* seems to be taking ascenders/descenders into account when
* centring. Ick.
*
- * - implement stronger modes of reasoning in nsolve, thus
- * enabling harder puzzles
- * + and having done that, supply configurable difficulty
- * levels
- *
* - it might still be nice to do some prioritisation on the
* removal of numbers from the grid
* + one possibility is to try to minimise the maximum number
* one thing is ever highlighted at a time, so there's no way
* to confuse the two.
* + `pencil marks' might be useful for more subtle forms of
- * deduction, once we implement creation of puzzles that
- * require it.
+ * deduction, now we can create puzzles that require them.
*/
/*
#include <ctype.h>
#include <math.h>
+#ifdef STANDALONE_SOLVER
+#include <stdarg.h>
+int solver_show_working;
+#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
enum { SYMM_NONE, SYMM_ROT2, SYMM_ROT4, SYMM_REF4 };
+enum { DIFF_BLOCK, DIFF_SIMPLE, DIFF_INTERSECT,
+ DIFF_SET, DIFF_RECURSIVE, DIFF_AMBIGUOUS, DIFF_IMPOSSIBLE };
+
enum {
COL_BACKGROUND,
COL_GRID,
};
struct game_params {
- int c, r, symm;
+ int c, r, symm, diff;
};
struct game_state {
ret->c = ret->r = 3;
ret->symm = SYMM_ROT2; /* a plausible default */
+ ret->diff = DIFF_SIMPLE; /* so is this */
return ret;
}
-static int game_fetch_preset(int i, char **name, game_params **params)
-{
- game_params *ret;
- int c, r;
- char buf[80];
-
- switch (i) {
- case 0: c = 2, r = 2; break;
- case 1: c = 2, r = 3; break;
- case 2: c = 3, r = 3; break;
- case 3: c = 3, r = 4; break;
- case 4: c = 4, r = 4; break;
- default: return FALSE;
- }
-
- sprintf(buf, "%dx%d", c, r);
- *name = dupstr(buf);
- *params = ret = snew(game_params);
- ret->c = c;
- ret->r = r;
- ret->symm = SYMM_ROT2;
- /* FIXME: difficulty presets? */
- return TRUE;
-}
-
static void free_params(game_params *params)
{
sfree(params);
return ret;
}
+static int game_fetch_preset(int i, char **name, game_params **params)
+{
+ static struct {
+ char *title;
+ game_params params;
+ } presets[] = {
+ { "2x2 Trivial", { 2, 2, SYMM_ROT2, DIFF_BLOCK } },
+ { "2x3 Basic", { 2, 3, SYMM_ROT2, DIFF_SIMPLE } },
+ { "3x3 Basic", { 3, 3, SYMM_ROT2, DIFF_SIMPLE } },
+ { "3x3 Intermediate", { 3, 3, SYMM_ROT2, DIFF_INTERSECT } },
+ { "3x3 Advanced", { 3, 3, SYMM_ROT2, DIFF_SET } },
+ { "3x4 Basic", { 3, 4, SYMM_ROT2, DIFF_SIMPLE } },
+ { "4x4 Basic", { 4, 4, SYMM_ROT2, DIFF_SIMPLE } },
+ };
+
+ if (i < 0 || i >= lenof(presets))
+ return FALSE;
+
+ *name = dupstr(presets[i].title);
+ *params = dup_params(&presets[i].params);
+
+ return TRUE;
+}
+
static game_params *decode_params(char const *string)
{
game_params *ret = default_params();
ret->r = atoi(string);
while (*string && isdigit((unsigned char)*string)) string++;
}
- if (*string == 'r' || *string == 'm' || *string == 'a') {
- int sn, sc;
- sc = *string++;
- sn = atoi(string);
- while (*string && isdigit((unsigned char)*string)) string++;
- if (sc == 'm' && sn == 4)
- ret->symm = SYMM_REF4;
- if (sc == 'r' && sn == 4)
- ret->symm = SYMM_ROT4;
- if (sc == 'r' && sn == 2)
- ret->symm = SYMM_ROT2;
- if (sc == 'a')
- ret->symm = SYMM_NONE;
+ while (*string) {
+ if (*string == 'r' || *string == 'm' || *string == 'a') {
+ int sn, sc;
+ sc = *string++;
+ sn = atoi(string);
+ while (*string && isdigit((unsigned char)*string)) string++;
+ if (sc == 'm' && sn == 4)
+ ret->symm = SYMM_REF4;
+ if (sc == 'r' && sn == 4)
+ ret->symm = SYMM_ROT4;
+ if (sc == 'r' && sn == 2)
+ ret->symm = SYMM_ROT2;
+ if (sc == 'a')
+ ret->symm = SYMM_NONE;
+ } else if (*string == 'd') {
+ string++;
+ if (*string == 't') /* trivial */
+ string++, ret->diff = DIFF_BLOCK;
+ else if (*string == 'b') /* basic */
+ string++, ret->diff = DIFF_SIMPLE;
+ else if (*string == 'i') /* intermediate */
+ string++, ret->diff = DIFF_INTERSECT;
+ else if (*string == 'a') /* advanced */
+ string++, ret->diff = DIFF_SET;
+ } else
+ string++; /* eat unknown character */
}
- /* FIXME: difficulty levels */
return ret;
}
ret[2].sval = ":None:2-way rotation:4-way rotation:4-way mirror";
ret[2].ival = params->symm;
- /*
- * FIXME: difficulty level.
- */
+ ret[3].name = "Difficulty";
+ ret[3].type = C_CHOICES;
+ ret[3].sval = ":Trivial:Basic:Intermediate:Advanced";
+ ret[3].ival = params->diff;
- ret[3].name = NULL;
- ret[3].type = C_END;
- ret[3].sval = NULL;
- ret[3].ival = 0;
+ ret[4].name = NULL;
+ ret[4].type = C_END;
+ ret[4].sval = NULL;
+ ret[4].ival = 0;
return ret;
}
ret->c = atoi(cfg[0].sval);
ret->r = atoi(cfg[1].sval);
ret->symm = cfg[2].ival;
+ ret->diff = cfg[3].ival;
return ret;
}
* in because all the other numbers that could go in it are
* ruled out.
*
- * More advanced modes of reasoning I'd like to support in future:
- *
- * - Intersectional elimination: given two domains which overlap
+ * - Intersectional analysis: given two domains which overlap
* (hence one must be a block, and the other can be a row or
* col), if the possible locations for a particular number in
* one of the domains can be narrowed down to the overlap, then
* that number can be ruled out everywhere but the overlap in
* the other domain too.
*
- * - Setwise numeric elimination: if there is a subset of the
- * empty squares within a domain such that the union of the
- * possible numbers in that subset has the same size as the
- * subset itself, then those numbers can be ruled out everywhere
- * else in the domain. (For example, if there are five empty
- * squares and the possible numbers in each are 12, 23, 13, 134
- * and 1345, then the first three empty squares form such a
- * subset: the numbers 1, 2 and 3 _must_ be in those three
- * squares in some permutation, and hence we can deduce none of
- * them can be in the fourth or fifth squares.)
- *
- * - Setwise positional elimination: if there is a subset of the
- * unplaced numbers within a domain such that the union of all
- * their possible positions has the same size as the subset
- * itself, then all other numbers can be ruled out for those
- * positions.
+ * - Set elimination: if there is a subset of the empty squares
+ * within a domain such that the union of the possible numbers
+ * in that subset has the same size as the subset itself, then
+ * those numbers can be ruled out everywhere else in the domain.
+ * (For example, if there are five empty squares and the
+ * possible numbers in each are 12, 23, 13, 134 and 1345, then
+ * the first three empty squares form such a subset: the numbers
+ * 1, 2 and 3 _must_ be in those three squares in some
+ * permutation, and hence we can deduce none of them can be in
+ * the fourth or fifth squares.)
+ * + You can also see this the other way round, concentrating
+ * on numbers rather than squares: if there is a subset of
+ * the unplaced numbers within a domain such that the union
+ * of all their possible positions has the same size as the
+ * subset itself, then all other numbers can be ruled out for
+ * those positions. However, it turns out that this is
+ * exactly equivalent to the first formulation at all times:
+ * there is a 1-1 correspondence between suitable subsets of
+ * the unplaced numbers and suitable subsets of the unfilled
+ * places, found by taking the _complement_ of the union of
+ * the numbers' possible positions (or the spaces' possible
+ * contents).
*/
/*
* in its row, its column and its block.
*/
usage->row[y*cr+n-1] = usage->col[x*cr+n-1] =
- usage->blk[((y/c)*c+(x/r))*cr+n-1] = TRUE;
+ 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 nsolve_elim(struct nsolve_usage *usage, int start, int step
+#ifdef STANDALONE_SOLVER
+ , char *fmt, ...
+#endif
+ )
{
int c = usage->c, r = usage->r, cr = c*r;
int fpos, m, i;
y %= cr;
if (!usage->grid[YUNTRANS(y)*cr+x]) {
+#ifdef STANDALONE_SOLVER
+ if (solver_show_working) {
+ va_list ap;
+ va_start(ap, fmt);
+ vprintf(fmt, ap);
+ va_end(ap);
+ printf(":\n placing %d at (%d,%d)\n",
+ n, 1+x, 1+YUNTRANS(y));
+ }
+#endif
nsolve_place(usage, x, y, n);
return TRUE;
}
return FALSE;
}
+static int nsolve_intersect(struct nsolve_usage *usage,
+ int start1, int step1, int start2, int step2
+#ifdef STANDALONE_SOLVER
+ , char *fmt, ...
+#endif
+ )
+{
+ int c = usage->c, r = usage->r, cr = c*r;
+ int ret, i;
+
+ /*
+ * Loop over the first domain and see if there's any set bit
+ * not also in the second.
+ */
+ for (i = 0; i < cr; i++) {
+ int p = start1+i*step1;
+ if (usage->cube[p] &&
+ !(p >= start2 && p < start2+cr*step2 &&
+ (p - start2) % step2 == 0))
+ return FALSE; /* 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.
+ */
+ ret = FALSE;
+ for (i = 0; i < cr; i++) {
+ int p = start2+i*step2;
+ if (usage->cube[p] &&
+ !(p >= start1 && p < start1+cr*step1 && (p - start1) % step1 == 0))
+ {
+#ifdef STANDALONE_SOLVER
+ if (solver_show_working) {
+ int px, py, pn;
+
+ if (!ret) {
+ va_list ap;
+ va_start(ap, fmt);
+ vprintf(fmt, ap);
+ va_end(ap);
+ printf(":\n");
+ }
+
+ pn = 1 + p % cr;
+ py = p / cr;
+ px = py / cr;
+ py %= cr;
+
+ printf(" ruling out %d at (%d,%d)\n",
+ pn, 1+px, 1+YUNTRANS(py));
+ }
+#endif
+ ret = TRUE; /* we did something */
+ usage->cube[p] = 0;
+ }
+ }
+
+ return ret;
+}
+
+static int nsolve_set(struct nsolve_usage *usage,
+ int start, int step1, int step2
+#ifdef STANDALONE_SOLVER
+ , char *fmt, ...
+#endif
+ )
+{
+ 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);
+
+ /*
+ * We are passed a cr-by-cr matrix of booleans. Our first job
+ * is to winnow it by finding any definite placements - i.e.
+ * any row with a solitary 1 - and discarding that row and the
+ * column containing the 1.
+ */
+ memset(rowidx, TRUE, cr);
+ memset(colidx, TRUE, cr);
+ for (i = 0; i < cr; i++) {
+ int count = 0, first = -1;
+ 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 == 1)
+ rowidx[i] = colidx[first] = FALSE;
+ }
+
+ /*
+ * Convert each of rowidx/colidx from a list of 0s and 1s to a
+ * list of the indices of the 1s.
+ */
+ for (i = j = 0; i < cr; i++)
+ if (rowidx[i])
+ rowidx[j++] = i;
+ n = j;
+ for (i = j = 0; i < cr; i++)
+ if (colidx[i])
+ colidx[j++] = i;
+ assert(n == j);
+
+ /*
+ * And create the smaller matrix.
+ */
+ for (i = 0; i < n; i++)
+ for (j = 0; j < n; j++)
+ grid[i*cr+j] = usage->cube[start+rowidx[i]*step1+colidx[j]*step2];
+
+ /*
+ * Having done that, we now have a matrix in which every row
+ * has at least two 1s in. Now we search to see if we can find
+ * a rectangle of zeroes (in the set-theoretic sense of
+ * `rectangle', i.e. a subset of rows crossed with a subset of
+ * columns) whose width and height add up to n.
+ */
+
+ memset(set, 0, n);
+ count = 0;
+ while (1) {
+ /*
+ * We have a candidate set. If its size is <=1 or >=n-1
+ * then we move on immediately.
+ */
+ if (count > 1 && count < n-1) {
+ /*
+ * The number of rows we need is n-count. See if we can
+ * find that many rows which each have a zero in all
+ * the positions listed in `set'.
+ */
+ int rows = 0;
+ for (i = 0; i < n; i++) {
+ int ok = TRUE;
+ for (j = 0; j < n; j++)
+ if (set[j] && grid[i*cr+j]) {
+ ok = FALSE;
+ break;
+ }
+ if (ok)
+ rows++;
+ }
+
+ /*
+ * We expect never to be able to get _more_ than
+ * n-count suitable rows: this would imply that (for
+ * example) there are four numbers which between them
+ * have at most three possible positions, and hence it
+ * indicates a faulty deduction before this point or
+ * even a bogus clue.
+ */
+ assert(rows <= n - count);
+ 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.
+ *
+ * This involves referring back through
+ * rowidx/colidx in order to work out which actual
+ * positions in the cube to meddle with.
+ */
+ for (i = 0; i < n; i++) {
+ int ok = TRUE;
+ for (j = 0; j < n; j++)
+ if (set[j] && grid[i*cr+j]) {
+ ok = FALSE;
+ break;
+ }
+ if (!ok) {
+ for (j = 0; j < n; j++)
+ if (!set[j] && grid[i*cr+j]) {
+ int fpos = (start+rowidx[i]*step1+
+ colidx[j]*step2);
+#ifdef STANDALONE_SOLVER
+ if (solver_show_working) {
+ int px, py, pn;
+
+ if (!progress) {
+ va_list ap;
+ va_start(ap, fmt);
+ vprintf(fmt, ap);
+ va_end(ap);
+ printf(":\n");
+ }
+
+ pn = 1 + fpos % cr;
+ py = fpos / cr;
+ px = py / cr;
+ py %= cr;
+
+ printf(" ruling out %d at (%d,%d)\n",
+ pn, 1+px, 1+YUNTRANS(py));
+ }
+#endif
+ progress = TRUE;
+ usage->cube[fpos] = FALSE;
+ }
+ }
+ }
+
+ if (progress) {
+ sfree(set);
+ sfree(colidx);
+ sfree(rowidx);
+ sfree(grid);
+ return 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 */
+ }
+
+ sfree(set);
+ sfree(colidx);
+ sfree(rowidx);
+ sfree(grid);
+
+ return FALSE;
+}
+
static int nsolve(int c, int r, digit *grid)
{
struct nsolve_usage *usage;
int cr = c*r;
int x, y, n;
+ int diff = DIFF_BLOCK;
/*
* Set up a usage structure as a clean slate (everything
* 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:
/*
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))
+ 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))
+ 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))
+ 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))
- goto cont;
+ 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, 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, 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, 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, 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, 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
for (x = 0; x < cr; x++)
for (y = 0; y < cr; y++)
if (!grid[y*cr+x])
- return FALSE;
- return TRUE;
+ return DIFF_IMPOSSIBLE;
+ return diff;
}
/* ----------------------------------------------------------------------
char *seed;
int coords[16], ncoords;
int xlim, ylim;
+ int maxdiff;
/*
- * Start the recursive solver with an empty grid to generate a
- * random solved state.
+ * Adjust the maximum difficulty level to be consistent with
+ * the puzzle size: all 2x2 puzzles appear to be Trivial
+ * (DIFF_BLOCK) so we cannot hold out for even a Basic
+ * (DIFF_SIMPLE) one.
*/
- grid = snewn(area, digit);
- memset(grid, 0, area);
- ret = rsolve(c, r, grid, rs, 1);
- assert(ret == 1);
- assert(check_valid(c, r, grid));
+ maxdiff = params->diff;
+ if (c == 2 && r == 2)
+ maxdiff = DIFF_BLOCK;
- /*
- * Now we have a solved grid, start removing things from it
- * while preserving solubility.
- */
+ grid = snewn(area, digit);
locs = snewn(area, struct xy);
grid2 = snewn(area, digit);
- symmetry_limit(params, &xlim, &ylim, params->symm);
- 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++;
- }
-
- /*
- * 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;
- }
- }
+ /*
+ * Loop until we get a grid of the required difficulty. This is
+ * nasty, but it seems to be unpleasantly hard to generate
+ * difficult grids otherwise.
+ */
+ do {
+ /*
+ * Start the recursive solver with an empty grid to generate a
+ * random solved state.
+ */
+ memset(grid, 0, area);
+ ret = rsolve(c, r, grid, rs, 1);
+ assert(ret == 1);
+ assert(check_valid(c, r, grid));
+
+ /*
+ * Now we have a solved grid, start removing things from it
+ * while preserving solubility.
+ */
+ symmetry_limit(params, &xlim, &ylim, params->symm);
+ 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++;
+ }
+
+ /*
+ * 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;
+ }
+ }
+
+ /*
+ * 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++) {
+ x = locs[i].x;
+ y = locs[i].y;
+
+ 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 (nsolve(c, r, grid2) <= maxdiff) {
+ 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.
+ */
+ break;
+ }
+ }
- /*
- * 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++) {
- x = locs[i].x;
- y = locs[i].y;
-
- 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 (nsolve(c, r, grid2)) {
- for (j = 0; j < ncoords; j++)
- grid[coords[2*j+1]*cr+coords[2*j]] = 0;
- break;
- }
- }
+ memcpy(grid2, grid, area);
+ } while (nsolve(c, r, grid2) != maxdiff);
- if (i == nlocs) {
- /*
- * There was nothing we could remove without destroying
- * solvability.
- */
- break;
- }
- }
sfree(grid2);
sfree(locs);
#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 start_draw(frontend *fe) {}
void draw_update(frontend *fe, int x, int y, int w, int h) {}
void end_draw(frontend *fe) {}
-
-#include <stdarg.h>
+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, ...)
{
{
game_params *p;
game_state *s;
- int recurse = FALSE;
+ int recurse = TRUE;
char *id = NULL, *seed, *err;
int y, x;
+ int grade = FALSE;
while (--argc > 0) {
char *p = *++argv;
recurse = TRUE;
} else if (!strcmp(p, "-n")) {
recurse = FALSE;
+ } else 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]);
return 1;
}
if (!id) {
- fprintf(stderr, "usage: %s [-n | -r] <game_id>\n", argv[0]);
+ fprintf(stderr, "usage: %s [-n | -r | -g | -v] <game_id>\n", argv[0]);
return 1;
}
if (recurse) {
int ret = rsolve(p->c, p->r, s->grid, NULL, 2);
if (ret > 1) {
- printf("multiple solutions detected; only first one output\n");
+ fprintf(stderr, "%s: rsolve: multiple solutions detected\n",
+ argv[0]);
}
} else {
- nsolve(p->c, p->r, s->grid);
+ 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 ? "blockwise positional elimination only":
+ ret==DIFF_SIMPLE ? "row/column/number elimination required":
+ ret==DIFF_INTERSECT ? "intersectional analysis required":
+ ret==DIFF_SET ? "set elimination required":
+ ret==DIFF_RECURSIVE ? "guesswork and backtracking required":
+ ret==DIFF_AMBIGUOUS ? "multiple solutions exist":
+ ret==DIFF_IMPOSSIBLE ? "no solution exists":
+ "INTERNAL ERROR: unrecognised difficulty code");
+ }
}
for (y = 0; y < p->c * p->r; y++) {
for (x = 0; x < p->c * p->r; x++) {
- printf("%2.0d", s->grid[y * p->c * p->r + x]);
+ int c = s->grid[y * p->c * p->r + x];
+ if (c == 0)
+ c = ' ';
+ else if (c <= 9)
+ c = '0' + c;
+ else
+ c = 'a' + c-10;
+ printf("%c", c);
+ if (x+1 < p->c * p->r) {
+ if ((x+1) % p->c)
+ printf(" ");
+ else
+ printf(" | ");
+ }
}
printf("\n");
+ if (y+1 < p->c * p->r && (y+1) % p->r == 0) {
+ for (x = 0; x < p->c * p->r; x++) {
+ printf("-");
+ if (x+1 < p->c * p->r) {
+ if ((x+1) % p->c)
+ printf("-");
+ else
+ printf("-+-");
+ }
+ }
+ printf("\n");
+ }
}
printf("\n");
~mdw / sgt / puzzles / commitdiff