params->h = params->w;
}
- /*
- * Assume a random generation scheme unless told otherwise, for the
- * sake of internal consistency.
- */
- params->type = TYPE_RANDOM;
for (i = 0; i < lenof(pegs_lowertypes); i++)
if (!strcmp(p, pegs_lowertypes[i]))
params->type = i;
case TYPE_OCTAGON:
cx = abs(x - w/2);
cy = abs(y - h/2);
- if (cx == 0 && cy == 0)
- v = GRID_HOLE;
- else if (cx + cy > 1 + max(w,h)/2)
+ if (cx + cy > 1 + max(w,h)/2)
v = GRID_OBST;
else
v = GRID_PEG;
}
grid[y*w+x] = v;
}
+
+ if (params->type == TYPE_OCTAGON) {
+ /*
+ * The octagonal (European) solitaire layout is
+ * actually _insoluble_ with the starting hole at the
+ * centre. Here's a proof:
+ *
+ * Colour the squares of the board diagonally in
+ * stripes of three different colours, which I'll call
+ * A, B and C. So the board looks like this:
+ *
+ * A B C
+ * A B C A B
+ * A B C A B C A
+ * B C A B C A B
+ * C A B C A B C
+ * B C A B C
+ * A B C
+ *
+ * Suppose we keep running track of the number of pegs
+ * occuping each colour of square. This colouring has
+ * the property that any valid move whatsoever changes
+ * all three of those counts by one (two of them go
+ * down and one goes up), which means that the _parity_
+ * of every count flips on every move.
+ *
+ * If the centre square starts off unoccupied, then
+ * there are twelve pegs on each colour and all three
+ * counts start off even; therefore, after 35 moves all
+ * three counts would have to be odd, which isn't
+ * possible if there's only one peg left. []
+ *
+ * This proof works just as well if the starting hole
+ * is _any_ of the thirteen positions labelled B. Also,
+ * we can stripe the board in the opposite direction
+ * and rule out any square labelled B in that colouring
+ * as well. This leaves:
+ *
+ * Y n Y
+ * n n Y n n
+ * Y n n Y n n Y
+ * n Y Y n Y Y n
+ * Y n n Y n n Y
+ * n n Y n n
+ * Y n Y
+ *
+ * where the ns are squares we've proved insoluble, and
+ * the Ys are the ones remaining.
+ *
+ * That doesn't prove all those starting positions to
+ * be soluble, of course; they're merely the ones we
+ * _haven't_ proved to be impossible. Nevertheless, it
+ * turns out that they are all soluble, so when the
+ * user requests an Octagon board the simplest thing is
+ * to pick one of these at random.
+ *
+ * Rather than picking equiprobably from those twelve
+ * positions, we'll pick equiprobably from the three
+ * equivalence classes
+ */
+ switch (random_upto(rs, 3)) {
+ case 0:
+ /* Remove a random corner piece. */
+ {
+ int dx, dy;
+
+ dx = random_upto(rs, 2) * 2 - 1; /* +1 or -1 */
+ dy = random_upto(rs, 2) * 2 - 1; /* +1 or -1 */
+ if (random_upto(rs, 2))
+ dy *= 3;
+ else
+ dx *= 3;
+ grid[(3+dy)*w+(3+dx)] = GRID_HOLE;
+ }
+ break;
+ case 1:
+ /* Remove a random piece two from the centre. */
+ {
+ int dx, dy;
+ dx = 2 * (random_upto(rs, 2) * 2 - 1);
+ if (random_upto(rs, 2))
+ dy = 0;
+ else
+ dy = dx, dx = 0;
+ grid[(3+dy)*w+(3+dx)] = GRID_HOLE;
+ }
+ break;
+ default /* case 2 */:
+ /* Remove a random piece one from the centre. */
+ {
+ int dx, dy;
+ dx = random_upto(rs, 2) * 2 - 1;
+ if (random_upto(rs, 2))
+ dy = 0;
+ else
+ dy = dx, dx = 0;
+ grid[(3+dy)*w+(3+dx)] = GRID_HOLE;
+ }
+ break;
+ }
+ }
}
/*
int sx, sy, tx, ty;
game_state *ret;
- if (sscanf(move, "%d,%d-%d,%d", &sx, &sy, &tx, &ty)) {
+ if (sscanf(move, "%d,%d-%d,%d", &sx, &sy, &tx, &ty) == 4) {
int mx, my, dx, dy;
if (sx < 0 || sx >= w || sy < 0 || sy >= h)
static float *game_colours(frontend *fe, game_state *state, int *ncolours)
{
float *ret = snewn(3 * NCOLOURS, float);
- int i;
- float max;
- frontend_default_colour(fe, &ret[COL_BACKGROUND * 3]);
-
- /*
- * Drop the background colour so that the highlight is
- * noticeably brighter than it while still being under 1.
- */
- max = ret[COL_BACKGROUND*3];
- for (i = 1; i < 3; i++)
- if (ret[COL_BACKGROUND*3+i] > max)
- max = ret[COL_BACKGROUND*3+i];
- if (max * 1.2F > 1.0F) {
- for (i = 0; i < 3; i++)
- ret[COL_BACKGROUND*3+i] /= (max * 1.2F);
- }
-
- for (i = 0; i < 3; i++) {
- ret[COL_HIGHLIGHT * 3 + i] = ret[COL_BACKGROUND * 3 + i] * 1.2F;
- ret[COL_LOWLIGHT * 3 + i] = ret[COL_BACKGROUND * 3 + i] * 0.8F;
- }
+ game_mkhighlight(fe, ret, COL_BACKGROUND, COL_HIGHLIGHT, COL_LOWLIGHT);
ret[COL_PEG * 3 + 0] = 0.0F;
ret[COL_PEG * 3 + 1] = 0.0F;
return FALSE;
}
-static int game_timing_state(game_state *state)
+static int game_timing_state(game_state *state, game_ui *ui)
{
return TRUE;
}