zakarya/puzzles
1
0
Fork 0
mirror of git://git.tartarus.org/simon/puzzles.git synced 2025-04-21 08:01:30 -07:00
Code Issues Packages Projects Releases Wiki Activity

Add a new deduction to Easy level, which is as small as I can make it

Browse Source 
to have the effect of enabling large Easy-level grids to be
constructed in all grid types. Without this, some generations at Easy
level (e.g. 'loopy --generate 1 7x7t9de') can spin forever because
_even with all clues filled in_ the generated grids can't be solved at
that level.

[originally from svn r9143]
This commit is contained in:
Simon Tatham
2011-04-03 07:59:35 +00:00
parent 2054da9f68
commit 9ece2832ce
1 changed files with 58 additions and 2 deletions
Download Patch File Download Diff File Expand all files Collapse all files

60
loopy.c
View File

@ -1834,7 +1834,6 @@ static char *new_game_desc(game_params *params, random_state *rs,
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++;
@ -1856,7 +1855,6 @@ static char *new_game_desc(game_params *params, random_state *rs,
* 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);
@ -2432,6 +2430,13 @@ static int trivial_deductions(solver_state *sstate)
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;
@ -2453,6 +2458,57 @@ static int trivial_deductions(solver_state *sstate)
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);