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