#ifdef STANDALONE_SOLVER
#include <stdarg.h>
-int solver_show_working;
+int solver_show_working, solver_recurse_depth;
#endif
#include "puzzles.h"
return ret;
}
-static char *validate_params(game_params *params)
+static char *validate_params(game_params *params, int full)
{
if (params->c < 2 || params->r < 2)
return "Both dimensions must be at least 2";
}
/* ----------------------------------------------------------------------
- * Full recursive Solo solver.
- *
- * The algorithm for this solver is shamelessly copied from a
- * Python solver written by Andrew Wilkinson (which is GPLed, but
- * I've reused only ideas and no code). It mostly just does the
- * obvious recursive thing: pick an empty square, put one of the
- * possible digits in it, recurse until all squares are filled,
- * backtrack and change some choices if necessary.
- *
- * The clever bit is that every time it chooses which square to
- * fill in next, it does so by counting the number of _possible_
- * numbers that can go in each square, and it prioritises so that
- * it picks a square with the _lowest_ number of possibilities. The
- * idea is that filling in lots of the obvious bits (particularly
- * any squares with only one possibility) will cut down on the list
- * of possibilities for other squares and hence reduce the enormous
- * search space as much as possible as early as possible.
- *
- * In practice the algorithm appeared to work very well; run on
- * sample problems from the Times it completed in well under a
- * second on my G5 even when written in Python, and given an empty
- * grid (so that in principle it would enumerate _all_ solved
- * grids!) it found the first valid solution just as quickly. So
- * with a bit more randomisation I see no reason not to use this as
- * my grid generator.
- */
-
-/*
- * Internal data structure used in solver to keep track of
- * progress.
- */
-struct rsolve_coord { int x, y, r; };
-struct rsolve_usage {
- int c, r, cr; /* cr == c*r */
- /* grid is a copy of the input grid, modified as we go along */
- digit *grid;
- /* row[y*cr+n-1] TRUE if digit n has been placed in row y */
- unsigned char *row;
- /* col[x*cr+n-1] TRUE if digit n has been placed in row x */
- unsigned char *col;
- /* blk[(y*c+x)*cr+n-1] TRUE if digit n has been placed in block (x,y) */
- unsigned char *blk;
- /* This lists all the empty spaces remaining in the grid. */
- struct rsolve_coord *spaces;
- int nspaces;
- /* If we need randomisation in the solve, this is our random state. */
- random_state *rs;
- /* Number of solutions so far found, and maximum number we care about. */
- int solns, maxsolns;
-};
-
-/*
- * The real recursive step in the solving function.
- */
-static void rsolve_real(struct rsolve_usage *usage, digit *grid)
-{
- int c = usage->c, r = usage->r, cr = usage->cr;
- int i, j, n, sx, sy, bestm, bestr;
- int *digits;
-
- /*
- * Firstly, check for completion! If there are no spaces left
- * in the grid, we have a solution.
- */
- if (usage->nspaces == 0) {
- if (!usage->solns) {
- /*
- * This is our first solution, so fill in the output grid.
- */
- memcpy(grid, usage->grid, cr * cr);
- }
- usage->solns++;
- return;
- }
-
- /*
- * Otherwise, there must be at least one space. Find the most
- * constrained space, using the `r' field as a tie-breaker.
- */
- bestm = cr+1; /* so that any space will beat it */
- bestr = 0;
- i = sx = sy = -1;
- for (j = 0; j < usage->nspaces; j++) {
- int x = usage->spaces[j].x, y = usage->spaces[j].y;
- int m;
-
- /*
- * Find the number of digits that could go in this space.
- */
- m = 0;
- for (n = 0; n < cr; n++)
- if (!usage->row[y*cr+n] && !usage->col[x*cr+n] &&
- !usage->blk[((y/c)*c+(x/r))*cr+n])
- m++;
-
- if (m < bestm || (m == bestm && usage->spaces[j].r < bestr)) {
- bestm = m;
- bestr = usage->spaces[j].r;
- sx = x;
- sy = y;
- i = j;
- }
- }
-
- /*
- * Swap that square into the final place in the spaces array,
- * so that decrementing nspaces will remove it from the list.
- */
- if (i != usage->nspaces-1) {
- struct rsolve_coord t;
- t = usage->spaces[usage->nspaces-1];
- usage->spaces[usage->nspaces-1] = usage->spaces[i];
- usage->spaces[i] = t;
- }
-
- /*
- * Now we've decided which square to start our recursion at,
- * simply go through all possible values, shuffling them
- * randomly first if necessary.
- */
- digits = snewn(bestm, int);
- j = 0;
- for (n = 0; n < cr; n++)
- if (!usage->row[sy*cr+n] && !usage->col[sx*cr+n] &&
- !usage->blk[((sy/c)*c+(sx/r))*cr+n]) {
- digits[j++] = n+1;
- }
-
- if (usage->rs) {
- /* shuffle */
- for (i = j; i > 1; i--) {
- int p = random_upto(usage->rs, i);
- if (p != i-1) {
- int t = digits[p];
- digits[p] = digits[i-1];
- digits[i-1] = t;
- }
- }
- }
-
- /* And finally, go through the digit list and actually recurse. */
- for (i = 0; i < j; i++) {
- n = digits[i];
-
- /* Update the usage structure to reflect the placing of this digit. */
- usage->row[sy*cr+n-1] = usage->col[sx*cr+n-1] =
- usage->blk[((sy/c)*c+(sx/r))*cr+n-1] = TRUE;
- usage->grid[sy*cr+sx] = n;
- usage->nspaces--;
-
- /* Call the solver recursively. */
- rsolve_real(usage, grid);
-
- /*
- * If we have seen as many solutions as we need, terminate
- * all processing immediately.
- */
- if (usage->solns >= usage->maxsolns)
- break;
-
- /* Revert the usage structure. */
- usage->row[sy*cr+n-1] = usage->col[sx*cr+n-1] =
- usage->blk[((sy/c)*c+(sx/r))*cr+n-1] = FALSE;
- usage->grid[sy*cr+sx] = 0;
- usage->nspaces++;
- }
-
- sfree(digits);
-}
-
-/*
- * Entry point to solver. You give it dimensions and a starting
- * grid, which is simply an array of N^4 digits. In that array, 0
- * means an empty square, and 1..N mean a clue square.
- *
- * Return value is the number of solutions found; searching will
- * stop after the provided `max'. (Thus, you can pass max==1 to
- * indicate that you only care about finding _one_ solution, or
- * max==2 to indicate that you want to know the difference between
- * a unique and non-unique solution.) The input parameter `grid' is
- * also filled in with the _first_ (or only) solution found by the
- * solver.
- */
-static int rsolve(int c, int r, digit *grid, random_state *rs, int max)
-{
- struct rsolve_usage *usage;
- int x, y, cr = c*r;
- int ret;
-
- /*
- * Create an rsolve_usage structure.
- */
- usage = snew(struct rsolve_usage);
-
- usage->c = c;
- usage->r = r;
- usage->cr = cr;
-
- usage->grid = snewn(cr * cr, digit);
- memcpy(usage->grid, grid, cr * cr);
-
- usage->row = snewn(cr * cr, unsigned char);
- usage->col = snewn(cr * cr, unsigned char);
- usage->blk = snewn(cr * cr, unsigned char);
- memset(usage->row, FALSE, cr * cr);
- memset(usage->col, FALSE, cr * cr);
- memset(usage->blk, FALSE, cr * cr);
-
- usage->spaces = snewn(cr * cr, struct rsolve_coord);
- usage->nspaces = 0;
-
- usage->solns = 0;
- usage->maxsolns = max;
-
- usage->rs = rs;
-
- /*
- * Now fill it in with data from the input grid.
- */
- for (y = 0; y < cr; y++) {
- for (x = 0; x < cr; x++) {
- int v = grid[y*cr+x];
- if (v == 0) {
- usage->spaces[usage->nspaces].x = x;
- usage->spaces[usage->nspaces].y = y;
- if (rs)
- usage->spaces[usage->nspaces].r = random_bits(rs, 31);
- else
- usage->spaces[usage->nspaces].r = usage->nspaces;
- usage->nspaces++;
- } else {
- usage->row[y*cr+v-1] = TRUE;
- usage->col[x*cr+v-1] = TRUE;
- usage->blk[((y/c)*c+(x/r))*cr+v-1] = TRUE;
- }
- }
- }
-
- /*
- * Run the real recursive solving function.
- */
- rsolve_real(usage, grid);
- ret = usage->solns;
-
- /*
- * Clean up the usage structure now we have our answer.
- */
- sfree(usage->spaces);
- sfree(usage->blk);
- sfree(usage->col);
- sfree(usage->row);
- sfree(usage->grid);
- sfree(usage);
-
- /*
- * And return.
- */
- return ret;
-}
-
-/* ----------------------------------------------------------------------
- * End of recursive solver code.
- */
-
-/* ----------------------------------------------------------------------
- * Less capable non-recursive solver. This one is used to check
- * solubility of a grid as we gradually remove numbers from it: by
- * verifying a grid using this solver we can ensure it isn't _too_
- * hard (e.g. does not actually require guessing and backtracking).
- *
+ * 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)
+ * + to check solubility of a grid as we gradually remove numbers
+ * from it
+ * + to solve an externally generated puzzle when the user selects
+ * `Solve'.
+ *
* It supports a variety of specific modes of reasoning. By
* enabling or disabling subsets of these modes we can arrange a
* range of difficulty levels.
* places, found by taking the _complement_ of the union of
* the numbers' possible positions (or the spaces' possible
* contents).
+ *
+ * - 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
+ * get any further.
*/
/*
#define YTRANS(y) (((y)%c)*r+(y)/c)
#define YUNTRANS(y) (((y)%r)*c+(y)/r)
-struct nsolve_usage {
+struct solver_usage {
int c, r, cr;
/*
* We set up a cubic array, indexed by x, y and digit; each
* a particular number in it. The y-coordinate passed in here is
* transformed.
*/
-static void nsolve_place(struct nsolve_usage *usage, int x, int y, int n)
+static void solver_place(struct solver_usage *usage, int x, int y, int n)
{
int c = usage->c, r = usage->r, cr = usage->cr;
int i, j, bx, by;
usage->blk[((y%r)*c+(x/r))*cr+n-1] = TRUE;
}
-static int nsolve_elim(struct nsolve_usage *usage, int start, int step
+static int solver_elim(struct solver_usage *usage, int start, int step
#ifdef STANDALONE_SOLVER
, char *fmt, ...
#endif
#ifdef STANDALONE_SOLVER
if (solver_show_working) {
va_list ap;
+ printf("%*s", solver_recurse_depth*4, "");
va_start(ap, fmt);
vprintf(fmt, ap);
va_end(ap);
- printf(":\n placing %d at (%d,%d)\n",
- n, 1+x, 1+YUNTRANS(y));
+ printf(":\n%*s placing %d at (%d,%d)\n",
+ solver_recurse_depth*4, "", n, 1+x, 1+YUNTRANS(y));
}
#endif
- nsolve_place(usage, x, y, n);
- return TRUE;
+ solver_place(usage, x, y, n);
+ return +1;
}
+ } else if (m == 0) {
+#ifdef STANDALONE_SOLVER
+ if (solver_show_working) {
+ va_list ap;
+ printf("%*s", solver_recurse_depth*4, "");
+ va_start(ap, fmt);
+ vprintf(fmt, ap);
+ va_end(ap);
+ printf(":\n%*s no possibilities available\n",
+ solver_recurse_depth*4, "");
+ }
+#endif
+ return -1;
}
- return FALSE;
+ return 0;
}
-static int nsolve_intersect(struct nsolve_usage *usage,
+static int solver_intersect(struct solver_usage *usage,
int start1, int step1, int start2, int step2
#ifdef STANDALONE_SOLVER
, char *fmt, ...
if (usage->cube[p] &&
!(p >= start2 && p < start2+cr*step2 &&
(p - start2) % step2 == 0))
- return FALSE; /* there is, so we can't deduce */
+ return 0; /* there is, so we can't deduce */
}
/*
* We have determined that all set bits in the first domain are
* within its overlap with the second. So loop over the second
* domain and remove all set bits that aren't also in that
- * overlap; return TRUE iff we actually _did_ anything.
+ * overlap; return +1 iff we actually _did_ anything.
*/
- ret = FALSE;
+ ret = 0;
for (i = 0; i < cr; i++) {
int p = start2+i*step2;
if (usage->cube[p] &&
if (!ret) {
va_list ap;
+ printf("%*s", solver_recurse_depth*4, "");
va_start(ap, fmt);
vprintf(fmt, ap);
va_end(ap);
px = py / cr;
py %= cr;
- printf(" ruling out %d at (%d,%d)\n",
- pn, 1+px, 1+YUNTRANS(py));
+ printf("%*s ruling out %d at (%d,%d)\n",
+ solver_recurse_depth*4, "", pn, 1+px, 1+YUNTRANS(py));
}
#endif
- ret = TRUE; /* we did something */
+ ret = +1; /* we did something */
usage->cube[p] = 0;
}
}
return ret;
}
-struct nsolve_scratch {
+struct solver_scratch {
unsigned char *grid, *rowidx, *colidx, *set;
};
-static int nsolve_set(struct nsolve_usage *usage,
- struct nsolve_scratch *scratch,
+static int solver_set(struct solver_usage *usage,
+ struct solver_scratch *scratch,
int start, int step1, int step2
#ifdef STANDALONE_SOLVER
, char *fmt, ...
for (j = 0; j < cr; j++)
if (usage->cube[start+i*step1+j*step2])
first = j, count++;
- if (count == 0) {
- /*
- * This condition actually marks a completely insoluble
- * (i.e. internally inconsistent) puzzle. We return and
- * report no progress made.
- */
- return FALSE;
- }
+
+ /*
+ * If count == 0, then there's a row with no 1s at all and
+ * the puzzle is internally inconsistent. However, we ought
+ * to have caught this already during the simpler reasoning
+ * methods, so we can safely fail an assertion if we reach
+ * this point here.
+ */
+ assert(count > 0);
if (count == 1)
rowidx[i] = colidx[first] = FALSE;
}
* indicates a faulty deduction before this point or
* even a bogus clue.
*/
- assert(rows <= n - count);
+ if (rows > n - count) {
+#ifdef STANDALONE_SOLVER
+ if (solver_show_working) {
+ va_list ap;
+ printf("%*s", solver_recurse_depth*4,
+ "");
+ va_start(ap, fmt);
+ vprintf(fmt, ap);
+ va_end(ap);
+ printf(":\n%*s contradiction reached\n",
+ solver_recurse_depth*4, "");
+ }
+#endif
+ return -1;
+ }
+
if (rows >= n - count) {
int progress = FALSE;
* We've got one! Now, for each row which _doesn't_
* satisfy the criterion, eliminate all its set
* bits in the positions _not_ listed in `set'.
- * Return TRUE (meaning progress has been made) if
- * we successfully eliminated anything at all.
+ * Return +1 (meaning progress has been made) if we
+ * successfully eliminated anything at all.
*
* This involves referring back through
* rowidx/colidx in order to work out which actual
#ifdef STANDALONE_SOLVER
if (solver_show_working) {
int px, py, pn;
-
+
if (!progress) {
va_list ap;
+ printf("%*s", solver_recurse_depth*4,
+ "");
va_start(ap, fmt);
vprintf(fmt, ap);
va_end(ap);
printf(":\n");
}
- pn = 1 + fpos % cr;
- py = fpos / cr;
- px = py / cr;
- py %= cr;
+ pn = 1 + fpos % cr;
+ py = fpos / cr;
+ px = py / cr;
+ py %= cr;
+
+ printf("%*s ruling out %d at (%d,%d)\n",
+ solver_recurse_depth*4, "",
+ pn, 1+px, 1+YUNTRANS(py));
+ }
+#endif
+ progress = TRUE;
+ usage->cube[fpos] = FALSE;
+ }
+ }
+ }
+
+ if (progress) {
+ return +1;
+ }
+ }
+ }
+
+ /*
+ * Binary increment: change the rightmost 0 to a 1, and
+ * change all 1s to the right of it to 0s.
+ */
+ i = n;
+ while (i > 0 && set[i-1])
+ set[--i] = 0, count--;
+ if (i > 0)
+ set[--i] = 1, count++;
+ else
+ break; /* done */
+ }
+
+ return 0;
+}
+
+static struct solver_scratch *solver_new_scratch(struct solver_usage *usage)
+{
+ struct solver_scratch *scratch = snew(struct solver_scratch);
+ int cr = usage->cr;
+ scratch->grid = snewn(cr*cr, unsigned char);
+ scratch->rowidx = snewn(cr, unsigned char);
+ scratch->colidx = snewn(cr, unsigned char);
+ scratch->set = snewn(cr, unsigned char);
+ return scratch;
+}
+
+static void solver_free_scratch(struct solver_scratch *scratch)
+{
+ sfree(scratch->set);
+ sfree(scratch->colidx);
+ sfree(scratch->rowidx);
+ sfree(scratch->grid);
+ sfree(scratch);
+}
+
+static int solver(int c, int r, digit *grid, random_state *rs, int maxdiff)
+{
+ struct solver_usage *usage;
+ struct solver_scratch *scratch;
+ int cr = c*r;
+ int x, y, n, ret;
+ int diff = DIFF_BLOCK;
+
+ /*
+ * Set up a usage structure as a clean slate (everything
+ * possible).
+ */
+ usage = snew(struct solver_usage);
+ usage->c = c;
+ usage->r = r;
+ usage->cr = cr;
+ usage->cube = snewn(cr*cr*cr, unsigned char);
+ usage->grid = grid; /* write straight back to the input */
+ memset(usage->cube, TRUE, cr*cr*cr);
+
+ usage->row = snewn(cr * cr, unsigned char);
+ usage->col = snewn(cr * cr, unsigned char);
+ usage->blk = snewn(cr * cr, unsigned char);
+ memset(usage->row, FALSE, cr * cr);
+ memset(usage->col, FALSE, cr * cr);
+ memset(usage->blk, FALSE, cr * cr);
+
+ scratch = solver_new_scratch(usage);
+
+ /*
+ * Place all the clue numbers we are given.
+ */
+ for (x = 0; x < cr; x++)
+ for (y = 0; y < cr; y++)
+ if (grid[y*cr+x])
+ solver_place(usage, x, YTRANS(y), grid[y*cr+x]);
+
+ /*
+ * Now loop over the grid repeatedly trying all permitted modes
+ * of reasoning. The loop terminates if we complete an
+ * iteration without making any progress; we then return
+ * failure or success depending on whether the grid is full or
+ * 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:
+
+ /*
+ * Blockwise positional elimination.
+ */
+ for (x = 0; x < cr; x += r)
+ for (y = 0; y < r; y++)
+ for (n = 1; n <= cr; n++)
+ if (!usage->blk[(y*c+(x/r))*cr+n-1]) {
+ ret = solver_elim(usage, cubepos(x,y,n), r*cr
+#ifdef STANDALONE_SOLVER
+ , "positional elimination,"
+ " %d in block (%d,%d)", n, 1+x/r, 1+y
+#endif
+ );
+ if (ret < 0) {
+ diff = DIFF_IMPOSSIBLE;
+ goto got_result;
+ } else if (ret > 0) {
+ diff = max(diff, DIFF_BLOCK);
+ goto cont;
+ }
+ }
+
+ if (maxdiff <= DIFF_BLOCK)
+ break;
+
+ /*
+ * Row-wise positional elimination.
+ */
+ for (y = 0; y < cr; y++)
+ for (n = 1; n <= cr; n++)
+ if (!usage->row[y*cr+n-1]) {
+ ret = solver_elim(usage, cubepos(0,y,n), cr*cr
+#ifdef STANDALONE_SOLVER
+ , "positional elimination,"
+ " %d in row %d", n, 1+YUNTRANS(y)
+#endif
+ );
+ if (ret < 0) {
+ diff = DIFF_IMPOSSIBLE;
+ goto got_result;
+ } else if (ret > 0) {
+ diff = max(diff, DIFF_SIMPLE);
+ goto cont;
+ }
+ }
+ /*
+ * Column-wise positional elimination.
+ */
+ for (x = 0; x < cr; x++)
+ for (n = 1; n <= cr; n++)
+ if (!usage->col[x*cr+n-1]) {
+ ret = solver_elim(usage, cubepos(x,0,n), cr
+#ifdef STANDALONE_SOLVER
+ , "positional elimination,"
+ " %d in column %d", n, 1+x
+#endif
+ );
+ if (ret < 0) {
+ diff = DIFF_IMPOSSIBLE;
+ goto got_result;
+ } else if (ret > 0) {
+ diff = max(diff, DIFF_SIMPLE);
+ goto cont;
+ }
+ }
+
+ /*
+ * Numeric elimination.
+ */
+ for (x = 0; x < cr; x++)
+ for (y = 0; y < cr; y++)
+ if (!usage->grid[YUNTRANS(y)*cr+x]) {
+ ret = solver_elim(usage, cubepos(x,y,1), 1
+#ifdef STANDALONE_SOLVER
+ , "numeric elimination at (%d,%d)", 1+x,
+ 1+YUNTRANS(y)
+#endif
+ );
+ 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;
+
+ /*
+ * Intersectional analysis, rows vs blocks.
+ */
+ for (y = 0; y < cr; y++)
+ for (x = 0; x < cr; x += r)
+ for (n = 1; n <= cr; n++)
+ /*
+ * solver_intersect() never returns -1.
+ */
+ if (!usage->row[y*cr+n-1] &&
+ !usage->blk[((y%r)*c+(x/r))*cr+n-1] &&
+ (solver_intersect(usage, cubepos(0,y,n), cr*cr,
+ cubepos(x,y%r,n), r*cr
+#ifdef STANDALONE_SOLVER
+ , "intersectional analysis,"
+ " %d in row %d vs block (%d,%d)",
+ n, 1+YUNTRANS(y), 1+x/r, 1+y%r
+#endif
+ ) ||
+ solver_intersect(usage, cubepos(x,y%r,n), r*cr,
+ cubepos(0,y,n), cr*cr
+#ifdef STANDALONE_SOLVER
+ , "intersectional analysis,"
+ " %d in block (%d,%d) vs row %d",
+ n, 1+x/r, 1+y%r, 1+YUNTRANS(y)
+#endif
+ ))) {
+ diff = max(diff, DIFF_INTERSECT);
+ goto cont;
+ }
+
+ /*
+ * Intersectional analysis, columns vs blocks.
+ */
+ for (x = 0; x < cr; x++)
+ for (y = 0; y < r; y++)
+ for (n = 1; n <= cr; n++)
+ if (!usage->col[x*cr+n-1] &&
+ !usage->blk[(y*c+(x/r))*cr+n-1] &&
+ (solver_intersect(usage, cubepos(x,0,n), cr,
+ cubepos((x/r)*r,y,n), r*cr
+#ifdef STANDALONE_SOLVER
+ , "intersectional analysis,"
+ " %d in column %d vs block (%d,%d)",
+ n, 1+x, 1+x/r, 1+y
+#endif
+ ) ||
+ solver_intersect(usage, cubepos((x/r)*r,y,n), r*cr,
+ cubepos(x,0,n), cr
+#ifdef STANDALONE_SOLVER
+ , "intersectional analysis,"
+ " %d in block (%d,%d) vs column %d",
+ n, 1+x/r, 1+y, 1+x
+#endif
+ ))) {
+ diff = max(diff, DIFF_INTERSECT);
+ goto cont;
+ }
+
+ if (maxdiff <= DIFF_INTERSECT)
+ break;
+
+ /*
+ * Blockwise set elimination.
+ */
+ for (x = 0; x < cr; x += r)
+ for (y = 0; y < r; y++) {
+ ret = solver_set(usage, scratch, cubepos(x,y,1), r*cr, 1
+#ifdef STANDALONE_SOLVER
+ , "set elimination, block (%d,%d)", 1+x/r, 1+y
+#endif
+ );
+ if (ret < 0) {
+ diff = DIFF_IMPOSSIBLE;
+ goto got_result;
+ } else if (ret > 0) {
+ diff = max(diff, DIFF_SET);
+ goto cont;
+ }
+ }
+
+ /*
+ * Row-wise set elimination.
+ */
+ for (y = 0; y < cr; y++) {
+ ret = solver_set(usage, scratch, cubepos(0,y,1), cr*cr, 1
+#ifdef STANDALONE_SOLVER
+ , "set elimination, row %d", 1+YUNTRANS(y)
+#endif
+ );
+ if (ret < 0) {
+ diff = DIFF_IMPOSSIBLE;
+ goto got_result;
+ } else if (ret > 0) {
+ diff = max(diff, DIFF_SET);
+ goto cont;
+ }
+ }
+
+ /*
+ * Column-wise set elimination.
+ */
+ for (x = 0; x < cr; x++) {
+ ret = solver_set(usage, scratch, cubepos(x,0,1), cr, 1
+#ifdef STANDALONE_SOLVER
+ , "set elimination, column %d", 1+x
+#endif
+ );
+ if (ret < 0) {
+ diff = DIFF_IMPOSSIBLE;
+ goto got_result;
+ } else if (ret > 0) {
+ diff = max(diff, DIFF_SET);
+ goto cont;
+ }
+ }
+
+ /*
+ * If we reach here, we have made no deductions in this
+ * iteration, so the algorithm terminates.
+ */
+ break;
+ }
+
+ /*
+ * Last chance: if we haven't fully solved the puzzle yet, try
+ * recursing based on guesses for a particular square. We pick
+ * one of the most constrained empty squares we can find, which
+ * has the effect of pruning the search tree as much as
+ * possible.
+ */
+ if (maxdiff >= DIFF_RECURSIVE) {
+ int best, bestcount, bestnumber;
+
+ best = -1;
+ bestcount = cr+1;
+ bestnumber = 0;
+
+ for (y = 0; y < cr; y++)
+ for (x = 0; x < cr; x++)
+ if (!grid[y*cr+x]) {
+ int count;
+
+ /*
+ * An unfilled square. Count the number of
+ * possible digits in it.
+ */
+ count = 0;
+ for (n = 1; n <= cr; n++)
+ if (cube(x,YTRANS(y),n))
+ count++;
+
+ /*
+ * We should have found any impossibilities
+ * already, so this can safely be an assert.
+ */
+ assert(count > 1);
+
+ if (count < bestcount) {
+ bestcount = count;
+ bestnumber = 0;
+ }
+
+ if (count == bestcount) {
+ bestnumber++;
+ if (bestnumber == 1 ||
+ (rs && random_upto(rs, bestnumber) == 0))
+ best = y*cr+x;
+ }
+ }
+
+ if (best != -1) {
+ int i, j;
+ digit *list, *ingrid, *outgrid;
+
+ diff = DIFF_IMPOSSIBLE; /* no solution found yet */
+
+ /*
+ * Attempt recursion.
+ */
+ y = best / cr;
+ x = best % cr;
+
+ list = snewn(cr, digit);
+ ingrid = snewn(cr * cr, digit);
+ outgrid = snewn(cr * cr, digit);
+ memcpy(ingrid, grid, cr * cr);
+
+ /* Make a list of the possible digits. */
+ for (j = 0, n = 1; n <= cr; n++)
+ if (cube(x,YTRANS(y),n))
+ list[j++] = n;
+
+#ifdef STANDALONE_SOLVER
+ if (solver_show_working) {
+ char *sep = "";
+ printf("%*srecursing on (%d,%d) [",
+ solver_recurse_depth*4, "", x, y);
+ for (i = 0; i < j; i++) {
+ printf("%s%d", sep, list[i]);
+ sep = " or ";
+ }
+ printf("]\n");
+ }
+#endif
+
+ /* Now shuffle the list. */
+ if (rs) {
+ for (i = j; i > 1; i--) {
+ int p = random_upto(rs, i);
+ if (p != i-1) {
+ int t = list[p];
+ list[p] = list[i-1];
+ list[i-1] = t;
+ }
+ }
+ }
+
+ /*
+ * And step along the list, recursing back into the
+ * main solver at every stage.
+ */
+ for (i = 0; i < j; i++) {
+ int ret;
+
+ memcpy(outgrid, ingrid, cr * cr);
+ outgrid[y*cr+x] = list[i];
+
+#ifdef STANDALONE_SOLVER
+ if (solver_show_working)
+ printf("%*sguessing %d at (%d,%d)\n",
+ solver_recurse_depth*4, "", list[i], x, y);
+ solver_recurse_depth++;
+#endif
+
+ ret = solver(c, r, outgrid, rs, maxdiff);
+
+#ifdef STANDALONE_SOLVER
+ solver_recurse_depth--;
+ if (solver_show_working) {
+ printf("%*sretracting %d at (%d,%d)\n",
+ solver_recurse_depth*4, "", list[i], x, y);
+ }
+#endif
+
+ /*
+ * If we have our first solution, copy it into the
+ * grid we will return.
+ */
+ if (diff == DIFF_IMPOSSIBLE && ret != DIFF_IMPOSSIBLE)
+ memcpy(grid, outgrid, cr*cr);
+
+ if (ret == DIFF_AMBIGUOUS)
+ diff = DIFF_AMBIGUOUS;
+ else if (ret == DIFF_IMPOSSIBLE)
+ /* do not change our return value */;
+ else {
+ /* the recursion turned up exactly one solution */
+ if (diff == DIFF_IMPOSSIBLE)
+ diff = DIFF_RECURSIVE;
+ else
+ diff = DIFF_AMBIGUOUS;
+ }
+
+ /*
+ * As soon as we've found more than one solution,
+ * give up immediately.
+ */
+ if (diff == DIFF_AMBIGUOUS)
+ break;
+ }
+
+ sfree(outgrid);
+ sfree(ingrid);
+ sfree(list);
+ }
+
+ } else {
+ /*
+ * We're forbidden to use recursion, so we just see whether
+ * our grid is fully solved, and return DIFF_IMPOSSIBLE
+ * otherwise.
+ */
+ for (y = 0; y < cr; y++)
+ for (x = 0; x < cr; x++)
+ if (!grid[y*cr+x])
+ diff = DIFF_IMPOSSIBLE;
+ }
+
+ got_result:;
+
+#ifdef STANDALONE_SOLVER
+ if (solver_show_working)
+ printf("%*s%s found\n",
+ solver_recurse_depth*4, "",
+ diff == DIFF_IMPOSSIBLE ? "no solution" :
+ diff == DIFF_AMBIGUOUS ? "multiple solutions" :
+ "one solution");
+#endif
+
+ sfree(usage->cube);
+ sfree(usage->row);
+ sfree(usage->col);
+ sfree(usage->blk);
+ sfree(usage);
+
+ solver_free_scratch(scratch);
+
+ return diff;
+}
+
+/* ----------------------------------------------------------------------
+ * End of solver code.
+ */
+
+/* ----------------------------------------------------------------------
+ * Solo filled-grid generator.
+ *
+ * This grid generator works by essentially trying to solve a grid
+ * starting from no clues, and not worrying that there's more than
+ * one possible solution. Unfortunately, it isn't computationally
+ * feasible to do this by calling the above solver with an empty
+ * grid, because that one needs to allocate a lot of scratch space
+ * at every recursion level. Instead, I have a much simpler
+ * algorithm which I shamelessly copied from a Python solver
+ * written by Andrew Wilkinson (which is GPLed, but I've reused
+ * only ideas and no code). It mostly just does the obvious
+ * recursive thing: pick an empty square, put one of the possible
+ * digits in it, recurse until all squares are filled, backtrack
+ * and change some choices if necessary.
+ *
+ * The clever bit is that every time it chooses which square to
+ * fill in next, it does so by counting the number of _possible_
+ * numbers that can go in each square, and it prioritises so that
+ * it picks a square with the _lowest_ number of possibilities. The
+ * idea is that filling in lots of the obvious bits (particularly
+ * any squares with only one possibility) will cut down on the list
+ * of possibilities for other squares and hence reduce the enormous
+ * search space as much as possible as early as possible.
+ */
+
+/*
+ * Internal data structure used in gridgen to keep track of
+ * progress.
+ */
+struct gridgen_coord { int x, y, r; };
+struct gridgen_usage {
+ int c, r, cr; /* cr == c*r */
+ /* grid is a copy of the input grid, modified as we go along */
+ digit *grid;
+ /* row[y*cr+n-1] TRUE if digit n has been placed in row y */
+ unsigned char *row;
+ /* col[x*cr+n-1] TRUE if digit n has been placed in row x */
+ unsigned char *col;
+ /* blk[(y*c+x)*cr+n-1] TRUE if digit n has been placed in block (x,y) */
+ unsigned char *blk;
+ /* This lists all the empty spaces remaining in the grid. */
+ struct gridgen_coord *spaces;
+ int nspaces;
+ /* If we need randomisation in the solve, this is our random state. */
+ random_state *rs;
+};
+
+/*
+ * The real recursive step in the generating function.
+ */
+static int gridgen_real(struct gridgen_usage *usage, digit *grid)
+{
+ int c = usage->c, r = usage->r, cr = usage->cr;
+ int i, j, n, sx, sy, bestm, bestr, ret;
+ int *digits;
+
+ /*
+ * Firstly, check for completion! If there are no spaces left
+ * in the grid, we have a solution.
+ */
+ if (usage->nspaces == 0) {
+ memcpy(grid, usage->grid, cr * cr);
+ return TRUE;
+ }
+
+ /*
+ * Otherwise, there must be at least one space. Find the most
+ * constrained space, using the `r' field as a tie-breaker.
+ */
+ bestm = cr+1; /* so that any space will beat it */
+ bestr = 0;
+ i = sx = sy = -1;
+ for (j = 0; j < usage->nspaces; j++) {
+ int x = usage->spaces[j].x, y = usage->spaces[j].y;
+ int m;
+
+ /*
+ * Find the number of digits that could go in this space.
+ */
+ m = 0;
+ for (n = 0; n < cr; n++)
+ if (!usage->row[y*cr+n] && !usage->col[x*cr+n] &&
+ !usage->blk[((y/c)*c+(x/r))*cr+n])
+ m++;
+
+ if (m < bestm || (m == bestm && usage->spaces[j].r < bestr)) {
+ bestm = m;
+ bestr = usage->spaces[j].r;
+ sx = x;
+ sy = y;
+ i = j;
+ }
+ }
+
+ /*
+ * Swap that square into the final place in the spaces array,
+ * so that decrementing nspaces will remove it from the list.
+ */
+ if (i != usage->nspaces-1) {
+ struct gridgen_coord t;
+ t = usage->spaces[usage->nspaces-1];
+ usage->spaces[usage->nspaces-1] = usage->spaces[i];
+ usage->spaces[i] = t;
+ }
+
+ /*
+ * Now we've decided which square to start our recursion at,
+ * simply go through all possible values, shuffling them
+ * randomly first if necessary.
+ */
+ digits = snewn(bestm, int);
+ j = 0;
+ for (n = 0; n < cr; n++)
+ if (!usage->row[sy*cr+n] && !usage->col[sx*cr+n] &&
+ !usage->blk[((sy/c)*c+(sx/r))*cr+n]) {
+ digits[j++] = n+1;
+ }
+
+ if (usage->rs) {
+ /* shuffle */
+ for (i = j; i > 1; i--) {
+ int p = random_upto(usage->rs, i);
+ if (p != i-1) {
+ int t = digits[p];
+ digits[p] = digits[i-1];
+ digits[i-1] = t;
+ }
+ }
+ }
+
+ /* And finally, go through the digit list and actually recurse. */
+ ret = FALSE;
+ for (i = 0; i < j; i++) {
+ n = digits[i];
- printf(" ruling out %d at (%d,%d)\n",
- pn, 1+px, 1+YUNTRANS(py));
- }
-#endif
- progress = TRUE;
- usage->cube[fpos] = FALSE;
- }
- }
- }
+ /* Update the usage structure to reflect the placing of this digit. */
+ usage->row[sy*cr+n-1] = usage->col[sx*cr+n-1] =
+ usage->blk[((sy/c)*c+(sx/r))*cr+n-1] = TRUE;
+ usage->grid[sy*cr+sx] = n;
+ usage->nspaces--;
- if (progress) {
- return TRUE;
- }
- }
- }
+ /* Call the solver recursively. Stop when we find a solution. */
+ if (gridgen_real(usage, grid))
+ ret = TRUE;
- /*
- * Binary increment: change the rightmost 0 to a 1, and
- * change all 1s to the right of it to 0s.
- */
- i = n;
- while (i > 0 && set[i-1])
- set[--i] = 0, count--;
- if (i > 0)
- set[--i] = 1, count++;
- else
- break; /* done */
- }
+ /* Revert the usage structure. */
+ usage->row[sy*cr+n-1] = usage->col[sx*cr+n-1] =
+ usage->blk[((sy/c)*c+(sx/r))*cr+n-1] = FALSE;
+ usage->grid[sy*cr+sx] = 0;
+ usage->nspaces++;
- return FALSE;
-}
+ if (ret)
+ break;
+ }
-static struct nsolve_scratch *nsolve_new_scratch(struct nsolve_usage *usage)
-{
- struct nsolve_scratch *scratch = snew(struct nsolve_scratch);
- int cr = usage->cr;
- scratch->grid = snewn(cr*cr, unsigned char);
- scratch->rowidx = snewn(cr, unsigned char);
- scratch->colidx = snewn(cr, unsigned char);
- scratch->set = snewn(cr, unsigned char);
- return scratch;
+ sfree(digits);
+ return ret;
}
-static void nsolve_free_scratch(struct nsolve_scratch *scratch)
+/*
+ * Entry point to generator. You give it dimensions and a starting
+ * grid, which is simply an array of cr*cr digits.
+ */
+static void gridgen(int c, int r, digit *grid, random_state *rs)
{
- sfree(scratch->set);
- sfree(scratch->colidx);
- sfree(scratch->rowidx);
- sfree(scratch->grid);
- sfree(scratch);
-}
+ struct gridgen_usage *usage;
+ int x, y, cr = c*r;
-static int nsolve(int c, int r, digit *grid)
-{
- struct nsolve_usage *usage;
- struct nsolve_scratch *scratch;
- int cr = c*r;
- int x, y, n;
- int diff = DIFF_BLOCK;
+ /*
+ * Clear the grid to start with.
+ */
+ memset(grid, 0, cr*cr);
/*
- * Set up a usage structure as a clean slate (everything
- * possible).
+ * Create a gridgen_usage structure.
*/
- usage = snew(struct nsolve_usage);
+ usage = snew(struct gridgen_usage);
+
usage->c = c;
usage->r = r;
usage->cr = cr;
- usage->cube = snewn(cr*cr*cr, unsigned char);
- usage->grid = grid; /* write straight back to the input */
- memset(usage->cube, TRUE, cr*cr*cr);
+
+ usage->grid = snewn(cr * cr, digit);
+ memcpy(usage->grid, grid, cr * cr);
usage->row = snewn(cr * cr, unsigned char);
usage->col = snewn(cr * cr, unsigned char);
memset(usage->col, FALSE, cr * cr);
memset(usage->blk, FALSE, cr * cr);
- scratch = nsolve_new_scratch(usage);
+ usage->spaces = snewn(cr * cr, struct gridgen_coord);
+ usage->nspaces = 0;
- /*
- * Place all the clue numbers we are given.
- */
- for (x = 0; x < cr; x++)
- for (y = 0; y < cr; y++)
- if (grid[y*cr+x])
- nsolve_place(usage, x, YTRANS(y), grid[y*cr+x]);
+ usage->rs = rs;
/*
- * Now loop over the grid repeatedly trying all permitted modes
- * of reasoning. The loop terminates if we complete an
- * iteration without making any progress; we then return
- * failure or success depending on whether the grid is full or
- * not.
+ * Initialise the list of grid spaces.
*/
- 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:
-
- /*
- * Blockwise positional elimination.
- */
- for (x = 0; x < cr; x += r)
- for (y = 0; y < r; y++)
- for (n = 1; n <= cr; n++)
- if (!usage->blk[(y*c+(x/r))*cr+n-1] &&
- nsolve_elim(usage, cubepos(x,y,n), r*cr
-#ifdef STANDALONE_SOLVER
- , "positional elimination,"
- " block (%d,%d)", 1+x/r, 1+y
-#endif
- )) {
- diff = max(diff, DIFF_BLOCK);
- goto cont;
- }
-
- /*
- * Row-wise positional elimination.
- */
- for (y = 0; y < cr; y++)
- for (n = 1; n <= cr; n++)
- if (!usage->row[y*cr+n-1] &&
- nsolve_elim(usage, cubepos(0,y,n), cr*cr
-#ifdef STANDALONE_SOLVER
- , "positional elimination,"
- " row %d", 1+YUNTRANS(y)
-#endif
- )) {
- diff = max(diff, DIFF_SIMPLE);
- goto cont;
- }
- /*
- * Column-wise positional elimination.
- */
- for (x = 0; x < cr; x++)
- for (n = 1; n <= cr; n++)
- if (!usage->col[x*cr+n-1] &&
- nsolve_elim(usage, cubepos(x,0,n), cr
-#ifdef STANDALONE_SOLVER
- , "positional elimination," " column %d", 1+x
-#endif
- )) {
- diff = max(diff, DIFF_SIMPLE);
- goto cont;
- }
-
- /*
- * Numeric elimination.
- */
- for (x = 0; x < cr; x++)
- for (y = 0; y < cr; y++)
- if (!usage->grid[YUNTRANS(y)*cr+x] &&
- nsolve_elim(usage, cubepos(x,y,1), 1
-#ifdef STANDALONE_SOLVER
- , "numeric elimination at (%d,%d)", 1+x,
- 1+YUNTRANS(y)
-#endif
- )) {
- diff = max(diff, DIFF_SIMPLE);
- goto cont;
- }
-
- /*
- * Intersectional analysis, rows vs blocks.
- */
- for (y = 0; y < cr; y++)
- for (x = 0; x < cr; x += r)
- for (n = 1; n <= cr; n++)
- if (!usage->row[y*cr+n-1] &&
- !usage->blk[((y%r)*c+(x/r))*cr+n-1] &&
- (nsolve_intersect(usage, cubepos(0,y,n), cr*cr,
- cubepos(x,y%r,n), r*cr
-#ifdef STANDALONE_SOLVER
- , "intersectional analysis,"
- " row %d vs block (%d,%d)",
- 1+YUNTRANS(y), 1+x/r, 1+y%r
-#endif
- ) ||
- nsolve_intersect(usage, cubepos(x,y%r,n), r*cr,
- cubepos(0,y,n), cr*cr
-#ifdef STANDALONE_SOLVER
- , "intersectional analysis,"
- " block (%d,%d) vs row %d",
- 1+x/r, 1+y%r, 1+YUNTRANS(y)
-#endif
- ))) {
- diff = max(diff, DIFF_INTERSECT);
- goto cont;
- }
-
- /*
- * Intersectional analysis, columns vs blocks.
- */
- for (x = 0; x < cr; x++)
- for (y = 0; y < r; y++)
- for (n = 1; n <= cr; n++)
- if (!usage->col[x*cr+n-1] &&
- !usage->blk[(y*c+(x/r))*cr+n-1] &&
- (nsolve_intersect(usage, cubepos(x,0,n), cr,
- cubepos((x/r)*r,y,n), r*cr
-#ifdef STANDALONE_SOLVER
- , "intersectional analysis,"
- " column %d vs block (%d,%d)",
- 1+x, 1+x/r, 1+y
-#endif
- ) ||
- nsolve_intersect(usage, cubepos((x/r)*r,y,n), r*cr,
- cubepos(x,0,n), cr
-#ifdef STANDALONE_SOLVER
- , "intersectional analysis,"
- " block (%d,%d) vs column %d",
- 1+x/r, 1+y, 1+x
-#endif
- ))) {
- diff = max(diff, DIFF_INTERSECT);
- goto cont;
- }
-
- /*
- * Blockwise set elimination.
- */
- for (x = 0; x < cr; x += r)
- for (y = 0; y < r; y++)
- if (nsolve_set(usage, scratch, cubepos(x,y,1), r*cr, 1
-#ifdef STANDALONE_SOLVER
- , "set elimination, block (%d,%d)", 1+x/r, 1+y
-#endif
- )) {
- diff = max(diff, DIFF_SET);
- goto cont;
- }
-
- /*
- * Row-wise set elimination.
- */
- for (y = 0; y < cr; y++)
- if (nsolve_set(usage, scratch, cubepos(0,y,1), cr*cr, 1
-#ifdef STANDALONE_SOLVER
- , "set elimination, row %d", 1+YUNTRANS(y)
-#endif
- )) {
- diff = max(diff, DIFF_SET);
- goto cont;
- }
-
- /*
- * Column-wise set elimination.
- */
- for (x = 0; x < cr; x++)
- if (nsolve_set(usage, scratch, cubepos(x,0,1), cr, 1
-#ifdef STANDALONE_SOLVER
- , "set elimination, column %d", 1+x
-#endif
- )) {
- diff = max(diff, DIFF_SET);
- goto cont;
- }
-
- /*
- * If we reach here, we have made no deductions in this
- * iteration, so the algorithm terminates.
- */
- break;
+ for (y = 0; y < cr; y++) {
+ for (x = 0; x < cr; x++) {
+ usage->spaces[usage->nspaces].x = x;
+ usage->spaces[usage->nspaces].y = y;
+ usage->spaces[usage->nspaces].r = random_bits(rs, 31);
+ usage->nspaces++;
+ }
}
- nsolve_free_scratch(scratch);
+ /*
+ * Run the real generator function.
+ */
+ gridgen_real(usage, grid);
- sfree(usage->cube);
- sfree(usage->row);
- sfree(usage->col);
+ /*
+ * Clean up the usage structure now we have our answer.
+ */
+ sfree(usage->spaces);
sfree(usage->blk);
+ sfree(usage->col);
+ sfree(usage->row);
+ sfree(usage->grid);
sfree(usage);
-
- for (x = 0; x < cr; x++)
- for (y = 0; y < cr; y++)
- if (!grid[y*cr+x])
- return DIFF_IMPOSSIBLE;
- return diff;
}
/* ----------------------------------------------------------------------
- * End of non-recursive solver code.
+ * End of grid generator code.
*/
/*
digit *grid, *grid2;
struct xy { int x, y; } *locs;
int nlocs;
- int ret;
char *desc;
int coords[16], ncoords;
int *symmclasses, nsymmclasses;
*/
do {
/*
- * Start the recursive solver with an empty grid to generate a
- * random solved state.
+ * Generate a random solved state.
*/
- memset(grid, 0, area);
- ret = rsolve(c, r, grid, rs, 1);
- assert(ret == 1);
+ gridgen(c, r, grid, rs);
assert(check_valid(c, r, grid));
/*
* Now loop over the shuffled list and, for each element,
* see whether removing that element (and its reflections)
* from the grid will still leave the grid soluble by
- * nsolve.
+ * solver.
*/
for (i = 0; i < nlocs; i++) {
int ret;
for (j = 0; j < ncoords; j++)
grid2[coords[2*j+1]*cr+coords[2*j]] = 0;
- if (recursing)
- ret = (rsolve(c, r, grid2, NULL, 2) == 1);
- else
- ret = (nsolve(c, r, grid2) <= maxdiff);
-
- if (ret) {
+ ret = solver(c, r, grid2, NULL, maxdiff);
+ if (ret != DIFF_IMPOSSIBLE && ret != DIFF_AMBIGUOUS) {
for (j = 0; j < ncoords; j++)
grid[coords[2*j+1]*cr+coords[2*j]] = 0;
break;
if (i == nlocs) {
/*
* There was nothing we could remove without
- * destroying solvability. If we're trying to
- * generate a recursion-only grid and haven't
- * switched over to rsolve yet, we now do;
- * otherwise we give up.
+ * destroying solvability. Give up.
*/
- if (maxdiff == DIFF_RECURSIVE && !recursing) {
- recursing = TRUE;
- } else {
- break;
- }
+ break;
}
}
memcpy(grid2, grid, area);
- } while (nsolve(c, r, grid2) < maxdiff);
+ } while (solver(c, r, grid2, NULL, maxdiff) < maxdiff);
sfree(grid2);
sfree(locs);
int c = state->c, r = state->r, cr = c*r;
char *ret;
digit *grid;
- int rsolve_ret;
+ int solve_ret;
/*
* If we already have the solution in ai, save ourselves some
grid = snewn(cr*cr, digit);
memcpy(grid, state->grid, cr*cr);
- rsolve_ret = rsolve(c, r, grid, NULL, 2);
+ solve_ret = solver(c, r, grid, NULL, DIFF_RECURSIVE);
+
+ *error = NULL;
- if (rsolve_ret != 1) {
+ if (solve_ret == DIFF_IMPOSSIBLE)
+ *error = "No solution exists for this puzzle";
+ else if (solve_ret == DIFF_AMBIGUOUS)
+ *error = "Multiple solutions exist for this puzzle";
+
+ if (*error) {
sfree(grid);
- if (rsolve_ret == 0)
- *error = "No solution exists for this puzzle";
- else
- *error = "Multiple solutions exist for this puzzle";
return NULL;
}
sfree(ui);
}
-char *encode_ui(game_ui *ui)
+static char *encode_ui(game_ui *ui)
{
return NULL;
}
-void decode_ui(game_ui *ui, char *encoding)
+static void decode_ui(game_ui *ui, char *encoding)
{
}
return NULL;
sprintf(buf, "%c%d,%d,%d",
- ui->hpencil && n > 0 ? 'P' : 'R', ui->hx, ui->hy, n);
+ (char)(ui->hpencil && n > 0 ? 'P' : 'R'), ui->hx, ui->hy, n);
ui->hx = ui->hy = -1;
*/
#define SIZE(cr) ((cr) * TILE_SIZE + 2*BORDER + 1)
-#define GETTILESIZE(cr, w) ( (w-1) / (cr+1) )
+#define GETTILESIZE(cr, w) ( (double)(w-1) / (double)(cr+1) )
-static void game_size(game_params *params, game_drawstate *ds,
- int *x, int *y, int expand)
+static void game_compute_size(game_params *params, int tilesize,
+ int *x, int *y)
{
- int c = params->c, r = params->r, cr = c*r;
- int ts;
+ /* Ick: fake up `ds->tilesize' for macro expansion purposes */
+ struct { int tilesize; } ads, *ds = &ads;
+ ads.tilesize = tilesize;
- ts = min(GETTILESIZE(cr, *x), GETTILESIZE(cr, *y));
- if (expand)
- ds->tilesize = ts;
- else
- ds->tilesize = min(ts, PREFERRED_TILE_SIZE);
+ *x = SIZE(params->c * params->r);
+ *y = SIZE(params->c * params->r);
+}
- *x = SIZE(cr);
- *y = SIZE(cr);
+static void game_set_size(game_drawstate *ds, game_params *params,
+ int tilesize)
+{
+ ds->tilesize = tilesize;
}
static float *game_colours(frontend *fe, game_state *state, int *ncolours)
coords[3] = cy;
coords[4] = cx;
coords[5] = cy+ch/2;
- draw_polygon(fe, coords, 3, TRUE, COL_HIGHLIGHT);
+ draw_polygon(fe, coords, 3, COL_HIGHLIGHT, COL_HIGHLIGHT);
}
/* new number needs drawing? */
return FALSE;
}
-static int game_timing_state(game_state *state)
+static int game_timing_state(game_state *state, game_ui *ui)
{
return TRUE;
}
game_changed_state,
interpret_move,
execute_move,
- game_size,
+ PREFERRED_TILE_SIZE, game_compute_size, game_set_size,
game_colours,
game_new_drawstate,
game_free_drawstate,
void draw_rect(frontend *fe, int x, int y, int w, int h, int colour) {}
void draw_line(frontend *fe, int x1, int y1, int x2, int y2, int colour) {}
void draw_polygon(frontend *fe, int *coords, int npoints,
- int fill, int colour) {}
+ int fillcolour, int outlinecolour) {}
void clip(frontend *fe, int x, int y, int w, int h) {}
void unclip(frontend *fe) {}
void start_draw(frontend *fe) {}
{
game_params *p;
game_state *s;
- int recurse = TRUE;
char *id = NULL, *desc, *err;
- int y, x;
int grade = FALSE;
+ int ret;
while (--argc > 0) {
char *p = *++argv;
- if (!strcmp(p, "-r")) {
- recurse = TRUE;
- } else if (!strcmp(p, "-n")) {
- recurse = FALSE;
- } else if (!strcmp(p, "-v")) {
+ if (!strcmp(p, "-v")) {
solver_show_working = TRUE;
- recurse = FALSE;
} else if (!strcmp(p, "-g")) {
grade = TRUE;
- recurse = FALSE;
} else if (*p == '-') {
- fprintf(stderr, "%s: unrecognised option `%s'\n", argv[0]);
+ fprintf(stderr, "%s: unrecognised option `%s'\n", argv[0], p);
return 1;
} else {
id = p;
}
if (!id) {
- fprintf(stderr, "usage: %s [-n | -r | -g | -v] <game_id>\n", argv[0]);
+ fprintf(stderr, "usage: %s [-g | -v] <game_id>\n", argv[0]);
return 1;
}
}
s = new_game(NULL, p, desc);
- if (recurse) {
- int ret = rsolve(p->c, p->r, s->grid, NULL, 2);
- if (ret > 1) {
- fprintf(stderr, "%s: rsolve: multiple solutions detected\n",
- argv[0]);
- }
+ ret = solver(p->c, p->r, s->grid, NULL, DIFF_RECURSIVE);
+ if (grade) {
+ printf("Difficulty rating: %s\n",
+ ret==DIFF_BLOCK ? "Trivial (blockwise positional elimination only)":
+ 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_RECURSIVE ? "Unreasonable (guesswork and backtracking required)":
+ ret==DIFF_AMBIGUOUS ? "Ambiguous (multiple solutions exist)":
+ ret==DIFF_IMPOSSIBLE ? "Impossible (no solution exists)":
+ "INTERNAL ERROR: unrecognised difficulty code");
} else {
- int ret = nsolve(p->c, p->r, s->grid);
- if (grade) {
- if (ret == DIFF_IMPOSSIBLE) {
- /*
- * Now resort to rsolve to determine whether it's
- * really soluble.
- */
- ret = rsolve(p->c, p->r, s->grid, NULL, 2);
- if (ret == 0)
- ret = DIFF_IMPOSSIBLE;
- else if (ret == 1)
- ret = DIFF_RECURSIVE;
- else
- ret = DIFF_AMBIGUOUS;
- }
- printf("Difficulty rating: %s\n",
- ret==DIFF_BLOCK ? "Trivial (blockwise positional elimination only)":
- 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_RECURSIVE ? "Unreasonable (guesswork and backtracking required)":
- ret==DIFF_AMBIGUOUS ? "Ambiguous (multiple solutions exist)":
- ret==DIFF_IMPOSSIBLE ? "Impossible (no solution exists)":
- "INTERNAL ERROR: unrecognised difficulty code");
- }
+ printf("%s\n", grid_text_format(p->c, p->r, s->grid));
}
- printf("%s\n", grid_text_format(p->c, p->r, s->grid));
-
return 0;
}