*
* Things still to do:
*
- * * write a recursive solver?
+ * - The solver's algorithmic design is not really ideal. It makes
+ * use of the same data representation as gameplay uses, which
+ * often looks like a tempting reuse of code but isn't always a
+ * good idea. In this case, it's unpleasant that each edge of the
+ * graph ends up represented as multiple squares on a grid, with
+ * flags indicating when edges and non-edges cross; that's useful
+ * when the result can be directly translated into positions of
+ * graphics on the display, but in purely internal work it makes
+ * even simple manipulations during solving more painful than they
+ * should be, and complex ones have no choice but to modify the
+ * data structures temporarily, test things, and put them back. I
+ * envisage a complete solver rewrite along the following lines:
+ * + We have a collection of vertices (islands) and edges
+ * (potential bridge locations, i.e. pairs of horizontal or
+ * vertical islands with no other island in between).
+ * + Each edge has an associated list of edges that cross it, and
+ * hence with which it is mutually exclusive.
+ * + For each edge, we track the min and max number of bridges we
+ * currently think possible.
+ * + For each vertex, we track the number of _liberties_ it has,
+ * i.e. its clue number minus the min bridge count for each edge
+ * out of it.
+ * + We also maintain a dsf that identifies sets of vertices which
+ * are connected components of the puzzle so far, and for each
+ * equivalence class we track the total number of liberties for
+ * that component. (The dsf mechanism will also already track
+ * the size of each component, i.e. number of islands.)
+ * + So incrementing the min for an edge requires processing along
+ * the lines of:
+ * - set the max for all edges crossing that one to zero
+ * - decrement the liberty count for the vertex at each end,
+ * and also for each vertex's equivalence class (NB they may
+ * be the same class)
+ * - unify the two equivalence classes if they're not already,
+ * and if so, set the liberty count for the new class to be
+ * the sum of the previous two.
+ * + Decrementing the max is much easier, however.
+ * + With this data structure the really fiddly stuff in stage3()
+ * becomes more or less trivial, because it's now a quick job to
+ * find out whether an island would form an isolated subgraph if
+ * connected to a given subset of its neighbours:
+ * - identify the connected components containing the test
+ * vertex and its putative new neighbours (but be careful not
+ * to count a component more than once if two or more of the
+ * vertices involved are already in the same one)
+ * - find the sum of those components' liberty counts, and also
+ * the total number of islands involved
+ * - if the total liberty count of the connected components is
+ * exactly equal to twice the number of edges we'd be adding
+ * (of course each edge destroys two liberties, one at each
+ * end) then these components would become a subgraph with
+ * zero liberties if connected together.
+ * - therefore, if that subgraph also contains fewer than the
+ * total number of islands, it's disallowed.
+ * - As mentioned in stage3(), once we've identified such a
+ * disallowed pattern, we have two choices for what to do
+ * with it: if the candidate set of neighbours has size 1 we
+ * can reduce the max for the edge to that one neighbour,
+ * whereas if its complement has size 1 we can increase the
+ * min for the edge to the _omitted_ neighbour.
+ *
+ * - write a recursive solver?
*/
#include <stdio.h>
idx = x;
s = e = -1;
bl = 0;
+ maxb = state->params.maxb; /* placate optimiser */
/* Unset possible flags until we find an island. */
for (y = 0; y < state->h; y++) {
is_s = IDX(state, gridi, idx);
idx = y*w;
s = e = -1;
bl = 0;
+ maxb = state->params.maxb; /* placate optimiser */
for (x = 0; x < state->w; x++) {
is_s = IDX(state, gridi, idx);
if (is_s) {
return 1;
}
-static int solve_island_subgroup(struct island *is, int direction, int n)
+static int solve_island_subgroup(struct island *is, int direction)
{
struct island *is_join;
int nislands, *dsf = is->state->solver->dsf;
debug(("..checking subgroups.\n"));
/* if is isn't full, return 0. */
- if (n < is->count) {
+ if (island_countbridges(is) < is->count) {
debug(("...orig island (%d,%d) not full.\n", is->x, is->y));
return 0;
}
/* we have a full subgroup that isn't the whole set.
* This isn't allowed. */
debug(("island at (%d,%d) makes full subgroup, disallowing.\n",
- is->x, is->y, n));
+ is->x, is->y));
return 1;
} else {
debug(("...has finished puzzle.\n"));
if (missing <= 0) return 1;
for (i = 0; i < is->adj.npoints; i++) {
- /* We only do right- or down-pointing bridges. */
- if (is->adj.points[i].dx == -1 ||
- is->adj.points[i].dy == -1) continue;
-
x = is->adj.points[i].x;
y = is->adj.points[i].y;
spc = island_adjspace(is, 1, missing, i);
solve_join(is, i, n, 0);
map_update_possibles(is->state);
- if (solve_island_subgroup(is, i, n) ||
+ if (solve_island_subgroup(is, i) ||
solve_island_impossible(is->state)) {
maxb = n-1;
debug(("island at (%d,%d) d(%d,%d) new max of %d bridges:\n",
}
map_update_possibles(is->state);
- if (solve_island_subgroup(is, -1, n))
+ if (solve_island_subgroup(is, -1))
got = 1;
for (j = 0; j < is->adj.npoints; j++)
int show_hints;
};
-static char *update_drag_dst(game_state *state, game_ui *ui, game_drawstate *ds,
- int nx, int ny)
+static char *update_drag_dst(game_state *state, game_ui *ui,
+ const game_drawstate *ds, int nx, int ny)
{
int ox, oy, dx, dy, i, currl, maxb;
struct island *is;
return dupstr(buf);
}
-static char *interpret_move(game_state *state, game_ui *ui, game_drawstate *ds,
+static char *interpret_move(game_state *state, game_ui *ui, const game_drawstate *ds,
int x, int y, int button)
{
int gx = FROMCOORD(x), gy = FROMCOORD(y);
~mdw / sgt / puzzles / blobdiff