13 #define MAXVERTICES 20
18 float vertices[MAXVERTICES * 3]; /* 3*npoints coordinates */
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 */
26 static const struct solid tetrahedron = {
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,
36 0,2,1, 3,1,2, 2,0,3, 1,3,0
39 -0.816496580928, -0.471404520791, 0.333333333334,
40 0.0, 0.942809041583, 0.333333333333,
41 0.816496580928, -0.471404520791, 0.333333333334,
47 static const struct solid cube = {
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,
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
58 -1,0,0, 0,0,+1, +1,0,0, 0,0,-1, 0,-1,0, 0,+1,0
63 static const struct solid octahedron = {
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,
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
78 -0.816496580928, -0.471404520791, -0.333333333334,
79 -0.816496580928, 0.471404520791, 0.333333333334,
80 0.0, -0.942809041583, 0.333333333333,
83 0.0, 0.942809041583, -0.333333333333,
84 0.816496580928, -0.471404520791, -0.333333333334,
85 0.816496580928, 0.471404520791, 0.333333333334,
90 static const struct solid icosahedron = {
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,
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,
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,
119 0.0, -0.666666666667, 0.745355992501,
120 0.0, 0.666666666667, -0.745355992501,
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,
139 TETRAHEDRON, CUBE, OCTAHEDRON, ICOSAHEDRON
141 static const struct solid *solids[] = {
142 &tetrahedron, &cube, &octahedron, &icosahedron
152 enum { LEFT, RIGHT, UP, DOWN };
154 #define GRID_SCALE 48
157 #define SQ(x) ( (x) * (x) )
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; \
167 #define APPROXEQ(x,y) ( SQ(x-y) < 0.1 )
172 float points[8]; /* maximum */
173 int directions[4]; /* bit masks showing point pairs */
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).
192 struct game_params params;
193 const struct solid *solid;
195 struct grid_square *squares;
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 */
208 game_params *default_params(void)
210 game_params *ret = snew(game_params);
219 void free_params(game_params *params)
224 static void enum_grid_squares(game_params *params,
225 void (*callback)(void *, struct grid_square *),
228 const struct solid *solid = solids[params->solid];
230 if (solid->order == 4) {
233 for (x = 0; x < params->d1; x++)
234 for (y = 0; y < params->d2; y++) {
235 struct grid_square sq;
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;
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 */
257 * This is supremely irrelevant, but just to avoid
258 * having any uninitialised structure members...
265 int row, rowlen, other, i, firstix = -1;
266 float theight = sqrt(3) / 2.0;
268 for (row = 0; row < params->d1 + params->d2; row++) {
269 if (row < params->d1) {
271 rowlen = row + params->d2;
274 rowlen = 2*params->d1 + params->d2 - row;
278 * There are `rowlen' down-pointing triangles.
280 for (i = 0; i < rowlen; i++) {
281 struct grid_square sq;
285 ix = (2 * i - (rowlen-1));
289 sq.y = y + theight / 3;
290 sq.points[0] = x - 0.5;
293 sq.points[3] = y + theight;
294 sq.points[4] = x + 0.5;
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 */
308 sq.tetra_class = ((row+(ix&1)) & 2) ^ (ix & 3);
314 * There are `rowlen+other' up-pointing triangles.
316 for (i = 0; i < rowlen+other; i++) {
317 struct grid_square sq;
321 ix = (2 * i - (rowlen+other-1));
325 sq.y = y + 2*theight / 3;
326 sq.points[0] = x + 0.5;
327 sq.points[1] = y + theight;
330 sq.points[4] = x - 0.5;
331 sq.points[5] = y + theight;
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 */
344 sq.tetra_class = ((row+(ix&1)) & 2) ^ (ix & 3);
352 static int grid_area(int d1, int d2, int order)
355 * An NxM grid of squares has NM squares in it.
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.)
367 return d1*d1 + d2*d2 + 4*d1*d2;
377 static void classify_grid_square_callback(void *ctx, struct grid_square *sq)
379 struct grid_data *data = (struct grid_data *)ctx;
382 if (data->nclasses == 4)
383 thisclass = sq->tetra_class;
384 else if (data->nclasses == 2)
385 thisclass = sq->flip;
389 data->gridptrs[thisclass][data->nsquares[thisclass]++] =
393 char *new_game_seed(game_params *params)
395 struct grid_data data;
396 int i, j, k, m, area, facesperclass;
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.)
407 area = grid_area(params->d1, params->d2, solids[params->solid]->order);
408 if (params->solid == TETRAHEDRON)
410 else if (params->solid == OCTAHEDRON)
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;
419 data.squareindex = 0;
420 enum_grid_squares(params, classify_grid_square_callback, &data);
422 facesperclass = solids[params->solid]->nfaces / data.nclasses;
424 for (i = 0; i < data.nclasses; i++)
425 assert(data.nsquares[i] >= facesperclass);
426 assert(data.squareindex == area);
429 * So now we know how many faces to allocate in each class. Get
432 flags = snewn(area, int);
433 for (i = 0; i < area; i++)
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];
448 assert(!flags[data.gridptrs[i][n]]);
449 flags[data.gridptrs[i][n]] = TRUE;
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.
456 while (n < data.nsquares[i]-1) {
457 data.gridptrs[i][n] = data.gridptrs[i][n+1];
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
470 seed = snewn(area / 4 + 40, char);
475 for (i = 0; i < area; i++) {
479 data.gridptrs[0][m++] = i;
483 *p++ = "0123456789ABCDEF"[j];
489 *p++ = "0123456789ABCDEF"[j];
492 * Choose a non-blue square for the polyhedron.
495 unsigned long divisor = RAND_MAX / m;
496 unsigned long max = divisor * m;
505 sprintf(p, ":%d", data.gridptrs[0][n]);
508 sfree(data.gridptrs[0]);
514 static void add_grid_square_callback(void *ctx, struct grid_square *sq)
516 game_state *state = (game_state *)ctx;
518 state->squares[state->nsquares] = *sq; /* structure copy */
519 state->squares[state->nsquares].blue = FALSE;
523 static int lowest_face(const struct solid *solid)
530 for (i = 0; i < solid->nfaces; i++) {
533 for (j = 0; j < solid->order; j++) {
534 int f = solid->faces[i*solid->order + j];
535 z += solid->vertices[f*3+2];
538 if (i == 0 || zmin > z) {
547 static int align_poly(const struct solid *solid, struct grid_square *sq,
552 int flip = (sq->flip ? -1 : +1);
555 * First, find the lowest z-coordinate present in the solid.
558 for (i = 0; i < solid->nvertices; i++)
559 if (zmin > solid->vertices[i*3+2])
560 zmin = solid->vertices[i*3+2];
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).
568 for (j = 0; j < sq->npoints; j++) {
574 for (i = 0; i < solid->nvertices; i++) {
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);
587 if (matches != 1 || index < 0)
595 static void flip_poly(struct solid *solid, int flip)
600 for (i = 0; i < solid->nvertices; i++) {
601 solid->vertices[i*3+0] *= -1;
602 solid->vertices[i*3+1] *= -1;
604 for (i = 0; i < solid->nfaces; i++) {
605 solid->normals[i*3+0] *= -1;
606 solid->normals[i*3+1] *= -1;
611 static struct solid *transform_poly(const struct solid *solid, int flip,
612 int key0, int key1, float angle)
614 struct solid *ret = snew(struct solid);
615 float vx, vy, ax, ay;
616 float vmatrix[9], amatrix[9], vmatrix2[9];
619 *ret = *solid; /* structure copy */
621 flip_poly(ret, flip);
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.
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));
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;
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;
644 memcpy(vmatrix2, vmatrix, sizeof(vmatrix));
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);
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);
662 game_state *new_game(game_params *params, char *seed)
664 game_state *state = snew(game_state);
667 state->params = *params; /* structure copy */
668 state->solid = solids[params->solid];
670 area = grid_area(params->d1, params->d2, state->solid->order);
671 state->squares = snewn(area, struct grid_square);
673 enum_grid_squares(params, add_grid_square_callback, state);
674 assert(state->nsquares == area);
676 state->facecolours = snewn(state->solid->nfaces, int);
677 memset(state->facecolours, 0, state->solid->nfaces * sizeof(int));
680 * Set up the blue squares and polyhedron position according to
689 for (i = 0; i < state->nsquares; i++) {
692 if (v >= '0' && v <= '9')
694 else if (v >= 'A' && v <= 'F')
696 else if (v >= 'a' && v <= 'f')
702 state->squares[i].blue = TRUE;
711 state->current = atoi(p);
712 if (state->current < 0 || state->current >= state->nsquares)
713 state->current = 0; /* got to do _something_ */
717 * Align the polyhedron with its grid square and determine
718 * initial key points.
724 ret = align_poly(state->solid, &state->squares[state->current], pkey);
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;
733 state->previous = state->current;
735 state->completed = FALSE;
736 state->movecount = 0;
741 game_state *dup_game(game_state *state)
743 game_state *ret = snew(game_state);
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;
770 void free_game(game_state *state)
775 game_state *make_move(game_state *from, int x, int y, int button)
778 int pkey[2], skey[2], dkey[2];
782 int i, j, dest, mask;
786 * All moves are made with the cursor keys.
788 if (button == CURSOR_UP)
790 else if (button == CURSOR_DOWN)
792 else if (button == CURSOR_LEFT)
794 else if (button == CURSOR_RIGHT)
800 * Find the two points in the current grid square which
801 * correspond to this move.
803 mask = from->squares[from->current].directions[direction];
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];
816 * Now find the other grid square which shares those points.
817 * This is our move destination.
820 for (i = 0; i < from->nsquares; i++)
821 if (i != from->current) {
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]));
830 dist = (SQ(from->squares[i].points[j*2] - points[2]) +
831 SQ(from->squares[i].points[j*2+1] - points[3]));
845 ret = dup_game(from);
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
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.
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]];
865 * Now pkey[0] corresponds to skey[0] and dkey[0], and
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.
879 for (i = 0; i < from->solid->nfaces; i++) {
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])
894 for (i = 0; i < 3; i++)
895 dp += (from->solid->normals[f[0]*3+i] *
896 from->solid->normals[f[1]*3+i]);
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!
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 :-/
917 if (from->solid->order == 4 && direction == UP)
918 angle = -angle; /* HACK */
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);
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);
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
946 int *newcolours = snewn(from->solid->nfaces, int);
948 for (i = 0; i < from->solid->nfaces; i++)
951 for (i = 0; i < from->solid->nfaces; i++) {
955 * Now go through the transformed polyhedron's faces
956 * and figure out which one's normal is approximately
959 for (j = 0; j < poly->nfaces; j++) {
965 for (k = 0; k < 3; k++)
966 dist += SQ(poly->normals[j*3+k] -
967 from->solid->normals[i*3+k]);
969 if (APPROXEQ(dist, 0)) {
971 newcolours[i] = ret->facecolours[j];
978 for (i = 0; i < from->solid->nfaces; i++)
979 assert(newcolours[i] != -1);
981 sfree(ret->facecolours);
982 ret->facecolours = newcolours;
986 * And finally, swap the colour between the bottom face of the
987 * polyhedron and the face we've just landed on.
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.
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;
1000 * Detect game completion.
1003 for (i = 0; i < ret->solid->nfaces; i++)
1004 if (ret->facecolours[i])
1006 if (j == ret->solid->nfaces)
1007 ret->completed = TRUE;
1013 * Align the normal polyhedron with its grid square, to get key
1014 * points for non-animated display.
1020 success = align_poly(ret->solid, &ret->squares[ret->current], pkey);
1023 ret->dpkey[0] = pkey[0];
1024 ret->dpkey[1] = pkey[1];
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;
1041 /* ----------------------------------------------------------------------
1049 struct game_drawstate {
1050 int ox, oy; /* pixel position of float origin */
1053 static void find_bbox_callback(void *ctx, struct grid_square *sq)
1055 struct bbox *bb = (struct bbox *)ctx;
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];
1066 static struct bbox find_bbox(game_params *params)
1071 * These should be hugely more than the real bounding box will
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);
1083 void game_size(game_params *params, int *x, int *y)
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;
1090 float *game_colours(frontend *fe, game_state *state, int *ncolours)
1092 float *ret = snewn(3 * NCOLOURS, float);
1094 frontend_default_colour(fe, &ret[COL_BACKGROUND * 3]);
1096 ret[COL_BORDER * 3 + 0] = 0.0;
1097 ret[COL_BORDER * 3 + 1] = 0.0;
1098 ret[COL_BORDER * 3 + 2] = 0.0;
1100 ret[COL_BLUE * 3 + 0] = 0.0;
1101 ret[COL_BLUE * 3 + 1] = 0.0;
1102 ret[COL_BLUE * 3 + 2] = 1.0;
1104 *ncolours = NCOLOURS;
1108 game_drawstate *game_new_drawstate(game_state *state)
1110 struct game_drawstate *ds = snew(struct game_drawstate);
1111 struct bbox bb = find_bbox(&state->params);
1113 ds->ox = -(bb.l - 1) * GRID_SCALE;
1114 ds->oy = -(bb.u - 1) * GRID_SCALE;
1119 void game_free_drawstate(game_drawstate *ds)
1124 void game_redraw(frontend *fe, game_drawstate *ds, game_state *oldstate,
1125 game_state *state, float animtime)
1128 struct bbox bb = find_bbox(&state->params);
1133 game_state *newstate;
1136 draw_rect(fe, 0, 0, (bb.r-bb.l+2) * GRID_SCALE,
1137 (bb.d-bb.u+2) * GRID_SCALE, COL_BACKGROUND);
1139 if (oldstate && oldstate->movecount > state->movecount) {
1143 * This is an Undo. So reverse the order of the states, and
1144 * run the roll timer backwards.
1150 animtime = ROLLTIME - animtime;
1156 square = state->current;
1157 pkey = state->dpkey;
1158 gkey = state->dgkey;
1160 angle = state->angle * animtime / ROLLTIME;
1161 square = state->previous;
1162 pkey = state->spkey;
1163 gkey = state->sgkey;
1168 for (i = 0; i < state->nsquares; i++) {
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;
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);
1184 * Now compute and draw the polyhedron.
1186 poly = transform_poly(state->solid, state->squares[square].flip,
1187 pkey[0], pkey[1], angle);
1190 * Compute the translation required to align the two key points
1191 * on the polyhedron with the same key points on the current
1194 for (i = 0; i < 3; i++) {
1197 for (j = 0; j < 2; j++) {
1202 state->squares[square].points[gkey[j]*2+i];
1207 tc += (grid_coord - poly->vertices[pkey[j]*3+i]);
1212 for (i = 0; i < poly->nvertices; i++)
1213 for (j = 0; j < 3; j++)
1214 poly->vertices[i*3+j] += t[j];
1217 * Now actually draw each face.
1219 for (i = 0; i < poly->nfaces; i++) {
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);
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;
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.
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).
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;
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);
1266 draw_update(fe, 0, 0, (bb.r-bb.l+2) * GRID_SCALE,
1267 (bb.d-bb.u+2) * GRID_SCALE);
1270 float game_anim_length(game_state *oldstate, game_state *newstate)