly(from->solid, &from->squares[from->current], all_pkey);
944 pkey[0] = all_pkey[skey[0]];
945 pkey[1] = all_pkey[skey[1]];
946 /*
947 * Now pkey[0] corresponds to skey[0] and dkey[0], and
948 * likewise [1].
949 */
950 }
951
952 /*
953 * Now find the angle through which to rotate the polyhedron.
954 * Do this by finding the two faces that share the two vertices
955 * we've found, and taking the dot product of their normals.
956 */
957 {
958 int f[2], nf = 0;
959 float dp;
960
961 for (i = 0; i < from->solid->nfaces; i++) {
962 int match = 0;
963 for (j = 0; j < from->solid->order; j++)
964 if (from->solid->faces[i*from->solid->order + j] == pkey[0] ||
965 from->solid->faces[i*from->solid->order + j] == pkey[1])
966 match++;
967 if (match == 2) {
968 assert(nf < 2);
969 f[nf++] = i;
970 }
971 }
972
973 assert(nf == 2);
974
975 dp = 0;
976 for (i = 0; i < 3; i++)
977 dp += (from->solid->normals[f[0]*3+i] *
978 from->solid->normals[f[1]*3+i]);
979 angle = (float)acos(dp);
980 }
981
982 /*
983 * Now transform the polyhedron. We aren't entirely sure
984 * whether we need to rotate through angle or -angle, and the
985 * simplest way round this is to try both and see which one
986 * aligns successfully!
987 *
988 * Unfortunately, _both_ will align successfully if this is a
989 * cube, which won't tell us anything much. So for that
990 * particular case, I resort to gross hackery: I simply negate
991 * the angle before trying the alignment, depending on the
992 * direction. Which directions work which way is determined by
993 * pure trial and error. I said it was gross :-/
994 */
995 {
996 int all_pkey[4];
997 int success;
998
999 if (from->solid->order == 4 && direction == UP)
1000 angle = -angle; /* HACK */
1001
1002 poly = transform_poly(from->solid,
1003 from->squares[from->current].flip,
1004 pkey[0], pkey[1], angle);
1005 flip_poly(poly, from->squares[ret->current].flip);
1006 success = align_poly(poly, &from->squares[ret->current], all_pkey);
1007
1008 if (!success) {
1009 angle = -angle;
1010 poly = transform_poly(from->solid,
1011 from->squares[from->current].flip,
1012 pkey[0], pkey[1], angle);
1013 flip_poly(poly, from->squares[ret->current].flip);
1014 success = align_poly(poly, &from->squares[ret->current], all_pkey);
1015 }
1016
1017 assert(success);
1018 }
1019
1020 /*
1021 * Now we have our rotated polyhedron, which we expect to be
1022 * exactly congruent to the one we started with - but with the
1023 * faces permuted. So we map that congruence and thereby figure
1024 * out how to permute the faces as a result of the polyhedron
1025 * having rolled.
1026 */
1027 {
1028 int *newcolours = snewn(from->solid->nfaces, int);
1029
1030 for (i = 0; i < from->solid->nfaces; i++)
1031 newcolours[i] = -1;
1032
1033 for (i = 0; i < from->solid->nfaces; i++) {
1034 int nmatch = 0;
1035
1036 /*
1037 * Now go through the transformed polyhedron's faces
1038 * and figure out which one's normal is approximately
1039 * equal to this one.
1040 */
1041 for (j = 0; j < poly->nfaces; j++) {
1042 float dist;
1043 int k;
1044
1045 dist = 0;
1046
1047 for (k = 0; k < 3; k++)
1048 dist += SQ(poly->normals[j*3+k] -
1049 from->solid->normals[i*3+k]);
1050
1051 if (APPROXEQ(dist, 0)) {
1052 nmatch++;
1053 newcolours[i] = ret->facecolours[j];
1054 }
1055 }
1056
1057 assert(nmatch == 1);
1058 }
1059
1060 for (i = 0; i < from->solid->nfaces; i++)
1061 assert(newcolours[i] != -1);
1062
1063 sfree(ret->facecolours);
1064 ret->facecolours = newcolours;
1065 }
1066
1067 /*
1068 * And finally, swap the colour between the bottom face of the
1069 * polyhedron and the face we've just landed on.
1070 *
1071 * We don't do this if the game is already complete, since we
1072 * allow the user to roll the fully blue polyhedron around the
1073 * grid as a feeble reward.
1074 */
1075 if (!ret->completed) {
1076 i = lowest_face(from->solid);
1077 j = ret->facecolours[i];
1078 ret->facecolours[i] = ret->squares[ret->current].blue;
1079 ret->squares[ret->current].blue = j;
1080
1081 /*
1082 * Detect game completion.
1083 */
1084 j = 0;
1085 for (i = 0; i < ret->solid->nfaces; i++)
1086 if (ret->facecolours[i])
1087 j++;
1088 if (j == ret->solid->nfaces)
1089 ret->completed = TRUE;
1090 }
1091
1092 sfree(poly);
1093
1094 /*
1095 * Align the normal polyhedron with its grid square, to get key
1096 * points for non-animated display.
1097 */
1098 {
1099 int pkey[4];
1100 int success;
1101
1102 success = align_poly(ret->solid, &ret->squares[ret->current], pkey);
1103 assert(success);
1104
1105 ret->dpkey[0] = pkey[0];
1106 ret->dpkey[1] = pkey[1];
1107 ret->dgkey[0] = 0;
1108 ret->dgkey[1] = 1;
1109 }
1110
1111
1112 ret->spkey[0] = pkey[0];
1113 ret->spkey[1] = pkey[1];
1114 ret->sgkey[0] = skey[0];
1115 ret->sgkey[1] = skey[1];
1116 ret->previous = from->current;
1117 ret->angle = angle;
1118 ret->movecount++;
1119
1120 return ret;
1121 }
1122
1123 /* ----------------------------------------------------------------------
1124 * Drawing routines.
1125 */
1126
1127 struct bbox {
1128 float l, r, u, d;
1129 };
1130
1131 struct game_drawstate {
1132 int ox, oy; /* pixel position of float origin */
1133 };
1134
1135 static void find_bbox_callback(void *ctx, struct grid_square *sq)
1136 {
1137 struct bbox *bb = (struct bbox *)ctx;
1138 int i;
1139
1140 for (i = 0; i < sq->npoints; i++) {
1141 if (bb->l > sq->points[i*2]) bb->l = sq->points[i*2];
1142 if (bb->r < sq->points[i*2]) bb->r = sq->points[i*2];
1143 if (bb->u > sq->points[i*2+1]) bb->u = sq->points[i*2+1];
1144 if (bb->d < sq->points[i*2+1]) bb->d = sq->points[i*2+1];
1145 }
1146 }
1147
1148 static struct bbox find_bbox(game_params *params)
1149 {
1150 struct bbox bb;
1151
1152 /*
1153 * These should be hugely more than the real bounding box will
1154 * be.
1155 */
1156 bb.l = 2.0F * (params->d1 + params->d2);
1157 bb.r = -2.0F * (params->d1 + params->d2);
1158 bb.u = 2.0F * (params->d1 + params->d2);
1159 bb.d = -2.0F * (params->d1 + params->d2);
1160 enum_grid_squares(params, find_bbox_callback, &bb);
1161
1162 return bb;
1163 }
1164
1165 void game_size(game_params *params, int *x, int *y)
1166 {
1167 struct bbox bb = find_bbox(params);
1168 *x = (int)((bb.r - bb.l + 2*solids[params->solid]->border) * GRID_SCALE);
1169 *y = (int)((bb.d - bb.u + 2*solids[params->solid]->border) * GRID_SCALE);
1170 }
1171
1172 float *game_colours(frontend *fe, game_state *state, int *ncolours)
1173 {
1174 float *ret = snewn(3 * NCOLOURS, float);
1175
1176 frontend_default_colour(fe, &ret[COL_BACKGROUND * 3]);
1177
1178 ret[COL_BORDER * 3 + 0] = 0.0;
1179 ret[COL_BORDER * 3 + 1] = 0.0;
1180 ret[COL_BORDER * 3 + 2] = 0.0;
1181
1182 ret[COL_BLUE * 3 + 0] = 0.0;
1183 ret[COL_BLUE * 3 + 1] = 0.0;
1184 ret[COL_BLUE * 3 + 2] = 1.0;
1185
1186 *ncolours = NCOLOURS;
1187 return ret;
1188 }
1189
1190 game_drawstate *game_new_drawstate(game_state *state)
1191 {
1192 struct game_drawstate *ds = snew(struct game_drawstate);
1193 struct bbox bb = find_bbox(&state->params);
1194
1195 ds->ox = (int)(-(bb.l - state->solid->border) * GRID_SCALE);
1196 ds->oy = (int)(-(bb.u - state->solid->border) * GRID_SCALE);
1197
1198 return ds;
1199 }
1200
1201 void game_free_drawstate(game_drawstate *ds)
1202 {
1203 sfree(ds);
1204 }
1205
1206 void game_redraw(frontend *fe, game_drawstate *ds, game_state *oldstate,
1207 game_state *state, float animtime)
1208 {
1209 int i, j;
1210 struct bbox bb = find_bbox(&state->params);
1211 struct solid *poly;
1212 int *pkey, *gkey;
1213 float t[3];
1214 float angle;
1215 game_state *newstate;
1216 int square;
1217
1218 draw_rect(fe, 0, 0, (int)((bb.r-bb.l+2.0F) * GRID_SCALE),
1219 (int)((bb.d-bb.u+2.0F) * GRID_SCALE), COL_BACKGROUND);
1220
1221 if (oldstate && oldstate->movecount > state->movecount) {
1222 game_state *t;
1223
1224 /*
1225 * This is an Undo. So reverse the order of the states, and
1226 * run the roll timer backwards.
1227 */
1228 t = oldstate;
1229 oldstate = state;
1230 state = t;
1231
1232 animtime = ROLLTIME - animtime;
1233 }
1234
1235 if (!oldstate) {
1236 oldstate = state;
1237 angle = 0.0;
1238 square = state->current;
1239 pkey = state->dpkey;
1240 gkey = state->dgkey;
1241 } else {
1242 angle = state->angle * animtime / ROLLTIME;
1243 square = state->previous;
1244 pkey = state->spkey;
1245 gkey = state->sgkey;
1246 }
1247 newstate = state;
1248 state = oldstate;
1249
1250 for (i = 0; i < state->nsquares; i++) {
1251 int coords[8];
1252
1253 for (j = 0; j < state->squares[i].npoints; j++) {
1254 coords[2*j] = ((int)(state->squares[i].points[2*j] * GRID_SCALE)
1255 + ds->ox);
1256 coords[2*j+1] = ((int)(state->squares[i].points[2*j+1]*GRID_SCALE)
1257 + ds->oy);
1258 }
1259
1260 draw_polygon(fe, coords, state->squares[i].npoints, TRUE,
1261 state->squares[i].blue ? COL_BLUE : COL_BACKGROUND);
1262 draw_polygon(fe, coords, state->squares[i].npoints, FALSE, COL_BORDER);
1263 }
1264
1265 /*
1266 * Now compute and draw the polyhedron.
1267 */
1268 poly = transform_poly(state->solid, state->squares[square].flip,
1269 pkey[0], pkey[1], angle);
1270
1271 /*
1272 * Compute the translation required to align the two key points
1273 * on the polyhedron with the same key points on the current
1274 * face.
1275 */
1276 for (i = 0; i < 3; i++) {
1277 float tc = 0.0;
1278
1279 for (j = 0; j < 2; j++) {
1280 float grid_coord;
1281
1282 if (i < 2) {
1283 grid_coord =
1284 state->squares[square].points[gkey[j]*2+i];
1285 } else {
1286 grid_coord = 0.0;
1287 }
1288
1289 tc += (grid_coord - poly->vertices[pkey[j]*3+i]);
1290 }
1291
1292 t[i] = tc / 2;
1293 }
1294 for (i = 0; i < poly->nvertices; i++)
1295 for (j = 0; j < 3; j++)
1296 poly->vertices[i*3+j] += t[j];
1297
1298 /*
1299 * Now actually draw each face.
1300 */
1301 for (i = 0; i < poly->nfaces; i++) {
1302 float points[8];
1303 int coords[8];
1304
1305 for (j = 0; j < poly->order; j++) {
1306 int f = poly->faces[i*poly->order + j];
1307 points[j*2] = (poly->vertices[f*3+0] -
1308 poly->vertices[f*3+2] * poly->shear);
1309 points[j*2+1] = (poly->vertices[f*3+1] -
1310 poly->vertices[f*3+2] * poly->shear);
1311 }
1312
1313 for (j = 0; j < poly->order; j++) {
1314 coords[j*2] = (int)(points[j*2] * GRID_SCALE) + ds->ox;
1315 coords[j*2+1] = (int)(points[j*2+1] * GRID_SCALE) + ds->oy;
1316 }
1317
1318 /*
1319 * Find out whether these points are in a clockwise or
1320 * anticlockwise arrangement. If the latter, discard the
1321 * face because it's facing away from the viewer.
1322 *
1323 * This would involve fiddly winding-number stuff for a
1324 * general polygon, but for the simple parallelograms we'll
1325 * be seeing here, all we have to do is check whether the
1326 * corners turn right or left. So we'll take the vector
1327 * from point 0 to point 1, turn it right 90 degrees,
1328 * and check the sign of the dot product with that and the
1329 * next vector (point 1 to point 2).
1330 */
1331 {
1332 float v1x = points[2]-points[0];
1333 float v1y = points[3]-points[1];
1334 float v2x = points[4]-points[2];
1335 float v2y = points[5]-points[3];
1336 float dp = v1x * v2y - v1y * v2x;
1337
1338 if (dp <= 0)
1339 continue;
1340 }
1341
1342 draw_polygon(fe, coords, poly->order, TRUE,
1343 state->facecolours[i] ? COL_BLUE : COL_BACKGROUND);
1344 draw_polygon(fe, coords, poly->order, FALSE, COL_BORDER);
1345 }
1346 sfree(poly);
1347
1348 draw_update(fe, 0, 0, (int)((bb.r-bb.l+2.0F) * GRID_SCALE),
1349 (int)((bb.d-bb.u+2.0F) * GRID_SCALE));
1350 }
1351
1352 float game_anim_length(game_state *oldstate, game_state *newstate)
1353 {
1354 return ROLLTIME;
1355 }