*
* TODO:
*
- * - can we do anything about nasty centring of text in GTK? It
- * 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
+ * - Jigsaw Sudoku is currently an undocumented feature enabled
+ * by setting r (`Rows of sub-blocks' in the GUI configurer) to
+ * 1. The reason it's undocumented is because they're rather
+ * erratic to generate, because gridgen tends to hang up for
+ * ages. I think this is because some jigsaw block layouts
+ * simply do not admit very many valid filled grids (and
+ * perhaps some have none at all).
+ * + To fix this, I think probably the solution is a change in
+ * grid generation policy: gridgen needs to have less of an
+ * all-or-nothing attitude and instead make only a limited
+ * amount of effort to construct a filled grid before giving
+ * up and trying a new layout. (Come to think of it, this
+ * same change might also make 5x5 standard Sudoku more
+ * practical to generate, if correctly tuned.)
+ * + If I get this fixed, other work needed on jigsaw mode is:
+ * * introduce a GUI config checkbox. game_configure()
+ * ticks this box iff r==1; if it's ticked in a call to
+ * custom_params(), we replace (c, r) with (c*r, 1).
+ * * document it.
*
+ * - reports from users are that `Trivial'-mode puzzles are still
+ * rather hard compared to newspapers' easy ones, so some better
+ * low-end difficulty grading would be nice
+ * + it's possible that really easy puzzles always have
+ * _several_ things you can do, so don't make you hunt too
+ * hard for the one deduction you can currently make
+ * + it's also possible that easy puzzles require fewer
+ * cross-eliminations: perhaps there's a higher incidence of
+ * things you can deduce by looking only at (say) rows,
+ * rather than things you have to check both rows and columns
+ * for
+ * + but really, what I need to do is find some really easy
+ * puzzles and _play_ them, to see what's actually easy about
+ * them
+ * + while I'm revamping this area, filling in the _last_
+ * number in a nearly-full row or column should certainly be
+ * permitted even at the lowest difficulty level.
+ * + also Owen noticed that `Basic' grids requiring numeric
+ * elimination are actually very hard, so I wonder if a
+ * difficulty gradation between that and positional-
+ * elimination-only might be in order
+ * + but it's not good to have _too_ many difficulty levels, or
+ * it'll take too long to randomly generate a given level.
+ *
* - 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
* of filled squares in any block, which in particular ought
* to enforce never leaving a completely filled block in the
* puzzle as presented.
- * + be careful of being too clever here, though, until after
- * I've tried implementing difficulty levels. It's not
- * impossible that those might impose much more important
- * constraints on this process.
*
* - alternative interface modes
* + sudoku.com's Windows program has a palette of possible
* click, _or_ you highlight a square and then type. At most
* 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.
+ * + then again, I don't actually like sudoku.com's interface;
+ * it's too much like a paint package whereas I prefer to
+ * think of Solo as a text editor.
+ * + another PDA-friendly possibility is a drag interface:
+ * _drag_ numbers from the palette into the grid squares.
+ * Thought experiments suggest I'd prefer that to the
+ * sudoku.com approach, but I haven't actually tried it.
*/
/*
#include <ctype.h>
#include <math.h>
+#ifdef STANDALONE_SOLVER
+#include <stdarg.h>
+int solver_show_working, solver_recurse_depth;
+#endif
+
#include "puzzles.h"
/*
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 GRIDEXTRA (TILE_SIZE / 32)
#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_EXTREME,
+ DIFF_RECURSIVE, DIFF_AMBIGUOUS, DIFF_IMPOSSIBLE };
enum {
COL_BACKGROUND,
+ COL_XDIAGONALS,
COL_GRID,
COL_CLUE,
COL_USER,
COL_HIGHLIGHT,
+ COL_ERROR,
+ COL_PENCIL,
NCOLOURS
};
struct game_params {
- int c, r, symm;
+ /*
+ * For a square puzzle, `c' and `r' indicate the puzzle
+ * parameters as described above.
+ *
+ * A jigsaw-style puzzle is indicated by r==1, in which case c
+ * can be whatever it likes (there is no constraint on
+ * compositeness - a 7x7 jigsaw sudoku makes perfect sense).
+ */
+ int c, r, symm, diff;
+ int xtype; /* require all digits in X-diagonals */
};
-struct game_state {
+struct block_structure {
+ int refcount;
+
+ /*
+ * For text formatting, we do need c and r here.
+ */
int c, r;
+
+ /*
+ * For any square index, whichblock[i] gives its block index.
+ *
+ * For 0 <= b,i < cr, blocks[b][i] gives the index of the ith
+ * square in block b.
+ *
+ * whichblock and blocks are each dynamically allocated in
+ * their own right, but the subarrays in blocks are appended
+ * to the whichblock array, so shouldn't be freed
+ * individually.
+ */
+ int *whichblock, **blocks;
+
+#ifdef STANDALONE_SOLVER
+ /*
+ * Textual descriptions of each block. For normal Sudoku these
+ * are of the form "(1,3)"; for jigsaw they are "starting at
+ * (5,7)". So the sensible usage in both cases is to say
+ * "elimination within block %s" with one of these strings.
+ *
+ * Only blocknames itself needs individually freeing; it's all
+ * one block.
+ */
+ char **blocknames;
+#endif
+};
+
+struct game_state {
+ /*
+ * For historical reasons, I use `cr' to denote the overall
+ * width/height of the puzzle. It was a natural notation when
+ * all puzzles were divided into blocks in a grid, but doesn't
+ * really make much sense given jigsaw puzzles. However, the
+ * obvious `n' is heavily used in the solver to describe the
+ * index of a number being placed, so `cr' will have to stay.
+ */
+ int cr;
+ struct block_structure *blocks;
+ int xtype;
digit *grid;
+ unsigned char *pencil; /* c*r*c*r elements */
unsigned char *immutable; /* marks which digits are clues */
- int completed;
+ int completed, cheated;
};
static game_params *default_params(void)
game_params *ret = snew(game_params);
ret->c = ret->r = 3;
+ ret->xtype = FALSE;
ret->symm = SYMM_ROT2; /* a plausible default */
+ ret->diff = DIFF_BLOCK; /* 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 game_params *decode_params(char const *string)
+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, FALSE } },
+ { "2x3 Basic", { 2, 3, SYMM_ROT2, DIFF_SIMPLE, FALSE } },
+ { "3x3 Trivial", { 3, 3, SYMM_ROT2, DIFF_BLOCK, FALSE } },
+ { "3x3 Basic", { 3, 3, SYMM_ROT2, DIFF_SIMPLE, FALSE } },
+ { "3x3 Basic X", { 3, 3, SYMM_ROT2, DIFF_SIMPLE, TRUE } },
+ { "3x3 Intermediate", { 3, 3, SYMM_ROT2, DIFF_INTERSECT, FALSE } },
+ { "3x3 Advanced", { 3, 3, SYMM_ROT2, DIFF_SET, FALSE } },
+ { "3x3 Advanced X", { 3, 3, SYMM_ROT2, DIFF_SET, TRUE } },
+ { "3x3 Extreme", { 3, 3, SYMM_ROT2, DIFF_EXTREME, FALSE } },
+ { "3x3 Unreasonable", { 3, 3, SYMM_ROT2, DIFF_RECURSIVE, FALSE } },
+#ifndef SLOW_SYSTEM
+ { "3x4 Basic", { 3, 4, SYMM_ROT2, DIFF_SIMPLE, FALSE } },
+ { "4x4 Basic", { 4, 4, SYMM_ROT2, DIFF_SIMPLE, FALSE } },
+#endif
+ };
+
+ if (i < 0 || i >= lenof(presets))
+ return FALSE;
+
+ *name = dupstr(presets[i].title);
+ *params = dup_params(&presets[i].params);
+
+ return TRUE;
+}
+
+static void decode_params(game_params *ret, char const *string)
{
- game_params *ret = default_params();
+ int seen_r = FALSE;
ret->c = ret->r = atoi(string);
- ret->symm = SYMM_ROT2;
+ ret->xtype = FALSE;
while (*string && isdigit((unsigned char)*string)) string++;
if (*string == 'x') {
string++;
ret->r = atoi(string);
+ seen_r = TRUE;
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 == 'j') {
+ string++;
+ if (seen_r)
+ ret->c *= ret->r;
+ ret->r = 1;
+ } else if (*string == 'x') {
+ string++;
+ ret->xtype = TRUE;
+ } else if (*string == 'r' || *string == 'm' || *string == 'a') {
+ int sn, sc, sd;
+ sc = *string++;
+ if (sc == 'm' && *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 = 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)
+ 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 if (*string == 'e') /* extreme */
+ string++, ret->diff = DIFF_EXTREME;
+ else if (*string == 'u') /* unreasonable */
+ string++, ret->diff = DIFF_RECURSIVE;
+ } else
+ string++; /* eat unknown character */
}
- /* FIXME: difficulty levels */
-
- return ret;
}
-static char *encode_params(game_params *params)
+static char *encode_params(game_params *params, int full)
{
char str[80];
- /*
- * Symmetry is a game generation preference and hence is left
- * out of the encoding. Users can add it back in as they see
- * fit.
- */
- sprintf(str, "%dx%d", params->c, params->r);
+ if (params->r > 1)
+ sprintf(str, "%dx%d", params->c, params->r);
+ else
+ sprintf(str, "%dj", params->c);
+ if (params->xtype)
+ strcat(str, "x");
+
+ 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;
+ }
+ switch (params->diff) {
+ /* case DIFF_BLOCK: strcat(str, "dt"); break; [default] */
+ 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;
+ }
+ }
return dupstr(str);
}
config_item *ret;
char buf[80];
- ret = snewn(5, config_item);
+ ret = snewn(6, config_item);
ret[0].name = "Columns of sub-blocks";
ret[0].type = C_STRING;
ret[1].sval = dupstr(buf);
ret[1].ival = 0;
- ret[2].name = "Symmetry";
- ret[2].type = C_CHOICES;
- ret[2].sval = ":None:2-way rotation:4-way rotation:4-way mirror";
- ret[2].ival = params->symm;
+ ret[2].name = "\"X\" (require every number in each main diagonal)";
+ ret[2].type = C_BOOLEAN;
+ ret[2].sval = NULL;
+ ret[2].ival = params->xtype;
- /*
- * FIXME: difficulty level.
- */
+ ret[3].name = "Symmetry";
+ ret[3].type = C_CHOICES;
+ ret[3].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[3].ival = params->symm;
- ret[3].name = NULL;
- ret[3].type = C_END;
- ret[3].sval = NULL;
- ret[3].ival = 0;
+ ret[4].name = "Difficulty";
+ ret[4].type = C_CHOICES;
+ ret[4].sval = ":Trivial:Basic:Intermediate:Advanced:Extreme:Unreasonable";
+ ret[4].ival = params->diff;
+
+ ret[5].name = NULL;
+ ret[5].type = C_END;
+ ret[5].sval = NULL;
+ ret[5].ival = 0;
return ret;
}
ret->c = atoi(cfg[0].sval);
ret->r = atoi(cfg[1].sval);
- ret->symm = cfg[2].ival;
+ ret->xtype = cfg[2].ival;
+ ret->symm = cfg[3].ival;
+ ret->diff = cfg[4].ival;
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)
+ if (params->c < 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) > 35)
+ return "Unable to support more than 35 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.
* 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.)
+ * - 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).
*
- * - 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.
- */
-
-/*
- * Within this solver, I'm going to transform all y-coordinates by
- * inverting the significance of the block number and the position
- * within the block. That is, we will start with the top row of
- * each block in order, then the second row of each block in order,
- * etc.
+ * - Forcing chains (see comment for solver_forcing().)
*
- * This transformation has the enormous advantage that it means
- * every row, column _and_ block is described by an arithmetic
- * progression of coordinates within the cubic array, so that I can
- * use the same very simple function to do blockwise, row-wise and
- * column-wise elimination.
+ * - 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 {
- int c, r, cr;
+struct solver_usage {
+ int cr;
+ struct block_structure *blocks;
/*
* We set up a cubic array, indexed by x, y and digit; each
* element of this array is TRUE or FALSE according to whether
* or not that digit _could_ in principle go in that position.
*
- * The way to index this array is cube[(x*cr+y)*cr+n-1].
- * y-coordinates in here are transformed.
+ * The way to index this array is cube[(y*cr+x)*cr+n-1]; there
+ * are macros below to help with this.
*/
unsigned char *cube;
/*
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) */
+ /* blk[i*cr+n-1] TRUE if digit n has been placed in block i */
unsigned char *blk;
+ /* diag[i*cr+n-1] TRUE if digit n has been placed in diagonal i */
+ unsigned char *diag; /* diag 0 is \, 1 is / */
};
-#define cubepos(x,y,n) (((x)*usage->cr+(y))*usage->cr+(n)-1)
+#define cubepos2(xy,n) ((xy)*usage->cr+(n)-1)
+#define cubepos(x,y,n) cubepos2((y)*usage->cr+(x),n)
#define cube(x,y,n) (usage->cube[cubepos(x,y,n)])
+#define cube2(xy,n) (usage->cube[cubepos2(xy,n)])
+
+#define ondiag0(xy) ((xy) % (cr+1) == 0)
+#define ondiag1(xy) ((xy) % (cr-1) == 0 && (xy) > 0 && (xy) < cr*cr-1)
+#define diag0(i) ((i) * (cr+1))
+#define diag1(i) ((i+1) * (cr-1))
/*
* Function called when we are certain that a particular square has
* 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;
+ int cr = usage->cr;
+ int sqindex = y*cr+x;
+ int i, bi;
assert(cube(x,y,n));
/*
* Rule out this number in all other positions in the block.
*/
- bx = (x/r)*r;
- by = y % r;
- for (i = 0; i < r; i++)
- for (j = 0; j < c; j++)
- if (bx+i != x || by+j*r != y)
- cube(bx+i,by+j*r,n) = FALSE;
+ bi = usage->blocks->whichblock[sqindex];
+ for (i = 0; i < cr; i++) {
+ int bp = usage->blocks->blocks[bi][i];
+ if (bp != sqindex)
+ cube2(bp,n) = FALSE;
+ }
/*
* Enter the number in the result grid.
*/
- usage->grid[YUNTRANS(y)*cr+x] = n;
+ usage->grid[sqindex] = n;
/*
* Cross out this number from the list of numbers left to place
* 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[bi*cr+n-1] = TRUE;
+
+ if (usage->diag) {
+ if (ondiag0(sqindex)) {
+ for (i = 0; i < cr; i++)
+ if (diag0(i) != sqindex)
+ cube2(diag0(i),n) = FALSE;
+ usage->diag[n-1] = TRUE;
+ }
+ if (ondiag1(sqindex)) {
+ for (i = 0; i < cr; i++)
+ if (diag1(i) != sqindex)
+ cube2(diag1(i),n) = FALSE;
+ usage->diag[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 *indices
+#ifdef STANDALONE_SOLVER
+ , char *fmt, ...
+#endif
+ )
{
- int c = usage->c, r = usage->r, cr = c*r;
+ int cr = usage->cr;
int fpos, m, i;
/*
m = 0;
fpos = -1;
for (i = 0; i < cr; i++)
- if (usage->cube[start+i*step]) {
- fpos = start+i*step;
+ if (usage->cube[indices[i]]) {
+ fpos = indices[i];
m++;
}
assert(fpos >= 0);
n = 1 + fpos % cr;
- y = fpos / cr;
- x = y / cr;
- y %= cr;
+ x = fpos / cr;
+ y = x / cr;
+ x %= cr;
- if (!usage->grid[YUNTRANS(y)*cr+x]) {
- nsolve_place(usage, x, y, n);
- return TRUE;
+ if (!usage->grid[y*cr+x]) {
+#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 placing %d at (%d,%d)\n",
+ solver_recurse_depth*4, "", n, 1+x, 1+y);
+ }
+#endif
+ solver_place(usage, x, y, n);
+ return +1;
}
- }
-
- return FALSE;
+ } 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 0;
}
-static int nsolve(int c, int r, digit *grid)
+static int solver_intersect(struct solver_usage *usage,
+ int *indices1, int *indices2
+#ifdef STANDALONE_SOLVER
+ , char *fmt, ...
+#endif
+ )
{
- struct nsolve_usage *usage;
- int cr = c*r;
- int x, y, n;
+ int cr = usage->cr;
+ int ret, i, j;
+
+ /*
+ * Loop over the first domain and see if there's any set bit
+ * not also in the second.
+ */
+ for (i = j = 0; i < cr; i++) {
+ int p = indices1[i];
+ while (j < cr && indices2[j] < p)
+ j++;
+ if (usage->cube[p]) {
+ if (j < cr && indices2[j] == p)
+ continue; /* both domains contain this index */
+ else
+ 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 +1 iff we actually _did_ anything.
+ */
+ ret = 0;
+ for (i = j = 0; i < cr; i++) {
+ int p = indices2[i];
+ while (j < cr && indices1[j] < p)
+ j++;
+ if (usage->cube[p] && (j >= cr || indices1[j] != p)) {
+#ifdef STANDALONE_SOLVER
+ if (solver_show_working) {
+ int px, py, pn;
+
+ if (!ret) {
+ va_list ap;
+ printf("%*s", solver_recurse_depth*4, "");
+ va_start(ap, fmt);
+ vprintf(fmt, ap);
+ va_end(ap);
+ printf(":\n");
+ }
+
+ pn = 1 + p % cr;
+ px = p / cr;
+ py = px / cr;
+ px %= cr;
+
+ printf("%*s ruling out %d at (%d,%d)\n",
+ solver_recurse_depth*4, "", pn, 1+px, 1+py);
+ }
+#endif
+ ret = +1; /* we did something */
+ usage->cube[p] = 0;
+ }
+ }
+
+ return ret;
+}
+
+struct solver_scratch {
+ unsigned char *grid, *rowidx, *colidx, *set;
+ int *neighbours, *bfsqueue;
+ int *indexlist, *indexlist2;
+#ifdef STANDALONE_SOLVER
+ int *bfsprev;
+#endif
+};
+
+static int solver_set(struct solver_usage *usage,
+ struct solver_scratch *scratch,
+ int *indices
+#ifdef STANDALONE_SOLVER
+ , char *fmt, ...
+#endif
+ )
+{
+ int cr = usage->cr;
+ int i, j, n, count;
+ 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
+ * 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[indices[i*cr+j]])
+ first = j, count++;
+
+ /*
+ * 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;
+ }
+
+ /*
+ * 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[indices[rowidx[i]*cr+colidx[j]]];
+
+ /*
+ * 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.
+ */
+ 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 +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
+ * 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 = indices[rowidx[i]*cr+colidx[j]];
+#ifdef STANDALONE_SOLVER
+ if (solver_show_working) {
+ int px, py, pn;
+
+ if (!progress) {
+ va_list ap;
+ printf("%*s", solver_recurse_depth*4,
+ "");
+ va_start(ap, fmt);
+ vprintf(fmt, ap);
+ va_end(ap);
+ printf(":\n");
+ }
+
+ pn = 1 + fpos % cr;
+ px = fpos / cr;
+ py = px / cr;
+ px %= cr;
+
+ printf("%*s ruling out %d at (%d,%d)\n",
+ solver_recurse_depth*4, "",
+ pn, 1+px, 1+py);
+ }
+#endif
+ progress = TRUE;
+ usage->cube[fpos] = FALSE;
+ }
+ }
+ }
+
+ if (progress) {
+ return +1;
+ }
+ }
+ }
+
+ /*
+ * Binary increment: change the rightmost 0 to a 1, and
+ * change all 1s to the right of it to 0s.
+ */
+ i = n;
+ while (i > 0 && set[i-1])
+ set[--i] = 0, count--;
+ if (i > 0)
+ set[--i] = 1, count++;
+ else
+ break; /* done */
+ }
+
+ return 0;
+}
+
+/*
+ * 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 the same
+ * 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 cr = usage->cr;
+ 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, 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;
+ xt = usage->blocks->whichblock[yy*cr+xx];
+ for (yt = 0; yt < cr; yt++)
+ neighbours[nneighbours++] = usage->blocks->blocks[xt][yt];
+ if (usage->diag) {
+ int sqindex = yy*cr+xx;
+ if (ondiag0(sqindex)) {
+ for (i = 0; i < cr; i++)
+ neighbours[nneighbours++] = diag0(i);
+ }
+ if (ondiag1(sqindex)) {
+ for (i = 0; i < cr; i++)
+ neighbours[nneighbours++] = diag1(i);
+ }
+ }
+
+ /*
+ * 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 ||
+ (usage->blocks->whichblock[yt*cr+xt] == usage->blocks->whichblock[y*cr+x]) ||
+ (usage->diag && ((ondiag0(yt*cr+xt) && ondiag0(y*cr+x)) ||
+ (ondiag1(yt*cr+xt) && ondiag1(y*cr+x)))))) {
+#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+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+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(5*cr, int);
+ scratch->bfsqueue = snewn(cr*cr, int);
+#ifdef STANDALONE_SOLVER
+ scratch->bfsprev = snewn(cr*cr, int);
+#endif
+ scratch->indexlist = snewn(cr*cr, int); /* used for set elimination */
+ scratch->indexlist2 = snewn(cr, int); /* only used for intersect() */
+ 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->indexlist);
+ sfree(scratch->indexlist2);
+ sfree(scratch);
+}
+
+static int solver(int cr, struct block_structure *blocks, int xtype,
+ digit *grid, int maxdiff)
+{
+ struct solver_usage *usage;
+ struct solver_scratch *scratch;
+ int x, y, b, i, n, ret;
+ int diff = DIFF_BLOCK;
/*
* Set up a usage structure as a clean slate (everything
* possible).
*/
- usage = snew(struct nsolve_usage);
- usage->c = c;
- usage->r = r;
+ usage = snew(struct solver_usage);
usage->cr = cr;
+ usage->blocks = blocks;
usage->cube = snewn(cr*cr*cr, unsigned char);
usage->grid = grid; /* write straight back to the input */
memset(usage->cube, TRUE, cr*cr*cr);
memset(usage->col, FALSE, cr * cr);
memset(usage->blk, FALSE, cr * cr);
+ if (xtype) {
+ usage->diag = snewn(cr * 2, unsigned char);
+ memset(usage->diag, FALSE, cr * 2);
+ } else
+ usage->diag = NULL;
+
+ 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, y, grid[y*cr+x]);
+
+ /*
+ * Now loop over the grid repeatedly trying all permitted modes
+ * of reasoning. The loop terminates if we complete an
+ * iteration without making any progress; we then return
+ * failure or success depending on whether the grid is full or
+ * not.
+ */
+ while (1) {
+ /*
+ * I'd like to write `continue;' inside each of the
+ * following loops, so that the solver returns here after
+ * making some progress. However, I can't specify that I
+ * want to continue an outer loop rather than the innermost
+ * one, so I'm apologetically resorting to a goto.
+ */
+ cont:
+
+ /*
+ * Blockwise positional elimination.
+ */
+ for (b = 0; b < cr; b++)
+ for (n = 1; n <= cr; n++)
+ if (!usage->blk[b*cr+n-1]) {
+ for (i = 0; i < cr; i++)
+ scratch->indexlist[i] = cubepos2(usage->blocks->blocks[b][i],n);
+ ret = solver_elim(usage, scratch->indexlist
+#ifdef STANDALONE_SOLVER
+ , "positional elimination,"
+ " %d in block %s", n,
+ usage->blocks->blocknames[b]
+#endif
+ );
+ if (ret < 0) {
+ diff = DIFF_IMPOSSIBLE;
+ goto got_result;
+ } else if (ret > 0) {
+ diff = max(diff, DIFF_BLOCK);
+ goto cont;
+ }
+ }
+
+ if (maxdiff <= DIFF_BLOCK)
+ break;
+
+ /*
+ * Row-wise positional elimination.
+ */
+ for (y = 0; y < cr; y++)
+ for (n = 1; n <= cr; n++)
+ if (!usage->row[y*cr+n-1]) {
+ for (x = 0; x < cr; x++)
+ scratch->indexlist[x] = cubepos(x, y, n);
+ ret = solver_elim(usage, scratch->indexlist
+#ifdef STANDALONE_SOLVER
+ , "positional elimination,"
+ " %d in row %d", n, 1+y
+#endif
+ );
+ if (ret < 0) {
+ diff = DIFF_IMPOSSIBLE;
+ goto got_result;
+ } else if (ret > 0) {
+ diff = max(diff, DIFF_SIMPLE);
+ goto cont;
+ }
+ }
+ /*
+ * Column-wise positional elimination.
+ */
+ for (x = 0; x < cr; x++)
+ for (n = 1; n <= cr; n++)
+ if (!usage->col[x*cr+n-1]) {
+ for (y = 0; y < cr; y++)
+ scratch->indexlist[y] = cubepos(x, y, n);
+ ret = solver_elim(usage, scratch->indexlist
+#ifdef STANDALONE_SOLVER
+ , "positional elimination,"
+ " %d in column %d", n, 1+x
+#endif
+ );
+ if (ret < 0) {
+ diff = DIFF_IMPOSSIBLE;
+ goto got_result;
+ } else if (ret > 0) {
+ diff = max(diff, DIFF_SIMPLE);
+ goto cont;
+ }
+ }
+
+ /*
+ * X-diagonal positional elimination.
+ */
+ if (usage->diag) {
+ for (n = 1; n <= cr; n++)
+ if (!usage->diag[n-1]) {
+ for (i = 0; i < cr; i++)
+ scratch->indexlist[i] = cubepos2(diag0(i), n);
+ ret = solver_elim(usage, scratch->indexlist
+#ifdef STANDALONE_SOLVER
+ , "positional elimination,"
+ " %d in \\-diagonal", n
+#endif
+ );
+ if (ret < 0) {
+ diff = DIFF_IMPOSSIBLE;
+ goto got_result;
+ } else if (ret > 0) {
+ diff = max(diff, DIFF_SIMPLE);
+ goto cont;
+ }
+ }
+ for (n = 1; n <= cr; n++)
+ if (!usage->diag[cr+n-1]) {
+ for (i = 0; i < cr; i++)
+ scratch->indexlist[i] = cubepos2(diag1(i), n);
+ ret = solver_elim(usage, scratch->indexlist
+#ifdef STANDALONE_SOLVER
+ , "positional elimination,"
+ " %d in /-diagonal", n
+#endif
+ );
+ if (ret < 0) {
+ diff = DIFF_IMPOSSIBLE;
+ goto got_result;
+ } else if (ret > 0) {
+ diff = max(diff, DIFF_SIMPLE);
+ goto cont;
+ }
+ }
+ }
+
+ /*
+ * Numeric elimination.
+ */
+ for (x = 0; x < cr; x++)
+ for (y = 0; y < cr; y++)
+ if (!usage->grid[y*cr+x]) {
+ for (n = 1; n <= cr; n++)
+ scratch->indexlist[n-1] = cubepos(x, y, n);
+ ret = solver_elim(usage, scratch->indexlist
+#ifdef STANDALONE_SOLVER
+ , "numeric elimination at (%d,%d)",
+ 1+x, 1+y
+#endif
+ );
+ if (ret < 0) {
+ diff = DIFF_IMPOSSIBLE;
+ goto got_result;
+ } else if (ret > 0) {
+ diff = max(diff, DIFF_SIMPLE);
+ goto cont;
+ }
+ }
+
+ if (maxdiff <= DIFF_SIMPLE)
+ break;
+
+ /*
+ * Intersectional analysis, rows vs blocks.
+ */
+ for (y = 0; y < cr; y++)
+ for (b = 0; b < cr; b++)
+ for (n = 1; n <= cr; n++) {
+ if (usage->row[y*cr+n-1] ||
+ usage->blk[b*cr+n-1])
+ continue;
+ for (i = 0; i < cr; i++) {
+ scratch->indexlist[i] = cubepos(i, y, n);
+ scratch->indexlist2[i] = cubepos2(usage->blocks->blocks[b][i], n);
+ }
+ /*
+ * solver_intersect() never returns -1.
+ */
+ if (solver_intersect(usage, scratch->indexlist,
+ scratch->indexlist2
+#ifdef STANDALONE_SOLVER
+ , "intersectional analysis,"
+ " %d in row %d vs block %s",
+ n, 1+y, usage->blocks->blocknames[b]
+#endif
+ ) ||
+ solver_intersect(usage, scratch->indexlist2,
+ scratch->indexlist
+#ifdef STANDALONE_SOLVER
+ , "intersectional analysis,"
+ " %d in block %s vs row %d",
+ n, usage->blocks->blocknames[b], 1+y
+#endif
+ )) {
+ diff = max(diff, DIFF_INTERSECT);
+ goto cont;
+ }
+ }
+
+ /*
+ * Intersectional analysis, columns vs blocks.
+ */
+ for (x = 0; x < cr; x++)
+ for (b = 0; b < cr; b++)
+ for (n = 1; n <= cr; n++) {
+ if (usage->col[x*cr+n-1] ||
+ usage->blk[b*cr+n-1])
+ continue;
+ for (i = 0; i < cr; i++) {
+ scratch->indexlist[i] = cubepos(x, i, n);
+ scratch->indexlist2[i] = cubepos2(usage->blocks->blocks[b][i], n);
+ }
+ if (solver_intersect(usage, scratch->indexlist,
+ scratch->indexlist2
+#ifdef STANDALONE_SOLVER
+ , "intersectional analysis,"
+ " %d in column %d vs block %s",
+ n, 1+x, usage->blocks->blocknames[b]
+#endif
+ ) ||
+ solver_intersect(usage, scratch->indexlist2,
+ scratch->indexlist
+#ifdef STANDALONE_SOLVER
+ , "intersectional analysis,"
+ " %d in block %s vs column %d",
+ n, usage->blocks->blocknames[b], 1+x
+#endif
+ )) {
+ diff = max(diff, DIFF_INTERSECT);
+ goto cont;
+ }
+ }
+
+ if (usage->diag) {
+ /*
+ * Intersectional analysis, \-diagonal vs blocks.
+ */
+ for (b = 0; b < cr; b++)
+ for (n = 1; n <= cr; n++) {
+ if (usage->diag[n-1] ||
+ usage->blk[b*cr+n-1])
+ continue;
+ for (i = 0; i < cr; i++) {
+ scratch->indexlist[i] = cubepos2(diag0(i), n);
+ scratch->indexlist2[i] = cubepos2(usage->blocks->blocks[b][i], n);
+ }
+ if (solver_intersect(usage, scratch->indexlist,
+ scratch->indexlist2
+#ifdef STANDALONE_SOLVER
+ , "intersectional analysis,"
+ " %d in \\-diagonal vs block %s",
+ n, 1+x, usage->blocks->blocknames[b]
+#endif
+ ) ||
+ solver_intersect(usage, scratch->indexlist2,
+ scratch->indexlist
+#ifdef STANDALONE_SOLVER
+ , "intersectional analysis,"
+ " %d in block %s vs \\-diagonal",
+ n, usage->blocks->blocknames[b], 1+x
+#endif
+ )) {
+ diff = max(diff, DIFF_INTERSECT);
+ goto cont;
+ }
+ }
+
+ /*
+ * Intersectional analysis, /-diagonal vs blocks.
+ */
+ for (b = 0; b < cr; b++)
+ for (n = 1; n <= cr; n++) {
+ if (usage->diag[cr+n-1] ||
+ usage->blk[b*cr+n-1])
+ continue;
+ for (i = 0; i < cr; i++) {
+ scratch->indexlist[i] = cubepos2(diag1(i), n);
+ scratch->indexlist2[i] = cubepos2(usage->blocks->blocks[b][i], n);
+ }
+ if (solver_intersect(usage, scratch->indexlist,
+ scratch->indexlist2
+#ifdef STANDALONE_SOLVER
+ , "intersectional analysis,"
+ " %d in /-diagonal vs block %s",
+ n, 1+x, usage->blocks->blocknames[b]
+#endif
+ ) ||
+ solver_intersect(usage, scratch->indexlist2,
+ scratch->indexlist
+#ifdef STANDALONE_SOLVER
+ , "intersectional analysis,"
+ " %d in block %s vs /-diagonal",
+ n, usage->blocks->blocknames[b], 1+x
+#endif
+ )) {
+ diff = max(diff, DIFF_INTERSECT);
+ goto cont;
+ }
+ }
+ }
+
+ if (maxdiff <= DIFF_INTERSECT)
+ break;
+
+ /*
+ * Blockwise set elimination.
+ */
+ for (b = 0; b < cr; b++) {
+ for (i = 0; i < cr; i++)
+ for (n = 1; n <= cr; n++)
+ scratch->indexlist[i*cr+n-1] = cubepos2(usage->blocks->blocks[b][i], n);
+ ret = solver_set(usage, scratch, scratch->indexlist
+#ifdef STANDALONE_SOLVER
+ , "set elimination, block %s",
+ usage->blocks->blocknames[b]
+#endif
+ );
+ if (ret < 0) {
+ diff = DIFF_IMPOSSIBLE;
+ goto got_result;
+ } else if (ret > 0) {
+ diff = max(diff, DIFF_SET);
+ goto cont;
+ }
+ }
+
+ /*
+ * Row-wise set elimination.
+ */
+ for (y = 0; y < cr; y++) {
+ for (x = 0; x < cr; x++)
+ for (n = 1; n <= cr; n++)
+ scratch->indexlist[x*cr+n-1] = cubepos(x, y, n);
+ ret = solver_set(usage, scratch, scratch->indexlist
+#ifdef STANDALONE_SOLVER
+ , "set elimination, row %d", 1+y
+#endif
+ );
+ if (ret < 0) {
+ diff = DIFF_IMPOSSIBLE;
+ goto got_result;
+ } else if (ret > 0) {
+ diff = max(diff, DIFF_SET);
+ goto cont;
+ }
+ }
+
+ /*
+ * Column-wise set elimination.
+ */
+ for (x = 0; x < cr; x++) {
+ for (y = 0; y < cr; y++)
+ for (n = 1; n <= cr; n++)
+ scratch->indexlist[y*cr+n-1] = cubepos(x, y, n);
+ ret = solver_set(usage, scratch, scratch->indexlist
+#ifdef STANDALONE_SOLVER
+ , "set elimination, column %d", 1+x
+#endif
+ );
+ if (ret < 0) {
+ diff = DIFF_IMPOSSIBLE;
+ goto got_result;
+ } else if (ret > 0) {
+ diff = max(diff, DIFF_SET);
+ goto cont;
+ }
+ }
+
+ if (usage->diag) {
+ /*
+ * \-diagonal set elimination.
+ */
+ for (i = 0; i < cr; i++)
+ for (n = 1; n <= cr; n++)
+ scratch->indexlist[i*cr+n-1] = cubepos2(diag0(i), n);
+ ret = solver_set(usage, scratch, scratch->indexlist
+#ifdef STANDALONE_SOLVER
+ , "set elimination, \\-diagonal"
+#endif
+ );
+ if (ret < 0) {
+ diff = DIFF_IMPOSSIBLE;
+ goto got_result;
+ } else if (ret > 0) {
+ diff = max(diff, DIFF_SET);
+ goto cont;
+ }
+
+ /*
+ * /-diagonal set elimination.
+ */
+ for (i = 0; i < cr; i++)
+ for (n = 1; n <= cr; n++)
+ scratch->indexlist[i*cr+n-1] = cubepos2(diag1(i), n);
+ ret = solver_set(usage, scratch, scratch->indexlist
+#ifdef STANDALONE_SOLVER
+ , "set elimination, \\-diagonal"
+#endif
+ );
+ if (ret < 0) {
+ diff = DIFF_IMPOSSIBLE;
+ goto got_result;
+ } else if (ret > 0) {
+ diff = max(diff, DIFF_SET);
+ goto cont;
+ }
+ }
+
+ if (maxdiff <= DIFF_SET)
+ break;
+
+ /*
+ * Row-vs-column set elimination on a single number.
+ */
+ for (n = 1; n <= cr; n++) {
+ for (y = 0; y < cr; y++)
+ for (x = 0; x < cr; x++)
+ scratch->indexlist[y*cr+x] = cubepos(x, y, n);
+ ret = solver_set(usage, scratch, scratch->indexlist
+#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;
+ }
+ }
+
+ /*
+ * 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,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,y,n))
+ list[j++] = n;
+
+#ifdef STANDALONE_SOLVER
+ if (solver_show_working) {
+ char *sep = "";
+ printf("%*srecursing on (%d,%d) [",
+ solver_recurse_depth*4, "", x + 1, y + 1);
+ 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 + 1, y + 1);
+ solver_recurse_depth++;
+#endif
+
+ ret = solver(cr, blocks, xtype, 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 + 1, y + 1);
+ }
+#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 cr;
+ struct block_structure *blocks;
+ /* 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;
+ /* diag[i*cr+n-1] TRUE if digit n has been placed in diagonal i */
+ unsigned char *diag;
+ /* 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.
+ *
+ * Return values: 1 means solution found, 0 means no solution
+ * found on this branch.
+ */
+static int gridgen_real(struct gridgen_usage *usage, digit *grid)
+{
+ int 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[usage->blocks->whichblock[y*cr+x]*cr+n] &&
+ (!usage->diag || ((!ondiag0(y*cr+x) || !usage->diag[n]) &&
+ (!ondiag1(y*cr+x) || !usage->diag[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[usage->blocks->whichblock[sy*cr+sx]*cr+n] &&
+ (!usage->diag || ((!ondiag0(sy*cr+sx) || !usage->diag[n]) &&
+ (!ondiag1(sy*cr+sx) || !usage->diag[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[usage->blocks->whichblock[sy*cr+sx]*cr+n-1] = TRUE;
+ if (usage->diag) {
+ if (ondiag0(sy*cr+sx))
+ usage->diag[n-1] = TRUE;
+ if (ondiag1(sy*cr+sx))
+ usage->diag[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[usage->blocks->whichblock[sy*cr+sx]*cr+n-1] = FALSE;
+ if (usage->diag) {
+ if (ondiag0(sy*cr+sx))
+ usage->diag[n-1] = FALSE;
+ if (ondiag1(sy*cr+sx))
+ usage->diag[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 parameters and a starting
+ * grid, which is simply an array of cr*cr digits.
+ */
+static int gridgen(int cr, struct block_structure *blocks, int xtype,
+ digit *grid, random_state *rs)
+{
+ struct gridgen_usage *usage;
+ int x, y, ret;
+
+ /*
+ * Clear the grid to start with.
+ */
+ memset(grid, 0, cr*cr);
+
+ /*
+ * Create a gridgen_usage structure.
+ */
+ usage = snew(struct gridgen_usage);
+
+ usage->cr = cr;
+ usage->blocks = blocks;
+
+ 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);
+
+ if (xtype) {
+ usage->diag = snewn(2 * cr, unsigned char);
+ memset(usage->diag, FALSE, 2 * cr);
+ } else {
+ usage->diag = NULL;
+ }
+
+ usage->spaces = snewn(cr * cr, struct gridgen_coord);
+ usage->nspaces = 0;
+
+ usage->rs = rs;
+
/*
- * Place all the clue numbers we are given.
+ * Initialise the list of grid spaces.
*/
- 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]);
+ 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++;
+ }
+ }
/*
- * 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.
+ * Run the real generator function.
*/
- while (1) {
- cont:
-
- /*
- * Blockwise positional elimination.
- */
- for (x = 0; x < cr; x += r)
- for (y = 0; y < r; y++)
- for (n = 1; n <= cr; n++)
- if (!usage->blk[(y*c+(x/r))*cr+n-1] &&
- nsolve_elim(usage, cubepos(x,y,n), r*cr))
- 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))
- 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))
- 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;
-
- /*
- * If we reach here, we have made no deductions in this
- * iteration, so the algorithm terminates.
- */
- break;
- }
+ ret = gridgen_real(usage, grid);
- sfree(usage->cube);
- sfree(usage->row);
- sfree(usage->col);
+ /*
+ * Clean up the usage structure now we have our answer.
+ */
+ sfree(usage->spaces);
sfree(usage->blk);
+ sfree(usage->col);
+ sfree(usage->row);
+ sfree(usage->grid);
sfree(usage);
- for (x = 0; x < cr; x++)
- for (y = 0; y < cr; y++)
- if (!grid[y*cr+x])
- return FALSE;
- return TRUE;
+ return ret;
}
/* ----------------------------------------------------------------------
- * End of non-recursive solver code.
+ * End of grid generator code.
*/
/*
* Check whether a grid contains a valid complete puzzle.
*/
-static int check_valid(int c, int r, digit *grid)
+static int check_valid(int cr, struct block_structure *blocks, int xtype,
+ digit *grid)
{
- int cr = c*r;
unsigned char *used;
- int x, y, n;
+ int x, y, i, j, n;
used = snewn(cr, unsigned char);
/*
* Check that each block contains precisely one of everything.
*/
- for (x = 0; x < cr; x += r) {
- for (y = 0; y < cr; y += c) {
- int xx, yy;
- memset(used, FALSE, cr);
- for (xx = x; xx < x+r; xx++)
- for (yy = 0; yy < y+c; yy++)
- if (grid[yy*cr+xx] > 0 && grid[yy*cr+xx] <= cr)
- used[grid[yy*cr+xx]-1] = TRUE;
- for (n = 0; n < cr; n++)
- if (!used[n]) {
- sfree(used);
- return FALSE;
- }
- }
+ for (i = 0; i < cr; i++) {
+ memset(used, FALSE, cr);
+ for (j = 0; j < cr; j++)
+ if (grid[blocks->blocks[i][j]] > 0 &&
+ grid[blocks->blocks[i][j]] <= cr)
+ used[grid[blocks->blocks[i][j]]-1] = TRUE;
+ for (n = 0; n < cr; n++)
+ if (!used[n]) {
+ sfree(used);
+ return FALSE;
+ }
+ }
+
+ /*
+ * Check that each diagonal contains precisely one of everything.
+ */
+ if (xtype) {
+ memset(used, FALSE, cr);
+ for (i = 0; i < cr; i++)
+ if (grid[diag0(i)] > 0 && grid[diag0(i)] <= cr)
+ used[grid[diag0(i)]-1] = TRUE;
+ for (n = 0; n < cr; n++)
+ if (!used[n]) {
+ sfree(used);
+ return FALSE;
+ }
+ for (i = 0; i < cr; i++)
+ if (grid[diag1(i)] > 0 && grid[diag1(i)] <= cr)
+ used[grid[diag1(i)]-1] = TRUE;
+ for (n = 0; n < cr; n++)
+ if (!used[n]) {
+ sfree(used);
+ return FALSE;
+ }
}
sfree(used);
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;
}
-static char *new_game_seed(game_params *params, random_state *rs)
+static char *new_game_desc(game_params *params, random_state *rs,
+ char **aux, int interactive)
{
int c = params->c, r = params->r, cr = c*r;
int area = cr*cr;
+ struct block_structure *blocks;
digit *grid, *grid2;
struct xy { int x, y; } *locs;
int nlocs;
- int ret;
- char *seed;
+ char *desc;
int coords[16], ncoords;
- int xlim, ylim;
+ int maxdiff;
+ int x, y, i, j;
/*
- * 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));
-
-#ifdef DEBUG
- memcpy(grid,
- "\x0\x1\x0\x0\x6\x0\x0\x0\x0"
- "\x5\x0\x0\x7\x0\x4\x0\x2\x0"
- "\x0\x0\x6\x1\x0\x0\x0\x0\x0"
- "\x8\x9\x7\x0\x0\x0\x0\x0\x0"
- "\x0\x0\x3\x0\x4\x0\x9\x0\x0"
- "\x0\x0\x0\x0\x0\x0\x8\x7\x6"
- "\x0\x0\x0\x0\x0\x9\x1\x0\x0"
- "\x0\x3\x0\x6\x0\x5\x0\x0\x7"
- "\x0\x0\x0\x0\x8\x0\x0\x5\x0"
- , area);
-
- {
- int y, x;
- for (y = 0; y < cr; y++) {
- for (x = 0; x < cr; x++) {
- printf("%2.0d", grid[y*cr+x]);
- }
- printf("\n");
- }
- printf("\n");
- }
+ maxdiff = params->diff;
+ if (c == 2 && r == 2)
+ maxdiff = DIFF_BLOCK;
- nsolve(c, r, grid);
+ grid = snewn(area, digit);
+ locs = snewn(area, struct xy);
+ grid2 = snewn(area, digit);
- {
- int y, x;
- for (y = 0; y < cr; y++) {
- for (x = 0; x < cr; x++) {
- printf("%2.0d", grid[y*cr+x]);
- }
- printf("\n");
- }
- printf("\n");
- }
+ blocks = snew(struct block_structure);
+ blocks->c = params->c; blocks->r = params->r;
+ blocks->whichblock = snewn(area*2, int);
+ blocks->blocks = snewn(cr, int *);
+ for (i = 0; i < cr; i++)
+ blocks->blocks[i] = blocks->whichblock + area + i*cr;
+#ifdef STANDALONE_SOLVER
+ assert(!"This should never happen, so we don't need to create blocknames");
#endif
/*
- * Now we have a solved grid, start removing things from it
- * while preserving solubility.
+ * 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.
*/
- 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;
+ /*
+ * Generate a random solved state, starting by
+ * constructing the block structure.
+ */
+ if (r == 1) { /* jigsaw mode */
+ int *dsf = divvy_rectangle(cr, cr, cr, rs);
+ int nb = 0;
+
+ for (i = 0; i < area; i++)
+ blocks->whichblock[i] = -1;
+ for (i = 0; i < area; i++) {
+ int j = dsf_canonify(dsf, i);
+ if (blocks->whichblock[j] < 0)
+ blocks->whichblock[j] = nb++;
+ blocks->whichblock[i] = blocks->whichblock[j];
}
+ assert(nb == cr);
+
+ sfree(dsf);
+ } else { /* basic Sudoku mode */
+ for (y = 0; y < cr; y++)
+ for (x = 0; x < cr; x++)
+ blocks->whichblock[y*cr+x] = (y/c) * c + (x/r);
}
+ for (i = 0; i < cr; i++)
+ blocks->blocks[i][cr-1] = 0;
+ for (i = 0; i < area; i++) {
+ int b = blocks->whichblock[i];
+ j = blocks->blocks[b][cr-1]++;
+ assert(j < cr);
+ blocks->blocks[b][j] = i;
+ }
+
+ if (!gridgen(cr, blocks, params->xtype, grid, rs))
+ continue; /* this might happen if the jigsaw is unsuitable */
+ assert(check_valid(cr, blocks, params->xtype, grid));
/*
- * Now loop over the shuffled list and, for each element,
- * see whether removing that element (and its reflections)
- * from the grid will still leave the grid soluble by
- * nsolve.
+ * Save the solved grid in aux.
*/
- 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;
- }
- }
-
- if (i == nlocs) {
+ {
/*
- * There was nothing we could remove without destroying
- * solvability.
+ * 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.
*/
- break;
+ if (*aux) {
+ sfree(*aux);
+ }
+
+ *aux = encode_solve_move(cr, grid);
}
+
+ /*
+ * Now we have a solved grid, start removing things from it
+ * while preserving solubility.
+ */
+
+ /*
+ * 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;
+
+ 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 shuffle that list.
+ */
+ shuffle(locs, nlocs, sizeof(*locs), rs);
+
+ /*
+ * 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;
+
+ 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;
+
+ ret = solver(cr, blocks, params->xtype, 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);
+
+ if (solver(cr, blocks, params->xtype, grid2, maxdiff) == maxdiff)
+ break; /* found one! */
}
+
sfree(grid2);
sfree(locs);
-#ifdef DEBUG
- {
- int y, x;
- for (y = 0; y < cr; y++) {
- for (x = 0; x < cr; x++) {
- printf("%2.0d", grid[y*cr+x]);
- }
- printf("\n");
- }
- printf("\n");
- }
-#endif
-
/*
* Now we have the grid as it will be presented to the user.
- * Encode it in a game seed.
+ * Encode it in a game desc.
*/
{
char *p;
int run, i;
- seed = snewn(5 * area, char);
- p = seed;
+ desc = snewn(7 * area, char);
+ p = desc;
run = 0;
for (i = 0; i <= area; i++) {
int n = (i < area ? grid[i] : -1);
* bottom right, there's no point putting an
* unnecessary _ before or after it.
*/
- if (p > seed && n > 0)
+ if (p > desc && n > 0)
*p++ = '_';
}
if (n > 0)
run = 0;
}
}
- assert(p - seed < 5 * area);
+
+ if (r == 1) {
+ int currrun = 0;
+
+ *p++ = ',';
+
+ /*
+ * Encode the block structure. We do this by encoding
+ * the pattern of dividing lines: first we iterate
+ * over the cr*(cr-1) internal vertical grid lines in
+ * ordinary reading order, then over the cr*(cr-1)
+ * internal horizontal ones in transposed reading
+ * order.
+ *
+ * We encode the number of non-lines between the
+ * lines; _ means zero (two adjacent divisions), a
+ * means 1, ..., y means 25, and z means 25 non-lines
+ * _and no following line_ (so that za means 26, zb 27
+ * etc).
+ */
+ for (i = 0; i <= 2*cr*(cr-1); i++) {
+ int p0, p1, edge;
+
+ if (i == 2*cr*(cr-1)) {
+ edge = TRUE; /* terminating virtual edge */
+ } else {
+ if (i < cr*(cr-1)) {
+ y = i/(cr-1);
+ x = i%(cr-1);
+ p0 = y*cr+x;
+ p1 = y*cr+x+1;
+ } else {
+ x = i/(cr-1) - cr;
+ y = i%(cr-1);
+ p0 = y*cr+x;
+ p1 = (y+1)*cr+x;
+ }
+ edge = (blocks->whichblock[p0] != blocks->whichblock[p1]);
+ }
+
+ if (edge) {
+ while (currrun > 25)
+ *p++ = 'z', currrun -= 25;
+ if (currrun)
+ *p++ = 'a'-1 + currrun;
+ else
+ *p++ = '_';
+ currrun = 0;
+ } else
+ currrun++;
+ }
+ }
+
+ assert(p - desc < 7 * area);
*p++ = '\0';
- seed = sresize(seed, p - seed, char);
+ desc = sresize(desc, p - desc, char);
}
sfree(grid);
- return seed;
+ return desc;
}
-static char *validate_seed(game_params *params, char *seed)
+static char *validate_desc(game_params *params, char *desc)
{
- int area = params->r * params->r * params->c * params->c;
+ int cr = params->c * params->r, area = cr*cr;
int squares = 0;
+ int *dsf;
- while (*seed) {
- int n = *seed++;
+ while (*desc && *desc != ',') {
+ int n = *desc++;
if (n >= 'a' && n <= 'z') {
squares += n - 'a' + 1;
} 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 (*seed >= '0' && *seed <= '9')
- seed++;
+ while (*desc >= '0' && *desc <= '9')
+ desc++;
} else
- return "Invalid character in game specification";
+ return "Invalid character in game description";
}
if (squares < area)
if (squares > area)
return "Too much data to fit in grid";
+ if (params->r == 1) {
+ int pos;
+
+ /*
+ * Now we expect a suffix giving the jigsaw block
+ * structure. Parse it and validate that it divides the
+ * grid into the right number of regions which are the
+ * right size.
+ */
+ if (*desc != ',')
+ return "Expected jigsaw block structure in game description";
+ pos = 0;
+
+ dsf = snew_dsf(area);
+ desc++;
+
+ while (*desc) {
+ int c, adv;
+
+ if (*desc == '_')
+ c = 0;
+ else if (*desc >= 'a' && *desc <= 'z')
+ c = *desc - 'a' + 1;
+ else {
+ sfree(dsf);
+ return "Invalid character in game description";
+ }
+ desc++;
+
+ adv = (c != 25); /* 'z' is a special case */
+
+ while (c-- > 0) {
+ int p0, p1;
+
+ /*
+ * Non-edge; merge the two dsf classes on either
+ * side of it.
+ */
+ if (pos >= 2*cr*(cr-1)) {
+ sfree(dsf);
+ return "Too much data in block structure specification";
+ } else if (pos < cr*(cr-1)) {
+ int y = pos/(cr-1);
+ int x = pos%(cr-1);
+ p0 = y*cr+x;
+ p1 = y*cr+x+1;
+ } else {
+ int x = pos/(cr-1) - cr;
+ int y = pos%(cr-1);
+ p0 = y*cr+x;
+ p1 = (y+1)*cr+x;
+ }
+ dsf_merge(dsf, p0, p1);
+
+ pos++;
+ }
+ if (adv)
+ pos++;
+ }
+
+ /*
+ * When desc is exhausted, we expect to have gone exactly
+ * one space _past_ the end of the grid, due to the dummy
+ * edge at the end.
+ */
+ if (pos != 2*cr*(cr-1)+1) {
+ sfree(dsf);
+ return "Not enough data in block structure specification";
+ }
+
+ /*
+ * Now we've got our dsf. Verify that it matches
+ * expectations.
+ */
+ {
+ int *canons, *counts;
+ int i, j, c, ncanons = 0;
+
+ canons = snewn(cr, int);
+ counts = snewn(cr, int);
+
+ for (i = 0; i < area; i++) {
+ j = dsf_canonify(dsf, i);
+
+ for (c = 0; c < ncanons; c++)
+ if (canons[c] == j) {
+ counts[c]++;
+ if (counts[c] > cr) {
+ sfree(dsf);
+ sfree(canons);
+ sfree(counts);
+ return "A jigsaw block is too big";
+ }
+ break;
+ }
+
+ if (c == ncanons) {
+ if (ncanons >= cr) {
+ sfree(dsf);
+ sfree(canons);
+ sfree(counts);
+ return "Too many distinct jigsaw blocks";
+ }
+ canons[ncanons] = j;
+ counts[ncanons] = 1;
+ ncanons++;
+ }
+ }
+
+ /*
+ * If we've managed to get through that loop without
+ * tripping either of the error conditions, then we
+ * must have partitioned the entire grid into at most
+ * cr blocks of at most cr squares each; therefore we
+ * must have _exactly_ cr blocks of _exactly_ cr
+ * squares each. I'll verify that by assertion just in
+ * case something has gone horribly wrong, but it
+ * shouldn't have been able to happen by duff input,
+ * only by a bug in the above code.
+ */
+ assert(ncanons == cr);
+ for (c = 0; c < ncanons; c++)
+ assert(counts[c] == cr);
+
+ sfree(canons);
+ sfree(counts);
+ }
+
+ sfree(dsf);
+ } else {
+ if (*desc)
+ return "Unexpected jigsaw block structure in game description";
+ }
+
return NULL;
}
-static game_state *new_game(game_params *params, char *seed)
+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;
int i;
- state->c = params->c;
- state->r = params->r;
+ state->cr = cr;
+ state->xtype = params->xtype;
state->grid = snewn(area, digit);
+ state->pencil = snewn(area * cr, unsigned char);
+ memset(state->pencil, 0, area * cr);
state->immutable = snewn(area, unsigned char);
memset(state->immutable, FALSE, area);
- state->completed = FALSE;
+ state->blocks = snew(struct block_structure);
+ state->blocks->c = c; state->blocks->r = r;
+ state->blocks->refcount = 1;
+ state->blocks->whichblock = snewn(area*2, int);
+ state->blocks->blocks = snewn(cr, int *);
+ for (i = 0; i < cr; i++)
+ state->blocks->blocks[i] = state->blocks->whichblock + area + i*cr;
+#ifdef STANDALONE_SOLVER
+ state->blocks->blocknames = (char **)smalloc(cr*(sizeof(char *)+80));
+#endif
+
+ state->completed = state->cheated = FALSE;
i = 0;
- while (*seed) {
- int n = *seed++;
+ while (*desc && *desc != ',') {
+ int n = *desc++;
if (n >= 'a' && n <= 'z') {
int run = n - 'a' + 1;
assert(i + run <= area);
} else if (n > '0' && n <= '9') {
assert(i < area);
state->immutable[i] = TRUE;
- state->grid[i++] = atoi(seed-1);
- while (*seed >= '0' && *seed <= '9')
- seed++;
+ state->grid[i++] = atoi(desc-1);
+ while (*desc >= '0' && *desc <= '9')
+ desc++;
} else {
assert(!"We can't get here");
}
}
- assert(i == area);
+ assert(i == area);
+
+ if (r == 1) {
+ int pos = 0;
+ int *dsf;
+ int nb;
+
+ assert(*desc == ',');
+
+ dsf = snew_dsf(area);
+ desc++;
+
+ while (*desc) {
+ int c, adv;
+
+ if (*desc == '_')
+ c = 0;
+ else if (*desc >= 'a' && *desc <= 'z')
+ c = *desc - 'a' + 1;
+ else
+ assert(!"Shouldn't get here");
+ desc++;
+
+ adv = (c != 25); /* 'z' is a special case */
+
+ while (c-- > 0) {
+ int p0, p1;
+
+ /*
+ * Non-edge; merge the two dsf classes on either
+ * side of it.
+ */
+ assert(pos < 2*cr*(cr-1));
+ if (pos < cr*(cr-1)) {
+ int y = pos/(cr-1);
+ int x = pos%(cr-1);
+ p0 = y*cr+x;
+ p1 = y*cr+x+1;
+ } else {
+ int x = pos/(cr-1) - cr;
+ int y = pos%(cr-1);
+ p0 = y*cr+x;
+ p1 = (y+1)*cr+x;
+ }
+ dsf_merge(dsf, p0, p1);
+
+ pos++;
+ }
+ if (adv)
+ pos++;
+ }
+
+ /*
+ * When desc is exhausted, we expect to have gone exactly
+ * one space _past_ the end of the grid, due to the dummy
+ * edge at the end.
+ */
+ assert(pos == 2*cr*(cr-1)+1);
+
+ /*
+ * Now we've got our dsf. Translate it into a block
+ * structure.
+ */
+ nb = 0;
+ for (i = 0; i < area; i++)
+ state->blocks->whichblock[i] = -1;
+ for (i = 0; i < area; i++) {
+ int j = dsf_canonify(dsf, i);
+ if (state->blocks->whichblock[j] < 0)
+ state->blocks->whichblock[j] = nb++;
+ state->blocks->whichblock[i] = state->blocks->whichblock[j];
+ }
+ assert(nb == cr);
+
+ sfree(dsf);
+ } else {
+ int x, y;
+
+ assert(!*desc);
+
+ for (y = 0; y < cr; y++)
+ for (x = 0; x < cr; x++)
+ state->blocks->whichblock[y*cr+x] = (y/c) * c + (x/r);
+ }
+
+ /*
+ * Having sorted out whichblock[], set up the block index arrays.
+ */
+ for (i = 0; i < cr; i++)
+ state->blocks->blocks[i][cr-1] = 0;
+ for (i = 0; i < area; i++) {
+ int b = state->blocks->whichblock[i];
+ int j = state->blocks->blocks[b][cr-1]++;
+ assert(j < cr);
+ state->blocks->blocks[b][j] = i;
+ }
+
+#ifdef STANDALONE_SOLVER
+ /*
+ * Set up the block names for solver diagnostic output.
+ */
+ {
+ char *p = (char *)(state->blocks->blocknames + cr);
+
+ if (r == 1) {
+ for (i = 0; i < cr; i++)
+ state->blocks->blocknames[i] = NULL;
+
+ for (i = 0; i < area; i++) {
+ int j = state->blocks->whichblock[i];
+ if (!state->blocks->blocknames[j]) {
+ state->blocks->blocknames[j] = p;
+ p += 1 + sprintf(p, "starting at (%d,%d)",
+ 1 + i%cr, 1 + i/cr);
+ }
+ }
+ } else {
+ int bx, by;
+ for (by = 0; by < r; by++)
+ for (bx = 0; bx < c; bx++) {
+ state->blocks->blocknames[by*c+bx] = p;
+ p += 1 + sprintf(p, "(%d,%d)", bx+1, by+1);
+ }
+ }
+ assert(p - (char *)state->blocks->blocknames < cr*(sizeof(char *)+80));
+ for (i = 0; i < cr; i++)
+ assert(state->blocks->blocknames[i]);
+ }
+#endif
return state;
}
static game_state *dup_game(game_state *state)
{
game_state *ret = snew(game_state);
- int c = state->c, r = state->r, cr = c*r, area = cr * cr;
+ int cr = state->cr, area = cr * cr;
+
+ ret->cr = state->cr;
+ ret->xtype = state->xtype;
- ret->c = state->c;
- ret->r = state->r;
+ ret->blocks = state->blocks;
+ ret->blocks->refcount++;
ret->grid = snewn(area, digit);
memcpy(ret->grid, state->grid, area);
+ ret->pencil = snewn(area * cr, unsigned char);
+ memcpy(ret->pencil, state->pencil, area * cr);
+
ret->immutable = snewn(area, unsigned char);
memcpy(ret->immutable, state->immutable, area);
ret->completed = state->completed;
+ ret->cheated = state->cheated;
return ret;
}
static void free_game(game_state *state)
{
+ if (--state->blocks->refcount == 0) {
+ sfree(state->blocks->whichblock);
+ sfree(state->blocks->blocks);
+#ifdef STANDALONE_SOLVER
+ sfree(state->blocks->blocknames);
+#endif
+ sfree(state->blocks);
+ }
sfree(state->immutable);
+ sfree(state->pencil);
sfree(state->grid);
sfree(state);
}
+static char *solve_game(game_state *state, game_state *currstate,
+ char *ai, char **error)
+{
+ int cr = state->cr;
+ char *ret;
+ digit *grid;
+ int solve_ret;
+
+ /*
+ * If we already have the solution in ai, save ourselves some
+ * time.
+ */
+ if (ai)
+ return dupstr(ai);
+
+ grid = snewn(cr*cr, digit);
+ memcpy(grid, state->grid, cr*cr);
+ solve_ret = solver(cr, state->blocks, state->xtype, grid, DIFF_RECURSIVE);
+
+ *error = 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;
+}
+
+static char *grid_text_format(int cr, struct block_structure *blocks,
+ int xtype, digit *grid)
+{
+ int vmod, hmod;
+ int x, y;
+ int totallen, linelen, nlines;
+ char *ret, *p, ch;
+
+ /*
+ * For non-jigsaw Sudoku, we format in the way we always have,
+ * by having the digits unevenly spaced so that the dividing
+ * lines can fit in:
+ *
+ * . . | . .
+ * . . | . .
+ * ----+----
+ * . . | . .
+ * . . | . .
+ *
+ * For jigsaw puzzles, however, we must leave space between
+ * _all_ pairs of digits for an optional dividing line, so we
+ * have to move to the rather ugly
+ *
+ * . . . .
+ * ------+------
+ * . . | . .
+ * +---+
+ * . . | . | .
+ * ------+ |
+ * . . . | .
+ *
+ * We deal with both cases using the same formatting code; we
+ * simply invent a vmod value such that there's a vertical
+ * dividing line before column i iff i is divisible by vmod
+ * (so it's r in the first case and 1 in the second), and hmod
+ * likewise for horizontal dividing lines.
+ */
+
+ if (blocks->r != 1) {
+ vmod = blocks->r;
+ hmod = blocks->c;
+ } else {
+ vmod = hmod = 1;
+ }
+
+ /*
+ * Line length: we have cr digits, each with a space after it,
+ * and (cr-1)/vmod dividing lines, each with a space after it.
+ * The final space is replaced by a newline, but that doesn't
+ * affect the length.
+ */
+ linelen = 2*(cr + (cr-1)/vmod);
+
+ /*
+ * Number of lines: we have cr rows of digits, and (cr-1)/hmod
+ * dividing rows.
+ */
+ nlines = cr + (cr-1)/hmod;
+
+ /*
+ * Allocate the space.
+ */
+ totallen = linelen * nlines;
+ ret = snewn(totallen+1, char); /* leave room for terminating NUL */
+
+ /*
+ * Write the text.
+ */
+ p = ret;
+ for (y = 0; y < cr; y++) {
+ /*
+ * Row of digits.
+ */
+ for (x = 0; x < cr; x++) {
+ /*
+ * Digit.
+ */
+ digit d = grid[y*cr+x];
+
+ if (d == 0) {
+ /*
+ * Empty space: we usually write a dot, but we'll
+ * highlight spaces on the X-diagonals (in X mode)
+ * by using underscores instead.
+ */
+ if (xtype && (ondiag0(y*cr+x) || ondiag1(y*cr+x)))
+ ch = '_';
+ else
+ ch = '.';
+ } else if (d <= 9) {
+ ch = '0' + d;
+ } else {
+ ch = 'a' + d-10;
+ }
+
+ *p++ = ch;
+ if (x == cr-1) {
+ *p++ = '\n';
+ continue;
+ }
+ *p++ = ' ';
+
+ if ((x+1) % vmod)
+ continue;
+
+ /*
+ * Optional dividing line.
+ */
+ if (blocks->whichblock[y*cr+x] != blocks->whichblock[y*cr+x+1])
+ ch = '|';
+ else
+ ch = ' ';
+ *p++ = ch;
+ *p++ = ' ';
+ }
+ if (y == cr-1 || (y+1) % hmod)
+ continue;
+
+ /*
+ * Dividing row.
+ */
+ for (x = 0; x < cr; x++) {
+ int dwid;
+ int tl, tr, bl, br;
+
+ /*
+ * Division between two squares. This varies
+ * complicatedly in length.
+ */
+ dwid = 2; /* digit and its following space */
+ if (x == cr-1)
+ dwid--; /* no following space at end of line */
+ if (x > 0 && x % vmod == 0)
+ dwid++; /* preceding space after a divider */
+
+ if (blocks->whichblock[y*cr+x] != blocks->whichblock[(y+1)*cr+x])
+ ch = '-';
+ else
+ ch = ' ';
+
+ while (dwid-- > 0)
+ *p++ = ch;
+
+ if (x == cr-1) {
+ *p++ = '\n';
+ break;
+ }
+
+ if ((x+1) % vmod)
+ continue;
+
+ /*
+ * Corner square. This is:
+ * - a space if all four surrounding squares are in
+ * the same block
+ * - a vertical line if the two left ones are in one
+ * block and the two right in another
+ * - a horizontal line if the two top ones are in one
+ * block and the two bottom in another
+ * - a plus sign in all other cases. (If we had a
+ * richer character set available we could break
+ * this case up further by doing fun things with
+ * line-drawing T-pieces.)
+ */
+ tl = blocks->whichblock[y*cr+x];
+ tr = blocks->whichblock[y*cr+x+1];
+ bl = blocks->whichblock[(y+1)*cr+x];
+ br = blocks->whichblock[(y+1)*cr+x+1];
+
+ if (tl == tr && tr == bl && bl == br)
+ ch = ' ';
+ else if (tl == bl && tr == br)
+ ch = '|';
+ else if (tl == tr && bl == br)
+ ch = '-';
+ else
+ ch = '+';
+
+ *p++ = ch;
+ }
+ }
+
+ assert(p - ret == totallen);
+ *p = '\0';
+ return ret;
+}
+
+static char *game_text_format(game_state *state)
+{
+ return grid_text_format(state->cr, state->blocks, state->xtype,
+ state->grid);
+}
+
struct game_ui {
/*
* These are the coordinates of the currently highlighted
* enter that number or letter in the grid.
*/
int hx, hy;
+ /*
+ * This indicates whether the current highlight is a
+ * pencil-mark one or a real one.
+ */
+ int hpencil;
};
static game_ui *new_ui(game_state *state)
game_ui *ui = snew(game_ui);
ui->hx = ui->hy = -1;
+ ui->hpencil = 0;
return ui;
}
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;
- int tx, ty;
- game_state *ret;
+ return NULL;
+}
+
+static void decode_ui(game_ui *ui, char *encoding)
+{
+}
- tx = (x - BORDER) / TILE_SIZE;
- ty = (y - BORDER) / TILE_SIZE;
+static void game_changed_state(game_ui *ui, game_state *oldstate,
+ game_state *newstate)
+{
+ int cr = newstate->cr;
+ /*
+ * 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;
+ }
+}
- if (tx >= 0 && tx < cr && ty >= 0 && ty < cr && button == LEFT_BUTTON) {
- if (tx == ui->hx && ty == ui->hy) {
- ui->hx = ui->hy = -1;
- } else {
- ui->hx = tx;
- ui->hy = ty;
- }
- return from; /* UI activity occurred */
+struct game_drawstate {
+ int started;
+ int cr, xtype;
+ 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 cr = state->cr;
+ int tx, ty;
+ 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) {
+ 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
+ * this shouldn't happen anyway because we should never
+ * have even been able to pencil-highlight the square, but
+ * it never hurts to be careful.
+ */
+ 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);
- ret = dup_game(from);
- ret->grid[ui->hy*cr+ui->hx] = n;
ui->hx = ui->hy = -1;
- /*
- * We've made a real change to the grid. Check to see
- * if the game has been completed.
- */
- if (!ret->completed && check_valid(c, r, ret->grid)) {
- ret->completed = TRUE;
- }
-
- return ret; /* made a valid move */
+ return dupstr(buf);
}
return NULL;
}
+static game_state *execute_move(game_state *from, char *move)
+{
+ int cr = from->cr;
+ 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 (move[0] == 'P' && n > 0) {
+ int index = (y*cr+x) * cr + (n-1);
+ ret->pencil[index] = !ret->pencil[index];
+ } else {
+ 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
+ * if the game has been completed.
+ */
+ if (!ret->completed && check_valid(cr, ret->blocks, ret->xtype,
+ ret->grid)) {
+ ret->completed = TRUE;
+ }
+ }
+ 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 *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);
frontend_default_colour(fe, &ret[COL_BACKGROUND * 3]);
+ ret[COL_XDIAGONALS * 3 + 0] = 0.9F * ret[COL_BACKGROUND * 3 + 0];
+ ret[COL_XDIAGONALS * 3 + 1] = 0.9F * ret[COL_BACKGROUND * 3 + 1];
+ ret[COL_XDIAGONALS * 3 + 2] = 0.9F * ret[COL_BACKGROUND * 3 + 2];
+
ret[COL_GRID * 3 + 0] = 0.0F;
ret[COL_GRID * 3 + 1] = 0.0F;
ret[COL_GRID * 3 + 2] = 0.0F;
ret[COL_USER * 3 + 1] = 0.6F * ret[COL_BACKGROUND * 3 + 1];
ret[COL_USER * 3 + 2] = 0.0F;
- ret[COL_HIGHLIGHT * 3 + 0] = 0.85F * ret[COL_BACKGROUND * 3 + 0];
- 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_HIGHLIGHT * 3 + 0] = 0.78F * ret[COL_BACKGROUND * 3 + 0];
+ ret[COL_HIGHLIGHT * 3 + 1] = 0.78F * ret[COL_BACKGROUND * 3 + 1];
+ ret[COL_HIGHLIGHT * 3 + 2] = 0.78F * 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];
*ncolours = NCOLOURS;
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;
+ int cr = state->cr;
ds->started = FALSE;
- ds->c = c;
- ds->r = r;
ds->cr = cr;
+ ds->xtype = state->xtype;
ds->grid = snewn(cr*cr, digit);
- memset(ds->grid, 0, cr*cr);
+ memset(ds->grid, cr+2, cr*cr);
+ ds->pencil = snewn(cr*cr*cr, digit);
+ 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;
+ int cr = state->cr;
int tx, ty;
int cx, cy, cw, ch;
char str[2];
- if (ds->grid[y*cr+x] == state->grid[y*cr+x] && ds->hl[y*cr+x] == hl)
+ if (ds->grid[y*cr+x] == state->grid[y*cr+x] &&
+ ds->hl[y*cr+x] == hl &&
+ !memcmp(ds->pencil+(y*cr+x)*cr, state->pencil+(y*cr+x)*cr, cr))
return; /* no change required */
- tx = BORDER + x * TILE_SIZE + 2;
- ty = BORDER + y * TILE_SIZE + 2;
+ tx = BORDER + x * TILE_SIZE + 1 + GRIDEXTRA;
+ ty = BORDER + y * TILE_SIZE + 1 + GRIDEXTRA;
cx = tx;
cy = ty;
- cw = TILE_SIZE-3;
- ch = TILE_SIZE-3;
-
- if (x % r)
- cx--, cw++;
- if ((x+1) % r)
- cw++;
- if (y % c)
- cy--, ch++;
- if ((y+1) % c)
- ch++;
+ cw = TILE_SIZE-1-2*GRIDEXTRA;
+ ch = TILE_SIZE-1-2*GRIDEXTRA;
+
+ if (x > 0 && state->blocks->whichblock[y*cr+x] == state->blocks->whichblock[y*cr+x-1])
+ cx -= GRIDEXTRA, cw += GRIDEXTRA;
+ if (x+1 < cr && state->blocks->whichblock[y*cr+x] == state->blocks->whichblock[y*cr+x+1])
+ cw += GRIDEXTRA;
+ if (y > 0 && state->blocks->whichblock[y*cr+x] == state->blocks->whichblock[(y-1)*cr+x])
+ cy -= GRIDEXTRA, ch += GRIDEXTRA;
+ if (y+1 < cr && state->blocks->whichblock[y*cr+x] == state->blocks->whichblock[(y+1)*cr+x])
+ ch += GRIDEXTRA;
+
+ clip(dr, cx, cy, cw, ch);
+
+ /* background needs erasing */
+ draw_rect(dr, cx, cy, cw, ch,
+ ((hl & 15) == 1 ? COL_HIGHLIGHT :
+ (ds->xtype && (ondiag0(y*cr+x) || ondiag1(y*cr+x))) ? COL_XDIAGONALS :
+ COL_BACKGROUND));
- clip(fe, cx, cy, cw, ch);
-
- /* background needs erasing? */
- if (ds->grid[y*cr+x] || ds->hl[y*cr+x] != hl)
- draw_rect(fe, cx, cy, cw, ch, hl ? COL_HIGHLIGHT : COL_BACKGROUND);
+ /*
+ * Draw the corners of thick lines in corner-adjacent squares,
+ * which jut into this square by one pixel.
+ */
+ if (x > 0 && y > 0 && state->blocks->whichblock[y*cr+x] != state->blocks->whichblock[(y-1)*cr+x-1])
+ draw_rect(dr, tx-GRIDEXTRA, ty-GRIDEXTRA, GRIDEXTRA, GRIDEXTRA, COL_GRID);
+ if (x+1 < cr && y > 0 && state->blocks->whichblock[y*cr+x] != state->blocks->whichblock[(y-1)*cr+x+1])
+ draw_rect(dr, tx+TILE_SIZE-1-2*GRIDEXTRA, ty-GRIDEXTRA, GRIDEXTRA, GRIDEXTRA, COL_GRID);
+ if (x > 0 && y+1 < cr && state->blocks->whichblock[y*cr+x] != state->blocks->whichblock[(y+1)*cr+x-1])
+ draw_rect(dr, tx-GRIDEXTRA, ty+TILE_SIZE-1-2*GRIDEXTRA, GRIDEXTRA, GRIDEXTRA, COL_GRID);
+ if (x+1 < cr && y+1 < cr && state->blocks->whichblock[y*cr+x] != state->blocks->whichblock[(y+1)*cr+x+1])
+ draw_rect(dr, tx+TILE_SIZE-1-2*GRIDEXTRA, ty+TILE_SIZE-1-2*GRIDEXTRA, GRIDEXTRA, GRIDEXTRA, COL_GRID);
+
+ /* pencil-mode highlight */
+ if ((hl & 15) == 2) {
+ int coords[6];
+ coords[0] = cx;
+ coords[1] = cy;
+ coords[2] = cx+cw/2;
+ coords[3] = cy;
+ coords[4] = cx;
+ coords[5] = cy+ch/2;
+ draw_polygon(dr, coords, 3, COL_HIGHLIGHT, COL_HIGHLIGHT);
+ }
/* new number needs drawing? */
if (state->grid[y*cr+x]) {
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 {
+ 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 % pw, dy = j / pw;
+
+ str[1] = '\0';
+ str[0] = i + '1';
+ if (str[0] > '9')
+ str[0] += 'a' - ('9'+1);
+ 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)
{
- int c = state->c, r = state->r, cr = c*r;
+ int cr = state->cr;
int x, y;
if (!ds->started) {
* 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.
+ * Draw the grid. We draw it as a big thick rectangle of
+ * COL_GRID initially; individual calls to draw_number()
+ * will poke the right-shaped holes in it.
*/
- for (x = 0; x <= cr; x++) {
- int thick = (x % r ? 0 : 1);
- draw_rect(fe, 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,
- cr*TILE_SIZE+3, 1+2*thick, COL_GRID);
- }
+ draw_rect(dr, BORDER-GRIDEXTRA, BORDER-GRIDEXTRA,
+ cr*TILE_SIZE+1+2*GRIDEXTRA, cr*TILE_SIZE+1+2*GRIDEXTRA,
+ 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 = state->blocks->whichblock[y*cr+x];
+ 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;
+ if (ds->xtype) {
+ if (ondiag0(y*cr+x))
+ ds->entered_items[d-1] |= ((ds->entered_items[d-1] & 64) << 1) | 64;
+ if (ondiag1(y*cr+x))
+ ds->entered_items[cr+d-1] |= ((ds->entered_items[cr+d-1] & 64) << 1) | 64;
+ }
+ }
+ }
+
+ /*
* Draw any numbers which need redrawing.
*/
for (x = 0; x < cr; x++) {
for (y = 0; y < cr; y++) {
- draw_number(fe, ds, state, x, y,
- (x == ui->hx && y == ui->hy) ||
- (flashtime > 0 &&
- (flashtime <= FLASH_TIME/3 ||
- flashtime >= FLASH_TIME*2/3)));
+ 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;
+
+ /* 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[state->blocks->whichblock[y*cr+x]*cr+d-1] & 32) ||
+ (ds->xtype && ((ondiag0(y*cr+x) && (ds->entered_items[d-1] & 128)) ||
+ (ondiag1(y*cr+x) && (ds->entered_items[cr+d-1] & 128))))))
+ 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;
}
}
static float game_anim_length(game_state *oldstate, game_state *newstate,
- int dir)
+ int dir, game_ui *ui)
{
return 0.0F;
}
static float game_flash_length(game_state *oldstate, game_state *newstate,
- int dir)
+ int dir, game_ui *ui)
{
- if (!oldstate->completed && newstate->completed)
+ if (!oldstate->completed && newstate->completed &&
+ !oldstate->cheated && !newstate->cheated)
return FLASH_TIME;
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 cr = state->cr;
+ 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);
+
+ /*
+ * Highlight X-diagonal squares.
+ */
+ if (state->xtype) {
+ int i;
+ int xhighlight = print_grey_colour(dr, 0.90F);
+
+ for (i = 0; i < cr; i++)
+ draw_rect(dr, BORDER + i*TILE_SIZE, BORDER + i*TILE_SIZE,
+ TILE_SIZE, TILE_SIZE, xhighlight);
+ for (i = 0; i < cr; i++)
+ if (i*2 != cr-1) /* avoid redoing centre square, just for fun */
+ draw_rect(dr, BORDER + i*TILE_SIZE,
+ BORDER + (cr-1-i)*TILE_SIZE,
+ TILE_SIZE, TILE_SIZE, xhighlight);
+ }
+
+ /*
+ * Main grid.
+ */
+ for (x = 1; x < cr; x++) {
+ print_line_width(dr, 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, TILE_SIZE / 40);
+ draw_line(dr, BORDER, BORDER+y*TILE_SIZE,
+ BORDER+cr*TILE_SIZE, BORDER+y*TILE_SIZE, ink);
+ }
+
+ /*
+ * Thick lines between cells. In order to do this using the
+ * line-drawing rather than rectangle-drawing API (so as to
+ * get line thicknesses to scale correctly) and yet have
+ * correctly mitred joins between lines, we must do this by
+ * tracing the boundary of each sub-block and drawing it in
+ * one go as a single polygon.
+ */
+ {
+ int *coords;
+ int bi, i, n;
+ int x, y, dx, dy, sx, sy, sdx, sdy;
+
+ print_line_width(dr, 3 * TILE_SIZE / 40);
+
+ /*
+ * Maximum perimeter of a k-omino is 2k+2. (Proof: start
+ * with k unconnected squares, with total perimeter 4k.
+ * Now repeatedly join two disconnected components
+ * together into a larger one; every time you do so you
+ * remove at least two unit edges, and you require k-1 of
+ * these operations to create a single connected piece, so
+ * you must have at most 4k-2(k-1) = 2k+2 unit edges left
+ * afterwards.)
+ */
+ coords = snewn(4*cr+4, int); /* 2k+2 points, 2 coords per point */
+
+ /*
+ * Iterate over all the blocks.
+ */
+ for (bi = 0; bi < cr; bi++) {
+
+ /*
+ * For each block, find a starting square within it
+ * which has a boundary at the left.
+ */
+ for (i = 0; i < cr; i++) {
+ int j = state->blocks->blocks[bi][i];
+ if (j % cr == 0 || state->blocks->whichblock[j-1] != bi)
+ break;
+ }
+ assert(i < cr); /* every block must have _some_ leftmost square */
+ x = state->blocks->blocks[bi][i] % cr;
+ y = state->blocks->blocks[bi][i] / cr;
+ dx = -1;
+ dy = 0;
+
+ /*
+ * Now begin tracing round the perimeter. At all
+ * times, (x,y) describes some square within the
+ * block, and (x+dx,y+dy) is some adjacent square
+ * outside it; so the edge between those two squares
+ * is always an edge of the block.
+ */
+ sx = x, sy = y, sdx = dx, sdy = dy; /* save starting position */
+ n = 0;
+ do {
+ int cx, cy, tx, ty, nin;
+
+ /*
+ * To begin with, record the point at one end of
+ * the edge. To do this, we translate (x,y) down
+ * and right by half a unit (so they're describing
+ * a point in the _centre_ of the square) and then
+ * translate back again in a manner rotated by dy
+ * and dx.
+ */
+ assert(n < 2*cr+2);
+ cx = ((2*x+1) + dy + dx) / 2;
+ cy = ((2*y+1) - dx + dy) / 2;
+ coords[2*n+0] = BORDER + cx * TILE_SIZE;
+ coords[2*n+1] = BORDER + cy * TILE_SIZE;
+ n++;
+
+ /*
+ * Now advance to the next edge, by looking at the
+ * two squares beyond it. If they're both outside
+ * the block, we turn right (by leaving x,y the
+ * same and rotating dx,dy clockwise); if they're
+ * both inside, we turn left (by rotating dx,dy
+ * anticlockwise and contriving to leave x+dx,y+dy
+ * unchanged); if one of each, we go straight on
+ * (and may enforce by assertion that they're one
+ * of each the _right_ way round).
+ */
+ nin = 0;
+ tx = x - dy + dx;
+ ty = y + dx + dy;
+ nin += (tx >= 0 && tx < cr && ty >= 0 && ty < cr &&
+ state->blocks->whichblock[ty*cr+tx] == bi);
+ tx = x - dy;
+ ty = y + dx;
+ nin += (tx >= 0 && tx < cr && ty >= 0 && ty < cr &&
+ state->blocks->whichblock[ty*cr+tx] == bi);
+ if (nin == 0) {
+ /*
+ * Turn right.
+ */
+ int tmp;
+ tmp = dx;
+ dx = -dy;
+ dy = tmp;
+ } else if (nin == 2) {
+ /*
+ * Turn left.
+ */
+ int tmp;
+
+ x += dx;
+ y += dy;
+
+ tmp = dx;
+ dx = dy;
+ dy = -tmp;
+
+ x -= dx;
+ y -= dy;
+ } else {
+ /*
+ * Go straight on.
+ */
+ x -= dy;
+ y += dx;
+ }
+
+ /*
+ * Now enforce by assertion that we ended up
+ * somewhere sensible.
+ */
+ assert(x >= 0 && x < cr && y >= 0 && y < cr &&
+ state->blocks->whichblock[y*cr+x] == bi);
+ assert(x+dx < 0 || x+dx >= cr || y+dy < 0 || y+dy >= cr ||
+ state->blocks->whichblock[(y+dy)*cr+(x+dx)] != bi);
+
+ } while (x != sx || y != sy || dx != sdx || dy != sdy);
+
+ /*
+ * That's our polygon; now draw it.
+ */
+ draw_polygon(dr, coords, n, -1, ink);
+ }
+
+ sfree(coords);
+ }
+
+ /*
+ * 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", TRUE,
+ "Solo", "games.solo", "solo",
default_params,
game_fetch_preset,
decode_params,
encode_params,
free_params,
dup_params,
- game_configure,
- custom_params,
+ TRUE, game_configure, custom_params,
validate_params,
- new_game_seed,
- validate_seed,
+ new_game_desc,
+ validate_desc,
new_game,
dup_game,
free_game,
+ TRUE, solve_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,
+ REQUIRE_RBUTTON | REQUIRE_NUMPAD, /* flags */
};
#ifdef STANDALONE_SOLVER
-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) {}
-
-#include <stdarg.h>
-
-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 = FALSE;
- char *id = NULL, *seed, *err;
- int y, x;
+ char *id = NULL, *desc, *err;
+ int grade = FALSE;
+ int ret;
while (--argc > 0) {
char *p = *++argv;
- if (!strcmp(p, "-r")) {
- recurse = TRUE;
- } else if (!strcmp(p, "-n")) {
- recurse = FALSE;
+ if (!strcmp(p, "-v")) {
+ solver_show_working = TRUE;
+ } else if (!strcmp(p, "-g")) {
+ grade = TRUE;
} 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] <game_id>\n", argv[0]);
+ fprintf(stderr, "usage: %s [-g | -v] <game_id>\n", argv[0]);
return 1;
}
- seed = strchr(id, ':');
- if (!seed) {
+ desc = strchr(id, ':');
+ if (!desc) {
fprintf(stderr, "%s: game id expects a colon in it\n", argv[0]);
return 1;
}
- *seed++ = '\0';
+ *desc++ = '\0';
- p = decode_params(id);
- err = validate_seed(p, seed);
+ p = default_params();
+ decode_params(p, id);
+ err = validate_desc(p, desc);
if (err) {
fprintf(stderr, "%s: %s\n", argv[0], err);
return 1;
}
- s = new_game(p, seed);
-
- 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");
- }
+ s = new_game(NULL, p, desc);
+
+ ret = solver(s->cr, s->blocks, s->xtype, 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 {
- nsolve(p->c, p->r, s->grid);
- }
-
- 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]);
- }
- printf("\n");
+ printf("%s\n", grid_text_format(s->cr, s->blocks, s->xtype, s->grid));
}
- printf("\n");
return 0;
}
~mdw / sgt / puzzles / blobdiff