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 */
24 float border
; /* border required around arena */
27 static const struct solid tetrahedron
= {
30 0.0F
, -0.57735026919F
, -0.20412414523F
,
31 -0.5F
, 0.28867513459F
, -0.20412414523F
,
32 0.0F
, -0.0F
, 0.6123724357F
,
33 0.5F
, 0.28867513459F
, -0.20412414523F
,
37 0,2,1, 3,1,2, 2,0,3, 1,3,0
40 -0.816496580928F
, -0.471404520791F
, 0.333333333334F
,
41 0.0F
, 0.942809041583F
, 0.333333333333F
,
42 0.816496580928F
, -0.471404520791F
, 0.333333333334F
,
48 static const struct solid cube
= {
51 -0.5F
,-0.5F
,-0.5F
, -0.5F
,-0.5F
,+0.5F
,
52 -0.5F
,+0.5F
,-0.5F
, -0.5F
,+0.5F
,+0.5F
,
53 +0.5F
,-0.5F
,-0.5F
, +0.5F
,-0.5F
,+0.5F
,
54 +0.5F
,+0.5F
,-0.5F
, +0.5F
,+0.5F
,+0.5F
,
58 0,1,3,2, 1,5,7,3, 5,4,6,7, 4,0,2,6, 0,4,5,1, 3,7,6,2
61 -1.0F
,0.0F
,0.0F
, 0.0F
,0.0F
,+1.0F
,
62 +1.0F
,0.0F
,0.0F
, 0.0F
,0.0F
,-1.0F
,
63 0.0F
,-1.0F
,0.0F
, 0.0F
,+1.0F
,0.0F
68 static const struct solid octahedron
= {
71 -0.5F
, -0.28867513459472505F
, 0.4082482904638664F
,
72 0.5F
, 0.28867513459472505F
, -0.4082482904638664F
,
73 -0.5F
, 0.28867513459472505F
, -0.4082482904638664F
,
74 0.5F
, -0.28867513459472505F
, 0.4082482904638664F
,
75 0.0F
, -0.57735026918945009F
, -0.4082482904638664F
,
76 0.0F
, 0.57735026918945009F
, 0.4082482904638664F
,
80 4,0,2, 0,5,2, 0,4,3, 5,0,3, 1,4,2, 5,1,2, 4,1,3, 1,5,3
83 -0.816496580928F
, -0.471404520791F
, -0.333333333334F
,
84 -0.816496580928F
, 0.471404520791F
, 0.333333333334F
,
85 0.0F
, -0.942809041583F
, 0.333333333333F
,
88 0.0F
, 0.942809041583F
, -0.333333333333F
,
89 0.816496580928F
, -0.471404520791F
, -0.333333333334F
,
90 0.816496580928F
, 0.471404520791F
, 0.333333333334F
,
95 static const struct solid icosahedron
= {
98 0.0F
, 0.57735026919F
, 0.75576131408F
,
99 0.0F
, -0.93417235896F
, 0.17841104489F
,
100 0.0F
, 0.93417235896F
, -0.17841104489F
,
101 0.0F
, -0.57735026919F
, -0.75576131408F
,
102 -0.5F
, -0.28867513459F
, 0.75576131408F
,
103 -0.5F
, 0.28867513459F
, -0.75576131408F
,
104 0.5F
, -0.28867513459F
, 0.75576131408F
,
105 0.5F
, 0.28867513459F
, -0.75576131408F
,
106 -0.80901699437F
, 0.46708617948F
, 0.17841104489F
,
107 0.80901699437F
, 0.46708617948F
, 0.17841104489F
,
108 -0.80901699437F
, -0.46708617948F
, -0.17841104489F
,
109 0.80901699437F
, -0.46708617948F
, -0.17841104489F
,
113 8,0,2, 0,9,2, 1,10,3, 11,1,3, 0,4,6,
114 4,1,6, 5,2,7, 3,5,7, 4,8,10, 8,5,10,
115 9,6,11, 7,9,11, 0,8,4, 9,0,6, 10,1,4,
116 1,11,6, 8,2,5, 2,9,7, 3,10,5, 11,3,7,
119 -0.356822089773F
, 0.87267799625F
, 0.333333333333F
,
120 0.356822089773F
, 0.87267799625F
, 0.333333333333F
,
121 -0.356822089773F
, -0.87267799625F
, -0.333333333333F
,
122 0.356822089773F
, -0.87267799625F
, -0.333333333333F
,
124 0.0F
, -0.666666666667F
, 0.745355992501F
,
125 0.0F
, 0.666666666667F
, -0.745355992501F
,
127 -0.934172358963F
, -0.12732200375F
, 0.333333333333F
,
128 -0.934172358963F
, 0.12732200375F
, -0.333333333333F
,
129 0.934172358963F
, -0.12732200375F
, 0.333333333333F
,
130 0.934172358963F
, 0.12732200375F
, -0.333333333333F
,
131 -0.57735026919F
, 0.333333333334F
, 0.745355992501F
,
132 0.57735026919F
, 0.333333333334F
, 0.745355992501F
,
133 -0.57735026919F
, -0.745355992501F
, 0.333333333334F
,
134 0.57735026919F
, -0.745355992501F
, 0.333333333334F
,
135 -0.57735026919F
, 0.745355992501F
, -0.333333333334F
,
136 0.57735026919F
, 0.745355992501F
, -0.333333333334F
,
137 -0.57735026919F
, -0.333333333334F
, -0.745355992501F
,
138 0.57735026919F
, -0.333333333334F
, -0.745355992501F
,
144 TETRAHEDRON
, CUBE
, OCTAHEDRON
, ICOSAHEDRON
146 static const struct solid
*solids
[] = {
147 &tetrahedron
, &cube
, &octahedron
, &icosahedron
157 enum { LEFT
, RIGHT
, UP
, DOWN
, UP_LEFT
, UP_RIGHT
, DOWN_LEFT
, DOWN_RIGHT
};
159 #define GRID_SCALE 48.0F
160 #define ROLLTIME 0.1F
162 #define SQ(x) ( (x) * (x) )
164 #define MATMUL(ra,m,a) do { \
165 float rx, ry, rz, xx = (a)[0], yy = (a)[1], zz = (a)[2], *mat = (m); \
166 rx = mat[0] * xx + mat[3] * yy + mat[6] * zz; \
167 ry = mat[1] * xx + mat[4] * yy + mat[7] * zz; \
168 rz = mat[2] * xx + mat[5] * yy + mat[8] * zz; \
169 (ra)[0] = rx; (ra)[1] = ry; (ra)[2] = rz; \
172 #define APPROXEQ(x,y) ( SQ(x-y) < 0.1 )
177 float points
[8]; /* maximum */
178 int directions
[8]; /* bit masks showing point pairs */
187 * Grid dimensions. For a square grid these are width and
188 * height respectively; otherwise the grid is a hexagon, with
189 * the top side and the two lower diagonals having length d1
190 * and the remaining three sides having length d2 (so that
191 * d1==d2 gives a regular hexagon, and d2==0 gives a triangle).
197 struct game_params params
;
198 const struct solid
*solid
;
200 struct grid_square
*squares
;
202 int current
; /* index of current grid square */
203 int sgkey
[2]; /* key-point indices into grid sq */
204 int dgkey
[2]; /* key-point indices into grid sq */
205 int spkey
[2]; /* key-point indices into polyhedron */
206 int dpkey
[2]; /* key-point indices into polyhedron */
213 game_params
*default_params(void)
215 game_params
*ret
= snew(game_params
);
224 int game_fetch_preset(int i
, char **name
, game_params
**params
)
226 game_params
*ret
= snew(game_params
);
238 ret
->solid
= TETRAHEDRON
;
244 ret
->solid
= OCTAHEDRON
;
250 ret
->solid
= ICOSAHEDRON
;
264 void free_params(game_params
*params
)
269 game_params
*dup_params(game_params
*params
)
271 game_params
*ret
= snew(game_params
);
272 *ret
= *params
; /* structure copy */
276 static void enum_grid_squares(game_params
*params
,
277 void (*callback
)(void *, struct grid_square
*),
280 const struct solid
*solid
= solids
[params
->solid
];
282 if (solid
->order
== 4) {
285 for (x
= 0; x
< params
->d1
; x
++)
286 for (y
= 0; y
< params
->d2
; y
++) {
287 struct grid_square sq
;
291 sq
.points
[0] = x
- 0.5F
;
292 sq
.points
[1] = y
- 0.5F
;
293 sq
.points
[2] = x
- 0.5F
;
294 sq
.points
[3] = y
+ 0.5F
;
295 sq
.points
[4] = x
+ 0.5F
;
296 sq
.points
[5] = y
+ 0.5F
;
297 sq
.points
[6] = x
+ 0.5F
;
298 sq
.points
[7] = y
- 0.5F
;
301 sq
.directions
[LEFT
] = 0x03; /* 0,1 */
302 sq
.directions
[RIGHT
] = 0x0C; /* 2,3 */
303 sq
.directions
[UP
] = 0x09; /* 0,3 */
304 sq
.directions
[DOWN
] = 0x06; /* 1,2 */
305 sq
.directions
[UP_LEFT
] = 0; /* no diagonals in a square */
306 sq
.directions
[UP_RIGHT
] = 0; /* no diagonals in a square */
307 sq
.directions
[DOWN_LEFT
] = 0; /* no diagonals in a square */
308 sq
.directions
[DOWN_RIGHT
] = 0; /* no diagonals in a square */
313 * This is supremely irrelevant, but just to avoid
314 * having any uninitialised structure members...
321 int row
, rowlen
, other
, i
, firstix
= -1;
322 float theight
= (float)(sqrt(3) / 2.0);
324 for (row
= 0; row
< params
->d1
+ params
->d2
; row
++) {
325 if (row
< params
->d1
) {
327 rowlen
= row
+ params
->d2
;
330 rowlen
= 2*params
->d1
+ params
->d2
- row
;
334 * There are `rowlen' down-pointing triangles.
336 for (i
= 0; i
< rowlen
; i
++) {
337 struct grid_square sq
;
341 ix
= (2 * i
- (rowlen
-1));
345 sq
.y
= y
+ theight
/ 3;
346 sq
.points
[0] = x
- 0.5F
;
349 sq
.points
[3] = y
+ theight
;
350 sq
.points
[4] = x
+ 0.5F
;
354 sq
.directions
[LEFT
] = 0x03; /* 0,1 */
355 sq
.directions
[RIGHT
] = 0x06; /* 1,2 */
356 sq
.directions
[UP
] = 0x05; /* 0,2 */
357 sq
.directions
[DOWN
] = 0; /* invalid move */
360 * Down-pointing triangle: both the up diagonals go
361 * up, and the down ones go left and right.
363 sq
.directions
[UP_LEFT
] = sq
.directions
[UP_RIGHT
] =
365 sq
.directions
[DOWN_LEFT
] = sq
.directions
[LEFT
];
366 sq
.directions
[DOWN_RIGHT
] = sq
.directions
[RIGHT
];
373 sq
.tetra_class
= ((row
+(ix
&1)) & 2) ^ (ix
& 3);
379 * There are `rowlen+other' up-pointing triangles.
381 for (i
= 0; i
< rowlen
+other
; i
++) {
382 struct grid_square sq
;
386 ix
= (2 * i
- (rowlen
+other
-1));
390 sq
.y
= y
+ 2*theight
/ 3;
391 sq
.points
[0] = x
+ 0.5F
;
392 sq
.points
[1] = y
+ theight
;
395 sq
.points
[4] = x
- 0.5F
;
396 sq
.points
[5] = y
+ theight
;
399 sq
.directions
[LEFT
] = 0x06; /* 1,2 */
400 sq
.directions
[RIGHT
] = 0x03; /* 0,1 */
401 sq
.directions
[DOWN
] = 0x05; /* 0,2 */
402 sq
.directions
[UP
] = 0; /* invalid move */
405 * Up-pointing triangle: both the down diagonals go
406 * down, and the up ones go left and right.
408 sq
.directions
[DOWN_LEFT
] = sq
.directions
[DOWN_RIGHT
] =
410 sq
.directions
[UP_LEFT
] = sq
.directions
[LEFT
];
411 sq
.directions
[UP_RIGHT
] = sq
.directions
[RIGHT
];
418 sq
.tetra_class
= ((row
+(ix
&1)) & 2) ^ (ix
& 3);
426 static int grid_area(int d1
, int d2
, int order
)
429 * An NxM grid of squares has NM squares in it.
431 * A grid of triangles with dimensions A and B has a total of
432 * A^2 + B^2 + 4AB triangles in it. (You can divide it up into
433 * a side-A triangle containing A^2 subtriangles, a side-B
434 * triangle containing B^2, and two congruent parallelograms,
435 * each with side lengths A and B, each therefore containing AB
436 * two-triangle rhombuses.)
441 return d1
*d1
+ d2
*d2
+ 4*d1
*d2
;
451 static void classify_grid_square_callback(void *ctx
, struct grid_square
*sq
)
453 struct grid_data
*data
= (struct grid_data
*)ctx
;
456 if (data
->nclasses
== 4)
457 thisclass
= sq
->tetra_class
;
458 else if (data
->nclasses
== 2)
459 thisclass
= sq
->flip
;
463 data
->gridptrs
[thisclass
][data
->nsquares
[thisclass
]++] =
467 char *new_game_seed(game_params
*params
)
469 struct grid_data data
;
470 int i
, j
, k
, m
, area
, facesperclass
;
475 * Enumerate the grid squares, dividing them into equivalence
476 * classes as appropriate. (For the tetrahedron, there is one
477 * equivalence class for each face; for the octahedron there
478 * are two classes; for the other two solids there's only one.)
481 area
= grid_area(params
->d1
, params
->d2
, solids
[params
->solid
]->order
);
482 if (params
->solid
== TETRAHEDRON
)
484 else if (params
->solid
== OCTAHEDRON
)
488 data
.gridptrs
[0] = snewn(data
.nclasses
* area
, int);
489 for (i
= 0; i
< data
.nclasses
; i
++) {
490 data
.gridptrs
[i
] = data
.gridptrs
[0] + i
* area
;
491 data
.nsquares
[i
] = 0;
493 data
.squareindex
= 0;
494 enum_grid_squares(params
, classify_grid_square_callback
, &data
);
496 facesperclass
= solids
[params
->solid
]->nfaces
/ data
.nclasses
;
498 for (i
= 0; i
< data
.nclasses
; i
++)
499 assert(data
.nsquares
[i
] >= facesperclass
);
500 assert(data
.squareindex
== area
);
503 * So now we know how many faces to allocate in each class. Get
506 flags
= snewn(area
, int);
507 for (i
= 0; i
< area
; i
++)
510 for (i
= 0; i
< data
.nclasses
; i
++) {
511 for (j
= 0; j
< facesperclass
; j
++) {
512 unsigned long divisor
= RAND_MAX
/ data
.nsquares
[i
];
513 unsigned long max
= divisor
* data
.nsquares
[i
];
522 assert(!flags
[data
.gridptrs
[i
][n
]]);
523 flags
[data
.gridptrs
[i
][n
]] = TRUE
;
526 * Move everything else up the array. I ought to use a
527 * better data structure for this, but for such small
528 * numbers it hardly seems worth the effort.
530 while ((int)n
< data
.nsquares
[i
]-1) {
531 data
.gridptrs
[i
][n
] = data
.gridptrs
[i
][n
+1];
539 * Now we know precisely which squares are blue. Encode this
540 * information in hex. While we're looping over this, collect
541 * the non-blue squares into a list in the now-unused gridptrs
544 seed
= snewn(area
/ 4 + 40, char);
549 for (i
= 0; i
< area
; i
++) {
553 data
.gridptrs
[0][m
++] = i
;
557 *p
++ = "0123456789ABCDEF"[j
];
563 *p
++ = "0123456789ABCDEF"[j
];
566 * Choose a non-blue square for the polyhedron.
569 unsigned long divisor
= RAND_MAX
/ m
;
570 unsigned long max
= divisor
* m
;
579 sprintf(p
, ":%d", data
.gridptrs
[0][n
]);
582 sfree(data
.gridptrs
[0]);
588 static void add_grid_square_callback(void *ctx
, struct grid_square
*sq
)
590 game_state
*state
= (game_state
*)ctx
;
592 state
->squares
[state
->nsquares
] = *sq
; /* structure copy */
593 state
->squares
[state
->nsquares
].blue
= FALSE
;
597 static int lowest_face(const struct solid
*solid
)
604 for (i
= 0; i
< solid
->nfaces
; i
++) {
607 for (j
= 0; j
< solid
->order
; j
++) {
608 int f
= solid
->faces
[i
*solid
->order
+ j
];
609 z
+= solid
->vertices
[f
*3+2];
612 if (i
== 0 || zmin
> z
) {
621 static int align_poly(const struct solid
*solid
, struct grid_square
*sq
,
626 int flip
= (sq
->flip ?
-1 : +1);
629 * First, find the lowest z-coordinate present in the solid.
632 for (i
= 0; i
< solid
->nvertices
; i
++)
633 if (zmin
> solid
->vertices
[i
*3+2])
634 zmin
= solid
->vertices
[i
*3+2];
637 * Now go round the grid square. For each point in the grid
638 * square, we're looking for a point of the polyhedron with the
639 * same x- and y-coordinates (relative to the square's centre),
640 * and z-coordinate equal to zmin (near enough).
642 for (j
= 0; j
< sq
->npoints
; j
++) {
648 for (i
= 0; i
< solid
->nvertices
; i
++) {
651 dist
+= SQ(solid
->vertices
[i
*3+0] * flip
- sq
->points
[j
*2+0] + sq
->x
);
652 dist
+= SQ(solid
->vertices
[i
*3+1] * flip
- sq
->points
[j
*2+1] + sq
->y
);
653 dist
+= SQ(solid
->vertices
[i
*3+2] - zmin
);
661 if (matches
!= 1 || index
< 0)
669 static void flip_poly(struct solid
*solid
, int flip
)
674 for (i
= 0; i
< solid
->nvertices
; i
++) {
675 solid
->vertices
[i
*3+0] *= -1;
676 solid
->vertices
[i
*3+1] *= -1;
678 for (i
= 0; i
< solid
->nfaces
; i
++) {
679 solid
->normals
[i
*3+0] *= -1;
680 solid
->normals
[i
*3+1] *= -1;
685 static struct solid
*transform_poly(const struct solid
*solid
, int flip
,
686 int key0
, int key1
, float angle
)
688 struct solid
*ret
= snew(struct solid
);
689 float vx
, vy
, ax
, ay
;
690 float vmatrix
[9], amatrix
[9], vmatrix2
[9];
693 *ret
= *solid
; /* structure copy */
695 flip_poly(ret
, flip
);
698 * Now rotate the polyhedron through the given angle. We must
699 * rotate about the Z-axis to bring the two vertices key0 and
700 * key1 into horizontal alignment, then rotate about the
701 * X-axis, then rotate back again.
703 vx
= ret
->vertices
[key1
*3+0] - ret
->vertices
[key0
*3+0];
704 vy
= ret
->vertices
[key1
*3+1] - ret
->vertices
[key0
*3+1];
705 assert(APPROXEQ(vx
*vx
+ vy
*vy
, 1.0));
707 vmatrix
[0] = vx
; vmatrix
[3] = vy
; vmatrix
[6] = 0;
708 vmatrix
[1] = -vy
; vmatrix
[4] = vx
; vmatrix
[7] = 0;
709 vmatrix
[2] = 0; vmatrix
[5] = 0; vmatrix
[8] = 1;
711 ax
= (float)cos(angle
);
712 ay
= (float)sin(angle
);
714 amatrix
[0] = 1; amatrix
[3] = 0; amatrix
[6] = 0;
715 amatrix
[1] = 0; amatrix
[4] = ax
; amatrix
[7] = ay
;
716 amatrix
[2] = 0; amatrix
[5] = -ay
; amatrix
[8] = ax
;
718 memcpy(vmatrix2
, vmatrix
, sizeof(vmatrix
));
722 for (i
= 0; i
< ret
->nvertices
; i
++) {
723 MATMUL(ret
->vertices
+ 3*i
, vmatrix
, ret
->vertices
+ 3*i
);
724 MATMUL(ret
->vertices
+ 3*i
, amatrix
, ret
->vertices
+ 3*i
);
725 MATMUL(ret
->vertices
+ 3*i
, vmatrix2
, ret
->vertices
+ 3*i
);
727 for (i
= 0; i
< ret
->nfaces
; i
++) {
728 MATMUL(ret
->normals
+ 3*i
, vmatrix
, ret
->normals
+ 3*i
);
729 MATMUL(ret
->normals
+ 3*i
, amatrix
, ret
->normals
+ 3*i
);
730 MATMUL(ret
->normals
+ 3*i
, vmatrix2
, ret
->normals
+ 3*i
);
736 game_state
*new_game(game_params
*params
, char *seed
)
738 game_state
*state
= snew(game_state
);
741 state
->params
= *params
; /* structure copy */
742 state
->solid
= solids
[params
->solid
];
744 area
= grid_area(params
->d1
, params
->d2
, state
->solid
->order
);
745 state
->squares
= snewn(area
, struct grid_square
);
747 enum_grid_squares(params
, add_grid_square_callback
, state
);
748 assert(state
->nsquares
== area
);
750 state
->facecolours
= snewn(state
->solid
->nfaces
, int);
751 memset(state
->facecolours
, 0, state
->solid
->nfaces
* sizeof(int));
754 * Set up the blue squares and polyhedron position according to
763 for (i
= 0; i
< state
->nsquares
; i
++) {
766 if (v
>= '0' && v
<= '9')
768 else if (v
>= 'A' && v
<= 'F')
770 else if (v
>= 'a' && v
<= 'f')
776 state
->squares
[i
].blue
= TRUE
;
785 state
->current
= atoi(p
);
786 if (state
->current
< 0 || state
->current
>= state
->nsquares
)
787 state
->current
= 0; /* got to do _something_ */
791 * Align the polyhedron with its grid square and determine
792 * initial key points.
798 ret
= align_poly(state
->solid
, &state
->squares
[state
->current
], pkey
);
801 state
->dpkey
[0] = state
->spkey
[0] = pkey
[0];
802 state
->dpkey
[1] = state
->spkey
[0] = pkey
[1];
803 state
->dgkey
[0] = state
->sgkey
[0] = 0;
804 state
->dgkey
[1] = state
->sgkey
[0] = 1;
807 state
->previous
= state
->current
;
809 state
->completed
= FALSE
;
810 state
->movecount
= 0;
815 game_state
*dup_game(game_state
*state
)
817 game_state
*ret
= snew(game_state
);
819 ret
->params
= state
->params
; /* structure copy */
820 ret
->solid
= state
->solid
;
821 ret
->facecolours
= snewn(ret
->solid
->nfaces
, int);
822 memcpy(ret
->facecolours
, state
->facecolours
,
823 ret
->solid
->nfaces
* sizeof(int));
824 ret
->nsquares
= state
->nsquares
;
825 ret
->squares
= snewn(ret
->nsquares
, struct grid_square
);
826 memcpy(ret
->squares
, state
->squares
,
827 ret
->nsquares
* sizeof(struct grid_square
));
828 ret
->dpkey
[0] = state
->dpkey
[0];
829 ret
->dpkey
[1] = state
->dpkey
[1];
830 ret
->dgkey
[0] = state
->dgkey
[0];
831 ret
->dgkey
[1] = state
->dgkey
[1];
832 ret
->spkey
[0] = state
->spkey
[0];
833 ret
->spkey
[1] = state
->spkey
[1];
834 ret
->sgkey
[0] = state
->sgkey
[0];
835 ret
->sgkey
[1] = state
->sgkey
[1];
836 ret
->previous
= state
->previous
;
837 ret
->angle
= state
->angle
;
838 ret
->completed
= state
->completed
;
839 ret
->movecount
= state
->movecount
;
844 void free_game(game_state
*state
)
849 game_state
*make_move(game_state
*from
, int x
, int y
, int button
)
852 int pkey
[2], skey
[2], dkey
[2];
856 int i
, j
, dest
, mask
;
860 * All moves are made with the cursor keys.
862 if (button
== CURSOR_UP
)
864 else if (button
== CURSOR_DOWN
)
866 else if (button
== CURSOR_LEFT
)
868 else if (button
== CURSOR_RIGHT
)
870 else if (button
== CURSOR_UP_LEFT
)
872 else if (button
== CURSOR_DOWN_LEFT
)
873 direction
= DOWN_LEFT
;
874 else if (button
== CURSOR_UP_RIGHT
)
875 direction
= UP_RIGHT
;
876 else if (button
== CURSOR_DOWN_RIGHT
)
877 direction
= DOWN_RIGHT
;
882 * Find the two points in the current grid square which
883 * correspond to this move.
885 mask
= from
->squares
[from
->current
].directions
[direction
];
888 for (i
= j
= 0; i
< from
->squares
[from
->current
].npoints
; i
++)
889 if (mask
& (1 << i
)) {
890 points
[j
*2] = from
->squares
[from
->current
].points
[i
*2];
891 points
[j
*2+1] = from
->squares
[from
->current
].points
[i
*2+1];
898 * Now find the other grid square which shares those points.
899 * This is our move destination.
902 for (i
= 0; i
< from
->nsquares
; i
++)
903 if (i
!= from
->current
) {
907 for (j
= 0; j
< from
->squares
[i
].npoints
; j
++) {
908 dist
= (SQ(from
->squares
[i
].points
[j
*2] - points
[0]) +
909 SQ(from
->squares
[i
].points
[j
*2+1] - points
[1]));
912 dist
= (SQ(from
->squares
[i
].points
[j
*2] - points
[2]) +
913 SQ(from
->squares
[i
].points
[j
*2+1] - points
[3]));
927 ret
= dup_game(from
);
931 * So we know what grid square we're aiming for, and we also
932 * know the two key points (as indices in both the source and
933 * destination grid squares) which are invariant between source
936 * Next we must roll the polyhedron on to that square. So we
937 * find the indices of the key points within the polyhedron's
938 * vertex array, then use those in a call to transform_poly,
939 * and align the result on the new grid square.
943 align_poly(from
->solid
, &from
->squares
[from
->current
], all_pkey
);
944 pkey
[0] = all_pkey
[skey
[0]];
945 pkey
[1] = all_pkey
[skey
[1]];
947 * Now pkey[0] corresponds to skey[0] and dkey[0], and
953 * Now find the angle through which to rotate the polyhedron.
954 * Do this by finding the two faces that share the two vertices
955 * we've found, and taking the dot product of their normals.
961 for (i
= 0; i
< from
->solid
->nfaces
; i
++) {
963 for (j
= 0; j
< from
->solid
->order
; j
++)
964 if (from
->solid
->faces
[i
*from
->solid
->order
+ j
] == pkey
[0] ||
965 from
->solid
->faces
[i
*from
->solid
->order
+ j
] == pkey
[1])
976 for (i
= 0; i
< 3; i
++)
977 dp
+= (from
->solid
->normals
[f
[0]*3+i
] *
978 from
->solid
->normals
[f
[1]*3+i
]);
979 angle
= (float)acos(dp
);
983 * Now transform the polyhedron. We aren't entirely sure
984 * whether we need to rotate through angle or -angle, and the
985 * simplest way round this is to try both and see which one
986 * aligns successfully!
988 * Unfortunately, _both_ will align successfully if this is a
989 * cube, which won't tell us anything much. So for that
990 * particular case, I resort to gross hackery: I simply negate
991 * the angle before trying the alignment, depending on the
992 * direction. Which directions work which way is determined by
993 * pure trial and error. I said it was gross :-/
999 if (from
->solid
->order
== 4 && direction
== UP
)
1000 angle
= -angle
; /* HACK */
1002 poly
= transform_poly(from
->solid
,
1003 from
->squares
[from
->current
].flip
,
1004 pkey
[0], pkey
[1], angle
);
1005 flip_poly(poly
, from
->squares
[ret
->current
].flip
);
1006 success
= align_poly(poly
, &from
->squares
[ret
->current
], all_pkey
);
1010 poly
= transform_poly(from
->solid
,
1011 from
->squares
[from
->current
].flip
,
1012 pkey
[0], pkey
[1], angle
);
1013 flip_poly(poly
, from
->squares
[ret
->current
].flip
);
1014 success
= align_poly(poly
, &from
->squares
[ret
->current
], all_pkey
);
1021 * Now we have our rotated polyhedron, which we expect to be
1022 * exactly congruent to the one we started with - but with the
1023 * faces permuted. So we map that congruence and thereby figure
1024 * out how to permute the faces as a result of the polyhedron
1028 int *newcolours
= snewn(from
->solid
->nfaces
, int);
1030 for (i
= 0; i
< from
->solid
->nfaces
; i
++)
1033 for (i
= 0; i
< from
->solid
->nfaces
; i
++) {
1037 * Now go through the transformed polyhedron's faces
1038 * and figure out which one's normal is approximately
1039 * equal to this one.
1041 for (j
= 0; j
< poly
->nfaces
; j
++) {
1047 for (k
= 0; k
< 3; k
++)
1048 dist
+= SQ(poly
->normals
[j
*3+k
] -
1049 from
->solid
->normals
[i
*3+k
]);
1051 if (APPROXEQ(dist
, 0)) {
1053 newcolours
[i
] = ret
->facecolours
[j
];
1057 assert(nmatch
== 1);
1060 for (i
= 0; i
< from
->solid
->nfaces
; i
++)
1061 assert(newcolours
[i
] != -1);
1063 sfree(ret
->facecolours
);
1064 ret
->facecolours
= newcolours
;
1068 * And finally, swap the colour between the bottom face of the
1069 * polyhedron and the face we've just landed on.
1071 * We don't do this if the game is already complete, since we
1072 * allow the user to roll the fully blue polyhedron around the
1073 * grid as a feeble reward.
1075 if (!ret
->completed
) {
1076 i
= lowest_face(from
->solid
);
1077 j
= ret
->facecolours
[i
];
1078 ret
->facecolours
[i
] = ret
->squares
[ret
->current
].blue
;
1079 ret
->squares
[ret
->current
].blue
= j
;
1082 * Detect game completion.
1085 for (i
= 0; i
< ret
->solid
->nfaces
; i
++)
1086 if (ret
->facecolours
[i
])
1088 if (j
== ret
->solid
->nfaces
)
1089 ret
->completed
= TRUE
;
1095 * Align the normal polyhedron with its grid square, to get key
1096 * points for non-animated display.
1102 success
= align_poly(ret
->solid
, &ret
->squares
[ret
->current
], pkey
);
1105 ret
->dpkey
[0] = pkey
[0];
1106 ret
->dpkey
[1] = pkey
[1];
1112 ret
->spkey
[0] = pkey
[0];
1113 ret
->spkey
[1] = pkey
[1];
1114 ret
->sgkey
[0] = skey
[0];
1115 ret
->sgkey
[1] = skey
[1];
1116 ret
->previous
= from
->current
;
1123 /* ----------------------------------------------------------------------
1131 struct game_drawstate
{
1132 int ox
, oy
; /* pixel position of float origin */
1135 static void find_bbox_callback(void *ctx
, struct grid_square
*sq
)
1137 struct bbox
*bb
= (struct bbox
*)ctx
;
1140 for (i
= 0; i
< sq
->npoints
; i
++) {
1141 if (bb
->l
> sq
->points
[i
*2]) bb
->l
= sq
->points
[i
*2];
1142 if (bb
->r
< sq
->points
[i
*2]) bb
->r
= sq
->points
[i
*2];
1143 if (bb
->u
> sq
->points
[i
*2+1]) bb
->u
= sq
->points
[i
*2+1];
1144 if (bb
->d
< sq
->points
[i
*2+1]) bb
->d
= sq
->points
[i
*2+1];
1148 static struct bbox
find_bbox(game_params
*params
)
1153 * These should be hugely more than the real bounding box will
1156 bb
.l
= 2.0F
* (params
->d1
+ params
->d2
);
1157 bb
.r
= -2.0F
* (params
->d1
+ params
->d2
);
1158 bb
.u
= 2.0F
* (params
->d1
+ params
->d2
);
1159 bb
.d
= -2.0F
* (params
->d1
+ params
->d2
);
1160 enum_grid_squares(params
, find_bbox_callback
, &bb
);
1165 void game_size(game_params
*params
, int *x
, int *y
)
1167 struct bbox bb
= find_bbox(params
);
1168 *x
= (int)((bb
.r
- bb
.l
+ 2*solids
[params
->solid
]->border
) * GRID_SCALE
);
1169 *y
= (int)((bb
.d
- bb
.u
+ 2*solids
[params
->solid
]->border
) * GRID_SCALE
);
1172 float *game_colours(frontend
*fe
, game_state
*state
, int *ncolours
)
1174 float *ret
= snewn(3 * NCOLOURS
, float);
1176 frontend_default_colour(fe
, &ret
[COL_BACKGROUND
* 3]);
1178 ret
[COL_BORDER
* 3 + 0] = 0.0;
1179 ret
[COL_BORDER
* 3 + 1] = 0.0;
1180 ret
[COL_BORDER
* 3 + 2] = 0.0;
1182 ret
[COL_BLUE
* 3 + 0] = 0.0;
1183 ret
[COL_BLUE
* 3 + 1] = 0.0;
1184 ret
[COL_BLUE
* 3 + 2] = 1.0;
1186 *ncolours
= NCOLOURS
;
1190 game_drawstate
*game_new_drawstate(game_state
*state
)
1192 struct game_drawstate
*ds
= snew(struct game_drawstate
);
1193 struct bbox bb
= find_bbox(&state
->params
);
1195 ds
->ox
= (int)(-(bb
.l
- state
->solid
->border
) * GRID_SCALE
);
1196 ds
->oy
= (int)(-(bb
.u
- state
->solid
->border
) * GRID_SCALE
);
1201 void game_free_drawstate(game_drawstate
*ds
)
1206 void game_redraw(frontend
*fe
, game_drawstate
*ds
, game_state
*oldstate
,
1207 game_state
*state
, float animtime
)
1210 struct bbox bb
= find_bbox(&state
->params
);
1215 game_state
*newstate
;
1218 draw_rect(fe
, 0, 0, (int)((bb
.r
-bb
.l
+2.0F
) * GRID_SCALE
),
1219 (int)((bb
.d
-bb
.u
+2.0F
) * GRID_SCALE
), COL_BACKGROUND
);
1221 if (oldstate
&& oldstate
->movecount
> state
->movecount
) {
1225 * This is an Undo. So reverse the order of the states, and
1226 * run the roll timer backwards.
1232 animtime
= ROLLTIME
- animtime
;
1238 square
= state
->current
;
1239 pkey
= state
->dpkey
;
1240 gkey
= state
->dgkey
;
1242 angle
= state
->angle
* animtime
/ ROLLTIME
;
1243 square
= state
->previous
;
1244 pkey
= state
->spkey
;
1245 gkey
= state
->sgkey
;
1250 for (i
= 0; i
< state
->nsquares
; i
++) {
1253 for (j
= 0; j
< state
->squares
[i
].npoints
; j
++) {
1254 coords
[2*j
] = ((int)(state
->squares
[i
].points
[2*j
] * GRID_SCALE
)
1256 coords
[2*j
+1] = ((int)(state
->squares
[i
].points
[2*j
+1]*GRID_SCALE
)
1260 draw_polygon(fe
, coords
, state
->squares
[i
].npoints
, TRUE
,
1261 state
->squares
[i
].blue ? COL_BLUE
: COL_BACKGROUND
);
1262 draw_polygon(fe
, coords
, state
->squares
[i
].npoints
, FALSE
, COL_BORDER
);
1266 * Now compute and draw the polyhedron.
1268 poly
= transform_poly(state
->solid
, state
->squares
[square
].flip
,
1269 pkey
[0], pkey
[1], angle
);
1272 * Compute the translation required to align the two key points
1273 * on the polyhedron with the same key points on the current
1276 for (i
= 0; i
< 3; i
++) {
1279 for (j
= 0; j
< 2; j
++) {
1284 state
->squares
[square
].points
[gkey
[j
]*2+i
];
1289 tc
+= (grid_coord
- poly
->vertices
[pkey
[j
]*3+i
]);
1294 for (i
= 0; i
< poly
->nvertices
; i
++)
1295 for (j
= 0; j
< 3; j
++)
1296 poly
->vertices
[i
*3+j
] += t
[j
];
1299 * Now actually draw each face.
1301 for (i
= 0; i
< poly
->nfaces
; i
++) {
1305 for (j
= 0; j
< poly
->order
; j
++) {
1306 int f
= poly
->faces
[i
*poly
->order
+ j
];
1307 points
[j
*2] = (poly
->vertices
[f
*3+0] -
1308 poly
->vertices
[f
*3+2] * poly
->shear
);
1309 points
[j
*2+1] = (poly
->vertices
[f
*3+1] -
1310 poly
->vertices
[f
*3+2] * poly
->shear
);
1313 for (j
= 0; j
< poly
->order
; j
++) {
1314 coords
[j
*2] = (int)(points
[j
*2] * GRID_SCALE
) + ds
->ox
;
1315 coords
[j
*2+1] = (int)(points
[j
*2+1] * GRID_SCALE
) + ds
->oy
;
1319 * Find out whether these points are in a clockwise or
1320 * anticlockwise arrangement. If the latter, discard the
1321 * face because it's facing away from the viewer.
1323 * This would involve fiddly winding-number stuff for a
1324 * general polygon, but for the simple parallelograms we'll
1325 * be seeing here, all we have to do is check whether the
1326 * corners turn right or left. So we'll take the vector
1327 * from point 0 to point 1, turn it right 90 degrees,
1328 * and check the sign of the dot product with that and the
1329 * next vector (point 1 to point 2).
1332 float v1x
= points
[2]-points
[0];
1333 float v1y
= points
[3]-points
[1];
1334 float v2x
= points
[4]-points
[2];
1335 float v2y
= points
[5]-points
[3];
1336 float dp
= v1x
* v2y
- v1y
* v2x
;
1342 draw_polygon(fe
, coords
, poly
->order
, TRUE
,
1343 state
->facecolours
[i
] ? COL_BLUE
: COL_BACKGROUND
);
1344 draw_polygon(fe
, coords
, poly
->order
, FALSE
, COL_BORDER
);
1348 draw_update(fe
, 0, 0, (int)((bb
.r
-bb
.l
+2.0F
) * GRID_SCALE
),
1349 (int)((bb
.d
-bb
.u
+2.0F
) * GRID_SCALE
));
1352 float game_anim_length(game_state
*oldstate
, game_state
*newstate
)