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 | |
0c490335 |
13 | const char *const game_name = "Cube"; |
14 | |
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15 | #define MAXVERTICES 20 |
16 | #define MAXFACES 20 |
17 | #define MAXORDER 4 |
18 | struct solid { |
19 | int nvertices; |
20 | float vertices[MAXVERTICES * 3]; /* 3*npoints coordinates */ |
21 | int order; |
22 | int nfaces; |
23 | int faces[MAXFACES * MAXORDER]; /* order*nfaces point indices */ |
24 | float normals[MAXFACES * 3]; /* 3*npoints vector components */ |
25 | float shear; /* isometric shear for nice drawing */ |
eb2ad6f1 |
26 | float border; /* border required around arena */ |
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27 | }; |
28 | |
29 | static const struct solid tetrahedron = { |
30 | 4, |
31 | { |
03f856c4 |
32 | 0.0F, -0.57735026919F, -0.20412414523F, |
33 | -0.5F, 0.28867513459F, -0.20412414523F, |
34 | 0.0F, -0.0F, 0.6123724357F, |
35 | 0.5F, 0.28867513459F, -0.20412414523F, |
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36 | }, |
37 | 3, 4, |
38 | { |
39 | 0,2,1, 3,1,2, 2,0,3, 1,3,0 |
40 | }, |
41 | { |
03f856c4 |
42 | -0.816496580928F, -0.471404520791F, 0.333333333334F, |
43 | 0.0F, 0.942809041583F, 0.333333333333F, |
44 | 0.816496580928F, -0.471404520791F, 0.333333333334F, |
45 | 0.0F, 0.0F, -1.0F, |
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46 | }, |
03f856c4 |
47 | 0.0F, 0.3F |
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48 | }; |
49 | |
50 | static const struct solid cube = { |
51 | 8, |
52 | { |
03f856c4 |
53 | -0.5F,-0.5F,-0.5F, -0.5F,-0.5F,+0.5F, |
54 | -0.5F,+0.5F,-0.5F, -0.5F,+0.5F,+0.5F, |
55 | +0.5F,-0.5F,-0.5F, +0.5F,-0.5F,+0.5F, |
56 | +0.5F,+0.5F,-0.5F, +0.5F,+0.5F,+0.5F, |
1482ee76 |
57 | }, |
58 | 4, 6, |
59 | { |
60 | 0,1,3,2, 1,5,7,3, 5,4,6,7, 4,0,2,6, 0,4,5,1, 3,7,6,2 |
61 | }, |
62 | { |
03f856c4 |
63 | -1.0F,0.0F,0.0F, 0.0F,0.0F,+1.0F, |
64 | +1.0F,0.0F,0.0F, 0.0F,0.0F,-1.0F, |
65 | 0.0F,-1.0F,0.0F, 0.0F,+1.0F,0.0F |
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66 | }, |
03f856c4 |
67 | 0.3F, 0.5F |
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68 | }; |
69 | |
70 | static const struct solid octahedron = { |
71 | 6, |
72 | { |
03f856c4 |
73 | -0.5F, -0.28867513459472505F, 0.4082482904638664F, |
74 | 0.5F, 0.28867513459472505F, -0.4082482904638664F, |
75 | -0.5F, 0.28867513459472505F, -0.4082482904638664F, |
76 | 0.5F, -0.28867513459472505F, 0.4082482904638664F, |
77 | 0.0F, -0.57735026918945009F, -0.4082482904638664F, |
78 | 0.0F, 0.57735026918945009F, 0.4082482904638664F, |
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79 | }, |
80 | 3, 8, |
81 | { |
82 | 4,0,2, 0,5,2, 0,4,3, 5,0,3, 1,4,2, 5,1,2, 4,1,3, 1,5,3 |
83 | }, |
84 | { |
03f856c4 |
85 | -0.816496580928F, -0.471404520791F, -0.333333333334F, |
86 | -0.816496580928F, 0.471404520791F, 0.333333333334F, |
87 | 0.0F, -0.942809041583F, 0.333333333333F, |
88 | 0.0F, 0.0F, 1.0F, |
89 | 0.0F, 0.0F, -1.0F, |
90 | 0.0F, 0.942809041583F, -0.333333333333F, |
91 | 0.816496580928F, -0.471404520791F, -0.333333333334F, |
92 | 0.816496580928F, 0.471404520791F, 0.333333333334F, |
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93 | }, |
03f856c4 |
94 | 0.0F, 0.5F |
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95 | }; |
96 | |
97 | static const struct solid icosahedron = { |
98 | 12, |
99 | { |
03f856c4 |
100 | 0.0F, 0.57735026919F, 0.75576131408F, |
101 | 0.0F, -0.93417235896F, 0.17841104489F, |
102 | 0.0F, 0.93417235896F, -0.17841104489F, |
103 | 0.0F, -0.57735026919F, -0.75576131408F, |
104 | -0.5F, -0.28867513459F, 0.75576131408F, |
105 | -0.5F, 0.28867513459F, -0.75576131408F, |
106 | 0.5F, -0.28867513459F, 0.75576131408F, |
107 | 0.5F, 0.28867513459F, -0.75576131408F, |
108 | -0.80901699437F, 0.46708617948F, 0.17841104489F, |
109 | 0.80901699437F, 0.46708617948F, 0.17841104489F, |
110 | -0.80901699437F, -0.46708617948F, -0.17841104489F, |
111 | 0.80901699437F, -0.46708617948F, -0.17841104489F, |
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112 | }, |
113 | 3, 20, |
114 | { |
115 | 8,0,2, 0,9,2, 1,10,3, 11,1,3, 0,4,6, |
116 | 4,1,6, 5,2,7, 3,5,7, 4,8,10, 8,5,10, |
117 | 9,6,11, 7,9,11, 0,8,4, 9,0,6, 10,1,4, |
118 | 1,11,6, 8,2,5, 2,9,7, 3,10,5, 11,3,7, |
119 | }, |
120 | { |
03f856c4 |
121 | -0.356822089773F, 0.87267799625F, 0.333333333333F, |
122 | 0.356822089773F, 0.87267799625F, 0.333333333333F, |
123 | -0.356822089773F, -0.87267799625F, -0.333333333333F, |
124 | 0.356822089773F, -0.87267799625F, -0.333333333333F, |
125 | -0.0F, 0.0F, 1.0F, |
126 | 0.0F, -0.666666666667F, 0.745355992501F, |
127 | 0.0F, 0.666666666667F, -0.745355992501F, |
128 | 0.0F, 0.0F, -1.0F, |
129 | -0.934172358963F, -0.12732200375F, 0.333333333333F, |
130 | -0.934172358963F, 0.12732200375F, -0.333333333333F, |
131 | 0.934172358963F, -0.12732200375F, 0.333333333333F, |
132 | 0.934172358963F, 0.12732200375F, -0.333333333333F, |
133 | -0.57735026919F, 0.333333333334F, 0.745355992501F, |
134 | 0.57735026919F, 0.333333333334F, 0.745355992501F, |
135 | -0.57735026919F, -0.745355992501F, 0.333333333334F, |
136 | 0.57735026919F, -0.745355992501F, 0.333333333334F, |
137 | -0.57735026919F, 0.745355992501F, -0.333333333334F, |
138 | 0.57735026919F, 0.745355992501F, -0.333333333334F, |
139 | -0.57735026919F, -0.333333333334F, -0.745355992501F, |
140 | 0.57735026919F, -0.333333333334F, -0.745355992501F, |
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141 | }, |
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142 | 0.0F, 0.8F |
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143 | }; |
144 | |
145 | enum { |
146 | TETRAHEDRON, CUBE, OCTAHEDRON, ICOSAHEDRON |
147 | }; |
148 | static const struct solid *solids[] = { |
149 | &tetrahedron, &cube, &octahedron, &icosahedron |
150 | }; |
151 | |
152 | enum { |
153 | COL_BACKGROUND, |
154 | COL_BORDER, |
155 | COL_BLUE, |
156 | NCOLOURS |
157 | }; |
158 | |
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159 | enum { LEFT, RIGHT, UP, DOWN, UP_LEFT, UP_RIGHT, DOWN_LEFT, DOWN_RIGHT }; |
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160 | |
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161 | #define GRID_SCALE 48.0F |
162 | #define ROLLTIME 0.1F |
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163 | |
164 | #define SQ(x) ( (x) * (x) ) |
165 | |
166 | #define MATMUL(ra,m,a) do { \ |
167 | float rx, ry, rz, xx = (a)[0], yy = (a)[1], zz = (a)[2], *mat = (m); \ |
168 | rx = mat[0] * xx + mat[3] * yy + mat[6] * zz; \ |
169 | ry = mat[1] * xx + mat[4] * yy + mat[7] * zz; \ |
170 | rz = mat[2] * xx + mat[5] * yy + mat[8] * zz; \ |
171 | (ra)[0] = rx; (ra)[1] = ry; (ra)[2] = rz; \ |
172 | } while (0) |
173 | |
174 | #define APPROXEQ(x,y) ( SQ(x-y) < 0.1 ) |
175 | |
176 | struct grid_square { |
177 | float x, y; |
178 | int npoints; |
179 | float points[8]; /* maximum */ |
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180 | int directions[8]; /* bit masks showing point pairs */ |
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181 | int flip; |
182 | int blue; |
183 | int tetra_class; |
184 | }; |
185 | |
186 | struct game_params { |
187 | int solid; |
188 | /* |
189 | * Grid dimensions. For a square grid these are width and |
190 | * height respectively; otherwise the grid is a hexagon, with |
191 | * the top side and the two lower diagonals having length d1 |
192 | * and the remaining three sides having length d2 (so that |
193 | * d1==d2 gives a regular hexagon, and d2==0 gives a triangle). |
194 | */ |
195 | int d1, d2; |
196 | }; |
197 | |
198 | struct game_state { |
199 | struct game_params params; |
200 | const struct solid *solid; |
201 | int *facecolours; |
202 | struct grid_square *squares; |
203 | int nsquares; |
204 | int current; /* index of current grid square */ |
205 | int sgkey[2]; /* key-point indices into grid sq */ |
206 | int dgkey[2]; /* key-point indices into grid sq */ |
207 | int spkey[2]; /* key-point indices into polyhedron */ |
208 | int dpkey[2]; /* key-point indices into polyhedron */ |
209 | int previous; |
210 | float angle; |
211 | int completed; |
212 | int movecount; |
213 | }; |
214 | |
215 | game_params *default_params(void) |
216 | { |
217 | game_params *ret = snew(game_params); |
218 | |
219 | ret->solid = CUBE; |
220 | ret->d1 = 4; |
221 | ret->d2 = 4; |
222 | |
223 | return ret; |
224 | } |
225 | |
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226 | int game_fetch_preset(int i, char **name, game_params **params) |
227 | { |
228 | game_params *ret = snew(game_params); |
229 | char *str; |
230 | |
231 | switch (i) { |
232 | case 0: |
233 | str = "Cube"; |
234 | ret->solid = CUBE; |
235 | ret->d1 = 4; |
236 | ret->d2 = 4; |
237 | break; |
238 | case 1: |
239 | str = "Tetrahedron"; |
240 | ret->solid = TETRAHEDRON; |
241 | ret->d1 = 2; |
242 | ret->d2 = 1; |
243 | break; |
244 | case 2: |
245 | str = "Octahedron"; |
246 | ret->solid = OCTAHEDRON; |
247 | ret->d1 = 2; |
248 | ret->d2 = 2; |
249 | break; |
250 | case 3: |
251 | str = "Icosahedron"; |
252 | ret->solid = ICOSAHEDRON; |
253 | ret->d1 = 3; |
254 | ret->d2 = 3; |
255 | break; |
256 | default: |
257 | sfree(ret); |
258 | return FALSE; |
259 | } |
260 | |
261 | *name = dupstr(str); |
262 | *params = ret; |
263 | return TRUE; |
264 | } |
265 | |
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266 | void free_params(game_params *params) |
267 | { |
268 | sfree(params); |
269 | } |
270 | |
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271 | game_params *dup_params(game_params *params) |
272 | { |
273 | game_params *ret = snew(game_params); |
274 | *ret = *params; /* structure copy */ |
275 | return ret; |
276 | } |
277 | |
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278 | static void enum_grid_squares(game_params *params, |
279 | void (*callback)(void *, struct grid_square *), |
280 | void *ctx) |
281 | { |
282 | const struct solid *solid = solids[params->solid]; |
283 | |
284 | if (solid->order == 4) { |
285 | int x, y; |
286 | |
287 | for (x = 0; x < params->d1; x++) |
288 | for (y = 0; y < params->d2; y++) { |
289 | struct grid_square sq; |
290 | |
03f856c4 |
291 | sq.x = (float)x; |
292 | sq.y = (float)y; |
293 | sq.points[0] = x - 0.5F; |
294 | sq.points[1] = y - 0.5F; |
295 | sq.points[2] = x - 0.5F; |
296 | sq.points[3] = y + 0.5F; |
297 | sq.points[4] = x + 0.5F; |
298 | sq.points[5] = y + 0.5F; |
299 | sq.points[6] = x + 0.5F; |
300 | sq.points[7] = y - 0.5F; |
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301 | sq.npoints = 4; |
302 | |
303 | sq.directions[LEFT] = 0x03; /* 0,1 */ |
304 | sq.directions[RIGHT] = 0x0C; /* 2,3 */ |
305 | sq.directions[UP] = 0x09; /* 0,3 */ |
306 | sq.directions[DOWN] = 0x06; /* 1,2 */ |
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307 | sq.directions[UP_LEFT] = 0; /* no diagonals in a square */ |
308 | sq.directions[UP_RIGHT] = 0; /* no diagonals in a square */ |
309 | sq.directions[DOWN_LEFT] = 0; /* no diagonals in a square */ |
310 | sq.directions[DOWN_RIGHT] = 0; /* no diagonals in a square */ |
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311 | |
312 | sq.flip = FALSE; |
313 | |
314 | /* |
315 | * This is supremely irrelevant, but just to avoid |
316 | * having any uninitialised structure members... |
317 | */ |
318 | sq.tetra_class = 0; |
319 | |
320 | callback(ctx, &sq); |
321 | } |
322 | } else { |
323 | int row, rowlen, other, i, firstix = -1; |
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324 | float theight = (float)(sqrt(3) / 2.0); |
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325 | |
326 | for (row = 0; row < params->d1 + params->d2; row++) { |
327 | if (row < params->d1) { |
328 | other = +1; |
329 | rowlen = row + params->d2; |
330 | } else { |
331 | other = -1; |
332 | rowlen = 2*params->d1 + params->d2 - row; |
333 | } |
334 | |
335 | /* |
336 | * There are `rowlen' down-pointing triangles. |
337 | */ |
338 | for (i = 0; i < rowlen; i++) { |
339 | struct grid_square sq; |
340 | int ix; |
341 | float x, y; |
342 | |
343 | ix = (2 * i - (rowlen-1)); |
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344 | x = ix * 0.5F; |
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345 | y = theight * row; |
346 | sq.x = x; |
347 | sq.y = y + theight / 3; |
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348 | sq.points[0] = x - 0.5F; |
1482ee76 |
349 | sq.points[1] = y; |
350 | sq.points[2] = x; |
351 | sq.points[3] = y + theight; |
03f856c4 |
352 | sq.points[4] = x + 0.5F; |
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353 | sq.points[5] = y; |
354 | sq.npoints = 3; |
355 | |
356 | sq.directions[LEFT] = 0x03; /* 0,1 */ |
357 | sq.directions[RIGHT] = 0x06; /* 1,2 */ |
358 | sq.directions[UP] = 0x05; /* 0,2 */ |
359 | sq.directions[DOWN] = 0; /* invalid move */ |
360 | |
c71454c0 |
361 | /* |
362 | * Down-pointing triangle: both the up diagonals go |
363 | * up, and the down ones go left and right. |
364 | */ |
365 | sq.directions[UP_LEFT] = sq.directions[UP_RIGHT] = |
366 | sq.directions[UP]; |
367 | sq.directions[DOWN_LEFT] = sq.directions[LEFT]; |
368 | sq.directions[DOWN_RIGHT] = sq.directions[RIGHT]; |
369 | |
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370 | sq.flip = TRUE; |
371 | |
372 | if (firstix < 0) |
373 | firstix = ix & 3; |
374 | ix -= firstix; |
375 | sq.tetra_class = ((row+(ix&1)) & 2) ^ (ix & 3); |
376 | |
377 | callback(ctx, &sq); |
378 | } |
379 | |
380 | /* |
381 | * There are `rowlen+other' up-pointing triangles. |
382 | */ |
383 | for (i = 0; i < rowlen+other; i++) { |
384 | struct grid_square sq; |
385 | int ix; |
386 | float x, y; |
387 | |
388 | ix = (2 * i - (rowlen+other-1)); |
03f856c4 |
389 | x = ix * 0.5F; |
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390 | y = theight * row; |
391 | sq.x = x; |
392 | sq.y = y + 2*theight / 3; |
03f856c4 |
393 | sq.points[0] = x + 0.5F; |
1482ee76 |
394 | sq.points[1] = y + theight; |
395 | sq.points[2] = x; |
396 | sq.points[3] = y; |
03f856c4 |
397 | sq.points[4] = x - 0.5F; |
1482ee76 |
398 | sq.points[5] = y + theight; |
399 | sq.npoints = 3; |
400 | |
401 | sq.directions[LEFT] = 0x06; /* 1,2 */ |
402 | sq.directions[RIGHT] = 0x03; /* 0,1 */ |
403 | sq.directions[DOWN] = 0x05; /* 0,2 */ |
404 | sq.directions[UP] = 0; /* invalid move */ |
405 | |
c71454c0 |
406 | /* |
407 | * Up-pointing triangle: both the down diagonals go |
408 | * down, and the up ones go left and right. |
409 | */ |
410 | sq.directions[DOWN_LEFT] = sq.directions[DOWN_RIGHT] = |
411 | sq.directions[DOWN]; |
412 | sq.directions[UP_LEFT] = sq.directions[LEFT]; |
413 | sq.directions[UP_RIGHT] = sq.directions[RIGHT]; |
414 | |
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415 | sq.flip = FALSE; |
416 | |
417 | if (firstix < 0) |
418 | firstix = ix; |
419 | ix -= firstix; |
420 | sq.tetra_class = ((row+(ix&1)) & 2) ^ (ix & 3); |
421 | |
422 | callback(ctx, &sq); |
423 | } |
424 | } |
425 | } |
426 | } |
427 | |
428 | static int grid_area(int d1, int d2, int order) |
429 | { |
430 | /* |
431 | * An NxM grid of squares has NM squares in it. |
432 | * |
433 | * A grid of triangles with dimensions A and B has a total of |
434 | * A^2 + B^2 + 4AB triangles in it. (You can divide it up into |
435 | * a side-A triangle containing A^2 subtriangles, a side-B |
436 | * triangle containing B^2, and two congruent parallelograms, |
437 | * each with side lengths A and B, each therefore containing AB |
438 | * two-triangle rhombuses.) |
439 | */ |
440 | if (order == 4) |
441 | return d1 * d2; |
442 | else |
443 | return d1*d1 + d2*d2 + 4*d1*d2; |
444 | } |
445 | |
446 | struct grid_data { |
447 | int *gridptrs[4]; |
448 | int nsquares[4]; |
449 | int nclasses; |
450 | int squareindex; |
451 | }; |
452 | |
453 | static void classify_grid_square_callback(void *ctx, struct grid_square *sq) |
454 | { |
455 | struct grid_data *data = (struct grid_data *)ctx; |
456 | int thisclass; |
457 | |
458 | if (data->nclasses == 4) |
459 | thisclass = sq->tetra_class; |
460 | else if (data->nclasses == 2) |
461 | thisclass = sq->flip; |
462 | else |
463 | thisclass = 0; |
464 | |
465 | data->gridptrs[thisclass][data->nsquares[thisclass]++] = |
466 | data->squareindex++; |
467 | } |
468 | |
469 | char *new_game_seed(game_params *params) |
470 | { |
471 | struct grid_data data; |
472 | int i, j, k, m, area, facesperclass; |
473 | int *flags; |
474 | char *seed, *p; |
475 | |
476 | /* |
477 | * Enumerate the grid squares, dividing them into equivalence |
478 | * classes as appropriate. (For the tetrahedron, there is one |
479 | * equivalence class for each face; for the octahedron there |
480 | * are two classes; for the other two solids there's only one.) |
481 | */ |
482 | |
483 | area = grid_area(params->d1, params->d2, solids[params->solid]->order); |
484 | if (params->solid == TETRAHEDRON) |
485 | data.nclasses = 4; |
486 | else if (params->solid == OCTAHEDRON) |
487 | data.nclasses = 2; |
488 | else |
489 | data.nclasses = 1; |
490 | data.gridptrs[0] = snewn(data.nclasses * area, int); |
491 | for (i = 0; i < data.nclasses; i++) { |
492 | data.gridptrs[i] = data.gridptrs[0] + i * area; |
493 | data.nsquares[i] = 0; |
494 | } |
495 | data.squareindex = 0; |
496 | enum_grid_squares(params, classify_grid_square_callback, &data); |
497 | |
498 | facesperclass = solids[params->solid]->nfaces / data.nclasses; |
499 | |
500 | for (i = 0; i < data.nclasses; i++) |
501 | assert(data.nsquares[i] >= facesperclass); |
502 | assert(data.squareindex == area); |
503 | |
504 | /* |
505 | * So now we know how many faces to allocate in each class. Get |
506 | * on with it. |
507 | */ |
508 | flags = snewn(area, int); |
509 | for (i = 0; i < area; i++) |
510 | flags[i] = FALSE; |
511 | |
512 | for (i = 0; i < data.nclasses; i++) { |
513 | for (j = 0; j < facesperclass; j++) { |
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514 | int n = rand_upto(data.nsquares[i]); |
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515 | |
516 | assert(!flags[data.gridptrs[i][n]]); |
517 | flags[data.gridptrs[i][n]] = TRUE; |
518 | |
519 | /* |
520 | * Move everything else up the array. I ought to use a |
521 | * better data structure for this, but for such small |
522 | * numbers it hardly seems worth the effort. |
523 | */ |
4efb3868 |
524 | while (n < data.nsquares[i]-1) { |
1482ee76 |
525 | data.gridptrs[i][n] = data.gridptrs[i][n+1]; |
526 | n++; |
527 | } |
528 | data.nsquares[i]--; |
529 | } |
530 | } |
531 | |
532 | /* |
533 | * Now we know precisely which squares are blue. Encode this |
534 | * information in hex. While we're looping over this, collect |
535 | * the non-blue squares into a list in the now-unused gridptrs |
536 | * array. |
537 | */ |
538 | seed = snewn(area / 4 + 40, char); |
539 | p = seed; |
540 | j = 0; |
541 | k = 8; |
542 | m = 0; |
543 | for (i = 0; i < area; i++) { |
544 | if (flags[i]) { |
545 | j |= k; |
546 | } else { |
547 | data.gridptrs[0][m++] = i; |
548 | } |
549 | k >>= 1; |
550 | if (!k) { |
551 | *p++ = "0123456789ABCDEF"[j]; |
552 | k = 8; |
553 | j = 0; |
554 | } |
555 | } |
556 | if (k != 8) |
557 | *p++ = "0123456789ABCDEF"[j]; |
558 | |
559 | /* |
560 | * Choose a non-blue square for the polyhedron. |
561 | */ |
b8c36486 |
562 | sprintf(p, ":%d", data.gridptrs[0][rand_upto(m)]); |
1482ee76 |
563 | |
564 | sfree(data.gridptrs[0]); |
565 | sfree(flags); |
566 | |
567 | return seed; |
568 | } |
569 | |
570 | static void add_grid_square_callback(void *ctx, struct grid_square *sq) |
571 | { |
572 | game_state *state = (game_state *)ctx; |
573 | |
574 | state->squares[state->nsquares] = *sq; /* structure copy */ |
575 | state->squares[state->nsquares].blue = FALSE; |
576 | state->nsquares++; |
577 | } |
578 | |
579 | static int lowest_face(const struct solid *solid) |
580 | { |
581 | int i, j, best; |
582 | float zmin; |
583 | |
584 | best = 0; |
585 | zmin = 0.0; |
586 | for (i = 0; i < solid->nfaces; i++) { |
587 | float z = 0; |
588 | |
589 | for (j = 0; j < solid->order; j++) { |
590 | int f = solid->faces[i*solid->order + j]; |
591 | z += solid->vertices[f*3+2]; |
592 | } |
593 | |
594 | if (i == 0 || zmin > z) { |
595 | zmin = z; |
596 | best = i; |
597 | } |
598 | } |
599 | |
600 | return best; |
601 | } |
602 | |
603 | static int align_poly(const struct solid *solid, struct grid_square *sq, |
604 | int *pkey) |
605 | { |
606 | float zmin; |
607 | int i, j; |
608 | int flip = (sq->flip ? -1 : +1); |
609 | |
610 | /* |
611 | * First, find the lowest z-coordinate present in the solid. |
612 | */ |
613 | zmin = 0.0; |
614 | for (i = 0; i < solid->nvertices; i++) |
615 | if (zmin > solid->vertices[i*3+2]) |
616 | zmin = solid->vertices[i*3+2]; |
617 | |
618 | /* |
619 | * Now go round the grid square. For each point in the grid |
620 | * square, we're looking for a point of the polyhedron with the |
621 | * same x- and y-coordinates (relative to the square's centre), |
622 | * and z-coordinate equal to zmin (near enough). |
623 | */ |
624 | for (j = 0; j < sq->npoints; j++) { |
625 | int matches, index; |
626 | |
627 | matches = 0; |
628 | index = -1; |
629 | |
630 | for (i = 0; i < solid->nvertices; i++) { |
631 | float dist = 0; |
632 | |
633 | dist += SQ(solid->vertices[i*3+0] * flip - sq->points[j*2+0] + sq->x); |
634 | dist += SQ(solid->vertices[i*3+1] * flip - sq->points[j*2+1] + sq->y); |
635 | dist += SQ(solid->vertices[i*3+2] - zmin); |
636 | |
637 | if (dist < 0.1) { |
638 | matches++; |
639 | index = i; |
640 | } |
641 | } |
642 | |
643 | if (matches != 1 || index < 0) |
644 | return FALSE; |
645 | pkey[j] = index; |
646 | } |
647 | |
648 | return TRUE; |
649 | } |
650 | |
651 | static void flip_poly(struct solid *solid, int flip) |
652 | { |
653 | int i; |
654 | |
655 | if (flip) { |
656 | for (i = 0; i < solid->nvertices; i++) { |
657 | solid->vertices[i*3+0] *= -1; |
658 | solid->vertices[i*3+1] *= -1; |
659 | } |
660 | for (i = 0; i < solid->nfaces; i++) { |
661 | solid->normals[i*3+0] *= -1; |
662 | solid->normals[i*3+1] *= -1; |
663 | } |
664 | } |
665 | } |
666 | |
667 | static struct solid *transform_poly(const struct solid *solid, int flip, |
668 | int key0, int key1, float angle) |
669 | { |
670 | struct solid *ret = snew(struct solid); |
671 | float vx, vy, ax, ay; |
672 | float vmatrix[9], amatrix[9], vmatrix2[9]; |
673 | int i; |
674 | |
675 | *ret = *solid; /* structure copy */ |
676 | |
677 | flip_poly(ret, flip); |
678 | |
679 | /* |
680 | * Now rotate the polyhedron through the given angle. We must |
681 | * rotate about the Z-axis to bring the two vertices key0 and |
682 | * key1 into horizontal alignment, then rotate about the |
683 | * X-axis, then rotate back again. |
684 | */ |
685 | vx = ret->vertices[key1*3+0] - ret->vertices[key0*3+0]; |
686 | vy = ret->vertices[key1*3+1] - ret->vertices[key0*3+1]; |
687 | assert(APPROXEQ(vx*vx + vy*vy, 1.0)); |
688 | |
689 | vmatrix[0] = vx; vmatrix[3] = vy; vmatrix[6] = 0; |
690 | vmatrix[1] = -vy; vmatrix[4] = vx; vmatrix[7] = 0; |
691 | vmatrix[2] = 0; vmatrix[5] = 0; vmatrix[8] = 1; |
692 | |
03f856c4 |
693 | ax = (float)cos(angle); |
694 | ay = (float)sin(angle); |
1482ee76 |
695 | |
696 | amatrix[0] = 1; amatrix[3] = 0; amatrix[6] = 0; |
697 | amatrix[1] = 0; amatrix[4] = ax; amatrix[7] = ay; |
698 | amatrix[2] = 0; amatrix[5] = -ay; amatrix[8] = ax; |
699 | |
700 | memcpy(vmatrix2, vmatrix, sizeof(vmatrix)); |
701 | vmatrix2[1] = vy; |
702 | vmatrix2[3] = -vy; |
703 | |
704 | for (i = 0; i < ret->nvertices; i++) { |
705 | MATMUL(ret->vertices + 3*i, vmatrix, ret->vertices + 3*i); |
706 | MATMUL(ret->vertices + 3*i, amatrix, ret->vertices + 3*i); |
707 | MATMUL(ret->vertices + 3*i, vmatrix2, ret->vertices + 3*i); |
708 | } |
709 | for (i = 0; i < ret->nfaces; i++) { |
710 | MATMUL(ret->normals + 3*i, vmatrix, ret->normals + 3*i); |
711 | MATMUL(ret->normals + 3*i, amatrix, ret->normals + 3*i); |
712 | MATMUL(ret->normals + 3*i, vmatrix2, ret->normals + 3*i); |
713 | } |
714 | |
715 | return ret; |
716 | } |
717 | |
718 | game_state *new_game(game_params *params, char *seed) |
719 | { |
720 | game_state *state = snew(game_state); |
721 | int area; |
722 | |
723 | state->params = *params; /* structure copy */ |
724 | state->solid = solids[params->solid]; |
725 | |
726 | area = grid_area(params->d1, params->d2, state->solid->order); |
727 | state->squares = snewn(area, struct grid_square); |
728 | state->nsquares = 0; |
729 | enum_grid_squares(params, add_grid_square_callback, state); |
730 | assert(state->nsquares == area); |
731 | |
732 | state->facecolours = snewn(state->solid->nfaces, int); |
733 | memset(state->facecolours, 0, state->solid->nfaces * sizeof(int)); |
734 | |
735 | /* |
736 | * Set up the blue squares and polyhedron position according to |
737 | * the game seed. |
738 | */ |
739 | { |
740 | char *p = seed; |
741 | int i, j, v; |
742 | |
743 | j = 8; |
744 | v = 0; |
745 | for (i = 0; i < state->nsquares; i++) { |
746 | if (j == 8) { |
747 | v = *p++; |
748 | if (v >= '0' && v <= '9') |
749 | v -= '0'; |
750 | else if (v >= 'A' && v <= 'F') |
751 | v -= 'A' - 10; |
752 | else if (v >= 'a' && v <= 'f') |
753 | v -= 'a' - 10; |
754 | else |
755 | break; |
756 | } |
757 | if (v & j) |
758 | state->squares[i].blue = TRUE; |
759 | j >>= 1; |
760 | if (j == 0) |
761 | j = 8; |
762 | } |
763 | |
764 | if (*p == ':') |
765 | p++; |
766 | |
767 | state->current = atoi(p); |
768 | if (state->current < 0 || state->current >= state->nsquares) |
769 | state->current = 0; /* got to do _something_ */ |
770 | } |
771 | |
772 | /* |
773 | * Align the polyhedron with its grid square and determine |
774 | * initial key points. |
775 | */ |
776 | { |
777 | int pkey[4]; |
778 | int ret; |
779 | |
780 | ret = align_poly(state->solid, &state->squares[state->current], pkey); |
781 | assert(ret); |
782 | |
783 | state->dpkey[0] = state->spkey[0] = pkey[0]; |
784 | state->dpkey[1] = state->spkey[0] = pkey[1]; |
785 | state->dgkey[0] = state->sgkey[0] = 0; |
786 | state->dgkey[1] = state->sgkey[0] = 1; |
787 | } |
788 | |
789 | state->previous = state->current; |
790 | state->angle = 0.0; |
fd1a1a2b |
791 | state->completed = 0; |
1482ee76 |
792 | state->movecount = 0; |
793 | |
794 | return state; |
795 | } |
796 | |
797 | game_state *dup_game(game_state *state) |
798 | { |
799 | game_state *ret = snew(game_state); |
800 | |
801 | ret->params = state->params; /* structure copy */ |
802 | ret->solid = state->solid; |
803 | ret->facecolours = snewn(ret->solid->nfaces, int); |
804 | memcpy(ret->facecolours, state->facecolours, |
805 | ret->solid->nfaces * sizeof(int)); |
806 | ret->nsquares = state->nsquares; |
807 | ret->squares = snewn(ret->nsquares, struct grid_square); |
808 | memcpy(ret->squares, state->squares, |
809 | ret->nsquares * sizeof(struct grid_square)); |
810 | ret->dpkey[0] = state->dpkey[0]; |
811 | ret->dpkey[1] = state->dpkey[1]; |
812 | ret->dgkey[0] = state->dgkey[0]; |
813 | ret->dgkey[1] = state->dgkey[1]; |
814 | ret->spkey[0] = state->spkey[0]; |
815 | ret->spkey[1] = state->spkey[1]; |
816 | ret->sgkey[0] = state->sgkey[0]; |
817 | ret->sgkey[1] = state->sgkey[1]; |
818 | ret->previous = state->previous; |
819 | ret->angle = state->angle; |
820 | ret->completed = state->completed; |
821 | ret->movecount = state->movecount; |
822 | |
823 | return ret; |
824 | } |
825 | |
826 | void free_game(game_state *state) |
827 | { |
828 | sfree(state); |
829 | } |
830 | |
831 | game_state *make_move(game_state *from, int x, int y, int button) |
832 | { |
833 | int direction; |
834 | int pkey[2], skey[2], dkey[2]; |
835 | float points[4]; |
836 | game_state *ret; |
837 | float angle; |
838 | int i, j, dest, mask; |
839 | struct solid *poly; |
840 | |
841 | /* |
842 | * All moves are made with the cursor keys. |
843 | */ |
844 | if (button == CURSOR_UP) |
845 | direction = UP; |
846 | else if (button == CURSOR_DOWN) |
847 | direction = DOWN; |
848 | else if (button == CURSOR_LEFT) |
849 | direction = LEFT; |
850 | else if (button == CURSOR_RIGHT) |
851 | direction = RIGHT; |
c71454c0 |
852 | else if (button == CURSOR_UP_LEFT) |
853 | direction = UP_LEFT; |
854 | else if (button == CURSOR_DOWN_LEFT) |
855 | direction = DOWN_LEFT; |
856 | else if (button == CURSOR_UP_RIGHT) |
857 | direction = UP_RIGHT; |
858 | else if (button == CURSOR_DOWN_RIGHT) |
859 | direction = DOWN_RIGHT; |
1482ee76 |
860 | else |
861 | return NULL; |
862 | |
863 | /* |
864 | * Find the two points in the current grid square which |
865 | * correspond to this move. |
866 | */ |
867 | mask = from->squares[from->current].directions[direction]; |
868 | if (mask == 0) |
869 | return NULL; |
870 | for (i = j = 0; i < from->squares[from->current].npoints; i++) |
871 | if (mask & (1 << i)) { |
872 | points[j*2] = from->squares[from->current].points[i*2]; |
873 | points[j*2+1] = from->squares[from->current].points[i*2+1]; |
874 | skey[j] = i; |
875 | j++; |
876 | } |
877 | assert(j == 2); |
878 | |
879 | /* |
880 | * Now find the other grid square which shares those points. |
881 | * This is our move destination. |
882 | */ |
883 | dest = -1; |
884 | for (i = 0; i < from->nsquares; i++) |
885 | if (i != from->current) { |
886 | int match = 0; |
887 | float dist; |
888 | |
889 | for (j = 0; j < from->squares[i].npoints; j++) { |
890 | dist = (SQ(from->squares[i].points[j*2] - points[0]) + |
891 | SQ(from->squares[i].points[j*2+1] - points[1])); |
892 | if (dist < 0.1) |
893 | dkey[match++] = j; |
894 | dist = (SQ(from->squares[i].points[j*2] - points[2]) + |
895 | SQ(from->squares[i].points[j*2+1] - points[3])); |
896 | if (dist < 0.1) |
897 | dkey[match++] = j; |
898 | } |
899 | |
900 | if (match == 2) { |
901 | dest = i; |
902 | break; |
903 | } |
904 | } |
905 | |
906 | if (dest < 0) |
907 | return NULL; |
908 | |
909 | ret = dup_game(from); |
910 | ret->current = i; |
911 | |
912 | /* |
913 | * So we know what grid square we're aiming for, and we also |
914 | * know the two key points (as indices in both the source and |
915 | * destination grid squares) which are invariant between source |
916 | * and destination. |
917 | * |
918 | * Next we must roll the polyhedron on to that square. So we |
919 | * find the indices of the key points within the polyhedron's |
920 | * vertex array, then use those in a call to transform_poly, |
921 | * and align the result on the new grid square. |
922 | */ |
923 | { |
924 | int all_pkey[4]; |
925 | align_poly(from->solid, &from->squares[from->current], all_pkey); |
926 | pkey[0] = all_pkey[skey[0]]; |
927 | pkey[1] = all_pkey[skey[1]]; |
928 | /* |
929 | * Now pkey[0] corresponds to skey[0] and dkey[0], and |
930 | * likewise [1]. |
931 | */ |
932 | } |
933 | |
934 | /* |
935 | * Now find the angle through which to rotate the polyhedron. |
936 | * Do this by finding the two faces that share the two vertices |
937 | * we've found, and taking the dot product of their normals. |
938 | */ |
939 | { |
940 | int f[2], nf = 0; |
941 | float dp; |
942 | |
943 | for (i = 0; i < from->solid->nfaces; i++) { |
944 | int match = 0; |
945 | for (j = 0; j < from->solid->order; j++) |
946 | if (from->solid->faces[i*from->solid->order + j] == pkey[0] || |
947 | from->solid->faces[i*from->solid->order + j] == pkey[1]) |
948 | match++; |
949 | if (match == 2) { |
950 | assert(nf < 2); |
951 | f[nf++] = i; |
952 | } |
953 | } |
954 | |
955 | assert(nf == 2); |
956 | |
957 | dp = 0; |
958 | for (i = 0; i < 3; i++) |
959 | dp += (from->solid->normals[f[0]*3+i] * |
960 | from->solid->normals[f[1]*3+i]); |
03f856c4 |
961 | angle = (float)acos(dp); |
1482ee76 |
962 | } |
963 | |
964 | /* |
965 | * Now transform the polyhedron. We aren't entirely sure |
966 | * whether we need to rotate through angle or -angle, and the |
967 | * simplest way round this is to try both and see which one |
968 | * aligns successfully! |
969 | * |
970 | * Unfortunately, _both_ will align successfully if this is a |
971 | * cube, which won't tell us anything much. So for that |
972 | * particular case, I resort to gross hackery: I simply negate |
973 | * the angle before trying the alignment, depending on the |
974 | * direction. Which directions work which way is determined by |
975 | * pure trial and error. I said it was gross :-/ |
976 | */ |
977 | { |
978 | int all_pkey[4]; |
979 | int success; |
980 | |
981 | if (from->solid->order == 4 && direction == UP) |
982 | angle = -angle; /* HACK */ |
983 | |
984 | poly = transform_poly(from->solid, |
985 | from->squares[from->current].flip, |
986 | pkey[0], pkey[1], angle); |
987 | flip_poly(poly, from->squares[ret->current].flip); |
988 | success = align_poly(poly, &from->squares[ret->current], all_pkey); |
989 | |
990 | if (!success) { |
991 | angle = -angle; |
992 | poly = transform_poly(from->solid, |
993 | from->squares[from->current].flip, |
994 | pkey[0], pkey[1], angle); |
995 | flip_poly(poly, from->squares[ret->current].flip); |
996 | success = align_poly(poly, &from->squares[ret->current], all_pkey); |
997 | } |
998 | |
999 | assert(success); |
1000 | } |
1001 | |
1002 | /* |
1003 | * Now we have our rotated polyhedron, which we expect to be |
1004 | * exactly congruent to the one we started with - but with the |
1005 | * faces permuted. So we map that congruence and thereby figure |
1006 | * out how to permute the faces as a result of the polyhedron |
1007 | * having rolled. |
1008 | */ |
1009 | { |
1010 | int *newcolours = snewn(from->solid->nfaces, int); |
1011 | |
1012 | for (i = 0; i < from->solid->nfaces; i++) |
1013 | newcolours[i] = -1; |
1014 | |
1015 | for (i = 0; i < from->solid->nfaces; i++) { |
1016 | int nmatch = 0; |
1017 | |
1018 | /* |
1019 | * Now go through the transformed polyhedron's faces |
1020 | * and figure out which one's normal is approximately |
1021 | * equal to this one. |
1022 | */ |
1023 | for (j = 0; j < poly->nfaces; j++) { |
1024 | float dist; |
1025 | int k; |
1026 | |
1027 | dist = 0; |
1028 | |
1029 | for (k = 0; k < 3; k++) |
1030 | dist += SQ(poly->normals[j*3+k] - |
1031 | from->solid->normals[i*3+k]); |
1032 | |
1033 | if (APPROXEQ(dist, 0)) { |
1034 | nmatch++; |
1035 | newcolours[i] = ret->facecolours[j]; |
1036 | } |
1037 | } |
1038 | |
1039 | assert(nmatch == 1); |
1040 | } |
1041 | |
1042 | for (i = 0; i < from->solid->nfaces; i++) |
1043 | assert(newcolours[i] != -1); |
1044 | |
1045 | sfree(ret->facecolours); |
1046 | ret->facecolours = newcolours; |
1047 | } |
1048 | |
ccd4e210 |
1049 | ret->movecount++; |
1050 | |
1482ee76 |
1051 | /* |
1052 | * And finally, swap the colour between the bottom face of the |
1053 | * polyhedron and the face we've just landed on. |
1054 | * |
1055 | * We don't do this if the game is already complete, since we |
1056 | * allow the user to roll the fully blue polyhedron around the |
1057 | * grid as a feeble reward. |
1058 | */ |
1059 | if (!ret->completed) { |
1060 | i = lowest_face(from->solid); |
1061 | j = ret->facecolours[i]; |
1062 | ret->facecolours[i] = ret->squares[ret->current].blue; |
1063 | ret->squares[ret->current].blue = j; |
1064 | |
1065 | /* |
1066 | * Detect game completion. |
1067 | */ |
1068 | j = 0; |
1069 | for (i = 0; i < ret->solid->nfaces; i++) |
1070 | if (ret->facecolours[i]) |
1071 | j++; |
1072 | if (j == ret->solid->nfaces) |
fd1a1a2b |
1073 | ret->completed = ret->movecount; |
1482ee76 |
1074 | } |
1075 | |
1076 | sfree(poly); |
1077 | |
1078 | /* |
1079 | * Align the normal polyhedron with its grid square, to get key |
1080 | * points for non-animated display. |
1081 | */ |
1082 | { |
1083 | int pkey[4]; |
1084 | int success; |
1085 | |
1086 | success = align_poly(ret->solid, &ret->squares[ret->current], pkey); |
1087 | assert(success); |
1088 | |
1089 | ret->dpkey[0] = pkey[0]; |
1090 | ret->dpkey[1] = pkey[1]; |
1091 | ret->dgkey[0] = 0; |
1092 | ret->dgkey[1] = 1; |
1093 | } |
1094 | |
1095 | |
1096 | ret->spkey[0] = pkey[0]; |
1097 | ret->spkey[1] = pkey[1]; |
1098 | ret->sgkey[0] = skey[0]; |
1099 | ret->sgkey[1] = skey[1]; |
1100 | ret->previous = from->current; |
1101 | ret->angle = angle; |
1482ee76 |
1102 | |
1103 | return ret; |
1104 | } |
1105 | |
1106 | /* ---------------------------------------------------------------------- |
1107 | * Drawing routines. |
1108 | */ |
1109 | |
1110 | struct bbox { |
1111 | float l, r, u, d; |
1112 | }; |
1113 | |
1114 | struct game_drawstate { |
1115 | int ox, oy; /* pixel position of float origin */ |
1116 | }; |
1117 | |
1118 | static void find_bbox_callback(void *ctx, struct grid_square *sq) |
1119 | { |
1120 | struct bbox *bb = (struct bbox *)ctx; |
1121 | int i; |
1122 | |
1123 | for (i = 0; i < sq->npoints; i++) { |
1124 | if (bb->l > sq->points[i*2]) bb->l = sq->points[i*2]; |
1125 | if (bb->r < sq->points[i*2]) bb->r = sq->points[i*2]; |
1126 | if (bb->u > sq->points[i*2+1]) bb->u = sq->points[i*2+1]; |
1127 | if (bb->d < sq->points[i*2+1]) bb->d = sq->points[i*2+1]; |
1128 | } |
1129 | } |
1130 | |
1131 | static struct bbox find_bbox(game_params *params) |
1132 | { |
1133 | struct bbox bb; |
1134 | |
1135 | /* |
1136 | * These should be hugely more than the real bounding box will |
1137 | * be. |
1138 | */ |
03f856c4 |
1139 | bb.l = 2.0F * (params->d1 + params->d2); |
1140 | bb.r = -2.0F * (params->d1 + params->d2); |
1141 | bb.u = 2.0F * (params->d1 + params->d2); |
1142 | bb.d = -2.0F * (params->d1 + params->d2); |
1482ee76 |
1143 | enum_grid_squares(params, find_bbox_callback, &bb); |
1144 | |
1145 | return bb; |
1146 | } |
1147 | |
1148 | void game_size(game_params *params, int *x, int *y) |
1149 | { |
1150 | struct bbox bb = find_bbox(params); |
03f856c4 |
1151 | *x = (int)((bb.r - bb.l + 2*solids[params->solid]->border) * GRID_SCALE); |
1152 | *y = (int)((bb.d - bb.u + 2*solids[params->solid]->border) * GRID_SCALE); |
1482ee76 |
1153 | } |
1154 | |
1155 | float *game_colours(frontend *fe, game_state *state, int *ncolours) |
1156 | { |
1157 | float *ret = snewn(3 * NCOLOURS, float); |
1158 | |
1159 | frontend_default_colour(fe, &ret[COL_BACKGROUND * 3]); |
1160 | |
1161 | ret[COL_BORDER * 3 + 0] = 0.0; |
1162 | ret[COL_BORDER * 3 + 1] = 0.0; |
1163 | ret[COL_BORDER * 3 + 2] = 0.0; |
1164 | |
1165 | ret[COL_BLUE * 3 + 0] = 0.0; |
1166 | ret[COL_BLUE * 3 + 1] = 0.0; |
1167 | ret[COL_BLUE * 3 + 2] = 1.0; |
1168 | |
1169 | *ncolours = NCOLOURS; |
1170 | return ret; |
1171 | } |
1172 | |
1173 | game_drawstate *game_new_drawstate(game_state *state) |
1174 | { |
1175 | struct game_drawstate *ds = snew(struct game_drawstate); |
1176 | struct bbox bb = find_bbox(&state->params); |
1177 | |
03f856c4 |
1178 | ds->ox = (int)(-(bb.l - state->solid->border) * GRID_SCALE); |
1179 | ds->oy = (int)(-(bb.u - state->solid->border) * GRID_SCALE); |
1482ee76 |
1180 | |
1181 | return ds; |
1182 | } |
1183 | |
1184 | void game_free_drawstate(game_drawstate *ds) |
1185 | { |
1186 | sfree(ds); |
1187 | } |
1188 | |
1189 | void game_redraw(frontend *fe, game_drawstate *ds, game_state *oldstate, |
87ed82be |
1190 | game_state *state, float animtime, float flashtime) |
1482ee76 |
1191 | { |
1192 | int i, j; |
1193 | struct bbox bb = find_bbox(&state->params); |
1194 | struct solid *poly; |
1195 | int *pkey, *gkey; |
1196 | float t[3]; |
1197 | float angle; |
1198 | game_state *newstate; |
1199 | int square; |
1200 | |
03f856c4 |
1201 | draw_rect(fe, 0, 0, (int)((bb.r-bb.l+2.0F) * GRID_SCALE), |
1202 | (int)((bb.d-bb.u+2.0F) * GRID_SCALE), COL_BACKGROUND); |
1482ee76 |
1203 | |
1204 | if (oldstate && oldstate->movecount > state->movecount) { |
1205 | game_state *t; |
1206 | |
1207 | /* |
1208 | * This is an Undo. So reverse the order of the states, and |
1209 | * run the roll timer backwards. |
1210 | */ |
1211 | t = oldstate; |
1212 | oldstate = state; |
1213 | state = t; |
1214 | |
1215 | animtime = ROLLTIME - animtime; |
1216 | } |
1217 | |
1218 | if (!oldstate) { |
1219 | oldstate = state; |
1220 | angle = 0.0; |
1221 | square = state->current; |
1222 | pkey = state->dpkey; |
1223 | gkey = state->dgkey; |
1224 | } else { |
1225 | angle = state->angle * animtime / ROLLTIME; |
1226 | square = state->previous; |
1227 | pkey = state->spkey; |
1228 | gkey = state->sgkey; |
1229 | } |
1230 | newstate = state; |
1231 | state = oldstate; |
1232 | |
1233 | for (i = 0; i < state->nsquares; i++) { |
1234 | int coords[8]; |
1235 | |
1236 | for (j = 0; j < state->squares[i].npoints; j++) { |
03f856c4 |
1237 | coords[2*j] = ((int)(state->squares[i].points[2*j] * GRID_SCALE) |
1238 | + ds->ox); |
1239 | coords[2*j+1] = ((int)(state->squares[i].points[2*j+1]*GRID_SCALE) |
1240 | + ds->oy); |
1482ee76 |
1241 | } |
1242 | |
1243 | draw_polygon(fe, coords, state->squares[i].npoints, TRUE, |
1244 | state->squares[i].blue ? COL_BLUE : COL_BACKGROUND); |
1245 | draw_polygon(fe, coords, state->squares[i].npoints, FALSE, COL_BORDER); |
1246 | } |
1247 | |
1248 | /* |
1249 | * Now compute and draw the polyhedron. |
1250 | */ |
1251 | poly = transform_poly(state->solid, state->squares[square].flip, |
1252 | pkey[0], pkey[1], angle); |
1253 | |
1254 | /* |
1255 | * Compute the translation required to align the two key points |
1256 | * on the polyhedron with the same key points on the current |
1257 | * face. |
1258 | */ |
1259 | for (i = 0; i < 3; i++) { |
1260 | float tc = 0.0; |
1261 | |
1262 | for (j = 0; j < 2; j++) { |
1263 | float grid_coord; |
1264 | |
1265 | if (i < 2) { |
1266 | grid_coord = |
1267 | state->squares[square].points[gkey[j]*2+i]; |
1268 | } else { |
1269 | grid_coord = 0.0; |
1270 | } |
1271 | |
1272 | tc += (grid_coord - poly->vertices[pkey[j]*3+i]); |
1273 | } |
1274 | |
1275 | t[i] = tc / 2; |
1276 | } |
1277 | for (i = 0; i < poly->nvertices; i++) |
1278 | for (j = 0; j < 3; j++) |
1279 | poly->vertices[i*3+j] += t[j]; |
1280 | |
1281 | /* |
1282 | * Now actually draw each face. |
1283 | */ |
1284 | for (i = 0; i < poly->nfaces; i++) { |
1285 | float points[8]; |
1286 | int coords[8]; |
1287 | |
1288 | for (j = 0; j < poly->order; j++) { |
1289 | int f = poly->faces[i*poly->order + j]; |
1290 | points[j*2] = (poly->vertices[f*3+0] - |
1291 | poly->vertices[f*3+2] * poly->shear); |
1292 | points[j*2+1] = (poly->vertices[f*3+1] - |
1293 | poly->vertices[f*3+2] * poly->shear); |
1294 | } |
1295 | |
1296 | for (j = 0; j < poly->order; j++) { |
962dcf9a |
1297 | coords[j*2] = (int)floor(points[j*2] * GRID_SCALE) + ds->ox; |
1298 | coords[j*2+1] = (int)floor(points[j*2+1] * GRID_SCALE) + ds->oy; |
1482ee76 |
1299 | } |
1300 | |
1301 | /* |
1302 | * Find out whether these points are in a clockwise or |
1303 | * anticlockwise arrangement. If the latter, discard the |
1304 | * face because it's facing away from the viewer. |
1305 | * |
1306 | * This would involve fiddly winding-number stuff for a |
1307 | * general polygon, but for the simple parallelograms we'll |
1308 | * be seeing here, all we have to do is check whether the |
1309 | * corners turn right or left. So we'll take the vector |
1310 | * from point 0 to point 1, turn it right 90 degrees, |
1311 | * and check the sign of the dot product with that and the |
1312 | * next vector (point 1 to point 2). |
1313 | */ |
1314 | { |
1315 | float v1x = points[2]-points[0]; |
1316 | float v1y = points[3]-points[1]; |
1317 | float v2x = points[4]-points[2]; |
1318 | float v2y = points[5]-points[3]; |
1319 | float dp = v1x * v2y - v1y * v2x; |
1320 | |
1321 | if (dp <= 0) |
1322 | continue; |
1323 | } |
1324 | |
1325 | draw_polygon(fe, coords, poly->order, TRUE, |
1326 | state->facecolours[i] ? COL_BLUE : COL_BACKGROUND); |
1327 | draw_polygon(fe, coords, poly->order, FALSE, COL_BORDER); |
1328 | } |
1329 | sfree(poly); |
1330 | |
03f856c4 |
1331 | draw_update(fe, 0, 0, (int)((bb.r-bb.l+2.0F) * GRID_SCALE), |
1332 | (int)((bb.d-bb.u+2.0F) * GRID_SCALE)); |
fd1a1a2b |
1333 | |
1334 | /* |
1335 | * Update the status bar. |
1336 | */ |
1337 | { |
1338 | char statusbuf[256]; |
1339 | |
1340 | sprintf(statusbuf, "%sMoves: %d", |
1341 | (state->completed ? "COMPLETED! " : ""), |
1342 | (state->completed ? state->completed : state->movecount)); |
1343 | |
1344 | status_bar(fe, statusbuf); |
1345 | } |
1482ee76 |
1346 | } |
1347 | |
1348 | float game_anim_length(game_state *oldstate, game_state *newstate) |
1349 | { |
1350 | return ROLLTIME; |
1351 | } |
87ed82be |
1352 | |
1353 | float game_flash_length(game_state *oldstate, game_state *newstate) |
1354 | { |
1355 | return 0.0F; |
1356 | } |
fd1a1a2b |
1357 | |
1358 | int game_wants_statusbar(void) |
1359 | { |
1360 | return TRUE; |
1361 | } |