13 const char *const game_name = "Cube";
15 #define MAXVERTICES 20
20 float vertices[MAXVERTICES * 3]; /* 3*npoints coordinates */
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 */
26 float border; /* border required around arena */
29 static const struct solid tetrahedron = {
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,
39 0,2,1, 3,1,2, 2,0,3, 1,3,0
42 -0.816496580928F, -0.471404520791F, 0.333333333334F,
43 0.0F, 0.942809041583F, 0.333333333333F,
44 0.816496580928F, -0.471404520791F, 0.333333333334F,
50 static const struct solid cube = {
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,
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
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
70 static const struct solid octahedron = {
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,
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
85 -0.816496580928F, -0.471404520791F, -0.333333333334F,
86 -0.816496580928F, 0.471404520791F, 0.333333333334F,
87 0.0F, -0.942809041583F, 0.333333333333F,
90 0.0F, 0.942809041583F, -0.333333333333F,
91 0.816496580928F, -0.471404520791F, -0.333333333334F,
92 0.816496580928F, 0.471404520791F, 0.333333333334F,
97 static const struct solid icosahedron = {
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,
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,
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,
126 0.0F, -0.666666666667F, 0.745355992501F,
127 0.0F, 0.666666666667F, -0.745355992501F,
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,
146 TETRAHEDRON, CUBE, OCTAHEDRON, ICOSAHEDRON
148 static const struct solid *solids[] = {
149 &tetrahedron, &cube, &octahedron, &icosahedron
159 enum { LEFT, RIGHT, UP, DOWN, UP_LEFT, UP_RIGHT, DOWN_LEFT, DOWN_RIGHT };
161 #define GRID_SCALE 48.0F
162 #define ROLLTIME 0.1F
164 #define SQ(x) ( (x) * (x) )
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; \
174 #define APPROXEQ(x,y) ( SQ(x-y) < 0.1 )
179 float points[8]; /* maximum */
180 int directions[8]; /* bit masks showing point pairs */
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).
199 struct game_params params;
200 const struct solid *solid;
202 struct grid_square *squares;
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 */
215 game_params *default_params(void)
217 game_params *ret = snew(game_params);
226 int game_fetch_preset(int i, char **name, game_params **params)
228 game_params *ret = snew(game_params);
240 ret->solid = TETRAHEDRON;
246 ret->solid = OCTAHEDRON;
252 ret->solid = ICOSAHEDRON;
266 void free_params(game_params *params)
271 game_params *dup_params(game_params *params)
273 game_params *ret = snew(game_params);
274 *ret = *params; /* structure copy */
278 static void enum_grid_squares(game_params *params,
279 void (*callback)(void *, struct grid_square *),
282 const struct solid *solid = solids[params->solid];
284 if (solid->order == 4) {
287 for (x = 0; x < params->d1; x++)
288 for (y = 0; y < params->d2; y++) {
289 struct grid_square sq;
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;
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 */
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 */
315 * This is supremely irrelevant, but just to avoid
316 * having any uninitialised structure members...
323 int row, rowlen, other, i, firstix = -1;
324 float theight = (float)(sqrt(3) / 2.0);
326 for (row = 0; row < params->d1 + params->d2; row++) {
327 if (row < params->d1) {
329 rowlen = row + params->d2;
332 rowlen = 2*params->d1 + params->d2 - row;
336 * There are `rowlen' down-pointing triangles.
338 for (i = 0; i < rowlen; i++) {
339 struct grid_square sq;
343 ix = (2 * i - (rowlen-1));
347 sq.y = y + theight / 3;
348 sq.points[0] = x - 0.5F;
351 sq.points[3] = y + theight;
352 sq.points[4] = x + 0.5F;
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 */
362 * Down-pointing triangle: both the up diagonals go
363 * up, and the down ones go left and right.
365 sq.directions[UP_LEFT] = sq.directions[UP_RIGHT] =
367 sq.directions[DOWN_LEFT] = sq.directions[LEFT];
368 sq.directions[DOWN_RIGHT] = sq.directions[RIGHT];
375 sq.tetra_class = ((row+(ix&1)) & 2) ^ (ix & 3);
381 * There are `rowlen+other' up-pointing triangles.
383 for (i = 0; i < rowlen+other; i++) {
384 struct grid_square sq;
388 ix = (2 * i - (rowlen+other-1));
392 sq.y = y + 2*theight / 3;
393 sq.points[0] = x + 0.5F;
394 sq.points[1] = y + theight;
397 sq.points[4] = x - 0.5F;
398 sq.points[5] = y + theight;
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 */
407 * Up-pointing triangle: both the down diagonals go
408 * down, and the up ones go left and right.
410 sq.directions[DOWN_LEFT] = sq.directions[DOWN_RIGHT] =
412 sq.directions[UP_LEFT] = sq.directions[LEFT];
413 sq.directions[UP_RIGHT] = sq.directions[RIGHT];
420 sq.tetra_class = ((row+(ix&1)) & 2) ^ (ix & 3);
428 static int grid_area(int d1, int d2, int order)
431 * An NxM grid of squares has NM squares in it.
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.)
443 return d1*d1 + d2*d2 + 4*d1*d2;
453 static void classify_grid_square_callback(void *ctx, struct grid_square *sq)
455 struct grid_data *data = (struct grid_data *)ctx;
458 if (data->nclasses == 4)
459 thisclass = sq->tetra_class;
460 else if (data->nclasses == 2)
461 thisclass = sq->flip;
465 data->gridptrs[thisclass][data->nsquares[thisclass]++] =
469 char *new_game_seed(game_params *params)
471 struct grid_data data;
472 int i, j, k, m, area, facesperclass;
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.)
483 area = grid_area(params->d1, params->d2, solids[params->solid]->order);
484 if (params->solid == TETRAHEDRON)
486 else if (params->solid == OCTAHEDRON)
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;
495 data.squareindex = 0;
496 enum_grid_squares(params, classify_grid_square_callback, &data);
498 facesperclass = solids[params->solid]->nfaces / data.nclasses;
500 for (i = 0; i < data.nclasses; i++)
501 assert(data.nsquares[i] >= facesperclass);
502 assert(data.squareindex == area);
505 * So now we know how many faces to allocate in each class. Get
508 flags = snewn(area, int);
509 for (i = 0; i < area; i++)
512 for (i = 0; i < data.nclasses; i++) {
513 for (j = 0; j < facesperclass; j++) {
514 int n = rand_upto(data.nsquares[i]);
516 assert(!flags[data.gridptrs[i][n]]);
517 flags[data.gridptrs[i][n]] = TRUE;
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.
524 while (n < data.nsquares[i]-1) {
525 data.gridptrs[i][n] = data.gridptrs[i][n+1];
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
538 seed = snewn(area / 4 + 40, char);
543 for (i = 0; i < area; i++) {
547 data.gridptrs[0][m++] = i;
551 *p++ = "0123456789ABCDEF"[j];
557 *p++ = "0123456789ABCDEF"[j];
560 * Choose a non-blue square for the polyhedron.
562 sprintf(p, ":%d", data.gridptrs[0][rand_upto(m)]);
564 sfree(data.gridptrs[0]);
570 static void add_grid_square_callback(void *ctx, struct grid_square *sq)
572 game_state *state = (game_state *)ctx;
574 state->squares[state->nsquares] = *sq; /* structure copy */
575 state->squares[state->nsquares].blue = FALSE;
579 static int lowest_face(const struct solid *solid)
586 for (i = 0; i < solid->nfaces; i++) {
589 for (j = 0; j < solid->order; j++) {
590 int f = solid->faces[i*solid->order + j];
591 z += solid->vertices[f*3+2];
594 if (i == 0 || zmin > z) {
603 static int align_poly(const struct solid *solid, struct grid_square *sq,
608 int flip = (sq->flip ? -1 : +1);
611 * First, find the lowest z-coordinate present in the solid.
614 for (i = 0; i < solid->nvertices; i++)
615 if (zmin > solid->vertices[i*3+2])
616 zmin = solid->vertices[i*3+2];
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).
624 for (j = 0; j < sq->npoints; j++) {
630 for (i = 0; i < solid->nvertices; i++) {
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);
643 if (matches != 1 || index < 0)
651 static void flip_poly(struct solid *solid, int flip)
656 for (i = 0; i < solid->nvertices; i++) {
657 solid->vertices[i*3+0] *= -1;
658 solid->vertices[i*3+1] *= -1;
660 for (i = 0; i < solid->nfaces; i++) {
661 solid->normals[i*3+0] *= -1;
662 solid->normals[i*3+1] *= -1;
667 static struct solid *transform_poly(const struct solid *solid, int flip,
668 int key0, int key1, float angle)
670 struct solid *ret = snew(struct solid);
671 float vx, vy, ax, ay;
672 float vmatrix[9], amatrix[9], vmatrix2[9];
675 *ret = *solid; /* structure copy */
677 flip_poly(ret, flip);
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.
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));
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;
693 ax = (float)cos(angle);
694 ay = (float)sin(angle);
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;
700 memcpy(vmatrix2, vmatrix, sizeof(vmatrix));
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);
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);
718 game_state *new_game(game_params *params, char *seed)
720 game_state *state = snew(game_state);
723 state->params = *params; /* structure copy */
724 state->solid = solids[params->solid];
726 area = grid_area(params->d1, params->d2, state->solid->order);
727 state->squares = snewn(area, struct grid_square);
729 enum_grid_squares(params, add_grid_square_callback, state);
730 assert(state->nsquares == area);
732 state->facecolours = snewn(state->solid->nfaces, int);
733 memset(state->facecolours, 0, state->solid->nfaces * sizeof(int));
736 * Set up the blue squares and polyhedron position according to
745 for (i = 0; i < state->nsquares; i++) {
748 if (v >= '0' && v <= '9')
750 else if (v >= 'A' && v <= 'F')
752 else if (v >= 'a' && v <= 'f')
758 state->squares[i].blue = TRUE;
767 state->current = atoi(p);
768 if (state->current < 0 || state->current >= state->nsquares)
769 state->current = 0; /* got to do _something_ */
773 * Align the polyhedron with its grid square and determine
774 * initial key points.
780 ret = align_poly(state->solid, &state->squares[state->current], pkey);
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;
789 state->previous = state->current;
791 state->completed = 0;
792 state->movecount = 0;
797 game_state *dup_game(game_state *state)
799 game_state *ret = snew(game_state);
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;
826 void free_game(game_state *state)
831 game_state *make_move(game_state *from, int x, int y, int button)
834 int pkey[2], skey[2], dkey[2];
838 int i, j, dest, mask;
842 * All moves are made with the cursor keys.
844 if (button == CURSOR_UP)
846 else if (button == CURSOR_DOWN)
848 else if (button == CURSOR_LEFT)
850 else if (button == CURSOR_RIGHT)
852 else if (button == CURSOR_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;
864 * Find the two points in the current grid square which
865 * correspond to this move.
867 mask = from->squares[from->current].directions[direction];
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];
880 * Now find the other grid square which shares those points.
881 * This is our move destination.
884 for (i = 0; i < from->nsquares; i++)
885 if (i != from->current) {
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]));
894 dist = (SQ(from->squares[i].points[j*2] - points[2]) +
895 SQ(from->squares[i].points[j*2+1] - points[3]));
909 ret = dup_game(from);
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
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.
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]];
929 * Now pkey[0] corresponds to skey[0] and dkey[0], and
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.
943 for (i = 0; i < from->solid->nfaces; i++) {
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])
958 for (i = 0; i < 3; i++)
959 dp += (from->solid->normals[f[0]*3+i] *
960 from->solid->normals[f[1]*3+i]);
961 angle = (float)acos(dp);
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!
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 :-/
981 if (from->solid->order == 4 && direction == UP)
982 angle = -angle; /* HACK */
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);
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);
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
1010 int *newcolours = snewn(from->solid->nfaces, int);
1012 for (i = 0; i < from->solid->nfaces; i++)
1015 for (i = 0; i < from->solid->nfaces; i++) {
1019 * Now go through the transformed polyhedron's faces
1020 * and figure out which one's normal is approximately
1021 * equal to this one.
1023 for (j = 0; j < poly->nfaces; j++) {
1029 for (k = 0; k < 3; k++)
1030 dist += SQ(poly->normals[j*3+k] -
1031 from->solid->normals[i*3+k]);
1033 if (APPROXEQ(dist, 0)) {
1035 newcolours[i] = ret->facecolours[j];
1039 assert(nmatch == 1);
1042 for (i = 0; i < from->solid->nfaces; i++)
1043 assert(newcolours[i] != -1);
1045 sfree(ret->facecolours);
1046 ret->facecolours = newcolours;
1050 * And finally, swap the colour between the bottom face of the
1051 * polyhedron and the face we've just landed on.
1053 * We don't do this if the game is already complete, since we
1054 * allow the user to roll the fully blue polyhedron around the
1055 * grid as a feeble reward.
1057 if (!ret->completed) {
1058 i = lowest_face(from->solid);
1059 j = ret->facecolours[i];
1060 ret->facecolours[i] = ret->squares[ret->current].blue;
1061 ret->squares[ret->current].blue = j;
1064 * Detect game completion.
1067 for (i = 0; i < ret->solid->nfaces; i++)
1068 if (ret->facecolours[i])
1070 if (j == ret->solid->nfaces)
1071 ret->completed = ret->movecount;
1077 * Align the normal polyhedron with its grid square, to get key
1078 * points for non-animated display.
1084 success = align_poly(ret->solid, &ret->squares[ret->current], pkey);
1087 ret->dpkey[0] = pkey[0];
1088 ret->dpkey[1] = pkey[1];
1094 ret->spkey[0] = pkey[0];
1095 ret->spkey[1] = pkey[1];
1096 ret->sgkey[0] = skey[0];
1097 ret->sgkey[1] = skey[1];
1098 ret->previous = from->current;
1105 /* ----------------------------------------------------------------------
1113 struct game_drawstate {
1114 int ox, oy; /* pixel position of float origin */
1117 static void find_bbox_callback(void *ctx, struct grid_square *sq)
1119 struct bbox *bb = (struct bbox *)ctx;
1122 for (i = 0; i < sq->npoints; i++) {
1123 if (bb->l > sq->points[i*2]) bb->l = sq->points[i*2];
1124 if (bb->r < sq->points[i*2]) bb->r = sq->points[i*2];
1125 if (bb->u > sq->points[i*2+1]) bb->u = sq->points[i*2+1];
1126 if (bb->d < sq->points[i*2+1]) bb->d = sq->points[i*2+1];
1130 static struct bbox find_bbox(game_params *params)
1135 * These should be hugely more than the real bounding box will
1138 bb.l = 2.0F * (params->d1 + params->d2);
1139 bb.r = -2.0F * (params->d1 + params->d2);
1140 bb.u = 2.0F * (params->d1 + params->d2);
1141 bb.d = -2.0F * (params->d1 + params->d2);
1142 enum_grid_squares(params, find_bbox_callback, &bb);
1147 void game_size(game_params *params, int *x, int *y)
1149 struct bbox bb = find_bbox(params);
1150 *x = (int)((bb.r - bb.l + 2*solids[params->solid]->border) * GRID_SCALE);
1151 *y = (int)((bb.d - bb.u + 2*solids[params->solid]->border) * GRID_SCALE);
1154 float *game_colours(frontend *fe, game_state *state, int *ncolours)
1156 float *ret = snewn(3 * NCOLOURS, float);
1158 frontend_default_colour(fe, &ret[COL_BACKGROUND * 3]);
1160 ret[COL_BORDER * 3 + 0] = 0.0;
1161 ret[COL_BORDER * 3 + 1] = 0.0;
1162 ret[COL_BORDER * 3 + 2] = 0.0;
1164 ret[COL_BLUE * 3 + 0] = 0.0;
1165 ret[COL_BLUE * 3 + 1] = 0.0;
1166 ret[COL_BLUE * 3 + 2] = 1.0;
1168 *ncolours = NCOLOURS;
1172 game_drawstate *game_new_drawstate(game_state *state)
1174 struct game_drawstate *ds = snew(struct game_drawstate);
1175 struct bbox bb = find_bbox(&state->params);
1177 ds->ox = (int)(-(bb.l - state->solid->border) * GRID_SCALE);
1178 ds->oy = (int)(-(bb.u - state->solid->border) * GRID_SCALE);
1183 void game_free_drawstate(game_drawstate *ds)
1188 void game_redraw(frontend *fe, game_drawstate *ds, game_state *oldstate,
1189 game_state *state, float animtime, float flashtime)
1192 struct bbox bb = find_bbox(&state->params);
1197 game_state *newstate;
1200 draw_rect(fe, 0, 0, (int)((bb.r-bb.l+2.0F) * GRID_SCALE),
1201 (int)((bb.d-bb.u+2.0F) * GRID_SCALE), COL_BACKGROUND);
1203 if (oldstate && oldstate->movecount > state->movecount) {
1207 * This is an Undo. So reverse the order of the states, and
1208 * run the roll timer backwards.
1214 animtime = ROLLTIME - animtime;
1220 square = state->current;
1221 pkey = state->dpkey;
1222 gkey = state->dgkey;
1224 angle = state->angle * animtime / ROLLTIME;
1225 square = state->previous;
1226 pkey = state->spkey;
1227 gkey = state->sgkey;
1232 for (i = 0; i < state->nsquares; i++) {
1235 for (j = 0; j < state->squares[i].npoints; j++) {
1236 coords[2*j] = ((int)(state->squares[i].points[2*j] * GRID_SCALE)
1238 coords[2*j+1] = ((int)(state->squares[i].points[2*j+1]*GRID_SCALE)
1242 draw_polygon(fe, coords, state->squares[i].npoints, TRUE,
1243 state->squares[i].blue ? COL_BLUE : COL_BACKGROUND);
1244 draw_polygon(fe, coords, state->squares[i].npoints, FALSE, COL_BORDER);
1248 * Now compute and draw the polyhedron.
1250 poly = transform_poly(state->solid, state->squares[square].flip,
1251 pkey[0], pkey[1], angle);
1254 * Compute the translation required to align the two key points
1255 * on the polyhedron with the same key points on the current
1258 for (i = 0; i < 3; i++) {
1261 for (j = 0; j < 2; j++) {
1266 state->squares[square].points[gkey[j]*2+i];
1271 tc += (grid_coord - poly->vertices[pkey[j]*3+i]);
1276 for (i = 0; i < poly->nvertices; i++)
1277 for (j = 0; j < 3; j++)
1278 poly->vertices[i*3+j] += t[j];
1281 * Now actually draw each face.
1283 for (i = 0; i < poly->nfaces; i++) {
1287 for (j = 0; j < poly->order; j++) {
1288 int f = poly->faces[i*poly->order + j];
1289 points[j*2] = (poly->vertices[f*3+0] -
1290 poly->vertices[f*3+2] * poly->shear);
1291 points[j*2+1] = (poly->vertices[f*3+1] -
1292 poly->vertices[f*3+2] * poly->shear);
1295 for (j = 0; j < poly->order; j++) {
1296 coords[j*2] = (int)(points[j*2] * GRID_SCALE) + ds->ox;
1297 coords[j*2+1] = (int)(points[j*2+1] * GRID_SCALE) + ds->oy;
1301 * Find out whether these points are in a clockwise or
1302 * anticlockwise arrangement. If the latter, discard the
1303 * face because it's facing away from the viewer.
1305 * This would involve fiddly winding-number stuff for a
1306 * general polygon, but for the simple parallelograms we'll
1307 * be seeing here, all we have to do is check whether the
1308 * corners turn right or left. So we'll take the vector
1309 * from point 0 to point 1, turn it right 90 degrees,
1310 * and check the sign of the dot product with that and the
1311 * next vector (point 1 to point 2).
1314 float v1x = points[2]-points[0];
1315 float v1y = points[3]-points[1];
1316 float v2x = points[4]-points[2];
1317 float v2y = points[5]-points[3];
1318 float dp = v1x * v2y - v1y * v2x;
1324 draw_polygon(fe, coords, poly->order, TRUE,
1325 state->facecolours[i] ? COL_BLUE : COL_BACKGROUND);
1326 draw_polygon(fe, coords, poly->order, FALSE, COL_BORDER);
1330 draw_update(fe, 0, 0, (int)((bb.r-bb.l+2.0F) * GRID_SCALE),
1331 (int)((bb.d-bb.u+2.0F) * GRID_SCALE));
1334 * Update the status bar.
1337 char statusbuf[256];
1339 sprintf(statusbuf, "%sMoves: %d",
1340 (state->completed ? "COMPLETED! " : ""),
1341 (state->completed ? state->completed : state->movecount));
1343 status_bar(fe, statusbuf);
1347 float game_anim_length(game_state *oldstate, game_state *newstate)
1352 float game_flash_length(game_state *oldstate, game_state *newstate)
1357 int game_wants_statusbar(void)