#ifdef STANDALONE_SOLVER
#include <stdarg.h>
-int solver_show_working;
+int solver_show_working, solver_recurse_depth;
#endif
#include "puzzles.h"
-#define max(x,y) ((x)>(y)?(x):(y))
-
/*
* To save space, I store digits internally as unsigned char. This
* imposes a hard limit of 255 on the order of the puzzle. Since
typedef unsigned char digit;
#define ORDER_MAX 255
-#define TILE_SIZE 32
-#define BORDER 18
+#define PREFERRED_TILE_SIZE 32
+#define TILE_SIZE (ds->tilesize)
+#define BORDER (TILE_SIZE / 2)
#define FLASH_TIME 0.4F
-enum { SYMM_NONE, SYMM_ROT2, SYMM_ROT4, SYMM_REF4 };
+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_EXTREME,
+ DIFF_RECURSIVE, DIFF_AMBIGUOUS, DIFF_IMPOSSIBLE };
enum {
COL_BACKGROUND,
COL_CLUE,
COL_USER,
COL_HIGHLIGHT,
+ COL_ERROR,
COL_PENCIL,
NCOLOURS
};
{ "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_EXTREME } },
{ "3x3 Unreasonable", { 3, 3, SYMM_ROT2, DIFF_RECURSIVE } },
+#ifndef SLOW_SYSTEM
{ "3x4 Basic", { 3, 4, SYMM_ROT2, DIFF_SIMPLE } },
{ "4x4 Basic", { 4, 4, SYMM_ROT2, DIFF_SIMPLE } },
+#endif
};
if (i < 0 || i >= lenof(presets))
}
while (*string) {
if (*string == 'r' || *string == 'm' || *string == 'a') {
- int sn, sc;
+ int sn, sc, sd;
sc = *string++;
+ if (*string == 'd') {
+ sd = TRUE;
+ string++;
+ } else {
+ sd = FALSE;
+ }
sn = atoi(string);
while (*string && isdigit((unsigned char)*string)) string++;
+ if (sc == 'm' && sn == 8)
+ ret->symm = SYMM_REF8;
if (sc == 'm' && sn == 4)
- ret->symm = SYMM_REF4;
+ ret->symm = sd ? SYMM_REF4D : SYMM_REF4;
+ if (sc == 'm' && sn == 2)
+ ret->symm = sd ? SYMM_REF2D : SYMM_REF2;
if (sc == 'r' && sn == 4)
ret->symm = SYMM_ROT4;
if (sc == 'r' && sn == 2)
string++, ret->diff = DIFF_INTERSECT;
else if (*string == 'a') /* advanced */
string++, ret->diff = DIFF_SET;
+ else if (*string == 'e') /* extreme */
+ string++, ret->diff = DIFF_EXTREME;
else if (*string == 'u') /* unreasonable */
string++, ret->diff = DIFF_RECURSIVE;
} else
sprintf(str, "%dx%d", params->c, params->r);
if (full) {
switch (params->symm) {
+ case SYMM_REF8: strcat(str, "m8"); break;
case SYMM_REF4: strcat(str, "m4"); break;
+ case SYMM_REF4D: strcat(str, "md4"); break;
+ case SYMM_REF2: strcat(str, "m2"); break;
+ case SYMM_REF2D: strcat(str, "md2"); break;
case SYMM_ROT4: strcat(str, "r4"); break;
/* case SYMM_ROT2: strcat(str, "r2"); break; [default] */
case SYMM_NONE: strcat(str, "a"); break;
case DIFF_SIMPLE: strcat(str, "db"); break;
case DIFF_INTERSECT: strcat(str, "di"); break;
case DIFF_SET: strcat(str, "da"); break;
+ case DIFF_EXTREME: strcat(str, "de"); break;
case DIFF_RECURSIVE: strcat(str, "du"); break;
}
}
ret[2].name = "Symmetry";
ret[2].type = C_CHOICES;
- ret[2].sval = ":None:2-way rotation:4-way rotation:4-way mirror";
+ ret[2].sval = ":None:2-way rotation:4-way rotation:2-way mirror:"
+ "2-way diagonal mirror:4-way mirror:4-way diagonal mirror:"
+ "8-way mirror";
ret[2].ival = params->symm;
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;
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";
if (params->c > ORDER_MAX || params->r > ORDER_MAX)
return "Dimensions greater than "STR(ORDER_MAX)" are not supported";
+ if ((params->c * params->r) > 36)
+ return "Unable to support more than 36 distinct symbols in a puzzle";
return NULL;
}
/* ----------------------------------------------------------------------
- * 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 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
+ * `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).
+ *
+ * - 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
+ * 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;
}
-static int nsolve_set(struct nsolve_usage *usage,
+struct solver_scratch {
+ unsigned char *grid, *rowidx, *colidx, *set;
+ int *neighbours, *bfsqueue;
+#ifdef STANDALONE_SOLVER
+ int *bfsprev;
+#endif
+};
+
+static int solver_set(struct solver_usage *usage,
+ struct solver_scratch *scratch,
int start, int step1, int step2
#ifdef STANDALONE_SOLVER
, char *fmt, ...
{
int c = usage->c, r = usage->r, cr = c*r;
int i, j, n, count;
- unsigned char *grid = snewn(cr*cr, unsigned char);
- unsigned char *rowidx = snewn(cr, unsigned char);
- unsigned char *colidx = snewn(cr, unsigned char);
- unsigned char *set = snewn(cr, unsigned char);
+ unsigned char *grid = scratch->grid;
+ unsigned char *rowidx = scratch->rowidx;
+ unsigned char *colidx = scratch->colidx;
+ unsigned char *set = scratch->set;
/*
* We are passed a cr-by-cr matrix of booleans. Our first job
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);
px = py / cr;
py %= cr;
- printf(" ruling out %d at (%d,%d)\n",
+ printf("%*s ruling out %d at (%d,%d)\n",
+ solver_recurse_depth*4, "",
pn, 1+px, 1+YUNTRANS(py));
}
#endif
}
if (progress) {
- sfree(set);
- sfree(colidx);
- sfree(rowidx);
- sfree(grid);
- return TRUE;
+ return +1;
}
}
}
break; /* done */
}
- sfree(set);
- sfree(colidx);
- sfree(rowidx);
- sfree(grid);
-
- return FALSE;
+ return 0;
}
-static int nsolve(int c, int r, digit *grid)
+/*
+ * 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)
{
- struct nsolve_usage *usage;
- int cr = c*r;
- int x, y, n;
- int diff = DIFF_BLOCK;
+ 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->neighbours;
+ nb[1] = scratch->neighbours + cr;
/*
- * Set up a usage structure as a clean slate (everything
- * possible).
+ * 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.
*/
- usage = snew(struct nsolve_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);
+ 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;
- 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);
+ assert(y1 % r != y2 % r);
- /*
- * 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]);
+ 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);
+ }
+ }
/*
- * 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.
+ * Right. Now loop over each possible number.
*/
- while (1) {
+ 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 */
+}
+
+/*
+ * 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);
+ 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);
+ 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)
+{
+#ifdef STANDALONE_SOLVER
+ sfree(scratch->bfsprev);
+#endif
+ sfree(scratch->bfsqueue);
+ sfree(scratch->neighbours);
+ sfree(scratch->set);
+ sfree(scratch->colidx);
+ sfree(scratch->rowidx);
+ sfree(scratch->grid);
+ sfree(scratch);
+}
+
+static int solver(int c, int r, digit *grid, int maxdiff)
+{
+ struct solver_usage *usage;
+ struct solver_scratch *scratch;
+ int cr = c*r;
+ int x, y, x2, y2, 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
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
+ 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,"
- " block (%d,%d)", 1+x/r, 1+y
+ , "positional elimination,"
+ " %d in block (%d,%d)", n, 1+x/r, 1+y
#endif
- )) {
- diff = max(diff, DIFF_BLOCK);
- goto cont;
+ );
+ 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] &&
- nsolve_elim(usage, cubepos(0,y,n), cr*cr
+ if (!usage->row[y*cr+n-1]) {
+ ret = solver_elim(usage, cubepos(0,y,n), cr*cr
#ifdef STANDALONE_SOLVER
- , "positional elimination,"
- " row %d", 1+YUNTRANS(y)
+ , "positional elimination,"
+ " %d in row %d", n, 1+YUNTRANS(y)
#endif
- )) {
- diff = max(diff, DIFF_SIMPLE);
- goto cont;
+ );
+ 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] &&
- nsolve_elim(usage, cubepos(x,0,n), cr
+ if (!usage->col[x*cr+n-1]) {
+ ret = solver_elim(usage, cubepos(x,0,n), cr
#ifdef STANDALONE_SOLVER
- , "positional elimination," " column %d", 1+x
+ , "positional elimination,"
+ " %d in column %d", n, 1+x
#endif
- )) {
- diff = max(diff, DIFF_SIMPLE);
- goto cont;
+ );
+ if (ret < 0) {
+ diff = DIFF_IMPOSSIBLE;
+ goto got_result;
+ } else if (ret > 0) {
+ diff = max(diff, DIFF_SIMPLE);
+ goto cont;
+ }
}
/*
*/
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
+ 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)
+ , "numeric elimination at (%d,%d)", 1+x,
+ 1+YUNTRANS(y)
#endif
- )) {
- diff = max(diff, DIFF_SIMPLE);
- goto cont;
+ );
+ 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] &&
- (nsolve_intersect(usage, cubepos(0,y,n), cr*cr,
+ (solver_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
+ " %d in row %d vs block (%d,%d)",
+ n, 1+YUNTRANS(y), 1+x/r, 1+y%r
#endif
) ||
- nsolve_intersect(usage, cubepos(x,y%r,n), r*cr,
+ solver_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)
+ " %d in block (%d,%d) vs row %d",
+ n, 1+x/r, 1+y%r, 1+YUNTRANS(y)
#endif
))) {
diff = max(diff, DIFF_INTERSECT);
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,
+ (solver_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
+ " %d in column %d vs block (%d,%d)",
+ n, 1+x, 1+x/r, 1+y
#endif
) ||
- nsolve_intersect(usage, cubepos((x/r)*r,y,n), r*cr,
+ solver_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
+ " %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++)
- if (nsolve_set(usage, cubepos(x,y,1), r*cr, 1
+ 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
+ , "set elimination, block (%d,%d)", 1+x/r, 1+y
#endif
- )) {
- diff = max(diff, DIFF_SET);
- goto cont;
- }
+ );
+ 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++)
- if (nsolve_set(usage, cubepos(0,y,1), cr*cr, 1
+ 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)
+ , "set elimination, row %d", 1+YUNTRANS(y)
#endif
- )) {
- diff = max(diff, DIFF_SET);
- goto cont;
- }
+ );
+ 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++)
- if (nsolve_set(usage, cubepos(x,0,1), cr, 1
+ 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
+ , "set elimination, column %d", 1+x
#endif
- )) {
- diff = max(diff, DIFF_SET);
- goto cont;
- }
+ );
+ if (ret < 0) {
+ diff = DIFF_IMPOSSIBLE;
+ goto got_result;
+ } else if (ret > 0) {
+ diff = max(diff, DIFF_SET);
+ goto cont;
+ }
+ }
+
+ /*
+ * 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.
+ */
+ 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_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_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.
+ */
+ 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;
+
+ best = -1;
+ bestcount = cr+1;
+
+ 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;
+ 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
+
+ /*
+ * 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, 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(digits, j, sizeof(*digits), usage->rs);
+
+ /* And finally, go through the digit list and actually recurse. */
+ ret = FALSE;
+ 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. Stop when we find a solution. */
+ if (gridgen_real(usage, grid))
+ ret = TRUE;
+
+ /* 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++;
+
+ if (ret)
+ break;
+ }
+
+ sfree(digits);
+ return ret;
+}
+
+/*
+ * 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)
+{
+ struct gridgen_usage *usage;
+ int x, y, cr = c*r;
+
+ /*
+ * Clear the grid to start with.
+ */
+ memset(grid, 0, cr*cr);
+
+ /*
+ * Create a gridgen_usage structure.
+ */
+ usage = snew(struct gridgen_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 gridgen_coord);
+ usage->nspaces = 0;
+
+ usage->rs = rs;
- /*
- * If we reach here, we have made no deductions in this
- * iteration, so the algorithm terminates.
- */
- break;
+ /*
+ * Initialise the list of grid spaces.
+ */
+ 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++;
+ }
}
- sfree(usage->cube);
- sfree(usage->row);
- sfree(usage->col);
+ /*
+ * Run the real generator function.
+ */
+ gridgen_real(usage, grid);
+
+ /*
+ * 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.
*/
/*
return TRUE;
}
-static void symmetry_limit(game_params *params, int *xlim, int *ylim, int s)
+static int symmetries(game_params *params, int x, int y, int *output, int s)
{
int c = params->c, r = params->r, cr = c*r;
+ int i = 0;
+
+#define ADD(x,y) (*output++ = (x), *output++ = (y), i++)
+
+ ADD(x, y);
switch (s) {
case SYMM_NONE:
- *xlim = *ylim = cr;
- break;
+ break; /* just x,y is all we need */
case SYMM_ROT2:
- *xlim = (cr+1) / 2;
- *ylim = cr;
- break;
- case SYMM_REF4:
+ ADD(cr - 1 - x, cr - 1 - y);
+ break;
case SYMM_ROT4:
- *xlim = *ylim = (cr+1) / 2;
- break;
+ ADD(cr - 1 - y, x);
+ ADD(y, cr - 1 - x);
+ ADD(cr - 1 - x, cr - 1 - y);
+ break;
+ case SYMM_REF2:
+ ADD(cr - 1 - x, y);
+ break;
+ case SYMM_REF2D:
+ ADD(y, x);
+ break;
+ case SYMM_REF4:
+ ADD(cr - 1 - x, y);
+ ADD(x, cr - 1 - y);
+ ADD(cr - 1 - x, cr - 1 - y);
+ break;
+ case SYMM_REF4D:
+ ADD(y, x);
+ ADD(cr - 1 - x, cr - 1 - y);
+ ADD(cr - 1 - y, cr - 1 - x);
+ break;
+ case SYMM_REF8:
+ ADD(cr - 1 - x, y);
+ ADD(x, cr - 1 - y);
+ ADD(cr - 1 - x, cr - 1 - y);
+ ADD(y, x);
+ ADD(y, cr - 1 - x);
+ ADD(cr - 1 - y, x);
+ ADD(cr - 1 - y, cr - 1 - x);
+ break;
}
+
+#undef ADD
+
+ return i;
}
-static int symmetries(game_params *params, int x, int y, int *output, int s)
+static char *encode_solve_move(int cr, digit *grid)
{
- int c = params->c, r = params->r, cr = c*r;
- int i = 0;
+ int i, len;
+ char *ret, *p, *sep;
- *output++ = x;
- *output++ = y;
- i++;
+ /*
+ * It's surprisingly easy to work out _exactly_ how long this
+ * string needs to be. To decimal-encode all the numbers from 1
+ * to n:
+ *
+ * - every number has a units digit; total is n.
+ * - all numbers above 9 have a tens digit; total is max(n-9,0).
+ * - all numbers above 99 have a hundreds digit; total is max(n-99,0).
+ * - and so on.
+ */
+ len = 0;
+ for (i = 1; i <= cr; i *= 10)
+ len += max(cr - i + 1, 0);
+ len += cr; /* don't forget the commas */
+ len *= cr; /* there are cr rows of these */
- switch (s) {
- case SYMM_NONE:
- break; /* just x,y is all we need */
- case SYMM_REF4:
- case SYMM_ROT4:
- switch (s) {
- case SYMM_REF4:
- *output++ = cr - 1 - x;
- *output++ = y;
- i++;
-
- *output++ = x;
- *output++ = cr - 1 - y;
- i++;
- break;
- case SYMM_ROT4:
- *output++ = cr - 1 - y;
- *output++ = x;
- i++;
-
- *output++ = y;
- *output++ = cr - 1 - x;
- i++;
- break;
- }
- /* fall through */
- case SYMM_ROT2:
- *output++ = cr - 1 - x;
- *output++ = cr - 1 - y;
- i++;
- break;
+ /*
+ * Now len is one bigger than the total size of the
+ * comma-separated numbers (because we counted an
+ * additional leading comma). We need to have a leading S
+ * and a trailing NUL, so we're off by one in total.
+ */
+ len++;
+
+ ret = snewn(len, char);
+ p = ret;
+ *p++ = 'S';
+ sep = "";
+ for (i = 0; i < cr*cr; i++) {
+ p += sprintf(p, "%s%d", sep, grid[i]);
+ sep = ",";
}
+ *p++ = '\0';
+ assert(p - ret == len);
- return i;
+ return ret;
}
-struct game_aux_info {
- int c, r;
- digit *grid;
-};
-
static char *new_game_desc(game_params *params, random_state *rs,
- game_aux_info **aux)
+ char **aux, int interactive)
{
int c = params->c, r = params->r, cr = c*r;
int area = cr*cr;
digit *grid, *grid2;
struct xy { int x, y; } *locs;
int nlocs;
- int ret;
char *desc;
int coords[16], ncoords;
- int xlim, ylim;
- int maxdiff, recursing;
+ int maxdiff;
+ int x, y, i, j;
/*
* Adjust the maximum difficulty level to be consistent with
*/
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));
/*
- * Save the solved grid in the aux_info.
+ * Save the solved grid in aux.
*/
{
- game_aux_info *ai = snew(game_aux_info);
- ai->c = c;
- ai->r = r;
- ai->grid = snewn(cr * cr, digit);
- memcpy(ai->grid, grid, cr * cr * sizeof(digit));
- *aux = ai;
+ /*
+ * We might already have written *aux the last time we
+ * went round this loop, in which case we should free
+ * the old aux before overwriting it with the new one.
+ */
+ if (*aux) {
+ sfree(*aux);
+ }
+
+ *aux = encode_solve_move(cr, grid);
}
/*
* Now we have a solved grid, start removing things from it
* while preserving solubility.
*/
- symmetry_limit(params, &xlim, &ylim, params->symm);
- recursing = FALSE;
- while (1) {
- int x, y, i, j;
- /*
- * Iterate over the grid and enumerate all the filled
- * squares we could empty.
- */
- nlocs = 0;
-
- for (x = 0; x < xlim; x++)
- for (y = 0; y < ylim; y++)
- if (grid[y*cr+x]) {
- locs[nlocs].x = x;
- locs[nlocs].y = y;
- nlocs++;
- }
+ /*
+ * Find the set of equivalence classes of squares permitted
+ * by the selected symmetry. We do this by enumerating all
+ * the grid squares which have no symmetric companion
+ * sorting lower than themselves.
+ */
+ nlocs = 0;
+ for (y = 0; y < cr; y++)
+ for (x = 0; x < cr; x++) {
+ int i = y*cr+x;
+ int j;
- /*
- * Now shuffle that list.
- */
- for (i = nlocs; i > 1; i--) {
- int p = random_upto(rs, i);
- if (p != i-1) {
- struct xy t = locs[p];
- locs[p] = locs[i-1];
- locs[i-1] = t;
+ ncoords = symmetries(params, x, y, coords, params->symm);
+ for (j = 0; j < ncoords; j++)
+ if (coords[2*j+1]*cr+coords[2*j] < i)
+ break;
+ if (j == ncoords) {
+ locs[nlocs].x = x;
+ locs[nlocs].y = y;
+ nlocs++;
}
}
- /*
- * 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.
- */
- for (i = 0; i < nlocs; i++) {
- int ret;
-
- x = locs[i].x;
- y = locs[i].y;
+ /*
+ * Now shuffle that list.
+ */
+ shuffle(locs, nlocs, sizeof(*locs), rs);
- memcpy(grid2, grid, area);
- ncoords = symmetries(params, x, y, coords, params->symm);
- for (j = 0; j < ncoords; j++)
- grid2[coords[2*j+1]*cr+coords[2*j]] = 0;
+ /*
+ * 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.
+ */
+ for (i = 0; i < nlocs; i++) {
+ int ret;
- if (recursing)
- ret = (rsolve(c, r, grid2, NULL, 2) == 1);
- else
- ret = (nsolve(c, r, grid2) <= maxdiff);
+ x = locs[i].x;
+ y = locs[i].y;
- if (ret) {
- for (j = 0; j < ncoords; j++)
- grid[coords[2*j+1]*cr+coords[2*j]] = 0;
- break;
- }
- }
+ memcpy(grid2, grid, area);
+ ncoords = symmetries(params, x, y, coords, params->symm);
+ for (j = 0; j < ncoords; j++)
+ grid2[coords[2*j+1]*cr+coords[2*j]] = 0;
- 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.
- */
- if (maxdiff == DIFF_RECURSIVE && !recursing) {
- recursing = TRUE;
- } else {
- break;
- }
+ ret = solver(c, r, grid2, maxdiff);
+ if (ret <= maxdiff) {
+ for (j = 0; j < ncoords; j++)
+ grid[coords[2*j+1]*cr+coords[2*j]] = 0;
}
}
memcpy(grid2, grid, area);
- } while (nsolve(c, r, grid2) < maxdiff);
+ } while (solver(c, r, grid2, maxdiff) < maxdiff);
sfree(grid2);
sfree(locs);
return desc;
}
-static void game_free_aux_info(game_aux_info *aux)
-{
- sfree(aux->grid);
- sfree(aux);
-}
-
static char *validate_desc(game_params *params, char *desc)
{
int area = params->r * params->r * params->c * params->c;
} else if (n == '_') {
/* do nothing */;
} else if (n > '0' && n <= '9') {
+ int val = atoi(desc-1);
+ if (val < 1 || val > params->c * params->r)
+ return "Out-of-range number in game description";
squares++;
while (*desc >= '0' && *desc <= '9')
desc++;
return NULL;
}
-static game_state *new_game(game_params *params, char *desc)
+static game_state *new_game(midend *me, game_params *params, char *desc)
{
game_state *state = snew(game_state);
int c = params->c, r = params->r, cr = c*r, area = cr * cr;
sfree(state);
}
-static game_state *solve_game(game_state *state, game_aux_info *ai,
- char **error)
+static char *solve_game(game_state *state, game_state *currstate,
+ char *ai, char **error)
{
- game_state *ret;
int c = state->c, r = state->r, cr = c*r;
- int rsolve_ret;
-
- ret = dup_game(state);
- ret->completed = ret->cheated = TRUE;
+ char *ret;
+ digit *grid;
+ int solve_ret;
/*
- * If we already have the solution in the aux_info, save
- * ourselves some time.
+ * If we already have the solution in ai, save ourselves some
+ * time.
*/
- if (ai) {
+ if (ai)
+ return dupstr(ai);
- assert(c == ai->c);
- assert(r == ai->r);
- memcpy(ret->grid, ai->grid, cr * cr * sizeof(digit));
+ grid = snewn(cr*cr, digit);
+ memcpy(grid, state->grid, cr*cr);
+ solve_ret = solver(c, r, grid, DIFF_RECURSIVE);
- } else {
- rsolve_ret = rsolve(c, r, ret->grid, NULL, 2);
+ *error = NULL;
- if (rsolve_ret != 1) {
- free_game(ret);
- if (rsolve_ret == 0)
- *error = "No solution exists for this puzzle";
- else
- *error = "Multiple solutions exist for this puzzle";
- return NULL;
- }
+ 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);
+ return NULL;
}
+ ret = encode_solve_move(cr, grid);
+
+ sfree(grid);
+
return ret;
}
for (x = 0; x < cr; x++) {
int ch = grid[y * cr + x];
if (ch == 0)
- ch = ' ';
+ ch = '.';
else if (ch <= 9)
ch = '0' + ch;
else
sfree(ui);
}
-static game_state *make_move(game_state *from, game_ui *ui, int x, int y,
- int button)
+static char *encode_ui(game_ui *ui)
{
- int c = from->c, r = from->r, cr = c*r;
+ return NULL;
+}
+
+static void decode_ui(game_ui *ui, char *encoding)
+{
+}
+
+static void game_changed_state(game_ui *ui, game_state *oldstate,
+ game_state *newstate)
+{
+ int c = newstate->c, r = newstate->r, cr = c*r;
+ /*
+ * We prevent pencil-mode highlighting of a filled square. So
+ * if the user has just filled in a square which we had a
+ * pencil-mode highlight in (by Undo, or by Redo, or by Solve),
+ * then we cancel the highlight.
+ */
+ if (ui->hx >= 0 && ui->hy >= 0 && ui->hpencil &&
+ newstate->grid[ui->hy * cr + ui->hx] != 0) {
+ ui->hx = ui->hy = -1;
+ }
+}
+
+struct game_drawstate {
+ int started;
+ int c, r, cr;
+ int tilesize;
+ digit *grid;
+ unsigned char *pencil;
+ unsigned char *hl;
+ /* This is scratch space used within a single call to game_redraw. */
+ int *entered_items;
+};
+
+static char *interpret_move(game_state *state, game_ui *ui, game_drawstate *ds,
+ int x, int y, int button)
+{
+ int c = state->c, r = state->r, cr = c*r;
int tx, ty;
- game_state *ret;
+ char buf[80];
button &= ~MOD_MASK;
tx = (x + TILE_SIZE - BORDER) / TILE_SIZE - 1;
ty = (y + TILE_SIZE - BORDER) / TILE_SIZE - 1;
- if (tx >= 0 && tx < cr && ty >= 0 && ty < cr &&
- (button == LEFT_BUTTON || button == RIGHT_BUTTON)) {
- /*
- * Prevent pencil-mode highlighting of a filled square.
- */
- if (button == RIGHT_BUTTON && from->grid[ty*cr+tx])
- return NULL;
-
- if (tx == ui->hx && ty == ui->hy) {
- ui->hx = ui->hy = -1;
- } else {
- ui->hx = tx;
- ui->hy = ty;
- }
- ui->hpencil = (button == RIGHT_BUTTON);
- return from; /* UI activity occurred */
+ if (tx >= 0 && tx < cr && ty >= 0 && ty < cr) {
+ if (button == LEFT_BUTTON) {
+ if (state->immutable[ty*cr+tx]) {
+ ui->hx = ui->hy = -1;
+ } else if (tx == ui->hx && ty == ui->hy && ui->hpencil == 0) {
+ ui->hx = ui->hy = -1;
+ } else {
+ ui->hx = tx;
+ ui->hy = ty;
+ ui->hpencil = 0;
+ }
+ return ""; /* UI activity occurred */
+ }
+ if (button == RIGHT_BUTTON) {
+ /*
+ * Pencil-mode highlighting for non filled squares.
+ */
+ if (state->grid[ty*cr+tx] == 0) {
+ if (tx == ui->hx && ty == ui->hy && ui->hpencil) {
+ ui->hx = ui->hy = -1;
+ } else {
+ ui->hpencil = 1;
+ ui->hx = tx;
+ ui->hy = ty;
+ }
+ } else {
+ ui->hx = ui->hy = -1;
+ }
+ return ""; /* UI activity occurred */
+ }
}
if (ui->hx != -1 && ui->hy != -1 &&
((button >= '1' && button <= '9' && button - '0' <= cr) ||
(button >= 'a' && button <= 'z' && button - 'a' + 10 <= cr) ||
(button >= 'A' && button <= 'Z' && button - 'A' + 10 <= cr) ||
- button == ' ')) {
+ button == ' ' || button == '\010' || button == '\177')) {
int n = button - '0';
if (button >= 'A' && button <= 'Z')
n = button - 'A' + 10;
if (button >= 'a' && button <= 'z')
n = button - 'a' + 10;
- if (button == ' ')
+ if (button == ' ' || button == '\010' || button == '\177')
n = 0;
- if (from->immutable[ui->hy*cr+ui->hx])
- return NULL; /* can't overwrite this square */
+ /*
+ * Can't overwrite this square. In principle this shouldn't
+ * happen anyway because we should never have even been
+ * able to highlight the square, but it never hurts to be
+ * careful.
+ */
+ if (state->immutable[ui->hy*cr+ui->hx])
+ return NULL;
/*
* Can't make pencil marks in a filled square. In principle
* have even been able to pencil-highlight the square, but
* it never hurts to be careful.
*/
- if (ui->hpencil && from->grid[ui->hy*cr+ui->hx])
+ if (ui->hpencil && state->grid[ui->hy*cr+ui->hx])
return NULL;
+ sprintf(buf, "%c%d,%d,%d",
+ (char)(ui->hpencil && n > 0 ? 'P' : 'R'), ui->hx, ui->hy, n);
+
+ ui->hx = ui->hy = -1;
+
+ return dupstr(buf);
+ }
+
+ return NULL;
+}
+
+static game_state *execute_move(game_state *from, char *move)
+{
+ int c = from->c, r = from->r, cr = c*r;
+ game_state *ret;
+ int x, y, n;
+
+ if (move[0] == 'S') {
+ char *p;
+
+ ret = dup_game(from);
+ ret->completed = ret->cheated = TRUE;
+
+ p = move+1;
+ for (n = 0; n < cr*cr; n++) {
+ ret->grid[n] = atoi(p);
+
+ if (!*p || ret->grid[n] < 1 || ret->grid[n] > cr) {
+ free_game(ret);
+ return NULL;
+ }
+
+ while (*p && isdigit((unsigned char)*p)) p++;
+ if (*p == ',') p++;
+ }
+
+ return ret;
+ } else if ((move[0] == 'P' || move[0] == 'R') &&
+ sscanf(move+1, "%d,%d,%d", &x, &y, &n) == 3 &&
+ x >= 0 && x < cr && y >= 0 && y < cr && n >= 0 && n <= cr) {
+
ret = dup_game(from);
- if (ui->hpencil && n > 0) {
- int index = (ui->hy*cr+ui->hx) * cr + (n-1);
+ if (move[0] == 'P' && n > 0) {
+ int index = (y*cr+x) * cr + (n-1);
ret->pencil[index] = !ret->pencil[index];
} else {
- ret->grid[ui->hy*cr+ui->hx] = n;
- memset(ret->pencil + (ui->hy*cr+ui->hx)*cr, 0, cr);
+ ret->grid[y*cr+x] = n;
+ memset(ret->pencil + (y*cr+x)*cr, 0, cr);
/*
* We've made a real change to the grid. Check to see
ret->completed = TRUE;
}
}
- ui->hx = ui->hy = -1;
-
- return ret; /* made a valid move */
- }
-
- return NULL;
+ return ret;
+ } else
+ return NULL; /* couldn't parse move string */
}
/* ----------------------------------------------------------------------
* Drawing routines.
*/
-struct game_drawstate {
- int started;
- int c, r, cr;
- digit *grid;
- unsigned char *pencil;
- unsigned char *hl;
-};
-
-#define XSIZE(cr) ((cr) * TILE_SIZE + 2*BORDER + 1)
-#define YSIZE(cr) ((cr) * TILE_SIZE + 2*BORDER + 1)
+#define SIZE(cr) ((cr) * TILE_SIZE + 2*BORDER + 1)
+#define GETTILESIZE(cr, w) ( (double)(w-1) / (double)(cr+1) )
-static void game_size(game_params *params, int *x, int *y)
+static void game_compute_size(game_params *params, int tilesize,
+ int *x, int *y)
{
- int c = params->c, r = params->r, cr = c*r;
+ /* Ick: fake up `ds->tilesize' for macro expansion purposes */
+ struct { int tilesize; } ads, *ds = &ads;
+ ads.tilesize = tilesize;
- *x = XSIZE(cr);
- *y = YSIZE(cr);
+ *x = SIZE(params->c * params->r);
+ *y = SIZE(params->c * params->r);
}
-static float *game_colours(frontend *fe, game_state *state, int *ncolours)
+static void game_set_size(drawing *dr, game_drawstate *ds,
+ game_params *params, int tilesize)
+{
+ ds->tilesize = tilesize;
+}
+
+static float *game_colours(frontend *fe, int *ncolours)
{
float *ret = snewn(3 * NCOLOURS, float);
ret[COL_HIGHLIGHT * 3 + 1] = 0.85F * ret[COL_BACKGROUND * 3 + 1];
ret[COL_HIGHLIGHT * 3 + 2] = 0.85F * ret[COL_BACKGROUND * 3 + 2];
+ ret[COL_ERROR * 3 + 0] = 1.0F;
+ ret[COL_ERROR * 3 + 1] = 0.0F;
+ ret[COL_ERROR * 3 + 2] = 0.0F;
+
ret[COL_PENCIL * 3 + 0] = 0.5F * ret[COL_BACKGROUND * 3 + 0];
ret[COL_PENCIL * 3 + 1] = 0.5F * ret[COL_BACKGROUND * 3 + 1];
ret[COL_PENCIL * 3 + 2] = ret[COL_BACKGROUND * 3 + 2];
return ret;
}
-static game_drawstate *game_new_drawstate(game_state *state)
+static game_drawstate *game_new_drawstate(drawing *dr, game_state *state)
{
struct game_drawstate *ds = snew(struct game_drawstate);
int c = state->c, r = state->r, cr = c*r;
memset(ds->pencil, 0, cr*cr*cr);
ds->hl = snewn(cr*cr, unsigned char);
memset(ds->hl, 0, cr*cr);
-
+ ds->entered_items = snewn(cr*cr, int);
+ ds->tilesize = 0; /* not decided yet */
return ds;
}
-static void game_free_drawstate(game_drawstate *ds)
+static void game_free_drawstate(drawing *dr, game_drawstate *ds)
{
sfree(ds->hl);
sfree(ds->pencil);
sfree(ds->grid);
+ sfree(ds->entered_items);
sfree(ds);
}
-static void draw_number(frontend *fe, game_drawstate *ds, game_state *state,
+static void draw_number(drawing *dr, game_drawstate *ds, game_state *state,
int x, int y, int hl)
{
int c = state->c, r = state->r, cr = c*r;
if ((y+1) % c)
ch++;
- clip(fe, cx, cy, cw, ch);
+ clip(dr, cx, cy, cw, ch);
/* background needs erasing */
- draw_rect(fe, cx, cy, cw, ch, hl == 1 ? COL_HIGHLIGHT : COL_BACKGROUND);
+ draw_rect(dr, cx, cy, cw, ch, (hl & 15) == 1 ? COL_HIGHLIGHT : COL_BACKGROUND);
/* pencil-mode highlight */
- if (hl == 2) {
+ if ((hl & 15) == 2) {
int coords[6];
coords[0] = cx;
coords[1] = cy;
coords[3] = cy;
coords[4] = cx;
coords[5] = cy+ch/2;
- draw_polygon(fe, coords, 3, TRUE, COL_HIGHLIGHT);
+ draw_polygon(dr, coords, 3, COL_HIGHLIGHT, COL_HIGHLIGHT);
}
/* new number needs drawing? */
str[0] = state->grid[y*cr+x] + '0';
if (str[0] > '9')
str[0] += 'a' - ('9'+1);
- draw_text(fe, tx + TILE_SIZE/2, ty + TILE_SIZE/2,
+ draw_text(dr, tx + TILE_SIZE/2, ty + TILE_SIZE/2,
FONT_VARIABLE, TILE_SIZE/2, ALIGN_VCENTRE | ALIGN_HCENTRE,
- state->immutable[y*cr+x] ? COL_CLUE : COL_USER, str);
+ state->immutable[y*cr+x] ? COL_CLUE : (hl & 16) ? COL_ERROR : COL_USER, str);
} else {
- /* pencil marks required? */
- int i, j;
+ int i, j, npencil;
+ int pw, ph, pmax, fontsize;
+
+ /* count the pencil marks required */
+ for (i = npencil = 0; i < cr; i++)
+ if (state->pencil[(y*cr+x)*cr+i])
+ npencil++;
+
+ /*
+ * It's not sensible to arrange pencil marks in the same
+ * layout as the squares within a block, because this leads
+ * to the font being too small. Instead, we arrange pencil
+ * marks in the nearest thing we can to a square layout,
+ * and we adjust the square layout depending on the number
+ * of pencil marks in the square.
+ */
+ for (pw = 1; pw * pw < npencil; pw++);
+ if (pw < 3) pw = 3; /* otherwise it just looks _silly_ */
+ ph = (npencil + pw - 1) / pw;
+ if (ph < 2) ph = 2; /* likewise */
+ pmax = max(pw, ph);
+ fontsize = TILE_SIZE/(pmax*(11-pmax)/8);
for (i = j = 0; i < cr; i++)
if (state->pencil[(y*cr+x)*cr+i]) {
- int dx = j % r, dy = j / r, crm = max(c, r);
+ int dx = j % pw, dy = j / pw;
+
str[1] = '\0';
str[0] = i + '1';
if (str[0] > '9')
str[0] += 'a' - ('9'+1);
- draw_text(fe, tx + (4*dx+3) * TILE_SIZE / (4*r+2),
- ty + (4*dy+3) * TILE_SIZE / (4*c+2),
- FONT_VARIABLE, TILE_SIZE/(crm*5/4),
+ draw_text(dr, tx + (4*dx+3) * TILE_SIZE / (4*pw+2),
+ ty + (4*dy+3) * TILE_SIZE / (4*ph+2),
+ FONT_VARIABLE, fontsize,
ALIGN_VCENTRE | ALIGN_HCENTRE, COL_PENCIL, str);
j++;
}
}
- unclip(fe);
+ unclip(dr);
- draw_update(fe, cx, cy, cw, ch);
+ draw_update(dr, cx, cy, cw, ch);
ds->grid[y*cr+x] = state->grid[y*cr+x];
memcpy(ds->pencil+(y*cr+x)*cr, state->pencil+(y*cr+x)*cr, cr);
ds->hl[y*cr+x] = hl;
}
-static void game_redraw(frontend *fe, game_drawstate *ds, game_state *oldstate,
+static void game_redraw(drawing *dr, game_drawstate *ds, game_state *oldstate,
game_state *state, int dir, game_ui *ui,
float animtime, float flashtime)
{
* all games should start by drawing a big
* background-colour rectangle covering the whole window.
*/
- draw_rect(fe, 0, 0, XSIZE(cr), YSIZE(cr), COL_BACKGROUND);
+ draw_rect(dr, 0, 0, SIZE(cr), SIZE(cr), COL_BACKGROUND);
/*
* Draw the grid.
*/
for (x = 0; x <= cr; x++) {
int thick = (x % r ? 0 : 1);
- draw_rect(fe, BORDER + x*TILE_SIZE - thick, BORDER-1,
+ draw_rect(dr, BORDER + x*TILE_SIZE - thick, BORDER-1,
1+2*thick, cr*TILE_SIZE+3, COL_GRID);
}
for (y = 0; y <= cr; y++) {
int thick = (y % c ? 0 : 1);
- draw_rect(fe, BORDER-1, BORDER + y*TILE_SIZE - thick,
+ draw_rect(dr, BORDER-1, BORDER + y*TILE_SIZE - thick,
cr*TILE_SIZE+3, 1+2*thick, COL_GRID);
}
}
/*
+ * This array is used to keep track of rows, columns and boxes
+ * which contain a number more than once.
+ */
+ for (x = 0; x < cr * cr; x++)
+ ds->entered_items[x] = 0;
+ for (x = 0; x < cr; x++)
+ for (y = 0; y < cr; y++) {
+ digit d = state->grid[y*cr+x];
+ if (d) {
+ int box = (x/r)+(y/c)*c;
+ ds->entered_items[x*cr+d-1] |= ((ds->entered_items[x*cr+d-1] & 1) << 1) | 1;
+ ds->entered_items[y*cr+d-1] |= ((ds->entered_items[y*cr+d-1] & 4) << 1) | 4;
+ ds->entered_items[box*cr+d-1] |= ((ds->entered_items[box*cr+d-1] & 16) << 1) | 16;
+ }
+ }
+
+ /*
* Draw any numbers which need redrawing.
*/
for (x = 0; x < cr; x++) {
for (y = 0; y < cr; y++) {
int highlight = 0;
+ digit d = state->grid[y*cr+x];
+
if (flashtime > 0 &&
(flashtime <= FLASH_TIME/3 ||
flashtime >= FLASH_TIME*2/3))
highlight = 1;
+
+ /* Highlight active input areas. */
if (x == ui->hx && y == ui->hy)
highlight = ui->hpencil ? 2 : 1;
- draw_number(fe, ds, state, x, y, highlight);
+
+ /* Mark obvious errors (ie, numbers which occur more than once
+ * in a single row, column, or box). */
+ if (d && ((ds->entered_items[x*cr+d-1] & 2) ||
+ (ds->entered_items[y*cr+d-1] & 8) ||
+ (ds->entered_items[((x/r)+(y/c)*c)*cr+d-1] & 32)))
+ highlight |= 16;
+
+ draw_number(dr, ds, state, x, y, highlight);
}
}
* Update the _entire_ grid if necessary.
*/
if (!ds->started) {
- draw_update(fe, 0, 0, XSIZE(cr), YSIZE(cr));
+ draw_update(dr, 0, 0, SIZE(cr), SIZE(cr));
ds->started = TRUE;
}
}
return 0.0F;
}
-static int game_wants_statusbar(void)
+static int game_timing_state(game_state *state, game_ui *ui)
+{
+ return TRUE;
+}
+
+static void game_print_size(game_params *params, float *x, float *y)
+{
+ int pw, ph;
+
+ /*
+ * I'll use 9mm squares by default. They should be quite big
+ * for this game, because players will want to jot down no end
+ * of pencil marks in the squares.
+ */
+ game_compute_size(params, 900, &pw, &ph);
+ *x = pw / 100.0;
+ *y = ph / 100.0;
+}
+
+static void game_print(drawing *dr, game_state *state, int tilesize)
{
- return FALSE;
+ int c = state->c, r = state->r, cr = c*r;
+ int ink = print_mono_colour(dr, 0);
+ int x, y;
+
+ /* Ick: fake up `ds->tilesize' for macro expansion purposes */
+ game_drawstate ads, *ds = &ads;
+ game_set_size(dr, ds, NULL, tilesize);
+
+ /*
+ * Border.
+ */
+ print_line_width(dr, 3 * TILE_SIZE / 40);
+ draw_rect_outline(dr, BORDER, BORDER, cr*TILE_SIZE, cr*TILE_SIZE, ink);
+
+ /*
+ * Grid.
+ */
+ for (x = 1; x < cr; x++) {
+ print_line_width(dr, (x % r ? 1 : 3) * TILE_SIZE / 40);
+ draw_line(dr, BORDER+x*TILE_SIZE, BORDER,
+ BORDER+x*TILE_SIZE, BORDER+cr*TILE_SIZE, ink);
+ }
+ for (y = 1; y < cr; y++) {
+ print_line_width(dr, (y % c ? 1 : 3) * TILE_SIZE / 40);
+ draw_line(dr, BORDER, BORDER+y*TILE_SIZE,
+ BORDER+cr*TILE_SIZE, BORDER+y*TILE_SIZE, ink);
+ }
+
+ /*
+ * Numbers.
+ */
+ for (y = 0; y < cr; y++)
+ for (x = 0; x < cr; x++)
+ if (state->grid[y*cr+x]) {
+ char str[2];
+ str[1] = '\0';
+ str[0] = state->grid[y*cr+x] + '0';
+ if (str[0] > '9')
+ str[0] += 'a' - ('9'+1);
+ draw_text(dr, BORDER + x*TILE_SIZE + TILE_SIZE/2,
+ BORDER + y*TILE_SIZE + TILE_SIZE/2,
+ FONT_VARIABLE, TILE_SIZE/2,
+ ALIGN_VCENTRE | ALIGN_HCENTRE, ink, str);
+ }
}
#ifdef COMBINED
#endif
const struct game thegame = {
- "Solo", "games.solo",
+ "Solo", "games.solo", "solo",
default_params,
game_fetch_preset,
decode_params,
TRUE, game_configure, custom_params,
validate_params,
new_game_desc,
- game_free_aux_info,
validate_desc,
new_game,
dup_game,
TRUE, game_text_format,
new_ui,
free_ui,
- make_move,
- game_size,
+ encode_ui,
+ decode_ui,
+ game_changed_state,
+ interpret_move,
+ execute_move,
+ PREFERRED_TILE_SIZE, game_compute_size, game_set_size,
game_colours,
game_new_drawstate,
game_free_drawstate,
game_redraw,
game_anim_length,
game_flash_length,
- game_wants_statusbar,
+ TRUE, FALSE, game_print_size, game_print,
+ FALSE, /* wants_statusbar */
+ FALSE, game_timing_state,
+ 0, /* flags */
};
#ifdef STANDALONE_SOLVER
-/*
- * gcc -DSTANDALONE_SOLVER -o solosolver solo.c malloc.c
- */
-
-void frontend_default_colour(frontend *fe, float *output) {}
-void draw_text(frontend *fe, int x, int y, int fonttype, int fontsize,
- int align, int colour, char *text) {}
-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) {}
-void clip(frontend *fe, int x, int y, int w, int h) {}
-void unclip(frontend *fe) {}
-void start_draw(frontend *fe) {}
-void draw_update(frontend *fe, int x, int y, int w, int h) {}
-void end_draw(frontend *fe) {}
-unsigned long random_bits(random_state *state, int bits)
-{ assert(!"Shouldn't get randomness"); return 0; }
-unsigned long random_upto(random_state *state, unsigned long limit)
-{ assert(!"Shouldn't get randomness"); return 0; }
-
-void fatal(char *fmt, ...)
-{
- va_list ap;
-
- fprintf(stderr, "fatal error: ");
-
- va_start(ap, fmt);
- vfprintf(stderr, fmt, ap);
- va_end(ap);
-
- fprintf(stderr, "\n");
- exit(1);
-}
-
int main(int argc, char **argv)
{
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;
}
fprintf(stderr, "%s: %s\n", argv[0], err);
return 1;
}
- s = new_game(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]);
- }
+ s = new_game(NULL, p, desc);
+
+ ret = solver(p->c, p->r, s->grid, 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_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)":
+ "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;
}
~mdw / sgt / puzzles / blobdiff