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