ref: d99e217cfb12f40dd7dbc26146ef287f9f9020fc
parent: 9867234e70002b8252a48c2bc023875ff87b8ca1
author: Simon Tatham <anakin@pobox.com>
date: Tue Apr 27 13:44:30 EDT 2004
Implemented Cube, in a sufficiently general way that it also handles the tetrahedron, octahedron and icosahedron. [originally from svn r4151]
--- a/Recipe
+++ b/Recipe
@@ -15,7 +15,7 @@
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
--- a/cube.c
+++ b/cube.c
@@ -1,3 +1,1273 @@
/*
* cube.c: Cube game.
*/
+
+#include <stdio.h>
+#include <stdlib.h>
+#include <string.h>
+#include <assert.h>
+#include <math.h>
+
+#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;
+}
--- a/gtk.c
+++ b/gtk.c
@@ -12,6 +12,7 @@
#include <stdarg.h>
#include <gtk/gtk.h>
+#include <gdk/gdkkeysyms.h>
#include "puzzles.h"
@@ -135,12 +136,26 @@
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;
--- a/midend.c
+++ b/midend.c
@@ -91,16 +91,22 @@
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 @@
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;
--- 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_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);