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_NEIGHBOUR,
+enum { DIFF_BLOCK, DIFF_SIMPLE, DIFF_INTERSECT, DIFF_SET, DIFF_EXTREME,
DIFF_RECURSIVE, DIFF_AMBIGUOUS, DIFF_IMPOSSIBLE };
enum {
{ "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 Extreme", { 3, 3, SYMM_ROT2, DIFF_EXTREME } },
{ "3x3 Unreasonable", { 3, 3, SYMM_ROT2, DIFF_RECURSIVE } },
#ifndef SLOW_SYSTEM
{ "3x4 Basic", { 3, 4, SYMM_ROT2, DIFF_SIMPLE } },
else if (*string == 'a') /* advanced */
string++, ret->diff = DIFF_SET;
else if (*string == 'e') /* extreme */
- string++, ret->diff = DIFF_NEIGHBOUR;
+ string++, ret->diff = DIFF_EXTREME;
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_EXTREME: strcat(str, "de"); break;
case DIFF_RECURSIVE: strcat(str, "du"); break;
}
}
struct solver_scratch {
unsigned char *grid, *rowidx, *colidx, *set;
- int *mne;
+ int *neighbours, *bfsqueue;
+#ifdef STANDALONE_SOLVER
+ int *bfsprev;
+#endif
};
static int solver_set(struct solver_usage *usage,
int nnb, count;
int i, j, n, nbi;
- nb[0] = scratch->mne;
- nb[1] = scratch->mne + cr;
+ nb[0] = scratch->neighbours;
+ nb[1] = scratch->neighbours + cr;
/*
* First, work out the mutual neighbour squares of the two. We
return 0; /* nothing found */
}
+/*
+ * Look for forcing chains. A forcing chain is a path of
+ * pairwise-exclusive squares (i.e. each pair of adjacent squares
+ * in the path are in the same row, column or block) with the
+ * following properties:
+ *
+ * (a) Each square on the path has precisely two possible numbers.
+ *
+ * (b) Each pair of squares which are adjacent on the path share
+ * at least one possible number in common.
+ *
+ * (c) Each square in the middle of the path shares _both_ of its
+ * numbers with at least one of its neighbours (not the same
+ * one with both neighbours).
+ *
+ * These together imply that at least one of the possible number
+ * choices at one end of the path forces _all_ the rest of the
+ * numbers along the path. In order to make real use of this, we
+ * need further properties:
+ *
+ * (c) Ruling out some number N from the square at one end
+ * of the path forces the square at the other end to
+ * take number N.
+ *
+ * (d) The two end squares are both in line with some third
+ * square.
+ *
+ * (e) That third square currently has N as a possibility.
+ *
+ * If we can find all of that lot, we can deduce that at least one
+ * of the two ends of the forcing chain has number N, and that
+ * therefore the mutually adjacent third square does not.
+ *
+ * To find forcing chains, we're going to start a bfs at each
+ * suitable square, once for each of its two possible numbers.
+ */
+static int solver_forcing(struct solver_usage *usage,
+ struct solver_scratch *scratch)
+{
+ int c = usage->c, r = usage->r, cr = c*r;
+ int *bfsqueue = scratch->bfsqueue;
+#ifdef STANDALONE_SOLVER
+ int *bfsprev = scratch->bfsprev;
+#endif
+ unsigned char *number = scratch->grid;
+ int *neighbours = scratch->neighbours;
+ int x, y;
+
+ for (y = 0; y < cr; y++)
+ for (x = 0; x < cr; x++) {
+ int count, t, n;
+
+ /*
+ * If this square doesn't have exactly two candidate
+ * numbers, don't try it.
+ *
+ * In this loop we also sum the candidate numbers,
+ * which is a nasty hack to allow us to quickly find
+ * `the other one' (since we will shortly know there
+ * are exactly two).
+ */
+ for (count = t = 0, n = 1; n <= cr; n++)
+ if (cube(x, y, n))
+ count++, t += n;
+ if (count != 2)
+ continue;
+
+ /*
+ * Now attempt a bfs for each candidate.
+ */
+ for (n = 1; n <= cr; n++)
+ if (cube(x, y, n)) {
+ int orign, currn, head, tail;
+
+ /*
+ * Begin a bfs.
+ */
+ orign = n;
+
+ memset(number, cr+1, cr*cr);
+ head = tail = 0;
+ bfsqueue[tail++] = y*cr+x;
+#ifdef STANDALONE_SOLVER
+ bfsprev[y*cr+x] = -1;
+#endif
+ number[y*cr+x] = t - n;
+
+ while (head < tail) {
+ int xx, yy, nneighbours, xt, yt, xblk, i;
+
+ xx = bfsqueue[head++];
+ yy = xx / cr;
+ xx %= cr;
+
+ currn = number[yy*cr+xx];
+
+ /*
+ * Find neighbours of yy,xx.
+ */
+ nneighbours = 0;
+ for (yt = 0; yt < cr; yt++)
+ neighbours[nneighbours++] = yt*cr+xx;
+ for (xt = 0; xt < cr; xt++)
+ neighbours[nneighbours++] = yy*cr+xt;
+ xblk = xx - (xx % r);
+ for (yt = yy % r; yt < cr; yt += r)
+ for (xt = xblk; xt < xblk+r; xt++)
+ neighbours[nneighbours++] = yt*cr+xt;
+
+ /*
+ * Try visiting each of those neighbours.
+ */
+ for (i = 0; i < nneighbours; i++) {
+ int cc, tt, nn;
+
+ xt = neighbours[i] % cr;
+ yt = neighbours[i] / cr;
+
+ /*
+ * We need this square to not be
+ * already visited, and to include
+ * currn as a possible number.
+ */
+ if (number[yt*cr+xt] <= cr)
+ continue;
+ if (!cube(xt, yt, currn))
+ continue;
+
+ /*
+ * Don't visit _this_ square a second
+ * time!
+ */
+ if (xt == xx && yt == yy)
+ continue;
+
+ /*
+ * To continue with the bfs, we need
+ * this square to have exactly two
+ * possible numbers.
+ */
+ for (cc = tt = 0, nn = 1; nn <= cr; nn++)
+ if (cube(xt, yt, nn))
+ cc++, tt += nn;
+ if (cc == 2) {
+ bfsqueue[tail++] = yt*cr+xt;
+#ifdef STANDALONE_SOLVER
+ bfsprev[yt*cr+xt] = yy*cr+xx;
+#endif
+ number[yt*cr+xt] = tt - currn;
+ }
+
+ /*
+ * One other possibility is that this
+ * might be the square in which we can
+ * make a real deduction: if it's
+ * adjacent to x,y, and currn is equal
+ * to the original number we ruled out.
+ */
+ if (currn == orign &&
+ (xt == x || yt == y ||
+ (xt / r == x / r && yt % r == y % r))) {
+#ifdef STANDALONE_SOLVER
+ if (solver_show_working) {
+ char *sep = "";
+ int xl, yl;
+ printf("%*sforcing chain, %d at ends of ",
+ solver_recurse_depth*4, "", orign);
+ xl = xx;
+ yl = yy;
+ while (1) {
+ printf("%s(%d,%d)", sep, 1+xl,
+ 1+YUNTRANS(yl));
+ xl = bfsprev[yl*cr+xl];
+ if (xl < 0)
+ break;
+ yl = xl / cr;
+ xl %= cr;
+ sep = "-";
+ }
+ printf("\n%*s ruling out %d at (%d,%d)\n",
+ solver_recurse_depth*4, "",
+ orign, 1+xt, 1+YUNTRANS(yt));
+ }
+#endif
+ cube(xt, yt, orign) = FALSE;
+ return 1;
+ }
+ }
+ }
+ }
+ }
+
+ return 0;
+}
+
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);
+ scratch->neighbours = snewn(3*cr, int);
+ scratch->bfsqueue = snewn(cr*cr, int);
+#ifdef STANDALONE_SOLVER
+ scratch->bfsprev = snewn(cr*cr, int);
+#endif
return scratch;
}
static void solver_free_scratch(struct solver_scratch *scratch)
{
- sfree(scratch->mne);
+#ifdef STANDALONE_SOLVER
+ sfree(scratch->bfsprev);
+#endif
+ sfree(scratch->bfsqueue);
+ sfree(scratch->neighbours);
sfree(scratch->set);
sfree(scratch->colidx);
sfree(scratch->rowidx);
}
}
+ /*
+ * Row-vs-column set elimination on a single number.
+ */
+ for (n = 1; n <= cr; n++) {
+ ret = solver_set(usage, scratch, cubepos(0,0,n), cr*cr, cr
+#ifdef STANDALONE_SOLVER
+ , "positional set elimination, number %d", n
+#endif
+ );
+ if (ret < 0) {
+ diff = DIFF_IMPOSSIBLE;
+ goto got_result;
+ } else if (ret > 0) {
+ diff = max(diff, DIFF_EXTREME);
+ goto cont;
+ }
+ }
+
/*
* Mutual neighbour elimination.
*/
!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);
+ diff = max(diff, DIFF_EXTREME);
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);
+ diff = max(diff, DIFF_EXTREME);
goto cont;
}
}
}
}
+ /*
+ * Forcing chains.
+ */
+ if (solver_forcing(usage, scratch)) {
+ diff = max(diff, DIFF_EXTREME);
+ 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_EXTREME ? "Extreme (complex non-recursive techniques 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