720a8fb7 |
1 | /* |
2 | * cube.c: Cube game. |
3 | */ |
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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 */ |
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24 | float border; /* border required around arena */ |
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25 | }; |
26 | |
27 | static const struct solid tetrahedron = { |
28 | 4, |
29 | { |
03f856c4 |
30 | 0.0F, -0.57735026919F, -0.20412414523F, |
31 | -0.5F, 0.28867513459F, -0.20412414523F, |
32 | 0.0F, -0.0F, 0.6123724357F, |
33 | 0.5F, 0.28867513459F, -0.20412414523F, |
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34 | }, |
35 | 3, 4, |
36 | { |
37 | 0,2,1, 3,1,2, 2,0,3, 1,3,0 |
38 | }, |
39 | { |
03f856c4 |
40 | -0.816496580928F, -0.471404520791F, 0.333333333334F, |
41 | 0.0F, 0.942809041583F, 0.333333333333F, |
42 | 0.816496580928F, -0.471404520791F, 0.333333333334F, |
43 | 0.0F, 0.0F, -1.0F, |
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44 | }, |
03f856c4 |
45 | 0.0F, 0.3F |
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46 | }; |
47 | |
48 | static const struct solid cube = { |
49 | 8, |
50 | { |
03f856c4 |
51 | -0.5F,-0.5F,-0.5F, -0.5F,-0.5F,+0.5F, |
52 | -0.5F,+0.5F,-0.5F, -0.5F,+0.5F,+0.5F, |
53 | +0.5F,-0.5F,-0.5F, +0.5F,-0.5F,+0.5F, |
54 | +0.5F,+0.5F,-0.5F, +0.5F,+0.5F,+0.5F, |
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55 | }, |
56 | 4, 6, |
57 | { |
58 | 0,1,3,2, 1,5,7,3, 5,4,6,7, 4,0,2,6, 0,4,5,1, 3,7,6,2 |
59 | }, |
60 | { |
03f856c4 |
61 | -1.0F,0.0F,0.0F, 0.0F,0.0F,+1.0F, |
62 | +1.0F,0.0F,0.0F, 0.0F,0.0F,-1.0F, |
63 | 0.0F,-1.0F,0.0F, 0.0F,+1.0F,0.0F |
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64 | }, |
03f856c4 |
65 | 0.3F, 0.5F |
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66 | }; |
67 | |
68 | static const struct solid octahedron = { |
69 | 6, |
70 | { |
03f856c4 |
71 | -0.5F, -0.28867513459472505F, 0.4082482904638664F, |
72 | 0.5F, 0.28867513459472505F, -0.4082482904638664F, |
73 | -0.5F, 0.28867513459472505F, -0.4082482904638664F, |
74 | 0.5F, -0.28867513459472505F, 0.4082482904638664F, |
75 | 0.0F, -0.57735026918945009F, -0.4082482904638664F, |
76 | 0.0F, 0.57735026918945009F, 0.4082482904638664F, |
1482ee76 |
77 | }, |
78 | 3, 8, |
79 | { |
80 | 4,0,2, 0,5,2, 0,4,3, 5,0,3, 1,4,2, 5,1,2, 4,1,3, 1,5,3 |
81 | }, |
82 | { |
03f856c4 |
83 | -0.816496580928F, -0.471404520791F, -0.333333333334F, |
84 | -0.816496580928F, 0.471404520791F, 0.333333333334F, |
85 | 0.0F, -0.942809041583F, 0.333333333333F, |
86 | 0.0F, 0.0F, 1.0F, |
87 | 0.0F, 0.0F, -1.0F, |
88 | 0.0F, 0.942809041583F, -0.333333333333F, |
89 | 0.816496580928F, -0.471404520791F, -0.333333333334F, |
90 | 0.816496580928F, 0.471404520791F, 0.333333333334F, |
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91 | }, |
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92 | 0.0F, 0.5F |
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93 | }; |
94 | |
95 | static const struct solid icosahedron = { |
96 | 12, |
97 | { |
03f856c4 |
98 | 0.0F, 0.57735026919F, 0.75576131408F, |
99 | 0.0F, -0.93417235896F, 0.17841104489F, |
100 | 0.0F, 0.93417235896F, -0.17841104489F, |
101 | 0.0F, -0.57735026919F, -0.75576131408F, |
102 | -0.5F, -0.28867513459F, 0.75576131408F, |
103 | -0.5F, 0.28867513459F, -0.75576131408F, |
104 | 0.5F, -0.28867513459F, 0.75576131408F, |
105 | 0.5F, 0.28867513459F, -0.75576131408F, |
106 | -0.80901699437F, 0.46708617948F, 0.17841104489F, |
107 | 0.80901699437F, 0.46708617948F, 0.17841104489F, |
108 | -0.80901699437F, -0.46708617948F, -0.17841104489F, |
109 | 0.80901699437F, -0.46708617948F, -0.17841104489F, |
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110 | }, |
111 | 3, 20, |
112 | { |
113 | 8,0,2, 0,9,2, 1,10,3, 11,1,3, 0,4,6, |
114 | 4,1,6, 5,2,7, 3,5,7, 4,8,10, 8,5,10, |
115 | 9,6,11, 7,9,11, 0,8,4, 9,0,6, 10,1,4, |
116 | 1,11,6, 8,2,5, 2,9,7, 3,10,5, 11,3,7, |
117 | }, |
118 | { |
03f856c4 |
119 | -0.356822089773F, 0.87267799625F, 0.333333333333F, |
120 | 0.356822089773F, 0.87267799625F, 0.333333333333F, |
121 | -0.356822089773F, -0.87267799625F, -0.333333333333F, |
122 | 0.356822089773F, -0.87267799625F, -0.333333333333F, |
123 | -0.0F, 0.0F, 1.0F, |
124 | 0.0F, -0.666666666667F, 0.745355992501F, |
125 | 0.0F, 0.666666666667F, -0.745355992501F, |
126 | 0.0F, 0.0F, -1.0F, |
127 | -0.934172358963F, -0.12732200375F, 0.333333333333F, |
128 | -0.934172358963F, 0.12732200375F, -0.333333333333F, |
129 | 0.934172358963F, -0.12732200375F, 0.333333333333F, |
130 | 0.934172358963F, 0.12732200375F, -0.333333333333F, |
131 | -0.57735026919F, 0.333333333334F, 0.745355992501F, |
132 | 0.57735026919F, 0.333333333334F, 0.745355992501F, |
133 | -0.57735026919F, -0.745355992501F, 0.333333333334F, |
134 | 0.57735026919F, -0.745355992501F, 0.333333333334F, |
135 | -0.57735026919F, 0.745355992501F, -0.333333333334F, |
136 | 0.57735026919F, 0.745355992501F, -0.333333333334F, |
137 | -0.57735026919F, -0.333333333334F, -0.745355992501F, |
138 | 0.57735026919F, -0.333333333334F, -0.745355992501F, |
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139 | }, |
03f856c4 |
140 | 0.0F, 0.8F |
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141 | }; |
142 | |
143 | enum { |
144 | TETRAHEDRON, CUBE, OCTAHEDRON, ICOSAHEDRON |
145 | }; |
146 | static const struct solid *solids[] = { |
147 | &tetrahedron, &cube, &octahedron, &icosahedron |
148 | }; |
149 | |
150 | enum { |
151 | COL_BACKGROUND, |
152 | COL_BORDER, |
153 | COL_BLUE, |
154 | NCOLOURS |
155 | }; |
156 | |
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157 | enum { LEFT, RIGHT, UP, DOWN, UP_LEFT, UP_RIGHT, DOWN_LEFT, DOWN_RIGHT }; |
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158 | |
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159 | #define GRID_SCALE 48.0F |
160 | #define ROLLTIME 0.1F |
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161 | |
162 | #define SQ(x) ( (x) * (x) ) |
163 | |
164 | #define MATMUL(ra,m,a) do { \ |
165 | float rx, ry, rz, xx = (a)[0], yy = (a)[1], zz = (a)[2], *mat = (m); \ |
166 | rx = mat[0] * xx + mat[3] * yy + mat[6] * zz; \ |
167 | ry = mat[1] * xx + mat[4] * yy + mat[7] * zz; \ |
168 | rz = mat[2] * xx + mat[5] * yy + mat[8] * zz; \ |
169 | (ra)[0] = rx; (ra)[1] = ry; (ra)[2] = rz; \ |
170 | } while (0) |
171 | |
172 | #define APPROXEQ(x,y) ( SQ(x-y) < 0.1 ) |
173 | |
174 | struct grid_square { |
175 | float x, y; |
176 | int npoints; |
177 | float points[8]; /* maximum */ |
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178 | int directions[8]; /* bit masks showing point pairs */ |
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179 | int flip; |
180 | int blue; |
181 | int tetra_class; |
182 | }; |
183 | |
184 | struct game_params { |
185 | int solid; |
186 | /* |
187 | * Grid dimensions. For a square grid these are width and |
188 | * height respectively; otherwise the grid is a hexagon, with |
189 | * the top side and the two lower diagonals having length d1 |
190 | * and the remaining three sides having length d2 (so that |
191 | * d1==d2 gives a regular hexagon, and d2==0 gives a triangle). |
192 | */ |
193 | int d1, d2; |
194 | }; |
195 | |
196 | struct game_state { |
197 | struct game_params params; |
198 | const struct solid *solid; |
199 | int *facecolours; |
200 | struct grid_square *squares; |
201 | int nsquares; |
202 | int current; /* index of current grid square */ |
203 | int sgkey[2]; /* key-point indices into grid sq */ |
204 | int dgkey[2]; /* key-point indices into grid sq */ |
205 | int spkey[2]; /* key-point indices into polyhedron */ |
206 | int dpkey[2]; /* key-point indices into polyhedron */ |
207 | int previous; |
208 | float angle; |
209 | int completed; |
210 | int movecount; |
211 | }; |
212 | |
213 | game_params *default_params(void) |
214 | { |
215 | game_params *ret = snew(game_params); |
216 | |
217 | ret->solid = CUBE; |
218 | ret->d1 = 4; |
219 | ret->d2 = 4; |
220 | |
221 | return ret; |
222 | } |
223 | |
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224 | int game_fetch_preset(int i, char **name, game_params **params) |
225 | { |
226 | game_params *ret = snew(game_params); |
227 | char *str; |
228 | |
229 | switch (i) { |
230 | case 0: |
231 | str = "Cube"; |
232 | ret->solid = CUBE; |
233 | ret->d1 = 4; |
234 | ret->d2 = 4; |
235 | break; |
236 | case 1: |
237 | str = "Tetrahedron"; |
238 | ret->solid = TETRAHEDRON; |
239 | ret->d1 = 2; |
240 | ret->d2 = 1; |
241 | break; |
242 | case 2: |
243 | str = "Octahedron"; |
244 | ret->solid = OCTAHEDRON; |
245 | ret->d1 = 2; |
246 | ret->d2 = 2; |
247 | break; |
248 | case 3: |
249 | str = "Icosahedron"; |
250 | ret->solid = ICOSAHEDRON; |
251 | ret->d1 = 3; |
252 | ret->d2 = 3; |
253 | break; |
254 | default: |
255 | sfree(ret); |
256 | return FALSE; |
257 | } |
258 | |
259 | *name = dupstr(str); |
260 | *params = ret; |
261 | return TRUE; |
262 | } |
263 | |
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264 | void free_params(game_params *params) |
265 | { |
266 | sfree(params); |
267 | } |
268 | |
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269 | game_params *dup_params(game_params *params) |
270 | { |
271 | game_params *ret = snew(game_params); |
272 | *ret = *params; /* structure copy */ |
273 | return ret; |
274 | } |
275 | |
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276 | static void enum_grid_squares(game_params *params, |
277 | void (*callback)(void *, struct grid_square *), |
278 | void *ctx) |
279 | { |
280 | const struct solid *solid = solids[params->solid]; |
281 | |
282 | if (solid->order == 4) { |
283 | int x, y; |
284 | |
285 | for (x = 0; x < params->d1; x++) |
286 | for (y = 0; y < params->d2; y++) { |
287 | struct grid_square sq; |
288 | |
03f856c4 |
289 | sq.x = (float)x; |
290 | sq.y = (float)y; |
291 | sq.points[0] = x - 0.5F; |
292 | sq.points[1] = y - 0.5F; |
293 | sq.points[2] = x - 0.5F; |
294 | sq.points[3] = y + 0.5F; |
295 | sq.points[4] = x + 0.5F; |
296 | sq.points[5] = y + 0.5F; |
297 | sq.points[6] = x + 0.5F; |
298 | sq.points[7] = y - 0.5F; |
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299 | sq.npoints = 4; |
300 | |
301 | sq.directions[LEFT] = 0x03; /* 0,1 */ |
302 | sq.directions[RIGHT] = 0x0C; /* 2,3 */ |
303 | sq.directions[UP] = 0x09; /* 0,3 */ |
304 | sq.directions[DOWN] = 0x06; /* 1,2 */ |
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305 | sq.directions[UP_LEFT] = 0; /* no diagonals in a square */ |
306 | sq.directions[UP_RIGHT] = 0; /* no diagonals in a square */ |
307 | sq.directions[DOWN_LEFT] = 0; /* no diagonals in a square */ |
308 | sq.directions[DOWN_RIGHT] = 0; /* no diagonals in a square */ |
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309 | |
310 | sq.flip = FALSE; |
311 | |
312 | /* |
313 | * This is supremely irrelevant, but just to avoid |
314 | * having any uninitialised structure members... |
315 | */ |
316 | sq.tetra_class = 0; |
317 | |
318 | callback(ctx, &sq); |
319 | } |
320 | } else { |
321 | int row, rowlen, other, i, firstix = -1; |
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322 | float theight = (float)(sqrt(3) / 2.0); |
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323 | |
324 | for (row = 0; row < params->d1 + params->d2; row++) { |
325 | if (row < params->d1) { |
326 | other = +1; |
327 | rowlen = row + params->d2; |
328 | } else { |
329 | other = -1; |
330 | rowlen = 2*params->d1 + params->d2 - row; |
331 | } |
332 | |
333 | /* |
334 | * There are `rowlen' down-pointing triangles. |
335 | */ |
336 | for (i = 0; i < rowlen; i++) { |
337 | struct grid_square sq; |
338 | int ix; |
339 | float x, y; |
340 | |
341 | ix = (2 * i - (rowlen-1)); |
03f856c4 |
342 | x = ix * 0.5F; |
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343 | y = theight * row; |
344 | sq.x = x; |
345 | sq.y = y + theight / 3; |
03f856c4 |
346 | sq.points[0] = x - 0.5F; |
1482ee76 |
347 | sq.points[1] = y; |
348 | sq.points[2] = x; |
349 | sq.points[3] = y + theight; |
03f856c4 |
350 | sq.points[4] = x + 0.5F; |
1482ee76 |
351 | sq.points[5] = y; |
352 | sq.npoints = 3; |
353 | |
354 | sq.directions[LEFT] = 0x03; /* 0,1 */ |
355 | sq.directions[RIGHT] = 0x06; /* 1,2 */ |
356 | sq.directions[UP] = 0x05; /* 0,2 */ |
357 | sq.directions[DOWN] = 0; /* invalid move */ |
358 | |
c71454c0 |
359 | /* |
360 | * Down-pointing triangle: both the up diagonals go |
361 | * up, and the down ones go left and right. |
362 | */ |
363 | sq.directions[UP_LEFT] = sq.directions[UP_RIGHT] = |
364 | sq.directions[UP]; |
365 | sq.directions[DOWN_LEFT] = sq.directions[LEFT]; |
366 | sq.directions[DOWN_RIGHT] = sq.directions[RIGHT]; |
367 | |
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368 | sq.flip = TRUE; |
369 | |
370 | if (firstix < 0) |
371 | firstix = ix & 3; |
372 | ix -= firstix; |
373 | sq.tetra_class = ((row+(ix&1)) & 2) ^ (ix & 3); |
374 | |
375 | callback(ctx, &sq); |
376 | } |
377 | |
378 | /* |
379 | * There are `rowlen+other' up-pointing triangles. |
380 | */ |
381 | for (i = 0; i < rowlen+other; i++) { |
382 | struct grid_square sq; |
383 | int ix; |
384 | float x, y; |
385 | |
386 | ix = (2 * i - (rowlen+other-1)); |
03f856c4 |
387 | x = ix * 0.5F; |
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388 | y = theight * row; |
389 | sq.x = x; |
390 | sq.y = y + 2*theight / 3; |
03f856c4 |
391 | sq.points[0] = x + 0.5F; |
1482ee76 |
392 | sq.points[1] = y + theight; |
393 | sq.points[2] = x; |
394 | sq.points[3] = y; |
03f856c4 |
395 | sq.points[4] = x - 0.5F; |
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396 | sq.points[5] = y + theight; |
397 | sq.npoints = 3; |
398 | |
399 | sq.directions[LEFT] = 0x06; /* 1,2 */ |
400 | sq.directions[RIGHT] = 0x03; /* 0,1 */ |
401 | sq.directions[DOWN] = 0x05; /* 0,2 */ |
402 | sq.directions[UP] = 0; /* invalid move */ |
403 | |
c71454c0 |
404 | /* |
405 | * Up-pointing triangle: both the down diagonals go |
406 | * down, and the up ones go left and right. |
407 | */ |
408 | sq.directions[DOWN_LEFT] = sq.directions[DOWN_RIGHT] = |
409 | sq.directions[DOWN]; |
410 | sq.directions[UP_LEFT] = sq.directions[LEFT]; |
411 | sq.directions[UP_RIGHT] = sq.directions[RIGHT]; |
412 | |
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413 | sq.flip = FALSE; |
414 | |
415 | if (firstix < 0) |
416 | firstix = ix; |
417 | ix -= firstix; |
418 | sq.tetra_class = ((row+(ix&1)) & 2) ^ (ix & 3); |
419 | |
420 | callback(ctx, &sq); |
421 | } |
422 | } |
423 | } |
424 | } |
425 | |
426 | static int grid_area(int d1, int d2, int order) |
427 | { |
428 | /* |
429 | * An NxM grid of squares has NM squares in it. |
430 | * |
431 | * A grid of triangles with dimensions A and B has a total of |
432 | * A^2 + B^2 + 4AB triangles in it. (You can divide it up into |
433 | * a side-A triangle containing A^2 subtriangles, a side-B |
434 | * triangle containing B^2, and two congruent parallelograms, |
435 | * each with side lengths A and B, each therefore containing AB |
436 | * two-triangle rhombuses.) |
437 | */ |
438 | if (order == 4) |
439 | return d1 * d2; |
440 | else |
441 | return d1*d1 + d2*d2 + 4*d1*d2; |
442 | } |
443 | |
444 | struct grid_data { |
445 | int *gridptrs[4]; |
446 | int nsquares[4]; |
447 | int nclasses; |
448 | int squareindex; |
449 | }; |
450 | |
451 | static void classify_grid_square_callback(void *ctx, struct grid_square *sq) |
452 | { |
453 | struct grid_data *data = (struct grid_data *)ctx; |
454 | int thisclass; |
455 | |
456 | if (data->nclasses == 4) |
457 | thisclass = sq->tetra_class; |
458 | else if (data->nclasses == 2) |
459 | thisclass = sq->flip; |
460 | else |
461 | thisclass = 0; |
462 | |
463 | data->gridptrs[thisclass][data->nsquares[thisclass]++] = |
464 | data->squareindex++; |
465 | } |
466 | |
467 | char *new_game_seed(game_params *params) |
468 | { |
469 | struct grid_data data; |
470 | int i, j, k, m, area, facesperclass; |
471 | int *flags; |
472 | char *seed, *p; |
473 | |
474 | /* |
475 | * Enumerate the grid squares, dividing them into equivalence |
476 | * classes as appropriate. (For the tetrahedron, there is one |
477 | * equivalence class for each face; for the octahedron there |
478 | * are two classes; for the other two solids there's only one.) |
479 | */ |
480 | |
481 | area = grid_area(params->d1, params->d2, solids[params->solid]->order); |
482 | if (params->solid == TETRAHEDRON) |
483 | data.nclasses = 4; |
484 | else if (params->solid == OCTAHEDRON) |
485 | data.nclasses = 2; |
486 | else |
487 | data.nclasses = 1; |
488 | data.gridptrs[0] = snewn(data.nclasses * area, int); |
489 | for (i = 0; i < data.nclasses; i++) { |
490 | data.gridptrs[i] = data.gridptrs[0] + i * area; |
491 | data.nsquares[i] = 0; |
492 | } |
493 | data.squareindex = 0; |
494 | enum_grid_squares(params, classify_grid_square_callback, &data); |
495 | |
496 | facesperclass = solids[params->solid]->nfaces / data.nclasses; |
497 | |
498 | for (i = 0; i < data.nclasses; i++) |
499 | assert(data.nsquares[i] >= facesperclass); |
500 | assert(data.squareindex == area); |
501 | |
502 | /* |
503 | * So now we know how many faces to allocate in each class. Get |
504 | * on with it. |
505 | */ |
506 | flags = snewn(area, int); |
507 | for (i = 0; i < area; i++) |
508 | flags[i] = FALSE; |
509 | |
510 | for (i = 0; i < data.nclasses; i++) { |
511 | for (j = 0; j < facesperclass; j++) { |
512 | unsigned long divisor = RAND_MAX / data.nsquares[i]; |
513 | unsigned long max = divisor * data.nsquares[i]; |
03f856c4 |
514 | unsigned long n; |
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515 | |
516 | do { |
517 | n = rand(); |
518 | } while (n >= max); |
519 | |
520 | n /= divisor; |
521 | |
522 | assert(!flags[data.gridptrs[i][n]]); |
523 | flags[data.gridptrs[i][n]] = TRUE; |
524 | |
525 | /* |
526 | * Move everything else up the array. I ought to use a |
527 | * better data structure for this, but for such small |
528 | * numbers it hardly seems worth the effort. |
529 | */ |
03f856c4 |
530 | while ((int)n < data.nsquares[i]-1) { |
1482ee76 |
531 | data.gridptrs[i][n] = data.gridptrs[i][n+1]; |
532 | n++; |
533 | } |
534 | data.nsquares[i]--; |
535 | } |
536 | } |
537 | |
538 | /* |
539 | * Now we know precisely which squares are blue. Encode this |
540 | * information in hex. While we're looping over this, collect |
541 | * the non-blue squares into a list in the now-unused gridptrs |
542 | * array. |
543 | */ |
544 | seed = snewn(area / 4 + 40, char); |
545 | p = seed; |
546 | j = 0; |
547 | k = 8; |
548 | m = 0; |
549 | for (i = 0; i < area; i++) { |
550 | if (flags[i]) { |
551 | j |= k; |
552 | } else { |
553 | data.gridptrs[0][m++] = i; |
554 | } |
555 | k >>= 1; |
556 | if (!k) { |
557 | *p++ = "0123456789ABCDEF"[j]; |
558 | k = 8; |
559 | j = 0; |
560 | } |
561 | } |
562 | if (k != 8) |
563 | *p++ = "0123456789ABCDEF"[j]; |
564 | |
565 | /* |
566 | * Choose a non-blue square for the polyhedron. |
567 | */ |
568 | { |
569 | unsigned long divisor = RAND_MAX / m; |
570 | unsigned long max = divisor * m; |
03f856c4 |
571 | unsigned long n; |
1482ee76 |
572 | |
573 | do { |
574 | n = rand(); |
575 | } while (n >= max); |
576 | |
577 | n /= divisor; |
578 | |
579 | sprintf(p, ":%d", data.gridptrs[0][n]); |
580 | } |
581 | |
582 | sfree(data.gridptrs[0]); |
583 | sfree(flags); |
584 | |
585 | return seed; |
586 | } |
587 | |
588 | static void add_grid_square_callback(void *ctx, struct grid_square *sq) |
589 | { |
590 | game_state *state = (game_state *)ctx; |
591 | |
592 | state->squares[state->nsquares] = *sq; /* structure copy */ |
593 | state->squares[state->nsquares].blue = FALSE; |
594 | state->nsquares++; |
595 | } |
596 | |
597 | static int lowest_face(const struct solid *solid) |
598 | { |
599 | int i, j, best; |
600 | float zmin; |
601 | |
602 | best = 0; |
603 | zmin = 0.0; |
604 | for (i = 0; i < solid->nfaces; i++) { |
605 | float z = 0; |
606 | |
607 | for (j = 0; j < solid->order; j++) { |
608 | int f = solid->faces[i*solid->order + j]; |
609 | z += solid->vertices[f*3+2]; |
610 | } |
611 | |
612 | if (i == 0 || zmin > z) { |
613 | zmin = z; |
614 | best = i; |
615 | } |
616 | } |
617 | |
618 | return best; |
619 | } |
620 | |
621 | static int align_poly(const struct solid *solid, struct grid_square *sq, |
622 | int *pkey) |
623 | { |
624 | float zmin; |
625 | int i, j; |
626 | int flip = (sq->flip ? -1 : +1); |
627 | |
628 | /* |
629 | * First, find the lowest z-coordinate present in the solid. |
630 | */ |
631 | zmin = 0.0; |
632 | for (i = 0; i < solid->nvertices; i++) |
633 | if (zmin > solid->vertices[i*3+2]) |
634 | zmin = solid->vertices[i*3+2]; |
635 | |
636 | /* |
637 | * Now go round the grid square. For each point in the grid |
638 | * square, we're looking for a point of the polyhedron with the |
639 | * same x- and y-coordinates (relative to the square's centre), |
640 | * and z-coordinate equal to zmin (near enough). |
641 | */ |
642 | for (j = 0; j < sq->npoints; j++) { |
643 | int matches, index; |
644 | |
645 | matches = 0; |
646 | index = -1; |
647 | |
648 | for (i = 0; i < solid->nvertices; i++) { |
649 | float dist = 0; |
650 | |
651 | dist += SQ(solid->vertices[i*3+0] * flip - sq->points[j*2+0] + sq->x); |
652 | dist += SQ(solid->vertices[i*3+1] * flip - sq->points[j*2+1] + sq->y); |
653 | dist += SQ(solid->vertices[i*3+2] - zmin); |
654 | |
655 | if (dist < 0.1) { |
656 | matches++; |
657 | index = i; |
658 | } |
659 | } |
660 | |
661 | if (matches != 1 || index < 0) |
662 | return FALSE; |
663 | pkey[j] = index; |
664 | } |
665 | |
666 | return TRUE; |
667 | } |
668 | |
669 | static void flip_poly(struct solid *solid, int flip) |
670 | { |
671 | int i; |
672 | |
673 | if (flip) { |
674 | for (i = 0; i < solid->nvertices; i++) { |
675 | solid->vertices[i*3+0] *= -1; |
676 | solid->vertices[i*3+1] *= -1; |
677 | } |
678 | for (i = 0; i < solid->nfaces; i++) { |
679 | solid->normals[i*3+0] *= -1; |
680 | solid->normals[i*3+1] *= -1; |
681 | } |
682 | } |
683 | } |
684 | |
685 | static struct solid *transform_poly(const struct solid *solid, int flip, |
686 | int key0, int key1, float angle) |
687 | { |
688 | struct solid *ret = snew(struct solid); |
689 | float vx, vy, ax, ay; |
690 | float vmatrix[9], amatrix[9], vmatrix2[9]; |
691 | int i; |
692 | |
693 | *ret = *solid; /* structure copy */ |
694 | |
695 | flip_poly(ret, flip); |
696 | |
697 | /* |
698 | * Now rotate the polyhedron through the given angle. We must |
699 | * rotate about the Z-axis to bring the two vertices key0 and |
700 | * key1 into horizontal alignment, then rotate about the |
701 | * X-axis, then rotate back again. |
702 | */ |
703 | vx = ret->vertices[key1*3+0] - ret->vertices[key0*3+0]; |
704 | vy = ret->vertices[key1*3+1] - ret->vertices[key0*3+1]; |
705 | assert(APPROXEQ(vx*vx + vy*vy, 1.0)); |
706 | |
707 | vmatrix[0] = vx; vmatrix[3] = vy; vmatrix[6] = 0; |
708 | vmatrix[1] = -vy; vmatrix[4] = vx; vmatrix[7] = 0; |
709 | vmatrix[2] = 0; vmatrix[5] = 0; vmatrix[8] = 1; |
710 | |
03f856c4 |
711 | ax = (float)cos(angle); |
712 | ay = (float)sin(angle); |
1482ee76 |
713 | |
714 | amatrix[0] = 1; amatrix[3] = 0; amatrix[6] = 0; |
715 | amatrix[1] = 0; amatrix[4] = ax; amatrix[7] = ay; |
716 | amatrix[2] = 0; amatrix[5] = -ay; amatrix[8] = ax; |
717 | |
718 | memcpy(vmatrix2, vmatrix, sizeof(vmatrix)); |
719 | vmatrix2[1] = vy; |
720 | vmatrix2[3] = -vy; |
721 | |
722 | for (i = 0; i < ret->nvertices; i++) { |
723 | MATMUL(ret->vertices + 3*i, vmatrix, ret->vertices + 3*i); |
724 | MATMUL(ret->vertices + 3*i, amatrix, ret->vertices + 3*i); |
725 | MATMUL(ret->vertices + 3*i, vmatrix2, ret->vertices + 3*i); |
726 | } |
727 | for (i = 0; i < ret->nfaces; i++) { |
728 | MATMUL(ret->normals + 3*i, vmatrix, ret->normals + 3*i); |
729 | MATMUL(ret->normals + 3*i, amatrix, ret->normals + 3*i); |
730 | MATMUL(ret->normals + 3*i, vmatrix2, ret->normals + 3*i); |
731 | } |
732 | |
733 | return ret; |
734 | } |
735 | |
736 | game_state *new_game(game_params *params, char *seed) |
737 | { |
738 | game_state *state = snew(game_state); |
739 | int area; |
740 | |
741 | state->params = *params; /* structure copy */ |
742 | state->solid = solids[params->solid]; |
743 | |
744 | area = grid_area(params->d1, params->d2, state->solid->order); |
745 | state->squares = snewn(area, struct grid_square); |
746 | state->nsquares = 0; |
747 | enum_grid_squares(params, add_grid_square_callback, state); |
748 | assert(state->nsquares == area); |
749 | |
750 | state->facecolours = snewn(state->solid->nfaces, int); |
751 | memset(state->facecolours, 0, state->solid->nfaces * sizeof(int)); |
752 | |
753 | /* |
754 | * Set up the blue squares and polyhedron position according to |
755 | * the game seed. |
756 | */ |
757 | { |
758 | char *p = seed; |
759 | int i, j, v; |
760 | |
761 | j = 8; |
762 | v = 0; |
763 | for (i = 0; i < state->nsquares; i++) { |
764 | if (j == 8) { |
765 | v = *p++; |
766 | if (v >= '0' && v <= '9') |
767 | v -= '0'; |
768 | else if (v >= 'A' && v <= 'F') |
769 | v -= 'A' - 10; |
770 | else if (v >= 'a' && v <= 'f') |
771 | v -= 'a' - 10; |
772 | else |
773 | break; |
774 | } |
775 | if (v & j) |
776 | state->squares[i].blue = TRUE; |
777 | j >>= 1; |
778 | if (j == 0) |
779 | j = 8; |
780 | } |
781 | |
782 | if (*p == ':') |
783 | p++; |
784 | |
785 | state->current = atoi(p); |
786 | if (state->current < 0 || state->current >= state->nsquares) |
787 | state->current = 0; /* got to do _something_ */ |
788 | } |
789 | |
790 | /* |
791 | * Align the polyhedron with its grid square and determine |
792 | * initial key points. |
793 | */ |
794 | { |
795 | int pkey[4]; |
796 | int ret; |
797 | |
798 | ret = align_poly(state->solid, &state->squares[state->current], pkey); |
799 | assert(ret); |
800 | |
801 | state->dpkey[0] = state->spkey[0] = pkey[0]; |
802 | state->dpkey[1] = state->spkey[0] = pkey[1]; |
803 | state->dgkey[0] = state->sgkey[0] = 0; |
804 | state->dgkey[1] = state->sgkey[0] = 1; |
805 | } |
806 | |
807 | state->previous = state->current; |
808 | state->angle = 0.0; |
809 | state->completed = FALSE; |
810 | state->movecount = 0; |
811 | |
812 | return state; |
813 | } |
814 | |
815 | game_state *dup_game(game_state *state) |
816 | { |
817 | game_state *ret = snew(game_state); |
818 | |
819 | ret->params = state->params; /* structure copy */ |
820 | ret->solid = state->solid; |
821 | ret->facecolours = snewn(ret->solid->nfaces, int); |
822 | memcpy(ret->facecolours, state->facecolours, |
823 | ret->solid->nfaces * sizeof(int)); |
824 | ret->nsquares = state->nsquares; |
825 | ret->squares = snewn(ret->nsquares, struct grid_square); |
826 | memcpy(ret->squares, state->squares, |
827 | ret->nsquares * sizeof(struct grid_square)); |
828 | ret->dpkey[0] = state->dpkey[0]; |
829 | ret->dpkey[1] = state->dpkey[1]; |
830 | ret->dgkey[0] = state->dgkey[0]; |
831 | ret->dgkey[1] = state->dgkey[1]; |
832 | ret->spkey[0] = state->spkey[0]; |
833 | ret->spkey[1] = state->spkey[1]; |
834 | ret->sgkey[0] = state->sgkey[0]; |
835 | ret->sgkey[1] = state->sgkey[1]; |
836 | ret->previous = state->previous; |
837 | ret->angle = state->angle; |
838 | ret->completed = state->completed; |
839 | ret->movecount = state->movecount; |
840 | |
841 | return ret; |
842 | } |
843 | |
844 | void free_game(game_state *state) |
845 | { |
846 | sfree(state); |
847 | } |
848 | |
849 | game_state *make_move(game_state *from, int x, int y, int button) |
850 | { |
851 | int direction; |
852 | int pkey[2], skey[2], dkey[2]; |
853 | float points[4]; |
854 | game_state *ret; |
855 | float angle; |
856 | int i, j, dest, mask; |
857 | struct solid *poly; |
858 | |
859 | /* |
860 | * All moves are made with the cursor keys. |
861 | */ |
862 | if (button == CURSOR_UP) |
863 | direction = UP; |
864 | else if (button == CURSOR_DOWN) |
865 | direction = DOWN; |
866 | else if (button == CURSOR_LEFT) |
867 | direction = LEFT; |
868 | else if (button == CURSOR_RIGHT) |
869 | direction = RIGHT; |
c71454c0 |
870 | else if (button == CURSOR_UP_LEFT) |
871 | direction = UP_LEFT; |
872 | else if (button == CURSOR_DOWN_LEFT) |
873 | direction = DOWN_LEFT; |
874 | else if (button == CURSOR_UP_RIGHT) |
875 | direction = UP_RIGHT; |
876 | else if (button == CURSOR_DOWN_RIGHT) |
877 | direction = DOWN_RIGHT; |
1482ee76 |
878 | else |
879 | return NULL; |
880 | |
881 | /* |
882 | * Find the two points in the current grid square which |
883 | * correspond to this move. |
884 | */ |
885 | mask = from->squares[from->current].directions[direction]; |
886 | if (mask == 0) |
887 | return NULL; |
888 | for (i = j = 0; i < from->squares[from->current].npoints; i++) |
889 | if (mask & (1 << i)) { |
890 | points[j*2] = from->squares[from->current].points[i*2]; |
891 | points[j*2+1] = from->squares[from->current].points[i*2+1]; |
892 | skey[j] = i; |
893 | j++; |
894 | } |
895 | assert(j == 2); |
896 | |
897 | /* |
898 | * Now find the other grid square which shares those points. |
899 | * This is our move destination. |
900 | */ |
901 | dest = -1; |
902 | for (i = 0; i < from->nsquares; i++) |
903 | if (i != from->current) { |
904 | int match = 0; |
905 | float dist; |
906 | |
907 | for (j = 0; j < from->squares[i].npoints; j++) { |
908 | dist = (SQ(from->squares[i].points[j*2] - points[0]) + |
909 | SQ(from->squares[i].points[j*2+1] - points[1])); |
910 | if (dist < 0.1) |
911 | dkey[match++] = j; |
912 | dist = (SQ(from->squares[i].points[j*2] - points[2]) + |
913 | SQ(from->squares[i].points[j*2+1] - points[3])); |
914 | if (dist < 0.1) |
915 | dkey[match++] = j; |
916 | } |
917 | |
918 | if (match == 2) { |
919 | dest = i; |
920 | break; |
921 | } |
922 | } |
923 | |
924 | if (dest < 0) |
925 | return NULL; |
926 | |
927 | ret = dup_game(from); |
928 | ret->current = i; |
929 | |
930 | /* |
931 | * So we know what grid square we're aiming for, and we also |
932 | * know the two key points (as indices in both the source and |
933 | * destination grid squares) which are invariant between source |
934 | * and destination. |
935 | * |
936 | * Next we must roll the polyhedron on to that square. So we |
937 | * find the indices of the key points within the polyhedron's |
938 | * vertex array, then use those in a call to transform_poly, |
939 | * and align the result on the new grid square. |
940 | */ |
941 | { |
942 | int all_pkey[4]; |
943 | align_poly(from->solid, &from->squares[from->current], all_pkey); |
944 | pkey[0] = all_pkey[skey[0]]; |
945 | pkey[1] = all_pkey[skey[1]]; |
946 | /* |
947 | * Now pkey[0] corresponds to skey[0] and dkey[0], and |
948 | * likewise [1]. |
949 | */ |
950 | } |
951 | |
952 | /* |
953 | * Now find the angle through which to rotate the polyhedron. |
954 | * Do this by finding the two faces that share the two vertices |
955 | * we've found, and taking the dot product of their normals. |
956 | */ |
957 | { |
958 | int f[2], nf = 0; |
959 | float dp; |
960 | |
961 | for (i = 0; i < from->solid->nfaces; i++) { |
962 | int match = 0; |
963 | for (j = 0; j < from->solid->order; j++) |
964 | if (from->solid->faces[i*from->solid->order + j] == pkey[0] || |
965 | from->solid->faces[i*from->solid->order + j] == pkey[1]) |
966 | match++; |
967 | if (match == 2) { |
968 | assert(nf < 2); |
969 | f[nf++] = i; |
970 | } |
971 | } |
972 | |
973 | assert(nf == 2); |
974 | |
975 | dp = 0; |
976 | for (i = 0; i < 3; i++) |
977 | dp += (from->solid->normals[f[0]*3+i] * |
978 | from->solid->normals[f[1]*3+i]); |
03f856c4 |
979 | angle = (float)acos(dp); |
1482ee76 |
980 | } |
981 | |
982 | /* |
983 | * Now transform the polyhedron. We aren't entirely sure |
984 | * whether we need to rotate through angle or -angle, and the |
985 | * simplest way round this is to try both and see which one |
986 | * aligns successfully! |
987 | * |
988 | * Unfortunately, _both_ will align successfully if this is a |
989 | * cube, which won't tell us anything much. So for that |
990 | * particular case, I resort to gross hackery: I simply negate |
991 | * the angle before trying the alignment, depending on the |
992 | * direction. Which directions work which way is determined by |
993 | * pure trial and error. I said it was gross :-/ |
994 | */ |
995 | { |
996 | int all_pkey[4]; |
997 | int success; |
998 | |
999 | if (from->solid->order == 4 && direction == UP) |
1000 | angle = -angle; /* HACK */ |
1001 | |
1002 | poly = transform_poly(from->solid, |
1003 | from->squares[from->current].flip, |
1004 | pkey[0], pkey[1], angle); |
1005 | flip_poly(poly, from->squares[ret->current].flip); |
1006 | success = align_poly(poly, &from->squares[ret->current], all_pkey); |
1007 | |
1008 | if (!success) { |
1009 | angle = -angle; |
1010 | poly = transform_poly(from->solid, |
1011 | from->squares[from->current].flip, |
1012 | pkey[0], pkey[1], angle); |
1013 | flip_poly(poly, from->squares[ret->current].flip); |
1014 | success = align_poly(poly, &from->squares[ret->current], all_pkey); |
1015 | } |
1016 | |
1017 | assert(success); |
1018 | } |
1019 | |
1020 | /* |
1021 | * Now we have our rotated polyhedron, which we expect to be |
1022 | * exactly congruent to the one we started with - but with the |
1023 | * faces permuted. So we map that congruence and thereby figure |
1024 | * out how to permute the faces as a result of the polyhedron |
1025 | * having rolled. |
1026 | */ |
1027 | { |
1028 | int *newcolours = snewn(from->solid->nfaces, int); |
1029 | |
1030 | for (i = 0; i < from->solid->nfaces; i++) |
1031 | newcolours[i] = -1; |
1032 | |
1033 | for (i = 0; i < from->solid->nfaces; i++) { |
1034 | int nmatch = 0; |
1035 | |
1036 | /* |
1037 | * Now go through the transformed polyhedron's faces |
1038 | * and figure out which one's normal is approximately |
1039 | * equal to this one. |
1040 | */ |
1041 | for (j = 0; j < poly->nfaces; j++) { |
1042 | float dist; |
1043 | int k; |
1044 | |
1045 | dist = 0; |
1046 | |
1047 | for (k = 0; k < 3; k++) |
1048 | dist += SQ(poly->normals[j*3+k] - |
1049 | from->solid->normals[i*3+k]); |
1050 | |
1051 | if (APPROXEQ(dist, 0)) { |
1052 | nmatch++; |
1053 | newcolours[i] = ret->facecolours[j]; |
1054 | } |
1055 | } |
1056 | |
1057 | assert(nmatch == 1); |
1058 | } |
1059 | |
1060 | for (i = 0; i < from->solid->nfaces; i++) |
1061 | assert(newcolours[i] != -1); |
1062 | |
1063 | sfree(ret->facecolours); |
1064 | ret->facecolours = newcolours; |
1065 | } |
1066 | |
1067 | /* |
1068 | * And finally, swap the colour between the bottom face of the |
1069 | * polyhedron and the face we've just landed on. |
1070 | * |
1071 | * We don't do this if the game is already complete, since we |
1072 | * allow the user to roll the fully blue polyhedron around the |
1073 | * grid as a feeble reward. |
1074 | */ |
1075 | if (!ret->completed) { |
1076 | i = lowest_face(from->solid); |
1077 | j = ret->facecolours[i]; |
1078 | ret->facecolours[i] = ret->squares[ret->current].blue; |
1079 | ret->squares[ret->current].blue = j; |
1080 | |
1081 | /* |
1082 | * Detect game completion. |
1083 | */ |
1084 | j = 0; |
1085 | for (i = 0; i < ret->solid->nfaces; i++) |
1086 | if (ret->facecolours[i]) |
1087 | j++; |
1088 | if (j == ret->solid->nfaces) |
1089 | ret->completed = TRUE; |
1090 | } |
1091 | |
1092 | sfree(poly); |
1093 | |
1094 | /* |
1095 | * Align the normal polyhedron with its grid square, to get key |
1096 | * points for non-animated display. |
1097 | */ |
1098 | { |
1099 | int pkey[4]; |
1100 | int success; |
1101 | |
1102 | success = align_poly(ret->solid, &ret->squares[ret->current], pkey); |
1103 | assert(success); |
1104 | |
1105 | ret->dpkey[0] = pkey[0]; |
1106 | ret->dpkey[1] = pkey[1]; |
1107 | ret->dgkey[0] = 0; |
1108 | ret->dgkey[1] = 1; |
1109 | } |
1110 | |
1111 | |
1112 | ret->spkey[0] = pkey[0]; |
1113 | ret->spkey[1] = pkey[1]; |
1114 | ret->sgkey[0] = skey[0]; |
1115 | ret->sgkey[1] = skey[1]; |
1116 | ret->previous = from->current; |
1117 | ret->angle = angle; |
1118 | ret->movecount++; |
1119 | |
1120 | return ret; |
1121 | } |
1122 | |
1123 | /* ---------------------------------------------------------------------- |
1124 | * Drawing routines. |
1125 | */ |
1126 | |
1127 | struct bbox { |
1128 | float l, r, u, d; |
1129 | }; |
1130 | |
1131 | struct game_drawstate { |
1132 | int ox, oy; /* pixel position of float origin */ |
1133 | }; |
1134 | |
1135 | static void find_bbox_callback(void *ctx, struct grid_square *sq) |
1136 | { |
1137 | struct bbox *bb = (struct bbox *)ctx; |
1138 | int i; |
1139 | |
1140 | for (i = 0; i < sq->npoints; i++) { |
1141 | if (bb->l > sq->points[i*2]) bb->l = sq->points[i*2]; |
1142 | if (bb->r < sq->points[i*2]) bb->r = sq->points[i*2]; |
1143 | if (bb->u > sq->points[i*2+1]) bb->u = sq->points[i*2+1]; |
1144 | if (bb->d < sq->points[i*2+1]) bb->d = sq->points[i*2+1]; |
1145 | } |
1146 | } |
1147 | |
1148 | static struct bbox find_bbox(game_params *params) |
1149 | { |
1150 | struct bbox bb; |
1151 | |
1152 | /* |
1153 | * These should be hugely more than the real bounding box will |
1154 | * be. |
1155 | */ |
03f856c4 |
1156 | bb.l = 2.0F * (params->d1 + params->d2); |
1157 | bb.r = -2.0F * (params->d1 + params->d2); |
1158 | bb.u = 2.0F * (params->d1 + params->d2); |
1159 | bb.d = -2.0F * (params->d1 + params->d2); |
1482ee76 |
1160 | enum_grid_squares(params, find_bbox_callback, &bb); |
1161 | |
1162 | return bb; |
1163 | } |
1164 | |
1165 | void game_size(game_params *params, int *x, int *y) |
1166 | { |
1167 | struct bbox bb = find_bbox(params); |
03f856c4 |
1168 | *x = (int)((bb.r - bb.l + 2*solids[params->solid]->border) * GRID_SCALE); |
1169 | *y = (int)((bb.d - bb.u + 2*solids[params->solid]->border) * GRID_SCALE); |
1482ee76 |
1170 | } |
1171 | |
1172 | float *game_colours(frontend *fe, game_state *state, int *ncolours) |
1173 | { |
1174 | float *ret = snewn(3 * NCOLOURS, float); |
1175 | |
1176 | frontend_default_colour(fe, &ret[COL_BACKGROUND * 3]); |
1177 | |
1178 | ret[COL_BORDER * 3 + 0] = 0.0; |
1179 | ret[COL_BORDER * 3 + 1] = 0.0; |
1180 | ret[COL_BORDER * 3 + 2] = 0.0; |
1181 | |
1182 | ret[COL_BLUE * 3 + 0] = 0.0; |
1183 | ret[COL_BLUE * 3 + 1] = 0.0; |
1184 | ret[COL_BLUE * 3 + 2] = 1.0; |
1185 | |
1186 | *ncolours = NCOLOURS; |
1187 | return ret; |
1188 | } |
1189 | |
1190 | game_drawstate *game_new_drawstate(game_state *state) |
1191 | { |
1192 | struct game_drawstate *ds = snew(struct game_drawstate); |
1193 | struct bbox bb = find_bbox(&state->params); |
1194 | |
03f856c4 |
1195 | ds->ox = (int)(-(bb.l - state->solid->border) * GRID_SCALE); |
1196 | ds->oy = (int)(-(bb.u - state->solid->border) * GRID_SCALE); |
1482ee76 |
1197 | |
1198 | return ds; |
1199 | } |
1200 | |
1201 | void game_free_drawstate(game_drawstate *ds) |
1202 | { |
1203 | sfree(ds); |
1204 | } |
1205 | |
1206 | void game_redraw(frontend *fe, game_drawstate *ds, game_state *oldstate, |
1207 | game_state *state, float animtime) |
1208 | { |
1209 | int i, j; |
1210 | struct bbox bb = find_bbox(&state->params); |
1211 | struct solid *poly; |
1212 | int *pkey, *gkey; |
1213 | float t[3]; |
1214 | float angle; |
1215 | game_state *newstate; |
1216 | int square; |
1217 | |
03f856c4 |
1218 | draw_rect(fe, 0, 0, (int)((bb.r-bb.l+2.0F) * GRID_SCALE), |
1219 | (int)((bb.d-bb.u+2.0F) * GRID_SCALE), COL_BACKGROUND); |
1482ee76 |
1220 | |
1221 | if (oldstate && oldstate->movecount > state->movecount) { |
1222 | game_state *t; |
1223 | |
1224 | /* |
1225 | * This is an Undo. So reverse the order of the states, and |
1226 | * run the roll timer backwards. |
1227 | */ |
1228 | t = oldstate; |
1229 | oldstate = state; |
1230 | state = t; |
1231 | |
1232 | animtime = ROLLTIME - animtime; |
1233 | } |
1234 | |
1235 | if (!oldstate) { |
1236 | oldstate = state; |
1237 | angle = 0.0; |
1238 | square = state->current; |
1239 | pkey = state->dpkey; |
1240 | gkey = state->dgkey; |
1241 | } else { |
1242 | angle = state->angle * animtime / ROLLTIME; |
1243 | square = state->previous; |
1244 | pkey = state->spkey; |
1245 | gkey = state->sgkey; |
1246 | } |
1247 | newstate = state; |
1248 | state = oldstate; |
1249 | |
1250 | for (i = 0; i < state->nsquares; i++) { |
1251 | int coords[8]; |
1252 | |
1253 | for (j = 0; j < state->squares[i].npoints; j++) { |
03f856c4 |
1254 | coords[2*j] = ((int)(state->squares[i].points[2*j] * GRID_SCALE) |
1255 | + ds->ox); |
1256 | coords[2*j+1] = ((int)(state->squares[i].points[2*j+1]*GRID_SCALE) |
1257 | + ds->oy); |
1482ee76 |
1258 | } |
1259 | |
1260 | draw_polygon(fe, coords, state->squares[i].npoints, TRUE, |
1261 | state->squares[i].blue ? COL_BLUE : COL_BACKGROUND); |
1262 | draw_polygon(fe, coords, state->squares[i].npoints, FALSE, COL_BORDER); |
1263 | } |
1264 | |
1265 | /* |
1266 | * Now compute and draw the polyhedron. |
1267 | */ |
1268 | poly = transform_poly(state->solid, state->squares[square].flip, |
1269 | pkey[0], pkey[1], angle); |
1270 | |
1271 | /* |
1272 | * Compute the translation required to align the two key points |
1273 | * on the polyhedron with the same key points on the current |
1274 | * face. |
1275 | */ |
1276 | for (i = 0; i < 3; i++) { |
1277 | float tc = 0.0; |
1278 | |
1279 | for (j = 0; j < 2; j++) { |
1280 | float grid_coord; |
1281 | |
1282 | if (i < 2) { |
1283 | grid_coord = |
1284 | state->squares[square].points[gkey[j]*2+i]; |
1285 | } else { |
1286 | grid_coord = 0.0; |
1287 | } |
1288 | |
1289 | tc += (grid_coord - poly->vertices[pkey[j]*3+i]); |
1290 | } |
1291 | |
1292 | t[i] = tc / 2; |
1293 | } |
1294 | for (i = 0; i < poly->nvertices; i++) |
1295 | for (j = 0; j < 3; j++) |
1296 | poly->vertices[i*3+j] += t[j]; |
1297 | |
1298 | /* |
1299 | * Now actually draw each face. |
1300 | */ |
1301 | for (i = 0; i < poly->nfaces; i++) { |
1302 | float points[8]; |
1303 | int coords[8]; |
1304 | |
1305 | for (j = 0; j < poly->order; j++) { |
1306 | int f = poly->faces[i*poly->order + j]; |
1307 | points[j*2] = (poly->vertices[f*3+0] - |
1308 | poly->vertices[f*3+2] * poly->shear); |
1309 | points[j*2+1] = (poly->vertices[f*3+1] - |
1310 | poly->vertices[f*3+2] * poly->shear); |
1311 | } |
1312 | |
1313 | for (j = 0; j < poly->order; j++) { |
03f856c4 |
1314 | coords[j*2] = (int)(points[j*2] * GRID_SCALE) + ds->ox; |
1315 | coords[j*2+1] = (int)(points[j*2+1] * GRID_SCALE) + ds->oy; |
1482ee76 |
1316 | } |
1317 | |
1318 | /* |
1319 | * Find out whether these points are in a clockwise or |
1320 | * anticlockwise arrangement. If the latter, discard the |
1321 | * face because it's facing away from the viewer. |
1322 | * |
1323 | * This would involve fiddly winding-number stuff for a |
1324 | * general polygon, but for the simple parallelograms we'll |
1325 | * be seeing here, all we have to do is check whether the |
1326 | * corners turn right or left. So we'll take the vector |
1327 | * from point 0 to point 1, turn it right 90 degrees, |
1328 | * and check the sign of the dot product with that and the |
1329 | * next vector (point 1 to point 2). |
1330 | */ |
1331 | { |
1332 | float v1x = points[2]-points[0]; |
1333 | float v1y = points[3]-points[1]; |
1334 | float v2x = points[4]-points[2]; |
1335 | float v2y = points[5]-points[3]; |
1336 | float dp = v1x * v2y - v1y * v2x; |
1337 | |
1338 | if (dp <= 0) |
1339 | continue; |
1340 | } |
1341 | |
1342 | draw_polygon(fe, coords, poly->order, TRUE, |
1343 | state->facecolours[i] ? COL_BLUE : COL_BACKGROUND); |
1344 | draw_polygon(fe, coords, poly->order, FALSE, COL_BORDER); |
1345 | } |
1346 | sfree(poly); |
1347 | |
03f856c4 |
1348 | draw_update(fe, 0, 0, (int)((bb.r-bb.l+2.0F) * GRID_SCALE), |
1349 | (int)((bb.d-bb.u+2.0F) * GRID_SCALE)); |
1482ee76 |
1350 | } |
1351 | |
1352 | float game_anim_length(game_state *oldstate, game_state *newstate) |
1353 | { |
1354 | return ROLLTIME; |
1355 | } |