From 1482ee76f9b9a587ea12a09d4c971f5ed92cb6fe Mon Sep 17 00:00:00 2001 From: simon Date: Tue, 27 Apr 2004 17:44:30 +0000 Subject: [PATCH] Implemented Cube, in a sufficiently general way that it also handles the tetrahedron, octahedron and icosahedron. git-svn-id: svn://svn.tartarus.org/sgt/puzzles@4151 cda61777-01e9-0310-a592-d414129be87e --- Recipe | 4 +- cube.c | 1270 +++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ gtk.c | 19 +- midend.c | 22 +- puzzles.h | 8 +- 5 files changed, 1309 insertions(+), 14 deletions(-) diff --git a/Recipe b/Recipe index 5a3567e..48f3e68 100644 --- a/Recipe +++ b/Recipe @@ -15,7 +15,7 @@ COMMON = midend malloc NET = net random tree234 net : [X] gtk COMMON NET -#cube : [X] gtk COMMON CUBE +cube : [X] gtk COMMON cube #net : [G] windows COMMON NET -#cube : [G] windows COMMON CUBE +#cube : [G] windows COMMON cube diff --git a/cube.c b/cube.c index 0ceef12..8a19ae7 100644 --- a/cube.c +++ b/cube.c @@ -1,3 +1,1273 @@ /* * cube.c: Cube game. */ + +#include +#include +#include +#include +#include + +#include "puzzles.h" + +#define MAXVERTICES 20 +#define MAXFACES 20 +#define MAXORDER 4 +struct solid { + int nvertices; + float vertices[MAXVERTICES * 3]; /* 3*npoints coordinates */ + int order; + int nfaces; + int faces[MAXFACES * MAXORDER]; /* order*nfaces point indices */ + float normals[MAXFACES * 3]; /* 3*npoints vector components */ + float shear; /* isometric shear for nice drawing */ +}; + +static const struct solid tetrahedron = { + 4, + { + 0.0, -0.57735026919, -0.20412414523, + -0.5, 0.28867513459, -0.20412414523, + 0.0, -0.0, 0.6123724357, + 0.5, 0.28867513459, -0.20412414523, + }, + 3, 4, + { + 0,2,1, 3,1,2, 2,0,3, 1,3,0 + }, + { + -0.816496580928, -0.471404520791, 0.333333333334, + 0.0, 0.942809041583, 0.333333333333, + 0.816496580928, -0.471404520791, 0.333333333334, + 0.0, 0.0, -1.0, + }, + 0.0 +}; + +static const struct solid cube = { + 8, + { + -0.5,-0.5,-0.5, -0.5,-0.5,+0.5, -0.5,+0.5,-0.5, -0.5,+0.5,+0.5, + +0.5,-0.5,-0.5, +0.5,-0.5,+0.5, +0.5,+0.5,-0.5, +0.5,+0.5,+0.5, + }, + 4, 6, + { + 0,1,3,2, 1,5,7,3, 5,4,6,7, 4,0,2,6, 0,4,5,1, 3,7,6,2 + }, + { + -1,0,0, 0,0,+1, +1,0,0, 0,0,-1, 0,-1,0, 0,+1,0 + }, + 0.3 +}; + +static const struct solid octahedron = { + 6, + { + -0.5, -0.28867513459472505, 0.4082482904638664, + 0.5, 0.28867513459472505, -0.4082482904638664, + -0.5, 0.28867513459472505, -0.4082482904638664, + 0.5, -0.28867513459472505, 0.4082482904638664, + 0.0, -0.57735026918945009, -0.4082482904638664, + 0.0, 0.57735026918945009, 0.4082482904638664, + }, + 3, 8, + { + 4,0,2, 0,5,2, 0,4,3, 5,0,3, 1,4,2, 5,1,2, 4,1,3, 1,5,3 + }, + { + -0.816496580928, -0.471404520791, -0.333333333334, + -0.816496580928, 0.471404520791, 0.333333333334, + 0.0, -0.942809041583, 0.333333333333, + 0.0, 0.0, 1.0, + 0.0, 0.0, -1.0, + 0.0, 0.942809041583, -0.333333333333, + 0.816496580928, -0.471404520791, -0.333333333334, + 0.816496580928, 0.471404520791, 0.333333333334, + }, + 0.0 +}; + +static const struct solid icosahedron = { + 12, + { + 0.0, 0.57735026919, 0.75576131408, + 0.0, -0.93417235896, 0.17841104489, + 0.0, 0.93417235896, -0.17841104489, + 0.0, -0.57735026919, -0.75576131408, + -0.5, -0.28867513459, 0.75576131408, + -0.5, 0.28867513459, -0.75576131408, + 0.5, -0.28867513459, 0.75576131408, + 0.5, 0.28867513459, -0.75576131408, + -0.80901699437, 0.46708617948, 0.17841104489, + 0.80901699437, 0.46708617948, 0.17841104489, + -0.80901699437, -0.46708617948, -0.17841104489, + 0.80901699437, -0.46708617948, -0.17841104489, + }, + 3, 20, + { + 8,0,2, 0,9,2, 1,10,3, 11,1,3, 0,4,6, + 4,1,6, 5,2,7, 3,5,7, 4,8,10, 8,5,10, + 9,6,11, 7,9,11, 0,8,4, 9,0,6, 10,1,4, + 1,11,6, 8,2,5, 2,9,7, 3,10,5, 11,3,7, + }, + { + -0.356822089773, 0.87267799625, 0.333333333333, + 0.356822089773, 0.87267799625, 0.333333333333, + -0.356822089773, -0.87267799625, -0.333333333333, + 0.356822089773, -0.87267799625, -0.333333333333, + -0.0, 0.0, 1.0, + 0.0, -0.666666666667, 0.745355992501, + 0.0, 0.666666666667, -0.745355992501, + 0.0, 0.0, -1.0, + -0.934172358963, -0.12732200375, 0.333333333333, + -0.934172358963, 0.12732200375, -0.333333333333, + 0.934172358963, -0.12732200375, 0.333333333333, + 0.934172358963, 0.12732200375, -0.333333333333, + -0.57735026919, 0.333333333334, 0.745355992501, + 0.57735026919, 0.333333333334, 0.745355992501, + -0.57735026919, -0.745355992501, 0.333333333334, + 0.57735026919, -0.745355992501, 0.333333333334, + -0.57735026919, 0.745355992501, -0.333333333334, + 0.57735026919, 0.745355992501, -0.333333333334, + -0.57735026919, -0.333333333334, -0.745355992501, + 0.57735026919, -0.333333333334, -0.745355992501, + }, + 0.0 +}; + +enum { + TETRAHEDRON, CUBE, OCTAHEDRON, ICOSAHEDRON +}; +static const struct solid *solids[] = { + &tetrahedron, &cube, &octahedron, &icosahedron +}; + +enum { + COL_BACKGROUND, + COL_BORDER, + COL_BLUE, + NCOLOURS +}; + +enum { LEFT, RIGHT, UP, DOWN }; + +#define GRID_SCALE 48 +#define ROLLTIME 0.1 + +#define SQ(x) ( (x) * (x) ) + +#define MATMUL(ra,m,a) do { \ + float rx, ry, rz, xx = (a)[0], yy = (a)[1], zz = (a)[2], *mat = (m); \ + rx = mat[0] * xx + mat[3] * yy + mat[6] * zz; \ + ry = mat[1] * xx + mat[4] * yy + mat[7] * zz; \ + rz = mat[2] * xx + mat[5] * yy + mat[8] * zz; \ + (ra)[0] = rx; (ra)[1] = ry; (ra)[2] = rz; \ +} while (0) + +#define APPROXEQ(x,y) ( SQ(x-y) < 0.1 ) + +struct grid_square { + float x, y; + int npoints; + float points[8]; /* maximum */ + int directions[4]; /* bit masks showing point pairs */ + int flip; + int blue; + int tetra_class; +}; + +struct game_params { + int solid; + /* + * Grid dimensions. For a square grid these are width and + * height respectively; otherwise the grid is a hexagon, with + * the top side and the two lower diagonals having length d1 + * and the remaining three sides having length d2 (so that + * d1==d2 gives a regular hexagon, and d2==0 gives a triangle). + */ + int d1, d2; +}; + +struct game_state { + struct game_params params; + const struct solid *solid; + int *facecolours; + struct grid_square *squares; + int nsquares; + int current; /* index of current grid square */ + int sgkey[2]; /* key-point indices into grid sq */ + int dgkey[2]; /* key-point indices into grid sq */ + int spkey[2]; /* key-point indices into polyhedron */ + int dpkey[2]; /* key-point indices into polyhedron */ + int previous; + float angle; + int completed; + int movecount; +}; + +game_params *default_params(void) +{ + game_params *ret = snew(game_params); + + ret->solid = CUBE; + ret->d1 = 4; + ret->d2 = 4; + + return ret; +} + +void free_params(game_params *params) +{ + sfree(params); +} + +static void enum_grid_squares(game_params *params, + void (*callback)(void *, struct grid_square *), + void *ctx) +{ + const struct solid *solid = solids[params->solid]; + + if (solid->order == 4) { + int x, y; + + for (x = 0; x < params->d1; x++) + for (y = 0; y < params->d2; y++) { + struct grid_square sq; + + sq.x = x; + sq.y = y; + sq.points[0] = x - 0.5; + sq.points[1] = y - 0.5; + sq.points[2] = x - 0.5; + sq.points[3] = y + 0.5; + sq.points[4] = x + 0.5; + sq.points[5] = y + 0.5; + sq.points[6] = x + 0.5; + sq.points[7] = y - 0.5; + sq.npoints = 4; + + sq.directions[LEFT] = 0x03; /* 0,1 */ + sq.directions[RIGHT] = 0x0C; /* 2,3 */ + sq.directions[UP] = 0x09; /* 0,3 */ + sq.directions[DOWN] = 0x06; /* 1,2 */ + + sq.flip = FALSE; + + /* + * This is supremely irrelevant, but just to avoid + * having any uninitialised structure members... + */ + sq.tetra_class = 0; + + callback(ctx, &sq); + } + } else { + int row, rowlen, other, i, firstix = -1; + float theight = sqrt(3) / 2.0; + + for (row = 0; row < params->d1 + params->d2; row++) { + if (row < params->d1) { + other = +1; + rowlen = row + params->d2; + } else { + other = -1; + rowlen = 2*params->d1 + params->d2 - row; + } + + /* + * There are `rowlen' down-pointing triangles. + */ + for (i = 0; i < rowlen; i++) { + struct grid_square sq; + int ix; + float x, y; + + ix = (2 * i - (rowlen-1)); + x = ix * 0.5; + y = theight * row; + sq.x = x; + sq.y = y + theight / 3; + sq.points[0] = x - 0.5; + sq.points[1] = y; + sq.points[2] = x; + sq.points[3] = y + theight; + sq.points[4] = x + 0.5; + sq.points[5] = y; + sq.npoints = 3; + + sq.directions[LEFT] = 0x03; /* 0,1 */ + sq.directions[RIGHT] = 0x06; /* 1,2 */ + sq.directions[UP] = 0x05; /* 0,2 */ + sq.directions[DOWN] = 0; /* invalid move */ + + sq.flip = TRUE; + + if (firstix < 0) + firstix = ix & 3; + ix -= firstix; + sq.tetra_class = ((row+(ix&1)) & 2) ^ (ix & 3); + + callback(ctx, &sq); + } + + /* + * There are `rowlen+other' up-pointing triangles. + */ + for (i = 0; i < rowlen+other; i++) { + struct grid_square sq; + int ix; + float x, y; + + ix = (2 * i - (rowlen+other-1)); + x = ix * 0.5; + y = theight * row; + sq.x = x; + sq.y = y + 2*theight / 3; + sq.points[0] = x + 0.5; + sq.points[1] = y + theight; + sq.points[2] = x; + sq.points[3] = y; + sq.points[4] = x - 0.5; + sq.points[5] = y + theight; + sq.npoints = 3; + + sq.directions[LEFT] = 0x06; /* 1,2 */ + sq.directions[RIGHT] = 0x03; /* 0,1 */ + sq.directions[DOWN] = 0x05; /* 0,2 */ + sq.directions[UP] = 0; /* invalid move */ + + sq.flip = FALSE; + + if (firstix < 0) + firstix = ix; + ix -= firstix; + sq.tetra_class = ((row+(ix&1)) & 2) ^ (ix & 3); + + callback(ctx, &sq); + } + } + } +} + +static int grid_area(int d1, int d2, int order) +{ + /* + * An NxM grid of squares has NM squares in it. + * + * A grid of triangles with dimensions A and B has a total of + * A^2 + B^2 + 4AB triangles in it. (You can divide it up into + * a side-A triangle containing A^2 subtriangles, a side-B + * triangle containing B^2, and two congruent parallelograms, + * each with side lengths A and B, each therefore containing AB + * two-triangle rhombuses.) + */ + if (order == 4) + return d1 * d2; + else + return d1*d1 + d2*d2 + 4*d1*d2; +} + +struct grid_data { + int *gridptrs[4]; + int nsquares[4]; + int nclasses; + int squareindex; +}; + +static void classify_grid_square_callback(void *ctx, struct grid_square *sq) +{ + struct grid_data *data = (struct grid_data *)ctx; + int thisclass; + + if (data->nclasses == 4) + thisclass = sq->tetra_class; + else if (data->nclasses == 2) + thisclass = sq->flip; + else + thisclass = 0; + + data->gridptrs[thisclass][data->nsquares[thisclass]++] = + data->squareindex++; +} + +char *new_game_seed(game_params *params) +{ + struct grid_data data; + int i, j, k, m, area, facesperclass; + int *flags; + char *seed, *p; + + /* + * Enumerate the grid squares, dividing them into equivalence + * classes as appropriate. (For the tetrahedron, there is one + * equivalence class for each face; for the octahedron there + * are two classes; for the other two solids there's only one.) + */ + + area = grid_area(params->d1, params->d2, solids[params->solid]->order); + if (params->solid == TETRAHEDRON) + data.nclasses = 4; + else if (params->solid == OCTAHEDRON) + data.nclasses = 2; + else + data.nclasses = 1; + data.gridptrs[0] = snewn(data.nclasses * area, int); + for (i = 0; i < data.nclasses; i++) { + data.gridptrs[i] = data.gridptrs[0] + i * area; + data.nsquares[i] = 0; + } + data.squareindex = 0; + enum_grid_squares(params, classify_grid_square_callback, &data); + + facesperclass = solids[params->solid]->nfaces / data.nclasses; + + for (i = 0; i < data.nclasses; i++) + assert(data.nsquares[i] >= facesperclass); + assert(data.squareindex == area); + + /* + * So now we know how many faces to allocate in each class. Get + * on with it. + */ + flags = snewn(area, int); + for (i = 0; i < area; i++) + flags[i] = FALSE; + + for (i = 0; i < data.nclasses; i++) { + for (j = 0; j < facesperclass; j++) { + unsigned long divisor = RAND_MAX / data.nsquares[i]; + unsigned long max = divisor * data.nsquares[i]; + int n; + + do { + n = rand(); + } while (n >= max); + + n /= divisor; + + assert(!flags[data.gridptrs[i][n]]); + flags[data.gridptrs[i][n]] = TRUE; + + /* + * Move everything else up the array. I ought to use a + * better data structure for this, but for such small + * numbers it hardly seems worth the effort. + */ + while (n < data.nsquares[i]-1) { + data.gridptrs[i][n] = data.gridptrs[i][n+1]; + n++; + } + data.nsquares[i]--; + } + } + + /* + * Now we know precisely which squares are blue. Encode this + * information in hex. While we're looping over this, collect + * the non-blue squares into a list in the now-unused gridptrs + * array. + */ + seed = snewn(area / 4 + 40, char); + p = seed; + j = 0; + k = 8; + m = 0; + for (i = 0; i < area; i++) { + if (flags[i]) { + j |= k; + } else { + data.gridptrs[0][m++] = i; + } + k >>= 1; + if (!k) { + *p++ = "0123456789ABCDEF"[j]; + k = 8; + j = 0; + } + } + if (k != 8) + *p++ = "0123456789ABCDEF"[j]; + + /* + * Choose a non-blue square for the polyhedron. + */ + { + unsigned long divisor = RAND_MAX / m; + unsigned long max = divisor * m; + int n; + + do { + n = rand(); + } while (n >= max); + + n /= divisor; + + sprintf(p, ":%d", data.gridptrs[0][n]); + } + + sfree(data.gridptrs[0]); + sfree(flags); + + return seed; +} + +static void add_grid_square_callback(void *ctx, struct grid_square *sq) +{ + game_state *state = (game_state *)ctx; + + state->squares[state->nsquares] = *sq; /* structure copy */ + state->squares[state->nsquares].blue = FALSE; + state->nsquares++; +} + +static int lowest_face(const struct solid *solid) +{ + int i, j, best; + float zmin; + + best = 0; + zmin = 0.0; + for (i = 0; i < solid->nfaces; i++) { + float z = 0; + + for (j = 0; j < solid->order; j++) { + int f = solid->faces[i*solid->order + j]; + z += solid->vertices[f*3+2]; + } + + if (i == 0 || zmin > z) { + zmin = z; + best = i; + } + } + + return best; +} + +static int align_poly(const struct solid *solid, struct grid_square *sq, + int *pkey) +{ + float zmin; + int i, j; + int flip = (sq->flip ? -1 : +1); + + /* + * First, find the lowest z-coordinate present in the solid. + */ + zmin = 0.0; + for (i = 0; i < solid->nvertices; i++) + if (zmin > solid->vertices[i*3+2]) + zmin = solid->vertices[i*3+2]; + + /* + * Now go round the grid square. For each point in the grid + * square, we're looking for a point of the polyhedron with the + * same x- and y-coordinates (relative to the square's centre), + * and z-coordinate equal to zmin (near enough). + */ + for (j = 0; j < sq->npoints; j++) { + int matches, index; + + matches = 0; + index = -1; + + for (i = 0; i < solid->nvertices; i++) { + float dist = 0; + + dist += SQ(solid->vertices[i*3+0] * flip - sq->points[j*2+0] + sq->x); + dist += SQ(solid->vertices[i*3+1] * flip - sq->points[j*2+1] + sq->y); + dist += SQ(solid->vertices[i*3+2] - zmin); + + if (dist < 0.1) { + matches++; + index = i; + } + } + + if (matches != 1 || index < 0) + return FALSE; + pkey[j] = index; + } + + return TRUE; +} + +static void flip_poly(struct solid *solid, int flip) +{ + int i; + + if (flip) { + for (i = 0; i < solid->nvertices; i++) { + solid->vertices[i*3+0] *= -1; + solid->vertices[i*3+1] *= -1; + } + for (i = 0; i < solid->nfaces; i++) { + solid->normals[i*3+0] *= -1; + solid->normals[i*3+1] *= -1; + } + } +} + +static struct solid *transform_poly(const struct solid *solid, int flip, + int key0, int key1, float angle) +{ + struct solid *ret = snew(struct solid); + float vx, vy, ax, ay; + float vmatrix[9], amatrix[9], vmatrix2[9]; + int i; + + *ret = *solid; /* structure copy */ + + flip_poly(ret, flip); + + /* + * Now rotate the polyhedron through the given angle. We must + * rotate about the Z-axis to bring the two vertices key0 and + * key1 into horizontal alignment, then rotate about the + * X-axis, then rotate back again. + */ + vx = ret->vertices[key1*3+0] - ret->vertices[key0*3+0]; + vy = ret->vertices[key1*3+1] - ret->vertices[key0*3+1]; + assert(APPROXEQ(vx*vx + vy*vy, 1.0)); + + vmatrix[0] = vx; vmatrix[3] = vy; vmatrix[6] = 0; + vmatrix[1] = -vy; vmatrix[4] = vx; vmatrix[7] = 0; + vmatrix[2] = 0; vmatrix[5] = 0; vmatrix[8] = 1; + + ax = cos(angle); + ay = sin(angle); + + amatrix[0] = 1; amatrix[3] = 0; amatrix[6] = 0; + amatrix[1] = 0; amatrix[4] = ax; amatrix[7] = ay; + amatrix[2] = 0; amatrix[5] = -ay; amatrix[8] = ax; + + memcpy(vmatrix2, vmatrix, sizeof(vmatrix)); + vmatrix2[1] = vy; + vmatrix2[3] = -vy; + + for (i = 0; i < ret->nvertices; i++) { + MATMUL(ret->vertices + 3*i, vmatrix, ret->vertices + 3*i); + MATMUL(ret->vertices + 3*i, amatrix, ret->vertices + 3*i); + MATMUL(ret->vertices + 3*i, vmatrix2, ret->vertices + 3*i); + } + for (i = 0; i < ret->nfaces; i++) { + MATMUL(ret->normals + 3*i, vmatrix, ret->normals + 3*i); + MATMUL(ret->normals + 3*i, amatrix, ret->normals + 3*i); + MATMUL(ret->normals + 3*i, vmatrix2, ret->normals + 3*i); + } + + return ret; +} + +game_state *new_game(game_params *params, char *seed) +{ + game_state *state = snew(game_state); + int area; + + state->params = *params; /* structure copy */ + state->solid = solids[params->solid]; + + area = grid_area(params->d1, params->d2, state->solid->order); + state->squares = snewn(area, struct grid_square); + state->nsquares = 0; + enum_grid_squares(params, add_grid_square_callback, state); + assert(state->nsquares == area); + + state->facecolours = snewn(state->solid->nfaces, int); + memset(state->facecolours, 0, state->solid->nfaces * sizeof(int)); + + /* + * Set up the blue squares and polyhedron position according to + * the game seed. + */ + { + char *p = seed; + int i, j, v; + + j = 8; + v = 0; + for (i = 0; i < state->nsquares; i++) { + if (j == 8) { + v = *p++; + if (v >= '0' && v <= '9') + v -= '0'; + else if (v >= 'A' && v <= 'F') + v -= 'A' - 10; + else if (v >= 'a' && v <= 'f') + v -= 'a' - 10; + else + break; + } + if (v & j) + state->squares[i].blue = TRUE; + j >>= 1; + if (j == 0) + j = 8; + } + + if (*p == ':') + p++; + + state->current = atoi(p); + if (state->current < 0 || state->current >= state->nsquares) + state->current = 0; /* got to do _something_ */ + } + + /* + * Align the polyhedron with its grid square and determine + * initial key points. + */ + { + int pkey[4]; + int ret; + + ret = align_poly(state->solid, &state->squares[state->current], pkey); + assert(ret); + + state->dpkey[0] = state->spkey[0] = pkey[0]; + state->dpkey[1] = state->spkey[0] = pkey[1]; + state->dgkey[0] = state->sgkey[0] = 0; + state->dgkey[1] = state->sgkey[0] = 1; + } + + state->previous = state->current; + state->angle = 0.0; + state->completed = FALSE; + state->movecount = 0; + + return state; +} + +game_state *dup_game(game_state *state) +{ + game_state *ret = snew(game_state); + + ret->params = state->params; /* structure copy */ + ret->solid = state->solid; + ret->facecolours = snewn(ret->solid->nfaces, int); + memcpy(ret->facecolours, state->facecolours, + ret->solid->nfaces * sizeof(int)); + ret->nsquares = state->nsquares; + ret->squares = snewn(ret->nsquares, struct grid_square); + memcpy(ret->squares, state->squares, + ret->nsquares * sizeof(struct grid_square)); + ret->dpkey[0] = state->dpkey[0]; + ret->dpkey[1] = state->dpkey[1]; + ret->dgkey[0] = state->dgkey[0]; + ret->dgkey[1] = state->dgkey[1]; + ret->spkey[0] = state->spkey[0]; + ret->spkey[1] = state->spkey[1]; + ret->sgkey[0] = state->sgkey[0]; + ret->sgkey[1] = state->sgkey[1]; + ret->previous = state->previous; + ret->angle = state->angle; + ret->completed = state->completed; + ret->movecount = state->movecount; + + return ret; +} + +void free_game(game_state *state) +{ + sfree(state); +} + +game_state *make_move(game_state *from, int x, int y, int button) +{ + int direction; + int pkey[2], skey[2], dkey[2]; + float points[4]; + game_state *ret; + float angle; + int i, j, dest, mask; + struct solid *poly; + + /* + * All moves are made with the cursor keys. + */ + if (button == CURSOR_UP) + direction = UP; + else if (button == CURSOR_DOWN) + direction = DOWN; + else if (button == CURSOR_LEFT) + direction = LEFT; + else if (button == CURSOR_RIGHT) + direction = RIGHT; + else + return NULL; + + /* + * Find the two points in the current grid square which + * correspond to this move. + */ + mask = from->squares[from->current].directions[direction]; + if (mask == 0) + return NULL; + for (i = j = 0; i < from->squares[from->current].npoints; i++) + if (mask & (1 << i)) { + points[j*2] = from->squares[from->current].points[i*2]; + points[j*2+1] = from->squares[from->current].points[i*2+1]; + skey[j] = i; + j++; + } + assert(j == 2); + + /* + * Now find the other grid square which shares those points. + * This is our move destination. + */ + dest = -1; + for (i = 0; i < from->nsquares; i++) + if (i != from->current) { + int match = 0; + float dist; + + for (j = 0; j < from->squares[i].npoints; j++) { + dist = (SQ(from->squares[i].points[j*2] - points[0]) + + SQ(from->squares[i].points[j*2+1] - points[1])); + if (dist < 0.1) + dkey[match++] = j; + dist = (SQ(from->squares[i].points[j*2] - points[2]) + + SQ(from->squares[i].points[j*2+1] - points[3])); + if (dist < 0.1) + dkey[match++] = j; + } + + if (match == 2) { + dest = i; + break; + } + } + + if (dest < 0) + return NULL; + + ret = dup_game(from); + ret->current = i; + + /* + * So we know what grid square we're aiming for, and we also + * know the two key points (as indices in both the source and + * destination grid squares) which are invariant between source + * and destination. + * + * Next we must roll the polyhedron on to that square. So we + * find the indices of the key points within the polyhedron's + * vertex array, then use those in a call to transform_poly, + * and align the result on the new grid square. + */ + { + int all_pkey[4]; + align_poly(from->solid, &from->squares[from->current], all_pkey); + pkey[0] = all_pkey[skey[0]]; + pkey[1] = all_pkey[skey[1]]; + /* + * Now pkey[0] corresponds to skey[0] and dkey[0], and + * likewise [1]. + */ + } + + /* + * Now find the angle through which to rotate the polyhedron. + * Do this by finding the two faces that share the two vertices + * we've found, and taking the dot product of their normals. + */ + { + int f[2], nf = 0; + float dp; + + for (i = 0; i < from->solid->nfaces; i++) { + int match = 0; + for (j = 0; j < from->solid->order; j++) + if (from->solid->faces[i*from->solid->order + j] == pkey[0] || + from->solid->faces[i*from->solid->order + j] == pkey[1]) + match++; + if (match == 2) { + assert(nf < 2); + f[nf++] = i; + } + } + + assert(nf == 2); + + dp = 0; + for (i = 0; i < 3; i++) + dp += (from->solid->normals[f[0]*3+i] * + from->solid->normals[f[1]*3+i]); + angle = acos(dp); + } + + /* + * Now transform the polyhedron. We aren't entirely sure + * whether we need to rotate through angle or -angle, and the + * simplest way round this is to try both and see which one + * aligns successfully! + * + * Unfortunately, _both_ will align successfully if this is a + * cube, which won't tell us anything much. So for that + * particular case, I resort to gross hackery: I simply negate + * the angle before trying the alignment, depending on the + * direction. Which directions work which way is determined by + * pure trial and error. I said it was gross :-/ + */ + { + int all_pkey[4]; + int success; + + if (from->solid->order == 4 && direction == UP) + angle = -angle; /* HACK */ + + poly = transform_poly(from->solid, + from->squares[from->current].flip, + pkey[0], pkey[1], angle); + flip_poly(poly, from->squares[ret->current].flip); + success = align_poly(poly, &from->squares[ret->current], all_pkey); + + if (!success) { + angle = -angle; + poly = transform_poly(from->solid, + from->squares[from->current].flip, + pkey[0], pkey[1], angle); + flip_poly(poly, from->squares[ret->current].flip); + success = align_poly(poly, &from->squares[ret->current], all_pkey); + } + + assert(success); + } + + /* + * Now we have our rotated polyhedron, which we expect to be + * exactly congruent to the one we started with - but with the + * faces permuted. So we map that congruence and thereby figure + * out how to permute the faces as a result of the polyhedron + * having rolled. + */ + { + int *newcolours = snewn(from->solid->nfaces, int); + + for (i = 0; i < from->solid->nfaces; i++) + newcolours[i] = -1; + + for (i = 0; i < from->solid->nfaces; i++) { + int nmatch = 0; + + /* + * Now go through the transformed polyhedron's faces + * and figure out which one's normal is approximately + * equal to this one. + */ + for (j = 0; j < poly->nfaces; j++) { + float dist; + int k; + + dist = 0; + + for (k = 0; k < 3; k++) + dist += SQ(poly->normals[j*3+k] - + from->solid->normals[i*3+k]); + + if (APPROXEQ(dist, 0)) { + nmatch++; + newcolours[i] = ret->facecolours[j]; + } + } + + assert(nmatch == 1); + } + + for (i = 0; i < from->solid->nfaces; i++) + assert(newcolours[i] != -1); + + sfree(ret->facecolours); + ret->facecolours = newcolours; + } + + /* + * And finally, swap the colour between the bottom face of the + * polyhedron and the face we've just landed on. + * + * We don't do this if the game is already complete, since we + * allow the user to roll the fully blue polyhedron around the + * grid as a feeble reward. + */ + if (!ret->completed) { + i = lowest_face(from->solid); + j = ret->facecolours[i]; + ret->facecolours[i] = ret->squares[ret->current].blue; + ret->squares[ret->current].blue = j; + + /* + * Detect game completion. + */ + j = 0; + for (i = 0; i < ret->solid->nfaces; i++) + if (ret->facecolours[i]) + j++; + if (j == ret->solid->nfaces) + ret->completed = TRUE; + } + + sfree(poly); + + /* + * Align the normal polyhedron with its grid square, to get key + * points for non-animated display. + */ + { + int pkey[4]; + int success; + + success = align_poly(ret->solid, &ret->squares[ret->current], pkey); + assert(success); + + ret->dpkey[0] = pkey[0]; + ret->dpkey[1] = pkey[1]; + ret->dgkey[0] = 0; + ret->dgkey[1] = 1; + } + + + ret->spkey[0] = pkey[0]; + ret->spkey[1] = pkey[1]; + ret->sgkey[0] = skey[0]; + ret->sgkey[1] = skey[1]; + ret->previous = from->current; + ret->angle = angle; + ret->movecount++; + + return ret; +} + +/* ---------------------------------------------------------------------- + * Drawing routines. + */ + +struct bbox { + float l, r, u, d; +}; + +struct game_drawstate { + int ox, oy; /* pixel position of float origin */ +}; + +static void find_bbox_callback(void *ctx, struct grid_square *sq) +{ + struct bbox *bb = (struct bbox *)ctx; + int i; + + for (i = 0; i < sq->npoints; i++) { + if (bb->l > sq->points[i*2]) bb->l = sq->points[i*2]; + if (bb->r < sq->points[i*2]) bb->r = sq->points[i*2]; + if (bb->u > sq->points[i*2+1]) bb->u = sq->points[i*2+1]; + if (bb->d < sq->points[i*2+1]) bb->d = sq->points[i*2+1]; + } +} + +static struct bbox find_bbox(game_params *params) +{ + struct bbox bb; + + /* + * These should be hugely more than the real bounding box will + * be. + */ + bb.l = 2 * (params->d1 + params->d2); + bb.r = -2 * (params->d1 + params->d2); + bb.u = 2 * (params->d1 + params->d2); + bb.d = -2 * (params->d1 + params->d2); + enum_grid_squares(params, find_bbox_callback, &bb); + + return bb; +} + +void game_size(game_params *params, int *x, int *y) +{ + struct bbox bb = find_bbox(params); + *x = (bb.r - bb.l + 2) * GRID_SCALE; + *y = (bb.d - bb.u + 2) * GRID_SCALE; +} + +float *game_colours(frontend *fe, game_state *state, int *ncolours) +{ + float *ret = snewn(3 * NCOLOURS, float); + + frontend_default_colour(fe, &ret[COL_BACKGROUND * 3]); + + ret[COL_BORDER * 3 + 0] = 0.0; + ret[COL_BORDER * 3 + 1] = 0.0; + ret[COL_BORDER * 3 + 2] = 0.0; + + ret[COL_BLUE * 3 + 0] = 0.0; + ret[COL_BLUE * 3 + 1] = 0.0; + ret[COL_BLUE * 3 + 2] = 1.0; + + *ncolours = NCOLOURS; + return ret; +} + +game_drawstate *game_new_drawstate(game_state *state) +{ + struct game_drawstate *ds = snew(struct game_drawstate); + struct bbox bb = find_bbox(&state->params); + + ds->ox = -(bb.l - 1) * GRID_SCALE; + ds->oy = -(bb.u - 1) * GRID_SCALE; + + return ds; +} + +void game_free_drawstate(game_drawstate *ds) +{ + sfree(ds); +} + +void game_redraw(frontend *fe, game_drawstate *ds, game_state *oldstate, + game_state *state, float animtime) +{ + int i, j; + struct bbox bb = find_bbox(&state->params); + struct solid *poly; + int *pkey, *gkey; + float t[3]; + float angle; + game_state *newstate; + int square; + + draw_rect(fe, 0, 0, (bb.r-bb.l+2) * GRID_SCALE, + (bb.d-bb.u+2) * GRID_SCALE, COL_BACKGROUND); + + if (oldstate && oldstate->movecount > state->movecount) { + game_state *t; + + /* + * This is an Undo. So reverse the order of the states, and + * run the roll timer backwards. + */ + t = oldstate; + oldstate = state; + state = t; + + animtime = ROLLTIME - animtime; + } + + if (!oldstate) { + oldstate = state; + angle = 0.0; + square = state->current; + pkey = state->dpkey; + gkey = state->dgkey; + } else { + angle = state->angle * animtime / ROLLTIME; + square = state->previous; + pkey = state->spkey; + gkey = state->sgkey; + } + newstate = state; + state = oldstate; + + for (i = 0; i < state->nsquares; i++) { + int coords[8]; + + for (j = 0; j < state->squares[i].npoints; j++) { + coords[2*j] = state->squares[i].points[2*j] + * GRID_SCALE + ds->ox; + coords[2*j+1] = state->squares[i].points[2*j+1] + * GRID_SCALE + ds->oy; + } + + draw_polygon(fe, coords, state->squares[i].npoints, TRUE, + state->squares[i].blue ? COL_BLUE : COL_BACKGROUND); + draw_polygon(fe, coords, state->squares[i].npoints, FALSE, COL_BORDER); + } + + /* + * Now compute and draw the polyhedron. + */ + poly = transform_poly(state->solid, state->squares[square].flip, + pkey[0], pkey[1], angle); + + /* + * Compute the translation required to align the two key points + * on the polyhedron with the same key points on the current + * face. + */ + for (i = 0; i < 3; i++) { + float tc = 0.0; + + for (j = 0; j < 2; j++) { + float grid_coord; + + if (i < 2) { + grid_coord = + state->squares[square].points[gkey[j]*2+i]; + } else { + grid_coord = 0.0; + } + + tc += (grid_coord - poly->vertices[pkey[j]*3+i]); + } + + t[i] = tc / 2; + } + for (i = 0; i < poly->nvertices; i++) + for (j = 0; j < 3; j++) + poly->vertices[i*3+j] += t[j]; + + /* + * Now actually draw each face. + */ + for (i = 0; i < poly->nfaces; i++) { + float points[8]; + int coords[8]; + + for (j = 0; j < poly->order; j++) { + int f = poly->faces[i*poly->order + j]; + points[j*2] = (poly->vertices[f*3+0] - + poly->vertices[f*3+2] * poly->shear); + points[j*2+1] = (poly->vertices[f*3+1] - + poly->vertices[f*3+2] * poly->shear); + } + + for (j = 0; j < poly->order; j++) { + coords[j*2] = points[j*2] * GRID_SCALE + ds->ox; + coords[j*2+1] = points[j*2+1] * GRID_SCALE + ds->oy; + } + + /* + * Find out whether these points are in a clockwise or + * anticlockwise arrangement. If the latter, discard the + * face because it's facing away from the viewer. + * + * This would involve fiddly winding-number stuff for a + * general polygon, but for the simple parallelograms we'll + * be seeing here, all we have to do is check whether the + * corners turn right or left. So we'll take the vector + * from point 0 to point 1, turn it right 90 degrees, + * and check the sign of the dot product with that and the + * next vector (point 1 to point 2). + */ + { + float v1x = points[2]-points[0]; + float v1y = points[3]-points[1]; + float v2x = points[4]-points[2]; + float v2y = points[5]-points[3]; + float dp = v1x * v2y - v1y * v2x; + + if (dp <= 0) + continue; + } + + draw_polygon(fe, coords, poly->order, TRUE, + state->facecolours[i] ? COL_BLUE : COL_BACKGROUND); + draw_polygon(fe, coords, poly->order, FALSE, COL_BORDER); + } + sfree(poly); + + draw_update(fe, 0, 0, (bb.r-bb.l+2) * GRID_SCALE, + (bb.d-bb.u+2) * GRID_SCALE); +} + +float game_anim_length(game_state *oldstate, game_state *newstate) +{ + return ROLLTIME; +} diff --git a/gtk.c b/gtk.c index ec086fc..0d63e58 100644 --- a/gtk.c +++ b/gtk.c @@ -12,6 +12,7 @@ #include #include +#include #include "puzzles.h" @@ -135,12 +136,26 @@ static void destroy(GtkWidget *widget, gpointer data) static gint key_event(GtkWidget *widget, GdkEventKey *event, gpointer data) { frontend *fe = (frontend *)data; + int keyval; if (!fe->pixmap) return TRUE; - if (event->string[0] && !event->string[1] && - !midend_process_key(fe->me, 0, 0, event->string[0])) + if (event->string[0] && !event->string[1]) + keyval = (unsigned char)event->string[0]; + else if (event->keyval == GDK_Up || event->keyval == GDK_KP_Up) + keyval = CURSOR_UP; + else if (event->keyval == GDK_Down || event->keyval == GDK_KP_Down) + keyval = CURSOR_DOWN; + else if (event->keyval == GDK_Left || event->keyval == GDK_KP_Left) + keyval = CURSOR_LEFT; + else if (event->keyval == GDK_Right || event->keyval == GDK_KP_Right) + keyval = CURSOR_RIGHT; + else + keyval = -1; + + if (keyval >= 0 && + !midend_process_key(fe->me, 0, 0, keyval)) gtk_widget_destroy(fe->window); return TRUE; diff --git a/midend.c b/midend.c index edde2e8..d309068 100644 --- a/midend.c +++ b/midend.c @@ -91,16 +91,22 @@ void midend_restart_game(midend_data *me) me->statepos = me->nstates; } -void midend_undo(midend_data *me) +static int midend_undo(midend_data *me) { - if (me->statepos > 1) + if (me->statepos > 1) { me->statepos--; + return 1; + } else + return 0; } -void midend_redo(midend_data *me) +static int midend_redo(midend_data *me) { - if (me->statepos < me->nstates) + if (me->statepos < me->nstates) { me->statepos++; + return 1; + } else + return 0; } int midend_process_key(midend_data *me, int x, int y, int button) @@ -126,10 +132,12 @@ int midend_process_key(midend_data *me, int x, int y, int button) midend_redraw(me); return 1; /* never animate */ } else if (button == 'u' || button == 'u' || - button == '\x1A' || button == '\x1F') { - midend_undo(me); + button == '\x1A' || button == '\x1F') { + if (!midend_undo(me)) + return 1; } else if (button == '\x12') { - midend_redo(me); + if (!midend_redo(me)) + return 1; } else if (button == 'q' || button == 'Q' || button == '\x11') { free_game(oldstate); return 0; diff --git a/puzzles.h b/puzzles.h index 79a938a..632f9b2 100644 --- a/puzzles.h +++ b/puzzles.h @@ -17,7 +17,11 @@ enum { LEFT_BUTTON = 0x1000, MIDDLE_BUTTON, - RIGHT_BUTTON + RIGHT_BUTTON, + CURSOR_UP, + CURSOR_DOWN, + CURSOR_LEFT, + CURSOR_RIGHT }; #define IGNORE(x) ( (x) = (x) ) @@ -53,8 +57,6 @@ void midend_set_params(midend_data *me, game_params *params); void midend_size(midend_data *me, int *x, int *y); void midend_new_game(midend_data *me, char *seed); void midend_restart_game(midend_data *me); -void midend_undo(midend_data *me); -void midend_redo(midend_data *me); int midend_process_key(midend_data *me, int x, int y, int button); void midend_redraw(midend_data *me); float *midend_colours(midend_data *me, int *ncolours); -- 2.11.0