int latin_solver_diff_set(struct latin_solver *solver,
struct latin_solver_scratch *scratch,
- int *extreme)
+ int extreme)
{
int x, y, n, ret, o = solver->o;
- /*
- * Row-wise set elimination.
- */
- for (y = 0; y < o; y++) {
- ret = latin_solver_set(solver, scratch, cubepos(0,y,1), o*o, 1
+
+ if (!extreme) {
+ /*
+ * Row-wise set elimination.
+ */
+ for (y = 0; y < o; y++) {
+ ret = latin_solver_set(solver, scratch, cubepos(0,y,1), o*o, 1
#ifdef STANDALONE_SOLVER
- , "set elimination, row %d", YUNTRANS(y)
+ , "set elimination, row %d", YUNTRANS(y)
#endif
- );
- if (ret > 0) *extreme = 0;
- if (ret != 0) return ret;
- }
-
- /*
- * Column-wise set elimination.
- */
- for (x = 0; x < o; x++) {
- ret = latin_solver_set(solver, scratch, cubepos(x,0,1), o, 1
+ );
+ if (ret != 0) return ret;
+ }
+ /*
+ * Column-wise set elimination.
+ */
+ for (x = 0; x < o; x++) {
+ ret = latin_solver_set(solver, scratch, cubepos(x,0,1), o, 1
#ifdef STANDALONE_SOLVER
- , "set elimination, column %d", x
+ , "set elimination, column %d", x
#endif
- );
- if (ret > 0) *extreme = 0;
- if (ret != 0) return ret;
- }
-
- /*
- * Row-vs-column set elimination on a single number.
- */
- for (n = 1; n <= o; n++) {
- ret = latin_solver_set(solver, scratch, cubepos(0,0,n), o*o, o
+ );
+ if (ret != 0) return ret;
+ }
+ } else {
+ /*
+ * Row-vs-column set elimination on a single number
+ * (much tricker for a human to do!)
+ */
+ for (n = 1; n <= o; n++) {
+ ret = latin_solver_set(solver, scratch, cubepos(0,0,n), o*o, o
#ifdef STANDALONE_SOLVER
- , "positional set elimination, number %d", n
+ , "positional set elimination, number %d", n
#endif
- );
- if (ret > 0) *extreme = 1;
- if (ret != 0) return ret;
+ );
+ if (ret != 0) return ret;
+ }
}
return 0;
}
-/* This uses our own diff_* internally, but doesn't require callers
- * to; this is so it can be used by games that want to rewrite
- * the solver so as to use a different set of difficulties.
- *
- * It returns:
+/*
+ * Returns:
* 0 for 'didn't do anything' implying it was already solved.
* -1 for 'impossible' (no solution)
* 1 for 'single solution'
*
* and this function may well assert if given an impossible board.
*/
-int latin_solver_recurse(struct latin_solver *solver, int recdiff,
- latin_solver_callback cb, void *ctx)
+static int latin_solver_recurse
+ (struct latin_solver *solver, int diff_simple, int diff_set_0,
+ int diff_set_1, int diff_forcing, int diff_recursive,
+ usersolver_t const *usersolvers, void *ctx,
+ ctxnew_t ctxnew, ctxfree_t ctxfree)
{
int best, bestcount;
int o = solver->o, x, y, n;
*/
for (i = 0; i < j; i++) {
int ret;
+ void *newctx;
memcpy(outgrid, ingrid, o*o);
outgrid[y*o+x] = list[i];
solver_recurse_depth++;
#endif
- ret = cb(outgrid, o, recdiff, ctx);
+ if (ctxnew) {
+ newctx = ctxnew(ctx);
+ } else {
+ newctx = ctx;
+ }
+ ret = latin_solver(outgrid, o, diff_recursive,
+ diff_simple, diff_set_0, diff_set_1,
+ diff_forcing, diff_recursive,
+ usersolvers, newctx, ctxnew, ctxfree);
+ if (ctxnew)
+ ctxfree(newctx);
#ifdef STANDALONE_SOLVER
solver_recurse_depth--;
else {
/* the recursion turned up exactly one solution */
if (diff == diff_impossible)
- diff = recdiff;
+ diff = diff_recursive;
else
diff = diff_ambiguous;
}
else if (diff == diff_ambiguous)
return 2;
else {
- assert(diff == recdiff);
+ assert(diff == diff_recursive);
return 1;
}
}
}
-enum { diff_simple = 1, diff_set, diff_extreme, diff_recursive };
-
-static int latin_solver_sub(struct latin_solver *solver, int maxdiff, void *ctx)
+int latin_solver_main(struct latin_solver *solver, int maxdiff,
+ int diff_simple, int diff_set_0, int diff_set_1,
+ int diff_forcing, int diff_recursive,
+ usersolver_t const *usersolvers, void *ctx,
+ ctxnew_t ctxnew, ctxfree_t ctxfree)
{
struct latin_solver_scratch *scratch = latin_solver_new_scratch(solver);
- int ret, diff = diff_simple, extreme;
+ int ret, diff = diff_simple;
assert(maxdiff <= diff_recursive);
/*
* not.
*/
while (1) {
- /*
- * I'd like to write `continue;' inside each of the
- * following loops, so that the solver returns here after
- * making some progress. However, I can't specify that I
- * want to continue an outer loop rather than the innermost
- * one, so I'm apologetically resorting to a goto.
- */
- cont:
- latin_solver_debug(solver->cube, solver->o);
-
- ret = latin_solver_diff_simple(solver);
- if (ret < 0) {
- diff = diff_impossible;
- goto got_result;
- } else if (ret > 0) {
- diff = max(diff, diff_simple);
- goto cont;
- }
-
- if (maxdiff <= diff_simple)
- break;
+ int i;
- ret = latin_solver_diff_set(solver, scratch, &extreme);
- if (ret < 0) {
- diff = diff_impossible;
- goto got_result;
- } else if (ret > 0) {
- diff = max(diff, extreme ? diff_extreme : diff_set);
- goto cont;
- }
+ cont:
- if (maxdiff <= diff_set)
- break;
+ latin_solver_debug(solver->cube, solver->o);
- /*
- * Forcing chains.
- */
- if (latin_solver_forcing(solver, scratch)) {
- diff = max(diff, diff_extreme);
- goto cont;
- }
+ for (i = 0; i <= maxdiff; i++) {
+ if (usersolvers[i])
+ ret = usersolvers[i](solver, ctx);
+ else
+ ret = 0;
+ if (ret == 0 && i == diff_simple)
+ ret = latin_solver_diff_simple(solver);
+ if (ret == 0 && i == diff_set_0)
+ ret = latin_solver_diff_set(solver, scratch, 0);
+ if (ret == 0 && i == diff_set_1)
+ ret = latin_solver_diff_set(solver, scratch, 1);
+ if (ret == 0 && i == diff_forcing)
+ ret = latin_solver_forcing(solver, scratch);
+
+ if (ret < 0) {
+ diff = diff_impossible;
+ goto got_result;
+ } else if (ret > 0) {
+ diff = max(diff, i);
+ goto cont;
+ }
+ }
/*
* If we reach here, we have made no deductions in this
* possible.
*/
if (maxdiff == diff_recursive) {
- int nsol = latin_solver_recurse(solver, diff_recursive, latin_solver, ctx);
+ int nsol = latin_solver_recurse(solver,
+ diff_simple, diff_set_0, diff_set_1,
+ diff_forcing, diff_recursive,
+ usersolvers, ctx, ctxnew, ctxfree);
if (nsol < 0) diff = diff_impossible;
else if (nsol == 1) diff = diff_recursive;
else if (nsol > 1) diff = diff_ambiguous;
return diff;
}
-int latin_solver(digit *grid, int o, int maxdiff, void *ctx)
+int latin_solver(digit *grid, int o, int maxdiff,
+ int diff_simple, int diff_set_0, int diff_set_1,
+ int diff_forcing, int diff_recursive,
+ usersolver_t const *usersolvers, void *ctx,
+ ctxnew_t ctxnew, ctxfree_t ctxfree)
{
struct latin_solver solver;
int diff;
latin_solver_alloc(&solver, grid, o);
- diff = latin_solver_sub(&solver, maxdiff, ctx);
+ diff = latin_solver_main(&solver, maxdiff,
+ diff_simple, diff_set_0, diff_set_1,
+ diff_forcing, diff_recursive,
+ usersolvers, ctx, ctxnew, ctxfree);
latin_solver_free(&solver);
return diff;
}
void latin_solver_debug(unsigned char *cube, int o)
{
#ifdef STANDALONE_SOLVER
- if (solver_show_working) {
+ if (solver_show_working > 1) {
struct latin_solver ls, *solver = &ls;
- unsigned char *dbg;
+ char *dbg;
int x, y, i, c = 0;
ls.cube = cube; ls.o = o; /* for cube() to work */
- dbg = snewn(3*o*o*o, unsigned char);
+ dbg = snewn(3*o*o*o, char);
for (y = 0; y < o; y++) {
for (x = 0; x < o; x++) {
for (i = 1; i <= o; i++) {
for (j = 0; j < o; j++)
col[j] = num[j] = j;
shuffle(col, j, sizeof(*col), rs);
+ shuffle(num, j, sizeof(*num), rs);
/* We need the num permutation in both forward and inverse forms. */
for (j = 0; j < o; j++)
numinv[num[j]] = j;
tree234 *dict = newtree234(latin_check_cmp);
int c, r;
int ret = 0;
- lcparams *lcp, lc;
+ lcparams *lcp, lc, *aret;
/* Use a tree234 as a simple hash table, go through the square
* adding elements as we go or incrementing their counts. */
lcp = snew(lcparams);
lcp->elt = ELT(sq, c, r);
lcp->count = 1;
- assert(add234(dict, lcp) == lcp);
+ aret = add234(dict, lcp);
+ assert(aret == lcp);
} else {
lcp->count++;
}
~mdw / sgt / puzzles / blobdiff