int started;
int tilesize;
int flashing;
+ int *textx, *texty;
char *lines;
char *clue_error;
char *clue_satisfied;
A(Great-Hexagonal,grid_new_greathexagonal,3,3) \
A(Octagonal,grid_new_octagonal,3,3) \
A(Kites,grid_new_kites,3,3) \
- A(Floret,grid_new_floret,1,2)
+ A(Floret,grid_new_floret,1,2) \
+ A(Dodecagonal,grid_new_dodecagonal,2,2) \
+ A(Great-Dodecagonal,grid_new_greatdodecagonal,2,2)
#define GRID_NAME(title,fn,amin,omin) #title,
#define GRID_CONFIG(title,fn,amin,omin) ":" #title
((field) &= ~(1<<(bit)), TRUE) : FALSE)
#define CLUE2CHAR(c) \
- ((c < 0) ? ' ' : c + '0')
+ ((c < 0) ? ' ' : c < 10 ? c + '0' : c - 10 + 'A')
/* ----------------------------------------------------------------------
* General struct manipulation and other straightforward code
{ 5, 5, DIFF_HARD, 6, NULL },
{ 5, 5, DIFF_HARD, 7, NULL },
{ 3, 3, DIFF_HARD, 8, NULL },
+ { 3, 3, DIFF_HARD, 9, NULL },
+ { 3, 3, DIFF_HARD, 10, NULL },
#else
{ 7, 7, DIFF_EASY, 0, NULL },
{ 10, 10, DIFF_EASY, 0, NULL },
{ 7, 7, DIFF_HARD, 6, NULL },
{ 5, 5, DIFF_HARD, 7, NULL },
{ 5, 5, DIFF_HARD, 8, NULL },
+ { 5, 4, DIFF_HARD, 9, NULL },
+ { 5, 4, DIFF_HARD, 10, NULL },
#endif
};
g = params->game_grid;
for (; *desc; ++desc) {
- if (*desc >= '0' && *desc <= '9') {
+ if ((*desc >= '0' && *desc <= '9') || (*desc >= 'A' && *desc <= 'Z')) {
count++;
continue;
}
ret[COL_FOREGROUND * 3 + 1] = 0.0F;
ret[COL_FOREGROUND * 3 + 2] = 0.0F;
- ret[COL_LINEUNKNOWN * 3 + 0] = 0.8F;
- ret[COL_LINEUNKNOWN * 3 + 1] = 0.8F;
+ /*
+ * We want COL_LINEUNKNOWN to be a yellow which is a bit darker
+ * than the background. (I previously set it to 0.8,0.8,0, but
+ * found that this went badly with the 0.8,0.8,0.8 favoured as a
+ * background by the Java frontend.)
+ */
+ ret[COL_LINEUNKNOWN * 3 + 0] = ret[COL_BACKGROUND * 3 + 0] * 0.9F;
+ ret[COL_LINEUNKNOWN * 3 + 1] = ret[COL_BACKGROUND * 3 + 1] * 0.9F;
ret[COL_LINEUNKNOWN * 3 + 2] = 0.0F;
ret[COL_HIGHLIGHT * 3 + 0] = 1.0F;
struct game_drawstate *ds = snew(struct game_drawstate);
int num_faces = state->game_grid->num_faces;
int num_edges = state->game_grid->num_edges;
+ int i;
ds->tilesize = 0;
ds->started = 0;
ds->lines = snewn(num_edges, char);
ds->clue_error = snewn(num_faces, char);
ds->clue_satisfied = snewn(num_faces, char);
+ ds->textx = snewn(num_faces, int);
+ ds->texty = snewn(num_faces, int);
ds->flashing = 0;
memset(ds->lines, LINE_UNKNOWN, num_edges);
memset(ds->clue_error, 0, num_faces);
memset(ds->clue_satisfied, 0, num_faces);
+ for (i = 0; i < num_faces; i++)
+ ds->textx[i] = ds->texty[i] = -1;
return ds;
}
int i;
game_state *state = snew(game_state);
int empties_to_make = 0;
- int n;
+ int n,n2;
const char *dp = desc;
grid *g;
int num_faces, num_edges;
assert(*dp);
n = *dp - '0';
+ n2 = *dp - 'A' + 10;
if (n >= 0 && n < 10) {
state->clues[i] = n;
+ } else if (n2 >= 10 && n2 < 36) {
+ state->clues[i] = n2;
} else {
n = *dp - 'a' + 1;
assert(n > 0);
if (state->clues[i] < 0)
continue;
+ /*
+ * This code checks whether the numeric clue on a face is so
+ * large as to permit all its remaining LINE_UNKNOWNs to be
+ * filled in as LINE_YES, or alternatively so small as to
+ * permit them all to be filled in as LINE_NO.
+ */
+
if (state->clues[i] < current_yes) {
sstate->solver_status = SOLVER_MISTAKE;
return DIFF_EASY;
sstate->face_solved[i] = TRUE;
continue;
}
+
+ if (f->order - state->clues[i] == current_no + 1 &&
+ f->order - current_yes - current_no > 2) {
+ /*
+ * One small refinement to the above: we also look for any
+ * adjacent pair of LINE_UNKNOWNs around the face with
+ * some LINE_YES incident on it from elsewhere. If we find
+ * one, then we know that pair of LINE_UNKNOWNs can't
+ * _both_ be LINE_YES, and hence that pushes us one line
+ * closer to being able to determine all the rest.
+ */
+ int j, k, e1, e2, e, d;
+
+ for (j = 0; j < f->order; j++) {
+ e1 = f->edges[j] - g->edges;
+ e2 = f->edges[j+1 < f->order ? j+1 : 0] - g->edges;
+
+ if (g->edges[e1].dot1 == g->edges[e2].dot1 ||
+ g->edges[e1].dot1 == g->edges[e2].dot2) {
+ d = g->edges[e1].dot1 - g->dots;
+ } else {
+ assert(g->edges[e1].dot2 == g->edges[e2].dot1 ||
+ g->edges[e1].dot2 == g->edges[e2].dot2);
+ d = g->edges[e1].dot2 - g->dots;
+ }
+
+ if (state->lines[e1] == LINE_UNKNOWN &&
+ state->lines[e2] == LINE_UNKNOWN) {
+ for (k = 0; k < g->dots[d].order; k++) {
+ int e = g->dots[d].edges[k] - g->edges;
+ if (state->lines[e] == LINE_YES)
+ goto found; /* multi-level break */
+ }
+ }
+ }
+ continue;
+
+ found:
+ /*
+ * If we get here, we've found such a pair of edges, and
+ * they're e1 and e2.
+ */
+ for (j = 0; j < f->order; j++) {
+ e = f->edges[j] - g->edges;
+ if (state->lines[e] == LINE_UNKNOWN && e != e1 && e != e2) {
+ int r = solver_set_line(sstate, e, LINE_YES);
+ assert(r);
+ diff = min(diff, DIFF_EASY);
+ }
+ }
+ }
}
check_caches(sstate);
* on that. We check this with an assertion, in case someone decides to
* make a grid which has larger faces than this. Note, this algorithm
* could get quite expensive if there are many large faces. */
-#define MAX_FACE_SIZE 8
+#define MAX_FACE_SIZE 12
for (i = 0; i < g->num_faces; i++) {
int maxs[MAX_FACE_SIZE][MAX_FACE_SIZE];
/* Returns (into x,y) position of centre of face for rendering the text clue.
*/
static void face_text_pos(const game_drawstate *ds, const grid *g,
- const grid_face *f, int *x, int *y)
+ const grid_face *f, int *xret, int *yret)
{
- int i;
+ int x, y, x0, y0, x1, y1, xbest, ybest, i, shift;
+ long bestdist;
+ int faceindex = f - g->faces;
- /* Simplest solution is the centroid. Might not work in some cases. */
+ /*
+ * Return the cached position for this face, if we've already
+ * worked it out.
+ */
+ if (ds->textx[faceindex] >= 0) {
+ *xret = ds->textx[faceindex];
+ *yret = ds->texty[faceindex];
+ return;
+ }
- /* Another algorithm to look into:
- * Find the midpoints of the sides, find the bounding-box,
- * then take the centre of that. */
+ /*
+ * Otherwise, try to find the point in the polygon with the
+ * maximum distance to any edge or corner.
+ *
+ * Start by working out the face's bounding box, in grid
+ * coordinates.
+ */
+ x0 = x1 = f->dots[0]->x;
+ y0 = y1 = f->dots[0]->y;
+ for (i = 1; i < f->order; i++) {
+ if (x0 > f->dots[i]->x) x0 = f->dots[i]->x;
+ if (x1 < f->dots[i]->x) x1 = f->dots[i]->x;
+ if (y0 > f->dots[i]->y) y0 = f->dots[i]->y;
+ if (y1 < f->dots[i]->y) y1 = f->dots[i]->y;
+ }
- /* Best solution probably involves incentres (inscribed circles) */
+ /*
+ * If the grid is at excessive resolution, decide on a scaling
+ * factor to bring it within reasonable bounds so we don't have to
+ * think too hard or suffer integer overflow.
+ */
+ shift = 0;
+ while (x1 - x0 > 128 || y1 - y0 > 128) {
+ shift++;
+ x0 >>= 1;
+ x1 >>= 1;
+ y0 >>= 1;
+ y1 >>= 1;
+ }
- int sx = 0, sy = 0; /* sums */
- for (i = 0; i < f->order; i++) {
- grid_dot *d = f->dots[i];
- sx += d->x;
- sy += d->y;
+ /*
+ * Now iterate over every point in that bounding box.
+ */
+ xbest = ybest = -1;
+ bestdist = -1;
+ for (y = y0; y <= y1; y++) {
+ for (x = x0; x <= x1; x++) {
+ /*
+ * First, disqualify the point if it's not inside the
+ * polygon, which we work out by counting the edges to the
+ * right of the point. (For tiebreaking purposes when
+ * edges start or end on our y-coordinate or go right
+ * through it, we consider our point to be offset by a
+ * small _positive_ epsilon in both the x- and
+ * y-direction.)
+ */
+ int in = 0;
+ for (i = 0; i < f->order; i++) {
+ int xs = f->edges[i]->dot1->x >> shift;
+ int xe = f->edges[i]->dot2->x >> shift;
+ int ys = f->edges[i]->dot1->y >> shift;
+ int ye = f->edges[i]->dot2->y >> shift;
+ if ((y >= ys && y < ye) || (y >= ye && y < ys)) {
+ /*
+ * The line goes past our y-position. Now we need
+ * to know if its x-coordinate when it does so is
+ * to our right.
+ *
+ * The x-coordinate in question is mathematically
+ * (y - ys) * (xe - xs) / (ye - ys), and we want
+ * to know whether (x - xs) >= that. Of course we
+ * avoid the division, so we can work in integers;
+ * to do this we must multiply both sides of the
+ * inequality by ye - ys, which means we must
+ * first check that's not negative.
+ */
+ int num = xe - xs, denom = ye - ys;
+ if (denom < 0) {
+ num = -num;
+ denom = -denom;
+ }
+ if ((x - xs) * denom >= (y - ys) * num)
+ in ^= 1;
+ }
+ }
+
+ if (in) {
+ long mindist = LONG_MAX;
+
+ /*
+ * This point is inside the polygon, so now we check
+ * its minimum distance to every edge and corner.
+ * First the corners ...
+ */
+ for (i = 0; i < f->order; i++) {
+ int xp = f->dots[i]->x >> shift;
+ int yp = f->dots[i]->y >> shift;
+ int dx = x - xp, dy = y - yp;
+ long dist = (long)dx*dx + (long)dy*dy;
+ if (mindist > dist)
+ mindist = dist;
+ }
+
+ /*
+ * ... and now also check the perpendicular distance
+ * to every edge, if the perpendicular lies between
+ * the edge's endpoints.
+ */
+ for (i = 0; i < f->order; i++) {
+ int xs = f->edges[i]->dot1->x >> shift;
+ int xe = f->edges[i]->dot2->x >> shift;
+ int ys = f->edges[i]->dot1->y >> shift;
+ int ye = f->edges[i]->dot2->y >> shift;
+
+ /*
+ * If s and e are our endpoints, and p our
+ * candidate circle centre, the foot of a
+ * perpendicular from p to the line se lies
+ * between s and e if and only if (p-s).(e-s) lies
+ * strictly between 0 and (e-s).(e-s).
+ */
+ int edx = xe - xs, edy = ye - ys;
+ int pdx = x - xs, pdy = y - ys;
+ long pde = (long)pdx * edx + (long)pdy * edy;
+ long ede = (long)edx * edx + (long)edy * edy;
+ if (0 < pde && pde < ede) {
+ /*
+ * Yes, the nearest point on this edge is
+ * closer than either endpoint, so we must
+ * take it into account by measuring the
+ * perpendicular distance to the edge and
+ * checking its square against mindist.
+ */
+
+ long pdre = (long)pdx * edy - (long)pdy * edx;
+ long sqlen = pdre * pdre / ede;
+
+ if (mindist > sqlen)
+ mindist = sqlen;
+ }
+ }
+
+ /*
+ * Right. Now we know the biggest circle around this
+ * point, so we can check it against bestdist.
+ */
+ if (bestdist < mindist) {
+ bestdist = mindist;
+ xbest = x;
+ ybest = y;
+ }
+ }
+ }
}
- sx /= f->order;
- sy /= f->order;
+
+ assert(bestdist >= 0);
/* convert to screen coordinates */
- grid_to_screen(ds, g, sx, sy, x, y);
+ grid_to_screen(ds, g, xbest << shift, ybest << shift,
+ &ds->textx[faceindex], &ds->texty[faceindex]);
+
+ *xret = ds->textx[faceindex];
+ *yret = ds->texty[faceindex];
+}
+
+static void face_text_bbox(game_drawstate *ds, grid *g, grid_face *f,
+ int *x, int *y, int *w, int *h)
+{
+ int xx, yy;
+ face_text_pos(ds, g, f, &xx, &yy);
+
+ /* There seems to be a certain amount of trial-and-error involved
+ * in working out the correct bounding-box for the text. */
+
+ *x = xx - ds->tilesize/4 - 1;
+ *y = yy - ds->tilesize/4 - 3;
+ *w = ds->tilesize/2 + 2;
+ *h = ds->tilesize/2 + 5;
}
static void game_redraw_clue(drawing *dr, game_drawstate *ds,
grid *g = state->game_grid;
grid_face *f = g->faces + i;
int x, y;
- char c[2];
+ char c[3];
- c[0] = CLUE2CHAR(state->clues[i]);
- c[1] = '\0';
+ if (state->clues[i] < 10) {
+ c[0] = CLUE2CHAR(state->clues[i]);
+ c[1] = '\0';
+ } else {
+ sprintf(c, "%d", state->clues[i]);
+ }
face_text_pos(ds, g, f, &x, &y);
draw_text(dr, x, y,
ds->clue_satisfied[i] ? COL_SATISFIED : COL_FOREGROUND, c);
}
+static void edge_bbox(game_drawstate *ds, grid *g, grid_edge *e,
+ int *x, int *y, int *w, int *h)
+{
+ int x1 = e->dot1->x;
+ int y1 = e->dot1->y;
+ int x2 = e->dot2->x;
+ int y2 = e->dot2->y;
+ int xmin, xmax, ymin, ymax;
+
+ grid_to_screen(ds, g, x1, y1, &x1, &y1);
+ grid_to_screen(ds, g, x2, y2, &x2, &y2);
+ /* Allow extra margin for dots, and thickness of lines */
+ xmin = min(x1, x2) - 2;
+ xmax = max(x1, x2) + 2;
+ ymin = min(y1, y2) - 2;
+ ymax = max(y1, y2) + 2;
+
+ *x = xmin;
+ *y = ymin;
+ *w = xmax - xmin + 1;
+ *h = ymax - ymin + 1;
+}
+
+static void dot_bbox(game_drawstate *ds, grid *g, grid_dot *d,
+ int *x, int *y, int *w, int *h)
+{
+ int x1, y1;
+
+ grid_to_screen(ds, g, d->x, d->y, &x1, &y1);
+
+ *x = x1 - 2;
+ *y = y1 - 2;
+ *w = 5;
+ *h = 5;
+}
+
+static const int loopy_line_redraw_phases[] = {
+ COL_FAINT, COL_LINEUNKNOWN, COL_FOREGROUND, COL_HIGHLIGHT, COL_MISTAKE
+};
+#define NPHASES lenof(loopy_line_redraw_phases)
+
static void game_redraw_line(drawing *dr, game_drawstate *ds,
- game_state *state, int i)
+ game_state *state, int i, int phase)
{
grid *g = state->game_grid;
grid_edge *e = g->edges + i;
line_colour = COL_HIGHLIGHT;
else
line_colour = COL_FOREGROUND;
+ if (line_colour != loopy_line_redraw_phases[phase])
+ return;
/* Convert from grid to screen coordinates */
grid_to_screen(ds, g, e->dot1->x, e->dot1->y, &x1, &y1);
draw_circle(dr, x, y, 2, COL_FOREGROUND, COL_FOREGROUND);
}
+static int boxes_intersect(int x0, int y0, int w0, int h0,
+ int x1, int y1, int w1, int h1)
+{
+ /*
+ * Two intervals intersect iff neither is wholly on one side of
+ * the other. Two boxes intersect iff their horizontal and
+ * vertical intervals both intersect.
+ */
+ return (x0 < x1+w1 && x1 < x0+w0 && y0 < y1+h1 && y1 < y0+h0);
+}
+
+static void game_redraw_in_rect(drawing *dr, game_drawstate *ds,
+ game_state *state, int x, int y, int w, int h)
+{
+ grid *g = state->game_grid;
+ int i, phase;
+ int bx, by, bw, bh;
+
+ clip(dr, x, y, w, h);
+ draw_rect(dr, x, y, w, h, COL_BACKGROUND);
+
+ for (i = 0; i < g->num_faces; i++) {
+ if (state->clues[i] >= 0) {
+ face_text_bbox(ds, g, &g->faces[i], &bx, &by, &bw, &bh);
+ if (boxes_intersect(x, y, w, h, bx, by, bw, bh))
+ game_redraw_clue(dr, ds, state, i);
+ }
+ }
+ for (phase = 0; phase < NPHASES; phase++) {
+ for (i = 0; i < g->num_edges; i++) {
+ edge_bbox(ds, g, &g->edges[i], &bx, &by, &bw, &bh);
+ if (boxes_intersect(x, y, w, h, bx, by, bw, bh))
+ game_redraw_line(dr, ds, state, i, phase);
+ }
+ }
+ for (i = 0; i < g->num_dots; i++) {
+ dot_bbox(ds, g, &g->dots[i], &bx, &by, &bw, &bh);
+ if (boxes_intersect(x, y, w, h, bx, by, bw, bh))
+ game_redraw_dot(dr, ds, state, i);
+ }
+
+ unclip(dr);
+ draw_update(dr, x, y, w, h);
+}
+
static void game_redraw(drawing *dr, game_drawstate *ds, game_state *oldstate,
game_state *state, int dir, game_ui *ui,
float animtime, float flashtime)
/* Pass one is now done. Now we do the actual drawing. */
if (redraw_everything) {
-
- /* This is the unsubtle version. */
-
int grid_width = g->highest_x - g->lowest_x;
int grid_height = g->highest_y - g->lowest_y;
int w = grid_width * ds->tilesize / g->tilesize;
int h = grid_height * ds->tilesize / g->tilesize;
- draw_rect(dr, 0, 0, w + 2*border + 1, h + 2*border + 1,
- COL_BACKGROUND);
-
- for (i = 0; i < g->num_faces; i++)
- game_redraw_clue(dr, ds, state, i);
- for (i = 0; i < g->num_edges; i++)
- game_redraw_line(dr, ds, state, i);
- for (i = 0; i < g->num_dots; i++)
- game_redraw_dot(dr, ds, state, i);
-
- draw_update(dr, 0, 0, w + 2*border + 1, h + 2*border + 1);
+ game_redraw_in_rect(dr, ds, state,
+ 0, 0, w + 2*border + 1, h + 2*border + 1);
} else {
/* Right. Now we roll up our sleeves. */
for (i = 0; i < nfaces; i++) {
grid_face *f = g->faces + faces[i];
- int xx, yy;
int x, y, w, h;
- int j;
-
- /* There seems to be a certain amount of trial-and-error
- * involved in working out the correct bounding-box for
- * the text. */
- face_text_pos(ds, g, f, &xx, &yy);
-
- x = xx - ds->tilesize/4 - 1; w = ds->tilesize/2 + 2;
- y = yy - ds->tilesize/4 - 3; h = ds->tilesize/2 + 5;
- clip(dr, x, y, w, h);
- draw_rect(dr, x, y, w, h, COL_BACKGROUND);
-
- game_redraw_clue(dr, ds, state, faces[i]);
- for (j = 0; j < f->order; j++)
- game_redraw_line(dr, ds, state, f->edges[j] - g->edges);
- for (j = 0; j < f->order; j++)
- game_redraw_dot(dr, ds, state, f->dots[j] - g->dots);
- unclip(dr);
- draw_update(dr, x, y, w, h);
+
+ face_text_bbox(ds, g, f, &x, &y, &w, &h);
+ game_redraw_in_rect(dr, ds, state, x, y, w, h);
}
for (i = 0; i < nedges; i++) {
- grid_edge *e = g->edges + edges[i], *ee;
- int x1 = e->dot1->x;
- int y1 = e->dot1->y;
- int x2 = e->dot2->x;
- int y2 = e->dot2->y;
- int xmin, xmax, ymin, ymax;
- int j;
-
- grid_to_screen(ds, g, x1, y1, &x1, &y1);
- grid_to_screen(ds, g, x2, y2, &x2, &y2);
- /* Allow extra margin for dots, and thickness of lines */
- xmin = min(x1, x2) - 2;
- xmax = max(x1, x2) + 2;
- ymin = min(y1, y2) - 2;
- ymax = max(y1, y2) + 2;
- /* For testing, I find it helpful to change COL_BACKGROUND
- * to COL_SATISFIED here. */
- clip(dr, xmin, ymin, xmax - xmin + 1, ymax - ymin + 1);
- draw_rect(dr, xmin, ymin, xmax - xmin + 1, ymax - ymin + 1,
- COL_BACKGROUND);
-
- if (e->face1)
- game_redraw_clue(dr, ds, state, e->face1 - g->faces);
- if (e->face2)
- game_redraw_clue(dr, ds, state, e->face2 - g->faces);
-
- game_redraw_line(dr, ds, state, edges[i]);
- for (j = 0; j < e->dot1->order; j++) {
- ee = e->dot1->edges[j];
- if (ee != e)
- game_redraw_line(dr, ds, state, ee - g->edges);
- }
- for (j = 0; j < e->dot2->order; j++) {
- ee = e->dot2->edges[j];
- if (ee != e)
- game_redraw_line(dr, ds, state, ee - g->edges);
- }
- game_redraw_dot(dr, ds, state, e->dot1 - g->dots);
- game_redraw_dot(dr, ds, state, e->dot2 - g->dots);
+ grid_edge *e = g->edges + edges[i];
+ int x, y, w, h;
- unclip(dr);
- draw_update(dr, xmin, ymin, xmax - xmin + 1, ymax - ymin + 1);
+ edge_bbox(ds, g, e, &x, &y, &w, &h);
+ game_redraw_in_rect(dr, ds, state, x, y, w, h);
}
}
return 0.0F;
}
+static int game_is_solved(game_state *state)
+{
+ return state->solved;
+}
+
static void game_print_size(game_params *params, float *x, float *y)
{
int pw, ph;
game_redraw,
game_anim_length,
game_flash_length,
+ game_is_solved,
TRUE, FALSE, game_print_size, game_print,
FALSE /* wants_statusbar */,
FALSE, game_timing_state,
~mdw / sgt / puzzles / blobdiff