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