Have each game declare a name which is used for window titles etc.
[sgt/puzzles] / cube.c
CommitLineData
720a8fb7 1/*
2 * cube.c: Cube game.
3 */
1482ee76 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
0c490335 13const char *const game_name = "Cube";
14
1482ee76 15#define MAXVERTICES 20
16#define MAXFACES 20
17#define MAXORDER 4
18struct solid {
19 int nvertices;
20 float vertices[MAXVERTICES * 3]; /* 3*npoints coordinates */
21 int order;
22 int nfaces;
23 int faces[MAXFACES * MAXORDER]; /* order*nfaces point indices */
24 float normals[MAXFACES * 3]; /* 3*npoints vector components */
25 float shear; /* isometric shear for nice drawing */
eb2ad6f1 26 float border; /* border required around arena */
1482ee76 27};
28
29static const struct solid tetrahedron = {
30 4,
31 {
03f856c4 32 0.0F, -0.57735026919F, -0.20412414523F,
33 -0.5F, 0.28867513459F, -0.20412414523F,
34 0.0F, -0.0F, 0.6123724357F,
35 0.5F, 0.28867513459F, -0.20412414523F,
1482ee76 36 },
37 3, 4,
38 {
39 0,2,1, 3,1,2, 2,0,3, 1,3,0
40 },
41 {
03f856c4 42 -0.816496580928F, -0.471404520791F, 0.333333333334F,
43 0.0F, 0.942809041583F, 0.333333333333F,
44 0.816496580928F, -0.471404520791F, 0.333333333334F,
45 0.0F, 0.0F, -1.0F,
1482ee76 46 },
03f856c4 47 0.0F, 0.3F
1482ee76 48};
49
50static const struct solid cube = {
51 8,
52 {
03f856c4 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 +0.5F,-0.5F,-0.5F, +0.5F,-0.5F,+0.5F,
56 +0.5F,+0.5F,-0.5F, +0.5F,+0.5F,+0.5F,
1482ee76 57 },
58 4, 6,
59 {
60 0,1,3,2, 1,5,7,3, 5,4,6,7, 4,0,2,6, 0,4,5,1, 3,7,6,2
61 },
62 {
03f856c4 63 -1.0F,0.0F,0.0F, 0.0F,0.0F,+1.0F,
64 +1.0F,0.0F,0.0F, 0.0F,0.0F,-1.0F,
65 0.0F,-1.0F,0.0F, 0.0F,+1.0F,0.0F
1482ee76 66 },
03f856c4 67 0.3F, 0.5F
1482ee76 68};
69
70static const struct solid octahedron = {
71 6,
72 {
03f856c4 73 -0.5F, -0.28867513459472505F, 0.4082482904638664F,
74 0.5F, 0.28867513459472505F, -0.4082482904638664F,
75 -0.5F, 0.28867513459472505F, -0.4082482904638664F,
76 0.5F, -0.28867513459472505F, 0.4082482904638664F,
77 0.0F, -0.57735026918945009F, -0.4082482904638664F,
78 0.0F, 0.57735026918945009F, 0.4082482904638664F,
1482ee76 79 },
80 3, 8,
81 {
82 4,0,2, 0,5,2, 0,4,3, 5,0,3, 1,4,2, 5,1,2, 4,1,3, 1,5,3
83 },
84 {
03f856c4 85 -0.816496580928F, -0.471404520791F, -0.333333333334F,
86 -0.816496580928F, 0.471404520791F, 0.333333333334F,
87 0.0F, -0.942809041583F, 0.333333333333F,
88 0.0F, 0.0F, 1.0F,
89 0.0F, 0.0F, -1.0F,
90 0.0F, 0.942809041583F, -0.333333333333F,
91 0.816496580928F, -0.471404520791F, -0.333333333334F,
92 0.816496580928F, 0.471404520791F, 0.333333333334F,
1482ee76 93 },
03f856c4 94 0.0F, 0.5F
1482ee76 95};
96
97static const struct solid icosahedron = {
98 12,
99 {
03f856c4 100 0.0F, 0.57735026919F, 0.75576131408F,
101 0.0F, -0.93417235896F, 0.17841104489F,
102 0.0F, 0.93417235896F, -0.17841104489F,
103 0.0F, -0.57735026919F, -0.75576131408F,
104 -0.5F, -0.28867513459F, 0.75576131408F,
105 -0.5F, 0.28867513459F, -0.75576131408F,
106 0.5F, -0.28867513459F, 0.75576131408F,
107 0.5F, 0.28867513459F, -0.75576131408F,
108 -0.80901699437F, 0.46708617948F, 0.17841104489F,
109 0.80901699437F, 0.46708617948F, 0.17841104489F,
110 -0.80901699437F, -0.46708617948F, -0.17841104489F,
111 0.80901699437F, -0.46708617948F, -0.17841104489F,
1482ee76 112 },
113 3, 20,
114 {
115 8,0,2, 0,9,2, 1,10,3, 11,1,3, 0,4,6,
116 4,1,6, 5,2,7, 3,5,7, 4,8,10, 8,5,10,
117 9,6,11, 7,9,11, 0,8,4, 9,0,6, 10,1,4,
118 1,11,6, 8,2,5, 2,9,7, 3,10,5, 11,3,7,
119 },
120 {
03f856c4 121 -0.356822089773F, 0.87267799625F, 0.333333333333F,
122 0.356822089773F, 0.87267799625F, 0.333333333333F,
123 -0.356822089773F, -0.87267799625F, -0.333333333333F,
124 0.356822089773F, -0.87267799625F, -0.333333333333F,
125 -0.0F, 0.0F, 1.0F,
126 0.0F, -0.666666666667F, 0.745355992501F,
127 0.0F, 0.666666666667F, -0.745355992501F,
128 0.0F, 0.0F, -1.0F,
129 -0.934172358963F, -0.12732200375F, 0.333333333333F,
130 -0.934172358963F, 0.12732200375F, -0.333333333333F,
131 0.934172358963F, -0.12732200375F, 0.333333333333F,
132 0.934172358963F, 0.12732200375F, -0.333333333333F,
133 -0.57735026919F, 0.333333333334F, 0.745355992501F,
134 0.57735026919F, 0.333333333334F, 0.745355992501F,
135 -0.57735026919F, -0.745355992501F, 0.333333333334F,
136 0.57735026919F, -0.745355992501F, 0.333333333334F,
137 -0.57735026919F, 0.745355992501F, -0.333333333334F,
138 0.57735026919F, 0.745355992501F, -0.333333333334F,
139 -0.57735026919F, -0.333333333334F, -0.745355992501F,
140 0.57735026919F, -0.333333333334F, -0.745355992501F,
1482ee76 141 },
03f856c4 142 0.0F, 0.8F
1482ee76 143};
144
145enum {
146 TETRAHEDRON, CUBE, OCTAHEDRON, ICOSAHEDRON
147};
148static const struct solid *solids[] = {
149 &tetrahedron, &cube, &octahedron, &icosahedron
150};
151
152enum {
153 COL_BACKGROUND,
154 COL_BORDER,
155 COL_BLUE,
156 NCOLOURS
157};
158
c71454c0 159enum { LEFT, RIGHT, UP, DOWN, UP_LEFT, UP_RIGHT, DOWN_LEFT, DOWN_RIGHT };
1482ee76 160
03f856c4 161#define GRID_SCALE 48.0F
162#define ROLLTIME 0.1F
1482ee76 163
164#define SQ(x) ( (x) * (x) )
165
166#define MATMUL(ra,m,a) do { \
167 float rx, ry, rz, xx = (a)[0], yy = (a)[1], zz = (a)[2], *mat = (m); \
168 rx = mat[0] * xx + mat[3] * yy + mat[6] * zz; \
169 ry = mat[1] * xx + mat[4] * yy + mat[7] * zz; \
170 rz = mat[2] * xx + mat[5] * yy + mat[8] * zz; \
171 (ra)[0] = rx; (ra)[1] = ry; (ra)[2] = rz; \
172} while (0)
173
174#define APPROXEQ(x,y) ( SQ(x-y) < 0.1 )
175
176struct grid_square {
177 float x, y;
178 int npoints;
179 float points[8]; /* maximum */
c71454c0 180 int directions[8]; /* bit masks showing point pairs */
1482ee76 181 int flip;
182 int blue;
183 int tetra_class;
184};
185
186struct game_params {
187 int solid;
188 /*
189 * Grid dimensions. For a square grid these are width and
190 * height respectively; otherwise the grid is a hexagon, with
191 * the top side and the two lower diagonals having length d1
192 * and the remaining three sides having length d2 (so that
193 * d1==d2 gives a regular hexagon, and d2==0 gives a triangle).
194 */
195 int d1, d2;
196};
197
198struct game_state {
199 struct game_params params;
200 const struct solid *solid;
201 int *facecolours;
202 struct grid_square *squares;
203 int nsquares;
204 int current; /* index of current grid square */
205 int sgkey[2]; /* key-point indices into grid sq */
206 int dgkey[2]; /* key-point indices into grid sq */
207 int spkey[2]; /* key-point indices into polyhedron */
208 int dpkey[2]; /* key-point indices into polyhedron */
209 int previous;
210 float angle;
211 int completed;
212 int movecount;
213};
214
215game_params *default_params(void)
216{
217 game_params *ret = snew(game_params);
218
219 ret->solid = CUBE;
220 ret->d1 = 4;
221 ret->d2 = 4;
222
223 return ret;
224}
225
eb2ad6f1 226int game_fetch_preset(int i, char **name, game_params **params)
227{
228 game_params *ret = snew(game_params);
229 char *str;
230
231 switch (i) {
232 case 0:
233 str = "Cube";
234 ret->solid = CUBE;
235 ret->d1 = 4;
236 ret->d2 = 4;
237 break;
238 case 1:
239 str = "Tetrahedron";
240 ret->solid = TETRAHEDRON;
241 ret->d1 = 2;
242 ret->d2 = 1;
243 break;
244 case 2:
245 str = "Octahedron";
246 ret->solid = OCTAHEDRON;
247 ret->d1 = 2;
248 ret->d2 = 2;
249 break;
250 case 3:
251 str = "Icosahedron";
252 ret->solid = ICOSAHEDRON;
253 ret->d1 = 3;
254 ret->d2 = 3;
255 break;
256 default:
257 sfree(ret);
258 return FALSE;
259 }
260
261 *name = dupstr(str);
262 *params = ret;
263 return TRUE;
264}
265
1482ee76 266void free_params(game_params *params)
267{
268 sfree(params);
269}
270
eb2ad6f1 271game_params *dup_params(game_params *params)
272{
273 game_params *ret = snew(game_params);
274 *ret = *params; /* structure copy */
275 return ret;
276}
277
1482ee76 278static void enum_grid_squares(game_params *params,
279 void (*callback)(void *, struct grid_square *),
280 void *ctx)
281{
282 const struct solid *solid = solids[params->solid];
283
284 if (solid->order == 4) {
285 int x, y;
286
287 for (x = 0; x < params->d1; x++)
288 for (y = 0; y < params->d2; y++) {
289 struct grid_square sq;
290
03f856c4 291 sq.x = (float)x;
292 sq.y = (float)y;
293 sq.points[0] = x - 0.5F;
294 sq.points[1] = y - 0.5F;
295 sq.points[2] = x - 0.5F;
296 sq.points[3] = y + 0.5F;
297 sq.points[4] = x + 0.5F;
298 sq.points[5] = y + 0.5F;
299 sq.points[6] = x + 0.5F;
300 sq.points[7] = y - 0.5F;
1482ee76 301 sq.npoints = 4;
302
303 sq.directions[LEFT] = 0x03; /* 0,1 */
304 sq.directions[RIGHT] = 0x0C; /* 2,3 */
305 sq.directions[UP] = 0x09; /* 0,3 */
306 sq.directions[DOWN] = 0x06; /* 1,2 */
c71454c0 307 sq.directions[UP_LEFT] = 0; /* no diagonals in a square */
308 sq.directions[UP_RIGHT] = 0; /* no diagonals in a square */
309 sq.directions[DOWN_LEFT] = 0; /* no diagonals in a square */
310 sq.directions[DOWN_RIGHT] = 0; /* no diagonals in a square */
1482ee76 311
312 sq.flip = FALSE;
313
314 /*
315 * This is supremely irrelevant, but just to avoid
316 * having any uninitialised structure members...
317 */
318 sq.tetra_class = 0;
319
320 callback(ctx, &sq);
321 }
322 } else {
323 int row, rowlen, other, i, firstix = -1;
03f856c4 324 float theight = (float)(sqrt(3) / 2.0);
1482ee76 325
326 for (row = 0; row < params->d1 + params->d2; row++) {
327 if (row < params->d1) {
328 other = +1;
329 rowlen = row + params->d2;
330 } else {
331 other = -1;
332 rowlen = 2*params->d1 + params->d2 - row;
333 }
334
335 /*
336 * There are `rowlen' down-pointing triangles.
337 */
338 for (i = 0; i < rowlen; i++) {
339 struct grid_square sq;
340 int ix;
341 float x, y;
342
343 ix = (2 * i - (rowlen-1));
03f856c4 344 x = ix * 0.5F;
1482ee76 345 y = theight * row;
346 sq.x = x;
347 sq.y = y + theight / 3;
03f856c4 348 sq.points[0] = x - 0.5F;
1482ee76 349 sq.points[1] = y;
350 sq.points[2] = x;
351 sq.points[3] = y + theight;
03f856c4 352 sq.points[4] = x + 0.5F;
1482ee76 353 sq.points[5] = y;
354 sq.npoints = 3;
355
356 sq.directions[LEFT] = 0x03; /* 0,1 */
357 sq.directions[RIGHT] = 0x06; /* 1,2 */
358 sq.directions[UP] = 0x05; /* 0,2 */
359 sq.directions[DOWN] = 0; /* invalid move */
360
c71454c0 361 /*
362 * Down-pointing triangle: both the up diagonals go
363 * up, and the down ones go left and right.
364 */
365 sq.directions[UP_LEFT] = sq.directions[UP_RIGHT] =
366 sq.directions[UP];
367 sq.directions[DOWN_LEFT] = sq.directions[LEFT];
368 sq.directions[DOWN_RIGHT] = sq.directions[RIGHT];
369
1482ee76 370 sq.flip = TRUE;
371
372 if (firstix < 0)
373 firstix = ix & 3;
374 ix -= firstix;
375 sq.tetra_class = ((row+(ix&1)) & 2) ^ (ix & 3);
376
377 callback(ctx, &sq);
378 }
379
380 /*
381 * There are `rowlen+other' up-pointing triangles.
382 */
383 for (i = 0; i < rowlen+other; i++) {
384 struct grid_square sq;
385 int ix;
386 float x, y;
387
388 ix = (2 * i - (rowlen+other-1));
03f856c4 389 x = ix * 0.5F;
1482ee76 390 y = theight * row;
391 sq.x = x;
392 sq.y = y + 2*theight / 3;
03f856c4 393 sq.points[0] = x + 0.5F;
1482ee76 394 sq.points[1] = y + theight;
395 sq.points[2] = x;
396 sq.points[3] = y;
03f856c4 397 sq.points[4] = x - 0.5F;
1482ee76 398 sq.points[5] = y + theight;
399 sq.npoints = 3;
400
401 sq.directions[LEFT] = 0x06; /* 1,2 */
402 sq.directions[RIGHT] = 0x03; /* 0,1 */
403 sq.directions[DOWN] = 0x05; /* 0,2 */
404 sq.directions[UP] = 0; /* invalid move */
405
c71454c0 406 /*
407 * Up-pointing triangle: both the down diagonals go
408 * down, and the up ones go left and right.
409 */
410 sq.directions[DOWN_LEFT] = sq.directions[DOWN_RIGHT] =
411 sq.directions[DOWN];
412 sq.directions[UP_LEFT] = sq.directions[LEFT];
413 sq.directions[UP_RIGHT] = sq.directions[RIGHT];
414
1482ee76 415 sq.flip = FALSE;
416
417 if (firstix < 0)
418 firstix = ix;
419 ix -= firstix;
420 sq.tetra_class = ((row+(ix&1)) & 2) ^ (ix & 3);
421
422 callback(ctx, &sq);
423 }
424 }
425 }
426}
427
428static int grid_area(int d1, int d2, int order)
429{
430 /*
431 * An NxM grid of squares has NM squares in it.
432 *
433 * A grid of triangles with dimensions A and B has a total of
434 * A^2 + B^2 + 4AB triangles in it. (You can divide it up into
435 * a side-A triangle containing A^2 subtriangles, a side-B
436 * triangle containing B^2, and two congruent parallelograms,
437 * each with side lengths A and B, each therefore containing AB
438 * two-triangle rhombuses.)
439 */
440 if (order == 4)
441 return d1 * d2;
442 else
443 return d1*d1 + d2*d2 + 4*d1*d2;
444}
445
446struct grid_data {
447 int *gridptrs[4];
448 int nsquares[4];
449 int nclasses;
450 int squareindex;
451};
452
453static void classify_grid_square_callback(void *ctx, struct grid_square *sq)
454{
455 struct grid_data *data = (struct grid_data *)ctx;
456 int thisclass;
457
458 if (data->nclasses == 4)
459 thisclass = sq->tetra_class;
460 else if (data->nclasses == 2)
461 thisclass = sq->flip;
462 else
463 thisclass = 0;
464
465 data->gridptrs[thisclass][data->nsquares[thisclass]++] =
466 data->squareindex++;
467}
468
469char *new_game_seed(game_params *params)
470{
471 struct grid_data data;
472 int i, j, k, m, area, facesperclass;
473 int *flags;
474 char *seed, *p;
475
476 /*
477 * Enumerate the grid squares, dividing them into equivalence
478 * classes as appropriate. (For the tetrahedron, there is one
479 * equivalence class for each face; for the octahedron there
480 * are two classes; for the other two solids there's only one.)
481 */
482
483 area = grid_area(params->d1, params->d2, solids[params->solid]->order);
484 if (params->solid == TETRAHEDRON)
485 data.nclasses = 4;
486 else if (params->solid == OCTAHEDRON)
487 data.nclasses = 2;
488 else
489 data.nclasses = 1;
490 data.gridptrs[0] = snewn(data.nclasses * area, int);
491 for (i = 0; i < data.nclasses; i++) {
492 data.gridptrs[i] = data.gridptrs[0] + i * area;
493 data.nsquares[i] = 0;
494 }
495 data.squareindex = 0;
496 enum_grid_squares(params, classify_grid_square_callback, &data);
497
498 facesperclass = solids[params->solid]->nfaces / data.nclasses;
499
500 for (i = 0; i < data.nclasses; i++)
501 assert(data.nsquares[i] >= facesperclass);
502 assert(data.squareindex == area);
503
504 /*
505 * So now we know how many faces to allocate in each class. Get
506 * on with it.
507 */
508 flags = snewn(area, int);
509 for (i = 0; i < area; i++)
510 flags[i] = FALSE;
511
512 for (i = 0; i < data.nclasses; i++) {
513 for (j = 0; j < facesperclass; j++) {
514 unsigned long divisor = RAND_MAX / data.nsquares[i];
515 unsigned long max = divisor * data.nsquares[i];
03f856c4 516 unsigned long n;
1482ee76 517
518 do {
519 n = rand();
520 } while (n >= max);
521
522 n /= divisor;
523
524 assert(!flags[data.gridptrs[i][n]]);
525 flags[data.gridptrs[i][n]] = TRUE;
526
527 /*
528 * Move everything else up the array. I ought to use a
529 * better data structure for this, but for such small
530 * numbers it hardly seems worth the effort.
531 */
03f856c4 532 while ((int)n < data.nsquares[i]-1) {
1482ee76 533 data.gridptrs[i][n] = data.gridptrs[i][n+1];
534 n++;
535 }
536 data.nsquares[i]--;
537 }
538 }
539
540 /*
541 * Now we know precisely which squares are blue. Encode this
542 * information in hex. While we're looping over this, collect
543 * the non-blue squares into a list in the now-unused gridptrs
544 * array.
545 */
546 seed = snewn(area / 4 + 40, char);
547 p = seed;
548 j = 0;
549 k = 8;
550 m = 0;
551 for (i = 0; i < area; i++) {
552 if (flags[i]) {
553 j |= k;
554 } else {
555 data.gridptrs[0][m++] = i;
556 }
557 k >>= 1;
558 if (!k) {
559 *p++ = "0123456789ABCDEF"[j];
560 k = 8;
561 j = 0;
562 }
563 }
564 if (k != 8)
565 *p++ = "0123456789ABCDEF"[j];
566
567 /*
568 * Choose a non-blue square for the polyhedron.
569 */
570 {
571 unsigned long divisor = RAND_MAX / m;
572 unsigned long max = divisor * m;
03f856c4 573 unsigned long n;
1482ee76 574
575 do {
576 n = rand();
577 } while (n >= max);
578
579 n /= divisor;
580
581 sprintf(p, ":%d", data.gridptrs[0][n]);
582 }
583
584 sfree(data.gridptrs[0]);
585 sfree(flags);
586
587 return seed;
588}
589
590static void add_grid_square_callback(void *ctx, struct grid_square *sq)
591{
592 game_state *state = (game_state *)ctx;
593
594 state->squares[state->nsquares] = *sq; /* structure copy */
595 state->squares[state->nsquares].blue = FALSE;
596 state->nsquares++;
597}
598
599static int lowest_face(const struct solid *solid)
600{
601 int i, j, best;
602 float zmin;
603
604 best = 0;
605 zmin = 0.0;
606 for (i = 0; i < solid->nfaces; i++) {
607 float z = 0;
608
609 for (j = 0; j < solid->order; j++) {
610 int f = solid->faces[i*solid->order + j];
611 z += solid->vertices[f*3+2];
612 }
613
614 if (i == 0 || zmin > z) {
615 zmin = z;
616 best = i;
617 }
618 }
619
620 return best;
621}
622
623static int align_poly(const struct solid *solid, struct grid_square *sq,
624 int *pkey)
625{
626 float zmin;
627 int i, j;
628 int flip = (sq->flip ? -1 : +1);
629
630 /*
631 * First, find the lowest z-coordinate present in the solid.
632 */
633 zmin = 0.0;
634 for (i = 0; i < solid->nvertices; i++)
635 if (zmin > solid->vertices[i*3+2])
636 zmin = solid->vertices[i*3+2];
637
638 /*
639 * Now go round the grid square. For each point in the grid
640 * square, we're looking for a point of the polyhedron with the
641 * same x- and y-coordinates (relative to the square's centre),
642 * and z-coordinate equal to zmin (near enough).
643 */
644 for (j = 0; j < sq->npoints; j++) {
645 int matches, index;
646
647 matches = 0;
648 index = -1;
649
650 for (i = 0; i < solid->nvertices; i++) {
651 float dist = 0;
652
653 dist += SQ(solid->vertices[i*3+0] * flip - sq->points[j*2+0] + sq->x);
654 dist += SQ(solid->vertices[i*3+1] * flip - sq->points[j*2+1] + sq->y);
655 dist += SQ(solid->vertices[i*3+2] - zmin);
656
657 if (dist < 0.1) {
658 matches++;
659 index = i;
660 }
661 }
662
663 if (matches != 1 || index < 0)
664 return FALSE;
665 pkey[j] = index;
666 }
667
668 return TRUE;
669}
670
671static void flip_poly(struct solid *solid, int flip)
672{
673 int i;
674
675 if (flip) {
676 for (i = 0; i < solid->nvertices; i++) {
677 solid->vertices[i*3+0] *= -1;
678 solid->vertices[i*3+1] *= -1;
679 }
680 for (i = 0; i < solid->nfaces; i++) {
681 solid->normals[i*3+0] *= -1;
682 solid->normals[i*3+1] *= -1;
683 }
684 }
685}
686
687static struct solid *transform_poly(const struct solid *solid, int flip,
688 int key0, int key1, float angle)
689{
690 struct solid *ret = snew(struct solid);
691 float vx, vy, ax, ay;
692 float vmatrix[9], amatrix[9], vmatrix2[9];
693 int i;
694
695 *ret = *solid; /* structure copy */
696
697 flip_poly(ret, flip);
698
699 /*
700 * Now rotate the polyhedron through the given angle. We must
701 * rotate about the Z-axis to bring the two vertices key0 and
702 * key1 into horizontal alignment, then rotate about the
703 * X-axis, then rotate back again.
704 */
705 vx = ret->vertices[key1*3+0] - ret->vertices[key0*3+0];
706 vy = ret->vertices[key1*3+1] - ret->vertices[key0*3+1];
707 assert(APPROXEQ(vx*vx + vy*vy, 1.0));
708
709 vmatrix[0] = vx; vmatrix[3] = vy; vmatrix[6] = 0;
710 vmatrix[1] = -vy; vmatrix[4] = vx; vmatrix[7] = 0;
711 vmatrix[2] = 0; vmatrix[5] = 0; vmatrix[8] = 1;
712
03f856c4 713 ax = (float)cos(angle);
714 ay = (float)sin(angle);
1482ee76 715
716 amatrix[0] = 1; amatrix[3] = 0; amatrix[6] = 0;
717 amatrix[1] = 0; amatrix[4] = ax; amatrix[7] = ay;
718 amatrix[2] = 0; amatrix[5] = -ay; amatrix[8] = ax;
719
720 memcpy(vmatrix2, vmatrix, sizeof(vmatrix));
721 vmatrix2[1] = vy;
722 vmatrix2[3] = -vy;
723
724 for (i = 0; i < ret->nvertices; i++) {
725 MATMUL(ret->vertices + 3*i, vmatrix, ret->vertices + 3*i);
726 MATMUL(ret->vertices + 3*i, amatrix, ret->vertices + 3*i);
727 MATMUL(ret->vertices + 3*i, vmatrix2, ret->vertices + 3*i);
728 }
729 for (i = 0; i < ret->nfaces; i++) {
730 MATMUL(ret->normals + 3*i, vmatrix, ret->normals + 3*i);
731 MATMUL(ret->normals + 3*i, amatrix, ret->normals + 3*i);
732 MATMUL(ret->normals + 3*i, vmatrix2, ret->normals + 3*i);
733 }
734
735 return ret;
736}
737
738game_state *new_game(game_params *params, char *seed)
739{
740 game_state *state = snew(game_state);
741 int area;
742
743 state->params = *params; /* structure copy */
744 state->solid = solids[params->solid];
745
746 area = grid_area(params->d1, params->d2, state->solid->order);
747 state->squares = snewn(area, struct grid_square);
748 state->nsquares = 0;
749 enum_grid_squares(params, add_grid_square_callback, state);
750 assert(state->nsquares == area);
751
752 state->facecolours = snewn(state->solid->nfaces, int);
753 memset(state->facecolours, 0, state->solid->nfaces * sizeof(int));
754
755 /*
756 * Set up the blue squares and polyhedron position according to
757 * the game seed.
758 */
759 {
760 char *p = seed;
761 int i, j, v;
762
763 j = 8;
764 v = 0;
765 for (i = 0; i < state->nsquares; i++) {
766 if (j == 8) {
767 v = *p++;
768 if (v >= '0' && v <= '9')
769 v -= '0';
770 else if (v >= 'A' && v <= 'F')
771 v -= 'A' - 10;
772 else if (v >= 'a' && v <= 'f')
773 v -= 'a' - 10;
774 else
775 break;
776 }
777 if (v & j)
778 state->squares[i].blue = TRUE;
779 j >>= 1;
780 if (j == 0)
781 j = 8;
782 }
783
784 if (*p == ':')
785 p++;
786
787 state->current = atoi(p);
788 if (state->current < 0 || state->current >= state->nsquares)
789 state->current = 0; /* got to do _something_ */
790 }
791
792 /*
793 * Align the polyhedron with its grid square and determine
794 * initial key points.
795 */
796 {
797 int pkey[4];
798 int ret;
799
800 ret = align_poly(state->solid, &state->squares[state->current], pkey);
801 assert(ret);
802
803 state->dpkey[0] = state->spkey[0] = pkey[0];
804 state->dpkey[1] = state->spkey[0] = pkey[1];
805 state->dgkey[0] = state->sgkey[0] = 0;
806 state->dgkey[1] = state->sgkey[0] = 1;
807 }
808
809 state->previous = state->current;
810 state->angle = 0.0;
811 state->completed = FALSE;
812 state->movecount = 0;
813
814 return state;
815}
816
817game_state *dup_game(game_state *state)
818{
819 game_state *ret = snew(game_state);
820
821 ret->params = state->params; /* structure copy */
822 ret->solid = state->solid;
823 ret->facecolours = snewn(ret->solid->nfaces, int);
824 memcpy(ret->facecolours, state->facecolours,
825 ret->solid->nfaces * sizeof(int));
826 ret->nsquares = state->nsquares;
827 ret->squares = snewn(ret->nsquares, struct grid_square);
828 memcpy(ret->squares, state->squares,
829 ret->nsquares * sizeof(struct grid_square));
830 ret->dpkey[0] = state->dpkey[0];
831 ret->dpkey[1] = state->dpkey[1];
832 ret->dgkey[0] = state->dgkey[0];
833 ret->dgkey[1] = state->dgkey[1];
834 ret->spkey[0] = state->spkey[0];
835 ret->spkey[1] = state->spkey[1];
836 ret->sgkey[0] = state->sgkey[0];
837 ret->sgkey[1] = state->sgkey[1];
838 ret->previous = state->previous;
839 ret->angle = state->angle;
840 ret->completed = state->completed;
841 ret->movecount = state->movecount;
842
843 return ret;
844}
845
846void free_game(game_state *state)
847{
848 sfree(state);
849}
850
851game_state *make_move(game_state *from, int x, int y, int button)
852{
853 int direction;
854 int pkey[2], skey[2], dkey[2];
855 float points[4];
856 game_state *ret;
857 float angle;
858 int i, j, dest, mask;
859 struct solid *poly;
860
861 /*
862 * All moves are made with the cursor keys.
863 */
864 if (button == CURSOR_UP)
865 direction = UP;
866 else if (button == CURSOR_DOWN)
867 direction = DOWN;
868 else if (button == CURSOR_LEFT)
869 direction = LEFT;
870 else if (button == CURSOR_RIGHT)
871 direction = RIGHT;
c71454c0 872 else if (button == CURSOR_UP_LEFT)
873 direction = UP_LEFT;
874 else if (button == CURSOR_DOWN_LEFT)
875 direction = DOWN_LEFT;
876 else if (button == CURSOR_UP_RIGHT)
877 direction = UP_RIGHT;
878 else if (button == CURSOR_DOWN_RIGHT)
879 direction = DOWN_RIGHT;
1482ee76 880 else
881 return NULL;
882
883 /*
884 * Find the two points in the current grid square which
885 * correspond to this move.
886 */
887 mask = from->squares[from->current].directions[direction];
888 if (mask == 0)
889 return NULL;
890 for (i = j = 0; i < from->squares[from->current].npoints; i++)
891 if (mask & (1 << i)) {
892 points[j*2] = from->squares[from->current].points[i*2];
893 points[j*2+1] = from->squares[from->current].points[i*2+1];
894 skey[j] = i;
895 j++;
896 }
897 assert(j == 2);
898
899 /*
900 * Now find the other grid square which shares those points.
901 * This is our move destination.
902 */
903 dest = -1;
904 for (i = 0; i < from->nsquares; i++)
905 if (i != from->current) {
906 int match = 0;
907 float dist;
908
909 for (j = 0; j < from->squares[i].npoints; j++) {
910 dist = (SQ(from->squares[i].points[j*2] - points[0]) +
911 SQ(from->squares[i].points[j*2+1] - points[1]));
912 if (dist < 0.1)
913 dkey[match++] = j;
914 dist = (SQ(from->squares[i].points[j*2] - points[2]) +
915 SQ(from->squares[i].points[j*2+1] - points[3]));
916 if (dist < 0.1)
917 dkey[match++] = j;
918 }
919
920 if (match == 2) {
921 dest = i;
922 break;
923 }
924 }
925
926 if (dest < 0)
927 return NULL;
928
929 ret = dup_game(from);
930 ret->current = i;
931
932 /*
933 * So we know what grid square we're aiming for, and we also
934 * know the two key points (as indices in both the source and
935 * destination grid squares) which are invariant between source
936 * and destination.
937 *
938 * Next we must roll the polyhedron on to that square. So we
939 * find the indices of the key points within the polyhedron's
940 * vertex array, then use those in a call to transform_poly,
941 * and align the result on the new grid square.
942 */
943 {
944 int all_pkey[4];
945 align_poly(from->solid, &from->squares[from->current], all_pkey);
946 pkey[0] = all_pkey[skey[0]];
947 pkey[1] = all_pkey[skey[1]];
948 /*
949 * Now pkey[0] corresponds to skey[0] and dkey[0], and
950 * likewise [1].
951 */
952 }
953
954 /*
955 * Now find the angle through which to rotate the polyhedron.
956 * Do this by finding the two faces that share the two vertices
957 * we've found, and taking the dot product of their normals.
958 */
959 {
960 int f[2], nf = 0;
961 float dp;
962
963 for (i = 0; i < from->solid->nfaces; i++) {
964 int match = 0;
965 for (j = 0; j < from->solid->order; j++)
966 if (from->solid->faces[i*from->solid->order + j] == pkey[0] ||
967 from->solid->faces[i*from->solid->order + j] == pkey[1])
968 match++;
969 if (match == 2) {
970 assert(nf < 2);
971 f[nf++] = i;
972 }
973 }
974
975 assert(nf == 2);
976
977 dp = 0;
978 for (i = 0; i < 3; i++)
979 dp += (from->solid->normals[f[0]*3+i] *
980 from->solid->normals[f[1]*3+i]);
03f856c4 981 angle = (float)acos(dp);
1482ee76 982 }
983
984 /*
985 * Now transform the polyhedron. We aren't entirely sure
986 * whether we need to rotate through angle or -angle, and the
987 * simplest way round this is to try both and see which one
988 * aligns successfully!
989 *
990 * Unfortunately, _both_ will align successfully if this is a
991 * cube, which won't tell us anything much. So for that
992 * particular case, I resort to gross hackery: I simply negate
993 * the angle before trying the alignment, depending on the
994 * direction. Which directions work which way is determined by
995 * pure trial and error. I said it was gross :-/
996 */
997 {
998 int all_pkey[4];
999 int success;
1000
1001 if (from->solid->order == 4 && direction == UP)
1002 angle = -angle; /* HACK */
1003
1004 poly = transform_poly(from->solid,
1005 from->squares[from->current].flip,
1006 pkey[0], pkey[1], angle);
1007 flip_poly(poly, from->squares[ret->current].flip);
1008 success = align_poly(poly, &from->squares[ret->current], all_pkey);
1009
1010 if (!success) {
1011 angle = -angle;
1012 poly = transform_poly(from->solid,
1013 from->squares[from->current].flip,
1014 pkey[0], pkey[1], angle);
1015 flip_poly(poly, from->squares[ret->current].flip);
1016 success = align_poly(poly, &from->squares[ret->current], all_pkey);
1017 }
1018
1019 assert(success);
1020 }
1021
1022 /*
1023 * Now we have our rotated polyhedron, which we expect to be
1024 * exactly congruent to the one we started with - but with the
1025 * faces permuted. So we map that congruence and thereby figure
1026 * out how to permute the faces as a result of the polyhedron
1027 * having rolled.
1028 */
1029 {
1030 int *newcolours = snewn(from->solid->nfaces, int);
1031
1032 for (i = 0; i < from->solid->nfaces; i++)
1033 newcolours[i] = -1;
1034
1035 for (i = 0; i < from->solid->nfaces; i++) {
1036 int nmatch = 0;
1037
1038 /*
1039 * Now go through the transformed polyhedron's faces
1040 * and figure out which one's normal is approximately
1041 * equal to this one.
1042 */
1043 for (j = 0; j < poly->nfaces; j++) {
1044 float dist;
1045 int k;
1046
1047 dist = 0;
1048
1049 for (k = 0; k < 3; k++)
1050 dist += SQ(poly->normals[j*3+k] -
1051 from->solid->normals[i*3+k]);
1052
1053 if (APPROXEQ(dist, 0)) {
1054 nmatch++;
1055 newcolours[i] = ret->facecolours[j];
1056 }
1057 }
1058
1059 assert(nmatch == 1);
1060 }
1061
1062 for (i = 0; i < from->solid->nfaces; i++)
1063 assert(newcolours[i] != -1);
1064
1065 sfree(ret->facecolours);
1066 ret->facecolours = newcolours;
1067 }
1068
1069 /*
1070 * And finally, swap the colour between the bottom face of the
1071 * polyhedron and the face we've just landed on.
1072 *
1073 * We don't do this if the game is already complete, since we
1074 * allow the user to roll the fully blue polyhedron around the
1075 * grid as a feeble reward.
1076 */
1077 if (!ret->completed) {
1078 i = lowest_face(from->solid);
1079 j = ret->facecolours[i];
1080 ret->facecolours[i] = ret->squares[ret->current].blue;
1081 ret->squares[ret->current].blue = j;
1082
1083 /*
1084 * Detect game completion.
1085 */
1086 j = 0;
1087 for (i = 0; i < ret->solid->nfaces; i++)
1088 if (ret->facecolours[i])
1089 j++;
1090 if (j == ret->solid->nfaces)
1091 ret->completed = TRUE;
1092 }
1093
1094 sfree(poly);
1095
1096 /*
1097 * Align the normal polyhedron with its grid square, to get key
1098 * points for non-animated display.
1099 */
1100 {
1101 int pkey[4];
1102 int success;
1103
1104 success = align_poly(ret->solid, &ret->squares[ret->current], pkey);
1105 assert(success);
1106
1107 ret->dpkey[0] = pkey[0];
1108 ret->dpkey[1] = pkey[1];
1109 ret->dgkey[0] = 0;
1110 ret->dgkey[1] = 1;
1111 }
1112
1113
1114 ret->spkey[0] = pkey[0];
1115 ret->spkey[1] = pkey[1];
1116 ret->sgkey[0] = skey[0];
1117 ret->sgkey[1] = skey[1];
1118 ret->previous = from->current;
1119 ret->angle = angle;
1120 ret->movecount++;
1121
1122 return ret;
1123}
1124
1125/* ----------------------------------------------------------------------
1126 * Drawing routines.
1127 */
1128
1129struct bbox {
1130 float l, r, u, d;
1131};
1132
1133struct game_drawstate {
1134 int ox, oy; /* pixel position of float origin */
1135};
1136
1137static void find_bbox_callback(void *ctx, struct grid_square *sq)
1138{
1139 struct bbox *bb = (struct bbox *)ctx;
1140 int i;
1141
1142 for (i = 0; i < sq->npoints; i++) {
1143 if (bb->l > sq->points[i*2]) bb->l = sq->points[i*2];
1144 if (bb->r < sq->points[i*2]) bb->r = sq->points[i*2];
1145 if (bb->u > sq->points[i*2+1]) bb->u = sq->points[i*2+1];
1146 if (bb->d < sq->points[i*2+1]) bb->d = sq->points[i*2+1];
1147 }
1148}
1149
1150static struct bbox find_bbox(game_params *params)
1151{
1152 struct bbox bb;
1153
1154 /*
1155 * These should be hugely more than the real bounding box will
1156 * be.
1157 */
03f856c4 1158 bb.l = 2.0F * (params->d1 + params->d2);
1159 bb.r = -2.0F * (params->d1 + params->d2);
1160 bb.u = 2.0F * (params->d1 + params->d2);
1161 bb.d = -2.0F * (params->d1 + params->d2);
1482ee76 1162 enum_grid_squares(params, find_bbox_callback, &bb);
1163
1164 return bb;
1165}
1166
1167void game_size(game_params *params, int *x, int *y)
1168{
1169 struct bbox bb = find_bbox(params);
03f856c4 1170 *x = (int)((bb.r - bb.l + 2*solids[params->solid]->border) * GRID_SCALE);
1171 *y = (int)((bb.d - bb.u + 2*solids[params->solid]->border) * GRID_SCALE);
1482ee76 1172}
1173
1174float *game_colours(frontend *fe, game_state *state, int *ncolours)
1175{
1176 float *ret = snewn(3 * NCOLOURS, float);
1177
1178 frontend_default_colour(fe, &ret[COL_BACKGROUND * 3]);
1179
1180 ret[COL_BORDER * 3 + 0] = 0.0;
1181 ret[COL_BORDER * 3 + 1] = 0.0;
1182 ret[COL_BORDER * 3 + 2] = 0.0;
1183
1184 ret[COL_BLUE * 3 + 0] = 0.0;
1185 ret[COL_BLUE * 3 + 1] = 0.0;
1186 ret[COL_BLUE * 3 + 2] = 1.0;
1187
1188 *ncolours = NCOLOURS;
1189 return ret;
1190}
1191
1192game_drawstate *game_new_drawstate(game_state *state)
1193{
1194 struct game_drawstate *ds = snew(struct game_drawstate);
1195 struct bbox bb = find_bbox(&state->params);
1196
03f856c4 1197 ds->ox = (int)(-(bb.l - state->solid->border) * GRID_SCALE);
1198 ds->oy = (int)(-(bb.u - state->solid->border) * GRID_SCALE);
1482ee76 1199
1200 return ds;
1201}
1202
1203void game_free_drawstate(game_drawstate *ds)
1204{
1205 sfree(ds);
1206}
1207
1208void game_redraw(frontend *fe, game_drawstate *ds, game_state *oldstate,
1209 game_state *state, float animtime)
1210{
1211 int i, j;
1212 struct bbox bb = find_bbox(&state->params);
1213 struct solid *poly;
1214 int *pkey, *gkey;
1215 float t[3];
1216 float angle;
1217 game_state *newstate;
1218 int square;
1219
03f856c4 1220 draw_rect(fe, 0, 0, (int)((bb.r-bb.l+2.0F) * GRID_SCALE),
1221 (int)((bb.d-bb.u+2.0F) * GRID_SCALE), COL_BACKGROUND);
1482ee76 1222
1223 if (oldstate && oldstate->movecount > state->movecount) {
1224 game_state *t;
1225
1226 /*
1227 * This is an Undo. So reverse the order of the states, and
1228 * run the roll timer backwards.
1229 */
1230 t = oldstate;
1231 oldstate = state;
1232 state = t;
1233
1234 animtime = ROLLTIME - animtime;
1235 }
1236
1237 if (!oldstate) {
1238 oldstate = state;
1239 angle = 0.0;
1240 square = state->current;
1241 pkey = state->dpkey;
1242 gkey = state->dgkey;
1243 } else {
1244 angle = state->angle * animtime / ROLLTIME;
1245 square = state->previous;
1246 pkey = state->spkey;
1247 gkey = state->sgkey;
1248 }
1249 newstate = state;
1250 state = oldstate;
1251
1252 for (i = 0; i < state->nsquares; i++) {
1253 int coords[8];
1254
1255 for (j = 0; j < state->squares[i].npoints; j++) {
03f856c4 1256 coords[2*j] = ((int)(state->squares[i].points[2*j] * GRID_SCALE)
1257 + ds->ox);
1258 coords[2*j+1] = ((int)(state->squares[i].points[2*j+1]*GRID_SCALE)
1259 + ds->oy);
1482ee76 1260 }
1261
1262 draw_polygon(fe, coords, state->squares[i].npoints, TRUE,
1263 state->squares[i].blue ? COL_BLUE : COL_BACKGROUND);
1264 draw_polygon(fe, coords, state->squares[i].npoints, FALSE, COL_BORDER);
1265 }
1266
1267 /*
1268 * Now compute and draw the polyhedron.
1269 */
1270 poly = transform_poly(state->solid, state->squares[square].flip,
1271 pkey[0], pkey[1], angle);
1272
1273 /*
1274 * Compute the translation required to align the two key points
1275 * on the polyhedron with the same key points on the current
1276 * face.
1277 */
1278 for (i = 0; i < 3; i++) {
1279 float tc = 0.0;
1280
1281 for (j = 0; j < 2; j++) {
1282 float grid_coord;
1283
1284 if (i < 2) {
1285 grid_coord =
1286 state->squares[square].points[gkey[j]*2+i];
1287 } else {
1288 grid_coord = 0.0;
1289 }
1290
1291 tc += (grid_coord - poly->vertices[pkey[j]*3+i]);
1292 }
1293
1294 t[i] = tc / 2;
1295 }
1296 for (i = 0; i < poly->nvertices; i++)
1297 for (j = 0; j < 3; j++)
1298 poly->vertices[i*3+j] += t[j];
1299
1300 /*
1301 * Now actually draw each face.
1302 */
1303 for (i = 0; i < poly->nfaces; i++) {
1304 float points[8];
1305 int coords[8];
1306
1307 for (j = 0; j < poly->order; j++) {
1308 int f = poly->faces[i*poly->order + j];
1309 points[j*2] = (poly->vertices[f*3+0] -
1310 poly->vertices[f*3+2] * poly->shear);
1311 points[j*2+1] = (poly->vertices[f*3+1] -
1312 poly->vertices[f*3+2] * poly->shear);
1313 }
1314
1315 for (j = 0; j < poly->order; j++) {
03f856c4 1316 coords[j*2] = (int)(points[j*2] * GRID_SCALE) + ds->ox;
1317 coords[j*2+1] = (int)(points[j*2+1] * GRID_SCALE) + ds->oy;
1482ee76 1318 }
1319
1320 /*
1321 * Find out whether these points are in a clockwise or
1322 * anticlockwise arrangement. If the latter, discard the
1323 * face because it's facing away from the viewer.
1324 *
1325 * This would involve fiddly winding-number stuff for a
1326 * general polygon, but for the simple parallelograms we'll
1327 * be seeing here, all we have to do is check whether the
1328 * corners turn right or left. So we'll take the vector
1329 * from point 0 to point 1, turn it right 90 degrees,
1330 * and check the sign of the dot product with that and the
1331 * next vector (point 1 to point 2).
1332 */
1333 {
1334 float v1x = points[2]-points[0];
1335 float v1y = points[3]-points[1];
1336 float v2x = points[4]-points[2];
1337 float v2y = points[5]-points[3];
1338 float dp = v1x * v2y - v1y * v2x;
1339
1340 if (dp <= 0)
1341 continue;
1342 }
1343
1344 draw_polygon(fe, coords, poly->order, TRUE,
1345 state->facecolours[i] ? COL_BLUE : COL_BACKGROUND);
1346 draw_polygon(fe, coords, poly->order, FALSE, COL_BORDER);
1347 }
1348 sfree(poly);
1349
03f856c4 1350 draw_update(fe, 0, 0, (int)((bb.r-bb.l+2.0F) * GRID_SCALE),
1351 (int)((bb.d-bb.u+2.0F) * GRID_SCALE));
1482ee76 1352}
1353
1354float game_anim_length(game_state *oldstate, game_state *newstate)
1355{
1356 return ROLLTIME;
1357}