enum { SYMM_NONE, SYMM_ROT2, SYMM_ROT4, SYMM_REF2, SYMM_REF2D, SYMM_REF4,
SYMM_REF4D, SYMM_REF8 };
-enum { DIFF_BLOCK, DIFF_SIMPLE, DIFF_INTERSECT,
- DIFF_SET, DIFF_RECURSIVE, DIFF_AMBIGUOUS, DIFF_IMPOSSIBLE };
+enum { DIFF_BLOCK, DIFF_SIMPLE, DIFF_INTERSECT, DIFF_SET, DIFF_NEIGHBOUR,
+ DIFF_RECURSIVE, DIFF_AMBIGUOUS, DIFF_IMPOSSIBLE };
enum {
COL_BACKGROUND,
{ "3x3 Basic", { 3, 3, SYMM_ROT2, DIFF_SIMPLE } },
{ "3x3 Intermediate", { 3, 3, SYMM_ROT2, DIFF_INTERSECT } },
{ "3x3 Advanced", { 3, 3, SYMM_ROT2, DIFF_SET } },
+ { "3x3 Extreme", { 3, 3, SYMM_ROT2, DIFF_NEIGHBOUR } },
{ "3x3 Unreasonable", { 3, 3, SYMM_ROT2, DIFF_RECURSIVE } },
#ifndef SLOW_SYSTEM
{ "3x4 Basic", { 3, 4, SYMM_ROT2, DIFF_SIMPLE } },
string++, ret->diff = DIFF_INTERSECT;
else if (*string == 'a') /* advanced */
string++, ret->diff = DIFF_SET;
+ else if (*string == 'e') /* extreme */
+ string++, ret->diff = DIFF_NEIGHBOUR;
else if (*string == 'u') /* unreasonable */
string++, ret->diff = DIFF_RECURSIVE;
} else
case DIFF_SIMPLE: strcat(str, "db"); break;
case DIFF_INTERSECT: strcat(str, "di"); break;
case DIFF_SET: strcat(str, "da"); break;
+ case DIFF_NEIGHBOUR: strcat(str, "de"); break;
case DIFF_RECURSIVE: strcat(str, "du"); break;
}
}
ret[3].name = "Difficulty";
ret[3].type = C_CHOICES;
- ret[3].sval = ":Trivial:Basic:Intermediate:Advanced:Unreasonable";
+ ret[3].sval = ":Trivial:Basic:Intermediate:Advanced:Extreme:Unreasonable";
ret[3].ival = params->diff;
ret[4].name = NULL;
/* ----------------------------------------------------------------------
* Solver.
*
- * This solver is used for several purposes:
- * + to generate filled grids as the basis for new puzzles (by
- * supplying no clue squares at all)
+ * This solver is used for two purposes:
* + to check solubility of a grid as we gradually remove numbers
* from it
* + to solve an externally generated puzzle when the user selects
* the numbers' possible positions (or the spaces' possible
* contents).
*
+ * - Mutual neighbour elimination: find two squares A,B and a
+ * number N in the possible set of A, such that putting N in A
+ * would rule out enough possibilities from the mutual
+ * neighbours of A and B that there would be no possibilities
+ * left for B. Thereby rule out N in A.
+ * + The simplest case of this is if B has two possibilities
+ * (wlog {1,2}), and there are two mutual neighbours of A and
+ * B which have possibilities {1,3} and {2,3}. Thus, if A
+ * were to be 3, then those neighbours would contain 1 and 2,
+ * and hence there would be nothing left which could go in B.
+ * + There can be more complex cases of it too: if A and B are
+ * in the same column of large blocks, then they can have
+ * more than two mutual neighbours, some of which can also be
+ * neighbours of one another. Suppose, for example, that B
+ * has possibilities {1,2,3}; there's one square P in the
+ * same column as B and the same block as A, with
+ * possibilities {1,4}; and there are _two_ squares Q,R in
+ * the same column as A and the same block as B with
+ * possibilities {2,3,4}. Then if A contained 4, P would
+ * contain 1, and Q and R would have to contain 2 and 3 in
+ * _some_ order; therefore, once again, B would have no
+ * remaining possibilities.
+ *
* - Recursion. If all else fails, we pick one of the currently
* most constrained empty squares and take a random guess at its
* contents, then continue solving on that basis and see if we
struct solver_scratch {
unsigned char *grid, *rowidx, *colidx, *set;
+ int *mne;
};
static int solver_set(struct solver_usage *usage,
return 0;
}
+/*
+ * Try to find a number in the possible set of (x1,y1) which can be
+ * ruled out because it would leave no possibilities for (x2,y2).
+ */
+static int solver_mne(struct solver_usage *usage,
+ struct solver_scratch *scratch,
+ int x1, int y1, int x2, int y2)
+{
+ int c = usage->c, r = usage->r, cr = c*r;
+ int *nb[2];
+ unsigned char *set = scratch->set;
+ unsigned char *numbers = scratch->rowidx;
+ unsigned char *numbersleft = scratch->colidx;
+ int nnb, count;
+ int i, j, n, nbi;
+
+ nb[0] = scratch->mne;
+ nb[1] = scratch->mne + cr;
+
+ /*
+ * First, work out the mutual neighbour squares of the two. We
+ * can assert that they're not actually in the same block,
+ * which leaves two possibilities: they're in different block
+ * rows _and_ different block columns (thus their mutual
+ * neighbours are precisely the other two corners of the
+ * rectangle), or they're in the same row (WLOG) and different
+ * columns, in which case their mutual neighbours are the
+ * column of each block aligned with the other square.
+ *
+ * We divide the mutual neighbours into two separate subsets
+ * nb[0] and nb[1]; squares in the same subset are not only
+ * adjacent to both our key squares, but are also always
+ * adjacent to one another.
+ */
+ if (x1 / r != x2 / r && y1 % r != y2 % r) {
+ /* Corners of the rectangle. */
+ nnb = 1;
+ nb[0][0] = cubepos(x2, y1, 1);
+ nb[1][0] = cubepos(x1, y2, 1);
+ } else if (x1 / r != x2 / r) {
+ /* Same row of blocks; different blocks within that row. */
+ int x1b = x1 - (x1 % r);
+ int x2b = x2 - (x2 % r);
+
+ nnb = r;
+ for (i = 0; i < r; i++) {
+ nb[0][i] = cubepos(x2b+i, y1, 1);
+ nb[1][i] = cubepos(x1b+i, y2, 1);
+ }
+ } else {
+ /* Same column of blocks; different blocks within that column. */
+ int y1b = y1 % r;
+ int y2b = y2 % r;
+
+ assert(y1 % r != y2 % r);
+
+ nnb = c;
+ for (i = 0; i < c; i++) {
+ nb[0][i] = cubepos(x2, y1b+i*r, 1);
+ nb[1][i] = cubepos(x1, y2b+i*r, 1);
+ }
+ }
+
+ /*
+ * Right. Now loop over each possible number.
+ */
+ for (n = 1; n <= cr; n++) {
+ if (!cube(x1, y1, n))
+ continue;
+ for (j = 0; j < cr; j++)
+ numbersleft[j] = cube(x2, y2, j+1);
+
+ /*
+ * Go over every possible subset of each neighbour list,
+ * and see if its union of possible numbers minus n has the
+ * same size as the subset. If so, add the numbers in that
+ * subset to the set of things which would be ruled out
+ * from (x2,y2) if n were placed at (x1,y1).
+ */
+ memset(set, 0, nnb);
+ count = 0;
+ while (1) {
+ /*
+ * Binary increment: change the rightmost 0 to a 1, and
+ * change all 1s to the right of it to 0s.
+ */
+ i = nnb;
+ while (i > 0 && set[i-1])
+ set[--i] = 0, count--;
+ if (i > 0)
+ set[--i] = 1, count++;
+ else
+ break; /* done */
+
+ /*
+ * Examine this subset of each neighbour set.
+ */
+ for (nbi = 0; nbi < 2; nbi++) {
+ int *nbs = nb[nbi];
+
+ memset(numbers, 0, cr);
+
+ for (i = 0; i < nnb; i++)
+ if (set[i])
+ for (j = 0; j < cr; j++)
+ if (j != n-1 && usage->cube[nbs[i] + j])
+ numbers[j] = 1;
+
+ for (i = j = 0; j < cr; j++)
+ i += numbers[j];
+
+ if (i == count) {
+ /*
+ * Got one. This subset of nbs, in the absence
+ * of n, would definitely contain all the
+ * numbers listed in `numbers'. Rule them out
+ * of `numbersleft'.
+ */
+ for (j = 0; j < cr; j++)
+ if (numbers[j])
+ numbersleft[j] = 0;
+ }
+ }
+ }
+
+ /*
+ * If we've got nothing left in `numbersleft', we have a
+ * successful mutual neighbour elimination.
+ */
+ for (j = 0; j < cr; j++)
+ if (numbersleft[j])
+ break;
+
+ if (j == cr) {
+#ifdef STANDALONE_SOLVER
+ if (solver_show_working) {
+ printf("%*smutual neighbour elimination, (%d,%d) vs (%d,%d):\n",
+ solver_recurse_depth*4, "",
+ 1+x1, 1+YUNTRANS(y1), 1+x2, 1+YUNTRANS(y2));
+ printf("%*s ruling out %d at (%d,%d)\n",
+ solver_recurse_depth*4, "",
+ n, 1+x1, 1+YUNTRANS(y1));
+ }
+#endif
+ cube(x1, y1, n) = FALSE;
+ return +1;
+ }
+ }
+
+ return 0; /* nothing found */
+}
+
static struct solver_scratch *solver_new_scratch(struct solver_usage *usage)
{
struct solver_scratch *scratch = snew(struct solver_scratch);
scratch->rowidx = snewn(cr, unsigned char);
scratch->colidx = snewn(cr, unsigned char);
scratch->set = snewn(cr, unsigned char);
+ scratch->mne = snewn(2*cr, int);
return scratch;
}
static void solver_free_scratch(struct solver_scratch *scratch)
{
+ sfree(scratch->mne);
sfree(scratch->set);
sfree(scratch->colidx);
sfree(scratch->rowidx);
struct solver_usage *usage;
struct solver_scratch *scratch;
int cr = c*r;
- int x, y, n, ret;
+ int x, y, x2, y2, n, ret;
int diff = DIFF_BLOCK;
/*
}
}
+ /*
+ * Mutual neighbour elimination.
+ */
+ for (y = 0; y+1 < cr; y++) {
+ for (x = 0; x+1 < cr; x++) {
+ for (y2 = y+1; y2 < cr; y2++) {
+ for (x2 = x+1; x2 < cr; x2++) {
+ /*
+ * Can't do mutual neighbour elimination
+ * between elements of the same actual
+ * block.
+ */
+ if (x/r == x2/r && y%r == y2%r)
+ continue;
+
+ /*
+ * Otherwise, try (x,y) vs (x2,y2) in both
+ * directions, and likewise (x2,y) vs
+ * (x,y2).
+ */
+ if (!usage->grid[YUNTRANS(y)*cr+x] &&
+ !usage->grid[YUNTRANS(y2)*cr+x2] &&
+ (solver_mne(usage, scratch, x, y, x2, y2) ||
+ solver_mne(usage, scratch, x2, y2, x, y))) {
+ diff = max(diff, DIFF_NEIGHBOUR);
+ goto cont;
+ }
+ if (!usage->grid[YUNTRANS(y)*cr+x2] &&
+ !usage->grid[YUNTRANS(y2)*cr+x] &&
+ (solver_mne(usage, scratch, x2, y, x, y2) ||
+ solver_mne(usage, scratch, x, y2, x2, y))) {
+ diff = max(diff, DIFF_NEIGHBOUR);
+ goto cont;
+ }
+ }
+ }
+ }
+ }
+
/*
* If we reach here, we have made no deductions in this
* iteration, so the algorithm terminates.
ret==DIFF_SIMPLE ? "Basic (row/column/number elimination required)":
ret==DIFF_INTERSECT ? "Intermediate (intersectional analysis required)":
ret==DIFF_SET ? "Advanced (set elimination required)":
+ ret==DIFF_NEIGHBOUR ? "Extreme (mutual neighbour elimination required)":
ret==DIFF_RECURSIVE ? "Unreasonable (guesswork and backtracking required)":
ret==DIFF_AMBIGUOUS ? "Ambiguous (multiple solutions exist)":
ret==DIFF_IMPOSSIBLE ? "Impossible (no solution exists)":
~ian / sgt-puzzles.git / commitdiff