int started;
int tilesize;
int flashing;
+ int *textx, *texty;
char *lines;
char *clue_error;
char *clue_satisfied;
A(Cairo,grid_new_cairo,3,4) \
A(Great-Hexagonal,grid_new_greathexagonal,3,3) \
A(Octagonal,grid_new_octagonal,3,3) \
- A(Kites,grid_new_kites,3,3)
+ A(Kites,grid_new_kites,3,3) \
+ 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, 4, DIFF_HARD, 5, NULL },
{ 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 },
{ 5, 4, DIFF_HARD, 5, 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;
}
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;
}
grid *g;
game_state *state = snew(game_state);
game_state *state_new;
+ int count = 0;
params_generate_grid(params);
state->game_grid = g = params->game_grid;
g->refcount++;
* preventing games smaller than 4x4 seems to stop this happening */
do {
add_full_clues(state, rs);
+ if (++count%100 == 0) printf("tried %d times to make a unique board\n", count);
} while (!game_has_unique_soln(state, params->diff));
state_new = remove_clues(state, rs, params->diff);
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);
* 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 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 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);
grid *g = state->game_grid;
int border = BORDER(ds->tilesize);
- int i;
+ int i, phase;
int flash_changed;
int redraw_everything = FALSE;
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 (phase = 0; phase < NPHASES; phase++)
+ for (i = 0; i < g->num_edges; i++)
+ game_redraw_line(dr, ds, state, i, phase);
for (i = 0; i < g->num_dots; i++)
game_redraw_dot(dr, ds, state, i);
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 (phase = 0; phase < NPHASES; phase++)
+ for (j = 0; j < f->order; j++)
+ game_redraw_line(dr, ds, state, f->edges[j] - g->edges,
+ phase);
for (j = 0; j < f->order; j++)
game_redraw_dot(dr, ds, state, f->dots[j] - g->dots);
unclip(dr);
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);
- }
+ for (phase = 0; phase < NPHASES; phase++) {
+ game_redraw_line(dr, ds, state, edges[i], phase);
+ 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, phase);
+ }
+ 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, phase);
+ }
+ }
game_redraw_dot(dr, ds, state, e->dot1 - g->dots);
game_redraw_dot(dr, ds, state, e->dot2 - g->dots);
~mdw / sgt / puzzles / blobdiff