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 | |
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 | } |