mirror of
git://git.tartarus.org/simon/puzzles.git
synced 2025-04-21 08:01:30 -07:00
Files

retired, and replaced with a simple string. Most of the games which use it simply encode the string in the same way that the Solve move will also be encoded, i.e. solve_game() simply returns dupstr(aux_info). Again, this is a better approach than writing separate game_aux_info serialise/deserialise functions because doing it this way is self-testing (the strings are created and parsed during the course of any Solve operation at all). [originally from svn r6029]
2556 lines
73 KiB
C
2556 lines
73 KiB
C
/*
|
|
* solo.c: the number-placing puzzle most popularly known as `Sudoku'.
|
|
*
|
|
* TODO:
|
|
*
|
|
* - 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.
|
|
*
|
|
* - alternative interface modes
|
|
* + sudoku.com's Windows program has a palette of possible
|
|
* entries; you select a palette entry first and then click
|
|
* on the square you want it to go in, thus enabling
|
|
* mouse-only play. Useful for PDAs! I don't think it's
|
|
* actually incompatible with the current highlight-then-type
|
|
* approach: you _either_ highlight a palette entry and then
|
|
* 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.
|
|
* + 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.
|
|
*/
|
|
|
|
/*
|
|
* Solo puzzles need to be square overall (since each row and each
|
|
* column must contain one of every digit), but they need not be
|
|
* subdivided the same way internally. I am going to adopt a
|
|
* convention whereby I _always_ refer to `r' as the number of rows
|
|
* of _big_ divisions, and `c' as the number of columns of _big_
|
|
* divisions. Thus, a 2c by 3r puzzle looks something like this:
|
|
*
|
|
* 4 5 1 | 2 6 3
|
|
* 6 3 2 | 5 4 1
|
|
* ------+------ (Of course, you can't subdivide it the other way
|
|
* 1 4 5 | 6 3 2 or you'll get clashes; observe that the 4 in the
|
|
* 3 2 6 | 4 1 5 top left would conflict with the 4 in the second
|
|
* ------+------ box down on the left-hand side.)
|
|
* 5 1 4 | 3 2 6
|
|
* 2 6 3 | 1 5 4
|
|
*
|
|
* The need for a strong naming convention should now be clear:
|
|
* each small box is two rows of digits by three columns, while the
|
|
* overall puzzle has three rows of small boxes by two columns. So
|
|
* I will (hopefully) consistently use `r' to denote the number of
|
|
* rows _of small boxes_ (here 3), which is also the number of
|
|
* columns of digits in each small box; and `c' vice versa (here
|
|
* 2).
|
|
*
|
|
* I'm also going to choose arbitrarily to list c first wherever
|
|
* possible: the above is a 2x3 puzzle, not a 3x2 one.
|
|
*/
|
|
|
|
#include <stdio.h>
|
|
#include <stdlib.h>
|
|
#include <string.h>
|
|
#include <assert.h>
|
|
#include <ctype.h>
|
|
#include <math.h>
|
|
|
|
#ifdef STANDALONE_SOLVER
|
|
#include <stdarg.h>
|
|
int solver_show_working;
|
|
#endif
|
|
|
|
#include "puzzles.h"
|
|
|
|
/*
|
|
* To save space, I store digits internally as unsigned char. This
|
|
* imposes a hard limit of 255 on the order of the puzzle. Since
|
|
* even a 5x5 takes unacceptably long to generate, I don't see this
|
|
* as a serious limitation unless something _really_ impressive
|
|
* happens in computing technology; but here's a typedef anyway for
|
|
* general good practice.
|
|
*/
|
|
typedef unsigned char digit;
|
|
#define ORDER_MAX 255
|
|
|
|
#define PREFERRED_TILE_SIZE 32
|
|
#define TILE_SIZE (ds->tilesize)
|
|
#define BORDER (TILE_SIZE / 2)
|
|
|
|
#define FLASH_TIME 0.4F
|
|
|
|
enum { SYMM_NONE, SYMM_ROT2, SYMM_ROT4, SYMM_REF2, SYMM_REF2D, SYMM_REF4,
|
|
SYMM_REF4D, SYMM_REF8 };
|
|
|
|
enum { DIFF_BLOCK, DIFF_SIMPLE, DIFF_INTERSECT,
|
|
DIFF_SET, DIFF_RECURSIVE, DIFF_AMBIGUOUS, DIFF_IMPOSSIBLE };
|
|
|
|
enum {
|
|
COL_BACKGROUND,
|
|
COL_GRID,
|
|
COL_CLUE,
|
|
COL_USER,
|
|
COL_HIGHLIGHT,
|
|
COL_ERROR,
|
|
COL_PENCIL,
|
|
NCOLOURS
|
|
};
|
|
|
|
struct game_params {
|
|
int c, r, symm, diff;
|
|
};
|
|
|
|
struct game_state {
|
|
int c, r;
|
|
digit *grid;
|
|
unsigned char *pencil; /* c*r*c*r elements */
|
|
unsigned char *immutable; /* marks which digits are clues */
|
|
int completed, cheated;
|
|
};
|
|
|
|
static game_params *default_params(void)
|
|
{
|
|
game_params *ret = snew(game_params);
|
|
|
|
ret->c = ret->r = 3;
|
|
ret->symm = SYMM_ROT2; /* a plausible default */
|
|
ret->diff = DIFF_BLOCK; /* so is this */
|
|
|
|
return ret;
|
|
}
|
|
|
|
static void free_params(game_params *params)
|
|
{
|
|
sfree(params);
|
|
}
|
|
|
|
static game_params *dup_params(game_params *params)
|
|
{
|
|
game_params *ret = snew(game_params);
|
|
*ret = *params; /* structure copy */
|
|
return ret;
|
|
}
|
|
|
|
static int game_fetch_preset(int i, char **name, game_params **params)
|
|
{
|
|
static struct {
|
|
char *title;
|
|
game_params params;
|
|
} presets[] = {
|
|
{ "2x2 Trivial", { 2, 2, SYMM_ROT2, DIFF_BLOCK } },
|
|
{ "2x3 Basic", { 2, 3, SYMM_ROT2, DIFF_SIMPLE } },
|
|
{ "3x3 Trivial", { 3, 3, SYMM_ROT2, DIFF_BLOCK } },
|
|
{ "3x3 Basic", { 3, 3, SYMM_ROT2, DIFF_SIMPLE } },
|
|
{ "3x3 Intermediate", { 3, 3, SYMM_ROT2, DIFF_INTERSECT } },
|
|
{ "3x3 Advanced", { 3, 3, SYMM_ROT2, DIFF_SET } },
|
|
{ "3x3 Unreasonable", { 3, 3, SYMM_ROT2, DIFF_RECURSIVE } },
|
|
#ifndef SLOW_SYSTEM
|
|
{ "3x4 Basic", { 3, 4, SYMM_ROT2, DIFF_SIMPLE } },
|
|
{ "4x4 Basic", { 4, 4, SYMM_ROT2, DIFF_SIMPLE } },
|
|
#endif
|
|
};
|
|
|
|
if (i < 0 || i >= lenof(presets))
|
|
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)
|
|
{
|
|
ret->c = ret->r = atoi(string);
|
|
while (*string && isdigit((unsigned char)*string)) string++;
|
|
if (*string == 'x') {
|
|
string++;
|
|
ret->r = atoi(string);
|
|
while (*string && isdigit((unsigned char)*string)) string++;
|
|
}
|
|
while (*string) {
|
|
if (*string == 'r' || *string == 'm' || *string == 'a') {
|
|
int sn, sc, sd;
|
|
sc = *string++;
|
|
if (*string == 'd') {
|
|
sd = TRUE;
|
|
string++;
|
|
} else {
|
|
sd = FALSE;
|
|
}
|
|
sn = atoi(string);
|
|
while (*string && isdigit((unsigned char)*string)) string++;
|
|
if (sc == 'm' && sn == 8)
|
|
ret->symm = SYMM_REF8;
|
|
if (sc == 'm' && sn == 4)
|
|
ret->symm = 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 == 'u') /* unreasonable */
|
|
string++, ret->diff = DIFF_RECURSIVE;
|
|
} else
|
|
string++; /* eat unknown character */
|
|
}
|
|
}
|
|
|
|
static char *encode_params(game_params *params, int full)
|
|
{
|
|
char str[80];
|
|
|
|
sprintf(str, "%dx%d", params->c, params->r);
|
|
if (full) {
|
|
switch (params->symm) {
|
|
case SYMM_REF8: strcat(str, "m8"); break;
|
|
case SYMM_REF4: strcat(str, "m4"); break;
|
|
case SYMM_REF4D: strcat(str, "md4"); break;
|
|
case SYMM_REF2: strcat(str, "m2"); break;
|
|
case SYMM_REF2D: strcat(str, "md2"); break;
|
|
case SYMM_ROT4: strcat(str, "r4"); break;
|
|
/* case SYMM_ROT2: strcat(str, "r2"); break; [default] */
|
|
case SYMM_NONE: strcat(str, "a"); break;
|
|
}
|
|
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_RECURSIVE: strcat(str, "du"); break;
|
|
}
|
|
}
|
|
return dupstr(str);
|
|
}
|
|
|
|
static config_item *game_configure(game_params *params)
|
|
{
|
|
config_item *ret;
|
|
char buf[80];
|
|
|
|
ret = snewn(5, config_item);
|
|
|
|
ret[0].name = "Columns of sub-blocks";
|
|
ret[0].type = C_STRING;
|
|
sprintf(buf, "%d", params->c);
|
|
ret[0].sval = dupstr(buf);
|
|
ret[0].ival = 0;
|
|
|
|
ret[1].name = "Rows of sub-blocks";
|
|
ret[1].type = C_STRING;
|
|
sprintf(buf, "%d", params->r);
|
|
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:2-way mirror:"
|
|
"2-way diagonal mirror:4-way mirror:4-way diagonal mirror:"
|
|
"8-way mirror";
|
|
ret[2].ival = params->symm;
|
|
|
|
ret[3].name = "Difficulty";
|
|
ret[3].type = C_CHOICES;
|
|
ret[3].sval = ":Trivial:Basic:Intermediate:Advanced:Unreasonable";
|
|
ret[3].ival = params->diff;
|
|
|
|
ret[4].name = NULL;
|
|
ret[4].type = C_END;
|
|
ret[4].sval = NULL;
|
|
ret[4].ival = 0;
|
|
|
|
return ret;
|
|
}
|
|
|
|
static game_params *custom_params(config_item *cfg)
|
|
{
|
|
game_params *ret = snew(game_params);
|
|
|
|
ret->c = atoi(cfg[0].sval);
|
|
ret->r = atoi(cfg[1].sval);
|
|
ret->symm = cfg[2].ival;
|
|
ret->diff = cfg[3].ival;
|
|
|
|
return ret;
|
|
}
|
|
|
|
static char *validate_params(game_params *params)
|
|
{
|
|
if (params->c < 2 || params->r < 2)
|
|
return "Both dimensions must be at least 2";
|
|
if (params->c > ORDER_MAX || params->r > ORDER_MAX)
|
|
return "Dimensions greater than "STR(ORDER_MAX)" are not supported";
|
|
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).
|
|
*
|
|
* 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.
|
|
*/
|
|
|
|
/*
|
|
* Modes of reasoning currently supported:
|
|
*
|
|
* - Positional elimination: a number must go in a particular
|
|
* square because all the other empty squares in a given
|
|
* row/col/blk are ruled out.
|
|
*
|
|
* - Numeric elimination: a square must have a particular number
|
|
* in because all the other numbers that could go in it are
|
|
* ruled out.
|
|
*
|
|
* - 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.
|
|
*
|
|
* - 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).
|
|
*/
|
|
|
|
/*
|
|
* 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.
|
|
*
|
|
* 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.
|
|
*/
|
|
#define YTRANS(y) (((y)%c)*r+(y)/c)
|
|
#define YUNTRANS(y) (((y)%r)*c+(y)/r)
|
|
|
|
struct nsolve_usage {
|
|
int c, r, cr;
|
|
/*
|
|
* 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.
|
|
*/
|
|
unsigned char *cube;
|
|
/*
|
|
* This is the grid in which we write down our final
|
|
* deductions. y-coordinates in here are _not_ transformed.
|
|
*/
|
|
digit *grid;
|
|
/*
|
|
* Now we keep track, at a slightly higher level, of what we
|
|
* have yet to work out, to prevent doing the same deduction
|
|
* many times.
|
|
*/
|
|
/* 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;
|
|
};
|
|
#define cubepos(x,y,n) (((x)*usage->cr+(y))*usage->cr+(n)-1)
|
|
#define cube(x,y,n) (usage->cube[cubepos(x,y,n)])
|
|
|
|
/*
|
|
* 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)
|
|
{
|
|
int c = usage->c, r = usage->r, cr = usage->cr;
|
|
int i, j, bx, by;
|
|
|
|
assert(cube(x,y,n));
|
|
|
|
/*
|
|
* Rule out all other numbers in this square.
|
|
*/
|
|
for (i = 1; i <= cr; i++)
|
|
if (i != n)
|
|
cube(x,y,i) = FALSE;
|
|
|
|
/*
|
|
* Rule out this number in all other positions in the row.
|
|
*/
|
|
for (i = 0; i < cr; i++)
|
|
if (i != y)
|
|
cube(x,i,n) = FALSE;
|
|
|
|
/*
|
|
* Rule out this number in all other positions in the column.
|
|
*/
|
|
for (i = 0; i < cr; i++)
|
|
if (i != x)
|
|
cube(i,y,n) = FALSE;
|
|
|
|
/*
|
|
* 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;
|
|
|
|
/*
|
|
* Enter the number in the result grid.
|
|
*/
|
|
usage->grid[YUNTRANS(y)*cr+x] = 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%r)*c+(x/r))*cr+n-1] = TRUE;
|
|
}
|
|
|
|
static int nsolve_elim(struct nsolve_usage *usage, int start, int step
|
|
#ifdef STANDALONE_SOLVER
|
|
, char *fmt, ...
|
|
#endif
|
|
)
|
|
{
|
|
int c = usage->c, r = usage->r, cr = c*r;
|
|
int fpos, m, i;
|
|
|
|
/*
|
|
* Count the number of set bits within this section of the
|
|
* cube.
|
|
*/
|
|
m = 0;
|
|
fpos = -1;
|
|
for (i = 0; i < cr; i++)
|
|
if (usage->cube[start+i*step]) {
|
|
fpos = start+i*step;
|
|
m++;
|
|
}
|
|
|
|
if (m == 1) {
|
|
int x, y, n;
|
|
assert(fpos >= 0);
|
|
|
|
n = 1 + fpos % cr;
|
|
y = fpos / cr;
|
|
x = y / cr;
|
|
y %= cr;
|
|
|
|
if (!usage->grid[YUNTRANS(y)*cr+x]) {
|
|
#ifdef STANDALONE_SOLVER
|
|
if (solver_show_working) {
|
|
va_list ap;
|
|
va_start(ap, fmt);
|
|
vprintf(fmt, ap);
|
|
va_end(ap);
|
|
printf(":\n placing %d at (%d,%d)\n",
|
|
n, 1+x, 1+YUNTRANS(y));
|
|
}
|
|
#endif
|
|
nsolve_place(usage, x, y, n);
|
|
return TRUE;
|
|
}
|
|
}
|
|
|
|
return FALSE;
|
|
}
|
|
|
|
static int nsolve_intersect(struct nsolve_usage *usage,
|
|
int start1, int step1, int start2, int step2
|
|
#ifdef STANDALONE_SOLVER
|
|
, char *fmt, ...
|
|
#endif
|
|
)
|
|
{
|
|
int c = usage->c, r = usage->r, cr = c*r;
|
|
int ret, i;
|
|
|
|
/*
|
|
* Loop over the first domain and see if there's any set bit
|
|
* not also in the second.
|
|
*/
|
|
for (i = 0; i < cr; i++) {
|
|
int p = start1+i*step1;
|
|
if (usage->cube[p] &&
|
|
!(p >= start2 && p < start2+cr*step2 &&
|
|
(p - start2) % step2 == 0))
|
|
return FALSE; /* there is, so we can't deduce */
|
|
}
|
|
|
|
/*
|
|
* We have determined that all set bits in the first domain are
|
|
* within its overlap with the second. So loop over the second
|
|
* domain and remove all set bits that aren't also in that
|
|
* overlap; return TRUE iff we actually _did_ anything.
|
|
*/
|
|
ret = FALSE;
|
|
for (i = 0; i < cr; i++) {
|
|
int p = start2+i*step2;
|
|
if (usage->cube[p] &&
|
|
!(p >= start1 && p < start1+cr*step1 && (p - start1) % step1 == 0))
|
|
{
|
|
#ifdef STANDALONE_SOLVER
|
|
if (solver_show_working) {
|
|
int px, py, pn;
|
|
|
|
if (!ret) {
|
|
va_list ap;
|
|
va_start(ap, fmt);
|
|
vprintf(fmt, ap);
|
|
va_end(ap);
|
|
printf(":\n");
|
|
}
|
|
|
|
pn = 1 + p % cr;
|
|
py = p / cr;
|
|
px = py / cr;
|
|
py %= cr;
|
|
|
|
printf(" ruling out %d at (%d,%d)\n",
|
|
pn, 1+px, 1+YUNTRANS(py));
|
|
}
|
|
#endif
|
|
ret = TRUE; /* we did something */
|
|
usage->cube[p] = 0;
|
|
}
|
|
}
|
|
|
|
return ret;
|
|
}
|
|
|
|
struct nsolve_scratch {
|
|
unsigned char *grid, *rowidx, *colidx, *set;
|
|
};
|
|
|
|
static int nsolve_set(struct nsolve_usage *usage,
|
|
struct nsolve_scratch *scratch,
|
|
int start, int step1, int step2
|
|
#ifdef STANDALONE_SOLVER
|
|
, char *fmt, ...
|
|
#endif
|
|
)
|
|
{
|
|
int c = usage->c, r = usage->r, cr = c*r;
|
|
int i, j, n, count;
|
|
unsigned char *grid = 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[start+i*step1+j*step2])
|
|
first = j, count++;
|
|
if (count == 0) {
|
|
/*
|
|
* This condition actually marks a completely insoluble
|
|
* (i.e. internally inconsistent) puzzle. We return and
|
|
* report no progress made.
|
|
*/
|
|
return FALSE;
|
|
}
|
|
if (count == 1)
|
|
rowidx[i] = colidx[first] = FALSE;
|
|
}
|
|
|
|
/*
|
|
* Convert each of rowidx/colidx from a list of 0s and 1s to a
|
|
* list of the indices of the 1s.
|
|
*/
|
|
for (i = j = 0; i < cr; i++)
|
|
if (rowidx[i])
|
|
rowidx[j++] = i;
|
|
n = j;
|
|
for (i = j = 0; i < cr; i++)
|
|
if (colidx[i])
|
|
colidx[j++] = i;
|
|
assert(n == j);
|
|
|
|
/*
|
|
* And create the smaller matrix.
|
|
*/
|
|
for (i = 0; i < n; i++)
|
|
for (j = 0; j < n; j++)
|
|
grid[i*cr+j] = usage->cube[start+rowidx[i]*step1+colidx[j]*step2];
|
|
|
|
/*
|
|
* Having done that, we now have a matrix in which every row
|
|
* has at least two 1s in. Now we search to see if we can find
|
|
* a rectangle of zeroes (in the set-theoretic sense of
|
|
* `rectangle', i.e. a subset of rows crossed with a subset of
|
|
* columns) whose width and height add up to n.
|
|
*/
|
|
|
|
memset(set, 0, n);
|
|
count = 0;
|
|
while (1) {
|
|
/*
|
|
* We have a candidate set. If its size is <=1 or >=n-1
|
|
* then we move on immediately.
|
|
*/
|
|
if (count > 1 && count < n-1) {
|
|
/*
|
|
* The number of rows we need is n-count. See if we can
|
|
* find that many rows which each have a zero in all
|
|
* the positions listed in `set'.
|
|
*/
|
|
int rows = 0;
|
|
for (i = 0; i < n; i++) {
|
|
int ok = TRUE;
|
|
for (j = 0; j < n; j++)
|
|
if (set[j] && grid[i*cr+j]) {
|
|
ok = FALSE;
|
|
break;
|
|
}
|
|
if (ok)
|
|
rows++;
|
|
}
|
|
|
|
/*
|
|
* We expect never to be able to get _more_ than
|
|
* n-count suitable rows: this would imply that (for
|
|
* example) there are four numbers which between them
|
|
* have at most three possible positions, and hence it
|
|
* indicates a faulty deduction before this point or
|
|
* even a bogus clue.
|
|
*/
|
|
assert(rows <= n - count);
|
|
if (rows >= n - count) {
|
|
int progress = FALSE;
|
|
|
|
/*
|
|
* We've got one! Now, for each row which _doesn't_
|
|
* satisfy the criterion, eliminate all its set
|
|
* bits in the positions _not_ listed in `set'.
|
|
* Return TRUE (meaning progress has been made) if
|
|
* we successfully eliminated anything at all.
|
|
*
|
|
* This involves referring back through
|
|
* rowidx/colidx in order to work out which actual
|
|
* positions in the cube to meddle with.
|
|
*/
|
|
for (i = 0; i < n; i++) {
|
|
int ok = TRUE;
|
|
for (j = 0; j < n; j++)
|
|
if (set[j] && grid[i*cr+j]) {
|
|
ok = FALSE;
|
|
break;
|
|
}
|
|
if (!ok) {
|
|
for (j = 0; j < n; j++)
|
|
if (!set[j] && grid[i*cr+j]) {
|
|
int fpos = (start+rowidx[i]*step1+
|
|
colidx[j]*step2);
|
|
#ifdef STANDALONE_SOLVER
|
|
if (solver_show_working) {
|
|
int px, py, pn;
|
|
|
|
if (!progress) {
|
|
va_list ap;
|
|
va_start(ap, fmt);
|
|
vprintf(fmt, ap);
|
|
va_end(ap);
|
|
printf(":\n");
|
|
}
|
|
|
|
pn = 1 + fpos % cr;
|
|
py = fpos / cr;
|
|
px = py / cr;
|
|
py %= cr;
|
|
|
|
printf(" ruling out %d at (%d,%d)\n",
|
|
pn, 1+px, 1+YUNTRANS(py));
|
|
}
|
|
#endif
|
|
progress = TRUE;
|
|
usage->cube[fpos] = FALSE;
|
|
}
|
|
}
|
|
}
|
|
|
|
if (progress) {
|
|
return TRUE;
|
|
}
|
|
}
|
|
}
|
|
|
|
/*
|
|
* Binary increment: change the rightmost 0 to a 1, and
|
|
* change all 1s to the right of it to 0s.
|
|
*/
|
|
i = n;
|
|
while (i > 0 && set[i-1])
|
|
set[--i] = 0, count--;
|
|
if (i > 0)
|
|
set[--i] = 1, count++;
|
|
else
|
|
break; /* done */
|
|
}
|
|
|
|
return FALSE;
|
|
}
|
|
|
|
static struct nsolve_scratch *nsolve_new_scratch(struct nsolve_usage *usage)
|
|
{
|
|
struct nsolve_scratch *scratch = snew(struct nsolve_scratch);
|
|
int cr = usage->cr;
|
|
scratch->grid = snewn(cr*cr, unsigned char);
|
|
scratch->rowidx = snewn(cr, unsigned char);
|
|
scratch->colidx = snewn(cr, unsigned char);
|
|
scratch->set = snewn(cr, unsigned char);
|
|
return scratch;
|
|
}
|
|
|
|
static void nsolve_free_scratch(struct nsolve_scratch *scratch)
|
|
{
|
|
sfree(scratch->set);
|
|
sfree(scratch->colidx);
|
|
sfree(scratch->rowidx);
|
|
sfree(scratch->grid);
|
|
sfree(scratch);
|
|
}
|
|
|
|
static int nsolve(int c, int r, digit *grid)
|
|
{
|
|
struct nsolve_usage *usage;
|
|
struct nsolve_scratch *scratch;
|
|
int cr = c*r;
|
|
int x, y, n;
|
|
int diff = DIFF_BLOCK;
|
|
|
|
/*
|
|
* Set up a usage structure as a clean slate (everything
|
|
* possible).
|
|
*/
|
|
usage = snew(struct nsolve_usage);
|
|
usage->c = c;
|
|
usage->r = r;
|
|
usage->cr = cr;
|
|
usage->cube = snewn(cr*cr*cr, unsigned char);
|
|
usage->grid = grid; /* write straight back to the input */
|
|
memset(usage->cube, TRUE, cr*cr*cr);
|
|
|
|
usage->row = snewn(cr * cr, unsigned char);
|
|
usage->col = snewn(cr * cr, unsigned char);
|
|
usage->blk = snewn(cr * cr, unsigned char);
|
|
memset(usage->row, FALSE, cr * cr);
|
|
memset(usage->col, FALSE, cr * cr);
|
|
memset(usage->blk, FALSE, cr * cr);
|
|
|
|
scratch = nsolve_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])
|
|
nsolve_place(usage, x, YTRANS(y), grid[y*cr+x]);
|
|
|
|
/*
|
|
* Now loop over the grid repeatedly trying all permitted modes
|
|
* of reasoning. The loop terminates if we complete an
|
|
* iteration without making any progress; we then return
|
|
* failure or success depending on whether the grid is full or
|
|
* not.
|
|
*/
|
|
while (1) {
|
|
/*
|
|
* I'd like to write `continue;' inside each of the
|
|
* following loops, so that the solver returns here after
|
|
* making some progress. However, I can't specify that I
|
|
* want to continue an outer loop rather than the innermost
|
|
* one, so I'm apologetically resorting to a goto.
|
|
*/
|
|
cont:
|
|
|
|
/*
|
|
* Blockwise positional elimination.
|
|
*/
|
|
for (x = 0; x < cr; x += r)
|
|
for (y = 0; y < r; y++)
|
|
for (n = 1; n <= cr; n++)
|
|
if (!usage->blk[(y*c+(x/r))*cr+n-1] &&
|
|
nsolve_elim(usage, cubepos(x,y,n), r*cr
|
|
#ifdef STANDALONE_SOLVER
|
|
, "positional elimination,"
|
|
" block (%d,%d)", 1+x/r, 1+y
|
|
#endif
|
|
)) {
|
|
diff = max(diff, DIFF_BLOCK);
|
|
goto cont;
|
|
}
|
|
|
|
/*
|
|
* Row-wise positional elimination.
|
|
*/
|
|
for (y = 0; y < cr; y++)
|
|
for (n = 1; n <= cr; n++)
|
|
if (!usage->row[y*cr+n-1] &&
|
|
nsolve_elim(usage, cubepos(0,y,n), cr*cr
|
|
#ifdef STANDALONE_SOLVER
|
|
, "positional elimination,"
|
|
" row %d", 1+YUNTRANS(y)
|
|
#endif
|
|
)) {
|
|
diff = max(diff, DIFF_SIMPLE);
|
|
goto cont;
|
|
}
|
|
/*
|
|
* Column-wise positional elimination.
|
|
*/
|
|
for (x = 0; x < cr; x++)
|
|
for (n = 1; n <= cr; n++)
|
|
if (!usage->col[x*cr+n-1] &&
|
|
nsolve_elim(usage, cubepos(x,0,n), cr
|
|
#ifdef STANDALONE_SOLVER
|
|
, "positional elimination," " column %d", 1+x
|
|
#endif
|
|
)) {
|
|
diff = max(diff, DIFF_SIMPLE);
|
|
goto cont;
|
|
}
|
|
|
|
/*
|
|
* Numeric elimination.
|
|
*/
|
|
for (x = 0; x < cr; x++)
|
|
for (y = 0; y < cr; y++)
|
|
if (!usage->grid[YUNTRANS(y)*cr+x] &&
|
|
nsolve_elim(usage, cubepos(x,y,1), 1
|
|
#ifdef STANDALONE_SOLVER
|
|
, "numeric elimination at (%d,%d)", 1+x,
|
|
1+YUNTRANS(y)
|
|
#endif
|
|
)) {
|
|
diff = max(diff, DIFF_SIMPLE);
|
|
goto cont;
|
|
}
|
|
|
|
/*
|
|
* Intersectional analysis, rows vs blocks.
|
|
*/
|
|
for (y = 0; y < cr; y++)
|
|
for (x = 0; x < cr; x += r)
|
|
for (n = 1; n <= cr; n++)
|
|
if (!usage->row[y*cr+n-1] &&
|
|
!usage->blk[((y%r)*c+(x/r))*cr+n-1] &&
|
|
(nsolve_intersect(usage, cubepos(0,y,n), cr*cr,
|
|
cubepos(x,y%r,n), r*cr
|
|
#ifdef STANDALONE_SOLVER
|
|
, "intersectional analysis,"
|
|
" row %d vs block (%d,%d)",
|
|
1+YUNTRANS(y), 1+x/r, 1+y%r
|
|
#endif
|
|
) ||
|
|
nsolve_intersect(usage, cubepos(x,y%r,n), r*cr,
|
|
cubepos(0,y,n), cr*cr
|
|
#ifdef STANDALONE_SOLVER
|
|
, "intersectional analysis,"
|
|
" block (%d,%d) vs row %d",
|
|
1+x/r, 1+y%r, 1+YUNTRANS(y)
|
|
#endif
|
|
))) {
|
|
diff = max(diff, DIFF_INTERSECT);
|
|
goto cont;
|
|
}
|
|
|
|
/*
|
|
* Intersectional analysis, columns vs blocks.
|
|
*/
|
|
for (x = 0; x < cr; x++)
|
|
for (y = 0; y < r; y++)
|
|
for (n = 1; n <= cr; n++)
|
|
if (!usage->col[x*cr+n-1] &&
|
|
!usage->blk[(y*c+(x/r))*cr+n-1] &&
|
|
(nsolve_intersect(usage, cubepos(x,0,n), cr,
|
|
cubepos((x/r)*r,y,n), r*cr
|
|
#ifdef STANDALONE_SOLVER
|
|
, "intersectional analysis,"
|
|
" column %d vs block (%d,%d)",
|
|
1+x, 1+x/r, 1+y
|
|
#endif
|
|
) ||
|
|
nsolve_intersect(usage, cubepos((x/r)*r,y,n), r*cr,
|
|
cubepos(x,0,n), cr
|
|
#ifdef STANDALONE_SOLVER
|
|
, "intersectional analysis,"
|
|
" block (%d,%d) vs column %d",
|
|
1+x/r, 1+y, 1+x
|
|
#endif
|
|
))) {
|
|
diff = max(diff, DIFF_INTERSECT);
|
|
goto cont;
|
|
}
|
|
|
|
/*
|
|
* Blockwise set elimination.
|
|
*/
|
|
for (x = 0; x < cr; x += r)
|
|
for (y = 0; y < r; y++)
|
|
if (nsolve_set(usage, scratch, cubepos(x,y,1), r*cr, 1
|
|
#ifdef STANDALONE_SOLVER
|
|
, "set elimination, block (%d,%d)", 1+x/r, 1+y
|
|
#endif
|
|
)) {
|
|
diff = max(diff, DIFF_SET);
|
|
goto cont;
|
|
}
|
|
|
|
/*
|
|
* Row-wise set elimination.
|
|
*/
|
|
for (y = 0; y < cr; y++)
|
|
if (nsolve_set(usage, scratch, cubepos(0,y,1), cr*cr, 1
|
|
#ifdef STANDALONE_SOLVER
|
|
, "set elimination, row %d", 1+YUNTRANS(y)
|
|
#endif
|
|
)) {
|
|
diff = max(diff, DIFF_SET);
|
|
goto cont;
|
|
}
|
|
|
|
/*
|
|
* Column-wise set elimination.
|
|
*/
|
|
for (x = 0; x < cr; x++)
|
|
if (nsolve_set(usage, scratch, cubepos(x,0,1), cr, 1
|
|
#ifdef STANDALONE_SOLVER
|
|
, "set elimination, column %d", 1+x
|
|
#endif
|
|
)) {
|
|
diff = max(diff, DIFF_SET);
|
|
goto cont;
|
|
}
|
|
|
|
/*
|
|
* If we reach here, we have made no deductions in this
|
|
* iteration, so the algorithm terminates.
|
|
*/
|
|
break;
|
|
}
|
|
|
|
nsolve_free_scratch(scratch);
|
|
|
|
sfree(usage->cube);
|
|
sfree(usage->row);
|
|
sfree(usage->col);
|
|
sfree(usage->blk);
|
|
sfree(usage);
|
|
|
|
for (x = 0; x < cr; x++)
|
|
for (y = 0; y < cr; y++)
|
|
if (!grid[y*cr+x])
|
|
return DIFF_IMPOSSIBLE;
|
|
return diff;
|
|
}
|
|
|
|
/* ----------------------------------------------------------------------
|
|
* End of non-recursive solver code.
|
|
*/
|
|
|
|
/*
|
|
* Check whether a grid contains a valid complete puzzle.
|
|
*/
|
|
static int check_valid(int c, int r, digit *grid)
|
|
{
|
|
int cr = c*r;
|
|
unsigned char *used;
|
|
int x, y, n;
|
|
|
|
used = snewn(cr, unsigned char);
|
|
|
|
/*
|
|
* Check that each row contains precisely one of everything.
|
|
*/
|
|
for (y = 0; y < cr; y++) {
|
|
memset(used, FALSE, cr);
|
|
for (x = 0; x < cr; x++)
|
|
if (grid[y*cr+x] > 0 && grid[y*cr+x] <= cr)
|
|
used[grid[y*cr+x]-1] = TRUE;
|
|
for (n = 0; n < cr; n++)
|
|
if (!used[n]) {
|
|
sfree(used);
|
|
return FALSE;
|
|
}
|
|
}
|
|
|
|
/*
|
|
* Check that each column contains precisely one of everything.
|
|
*/
|
|
for (x = 0; x < cr; x++) {
|
|
memset(used, FALSE, cr);
|
|
for (y = 0; y < cr; y++)
|
|
if (grid[y*cr+x] > 0 && grid[y*cr+x] <= cr)
|
|
used[grid[y*cr+x]-1] = TRUE;
|
|
for (n = 0; n < cr; n++)
|
|
if (!used[n]) {
|
|
sfree(used);
|
|
return FALSE;
|
|
}
|
|
}
|
|
|
|
/*
|
|
* 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;
|
|
}
|
|
}
|
|
}
|
|
|
|
sfree(used);
|
|
return TRUE;
|
|
}
|
|
|
|
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:
|
|
break; /* just x,y is all we need */
|
|
case SYMM_ROT2:
|
|
ADD(cr - 1 - x, cr - 1 - y);
|
|
break;
|
|
case SYMM_ROT4:
|
|
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 char *encode_solve_move(int cr, digit *grid)
|
|
{
|
|
int i, len;
|
|
char *ret, *p, *sep;
|
|
|
|
/*
|
|
* 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 */
|
|
|
|
/*
|
|
* 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 ret;
|
|
}
|
|
|
|
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;
|
|
digit *grid, *grid2;
|
|
struct xy { int x, y; } *locs;
|
|
int nlocs;
|
|
int ret;
|
|
char *desc;
|
|
int coords[16], ncoords;
|
|
int *symmclasses, nsymmclasses;
|
|
int maxdiff, recursing;
|
|
|
|
/*
|
|
* 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.
|
|
*/
|
|
maxdiff = params->diff;
|
|
if (c == 2 && r == 2)
|
|
maxdiff = DIFF_BLOCK;
|
|
|
|
grid = snewn(area, digit);
|
|
locs = snewn(area, struct xy);
|
|
grid2 = snewn(area, digit);
|
|
|
|
/*
|
|
* 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.
|
|
*/
|
|
nsymmclasses = 0;
|
|
symmclasses = snewn(cr * cr, int);
|
|
{
|
|
int x, y;
|
|
|
|
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)
|
|
symmclasses[nsymmclasses++] = i;
|
|
}
|
|
}
|
|
|
|
/*
|
|
* Loop until we get a grid of the required difficulty. This is
|
|
* nasty, but it seems to be unpleasantly hard to generate
|
|
* difficult grids otherwise.
|
|
*/
|
|
do {
|
|
/*
|
|
* Start the recursive solver with an empty grid to generate a
|
|
* random solved state.
|
|
*/
|
|
memset(grid, 0, area);
|
|
ret = rsolve(c, r, grid, rs, 1);
|
|
assert(ret == 1);
|
|
assert(check_valid(c, r, grid));
|
|
|
|
/*
|
|
* Save the solved grid in aux.
|
|
*/
|
|
{
|
|
/*
|
|
* We might already have written *aux the last time we
|
|
* went round this loop, in which case we should free
|
|
* the old aux before overwriting it with the new one.
|
|
*/
|
|
if (*aux) {
|
|
sfree(*aux);
|
|
}
|
|
|
|
*aux = encode_solve_move(cr, grid);
|
|
}
|
|
|
|
/*
|
|
* Now we have a solved grid, start removing things from it
|
|
* while preserving solubility.
|
|
*/
|
|
recursing = FALSE;
|
|
while (1) {
|
|
int x, y, i, j;
|
|
|
|
/*
|
|
* Iterate over the grid and enumerate all the filled
|
|
* squares we could empty.
|
|
*/
|
|
nlocs = 0;
|
|
|
|
for (i = 0; i < nsymmclasses; i++) {
|
|
x = symmclasses[i] % cr;
|
|
y = symmclasses[i] / cr;
|
|
if (grid[y*cr+x]) {
|
|
locs[nlocs].x = x;
|
|
locs[nlocs].y = y;
|
|
nlocs++;
|
|
}
|
|
}
|
|
|
|
/*
|
|
* Now shuffle that list.
|
|
*/
|
|
for (i = nlocs; i > 1; i--) {
|
|
int p = random_upto(rs, i);
|
|
if (p != i-1) {
|
|
struct xy t = locs[p];
|
|
locs[p] = locs[i-1];
|
|
locs[i-1] = t;
|
|
}
|
|
}
|
|
|
|
/*
|
|
* Now loop over the shuffled list and, for each element,
|
|
* see whether removing that element (and its reflections)
|
|
* from the grid will still leave the grid soluble by
|
|
* nsolve.
|
|
*/
|
|
for (i = 0; i < nlocs; i++) {
|
|
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;
|
|
|
|
if (recursing)
|
|
ret = (rsolve(c, r, grid2, NULL, 2) == 1);
|
|
else
|
|
ret = (nsolve(c, r, grid2) <= maxdiff);
|
|
|
|
if (ret) {
|
|
for (j = 0; j < ncoords; j++)
|
|
grid[coords[2*j+1]*cr+coords[2*j]] = 0;
|
|
break;
|
|
}
|
|
}
|
|
|
|
if (i == nlocs) {
|
|
/*
|
|
* There was nothing we could remove without
|
|
* destroying solvability. If we're trying to
|
|
* generate a recursion-only grid and haven't
|
|
* switched over to rsolve yet, we now do;
|
|
* otherwise we give up.
|
|
*/
|
|
if (maxdiff == DIFF_RECURSIVE && !recursing) {
|
|
recursing = TRUE;
|
|
} else {
|
|
break;
|
|
}
|
|
}
|
|
}
|
|
|
|
memcpy(grid2, grid, area);
|
|
} while (nsolve(c, r, grid2) < maxdiff);
|
|
|
|
sfree(grid2);
|
|
sfree(locs);
|
|
|
|
sfree(symmclasses);
|
|
|
|
/*
|
|
* Now we have the grid as it will be presented to the user.
|
|
* Encode it in a game desc.
|
|
*/
|
|
{
|
|
char *p;
|
|
int run, i;
|
|
|
|
desc = snewn(5 * area, char);
|
|
p = desc;
|
|
run = 0;
|
|
for (i = 0; i <= area; i++) {
|
|
int n = (i < area ? grid[i] : -1);
|
|
|
|
if (!n)
|
|
run++;
|
|
else {
|
|
if (run) {
|
|
while (run > 0) {
|
|
int c = 'a' - 1 + run;
|
|
if (run > 26)
|
|
c = 'z';
|
|
*p++ = c;
|
|
run -= c - ('a' - 1);
|
|
}
|
|
} else {
|
|
/*
|
|
* If there's a number in the very top left or
|
|
* bottom right, there's no point putting an
|
|
* unnecessary _ before or after it.
|
|
*/
|
|
if (p > desc && n > 0)
|
|
*p++ = '_';
|
|
}
|
|
if (n > 0)
|
|
p += sprintf(p, "%d", n);
|
|
run = 0;
|
|
}
|
|
}
|
|
assert(p - desc < 5 * area);
|
|
*p++ = '\0';
|
|
desc = sresize(desc, p - desc, char);
|
|
}
|
|
|
|
sfree(grid);
|
|
|
|
return desc;
|
|
}
|
|
|
|
static char *validate_desc(game_params *params, char *desc)
|
|
{
|
|
int area = params->r * params->r * params->c * params->c;
|
|
int squares = 0;
|
|
|
|
while (*desc) {
|
|
int n = *desc++;
|
|
if (n >= 'a' && n <= 'z') {
|
|
squares += n - 'a' + 1;
|
|
} else if (n == '_') {
|
|
/* do nothing */;
|
|
} else if (n > '0' && n <= '9') {
|
|
squares++;
|
|
while (*desc >= '0' && *desc <= '9')
|
|
desc++;
|
|
} else
|
|
return "Invalid character in game description";
|
|
}
|
|
|
|
if (squares < area)
|
|
return "Not enough data to fill grid";
|
|
|
|
if (squares > area)
|
|
return "Too much data to fit in grid";
|
|
|
|
return NULL;
|
|
}
|
|
|
|
static game_state *new_game(midend_data *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->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 = state->cheated = FALSE;
|
|
|
|
i = 0;
|
|
while (*desc) {
|
|
int n = *desc++;
|
|
if (n >= 'a' && n <= 'z') {
|
|
int run = n - 'a' + 1;
|
|
assert(i + run <= area);
|
|
while (run-- > 0)
|
|
state->grid[i++] = 0;
|
|
} else if (n == '_') {
|
|
/* do nothing */;
|
|
} else if (n > '0' && n <= '9') {
|
|
assert(i < area);
|
|
state->immutable[i] = TRUE;
|
|
state->grid[i++] = atoi(desc-1);
|
|
while (*desc >= '0' && *desc <= '9')
|
|
desc++;
|
|
} else {
|
|
assert(!"We can't get here");
|
|
}
|
|
}
|
|
assert(i == area);
|
|
|
|
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;
|
|
|
|
ret->c = state->c;
|
|
ret->r = state->r;
|
|
|
|
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)
|
|
{
|
|
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 c = state->c, r = state->r, cr = c*r;
|
|
char *ret;
|
|
digit *grid;
|
|
int rsolve_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);
|
|
rsolve_ret = rsolve(c, r, grid, NULL, 2);
|
|
|
|
if (rsolve_ret != 1) {
|
|
sfree(grid);
|
|
if (rsolve_ret == 0)
|
|
*error = "No solution exists for this puzzle";
|
|
else
|
|
*error = "Multiple solutions exist for this puzzle";
|
|
return NULL;
|
|
}
|
|
|
|
ret = encode_solve_move(cr, grid);
|
|
|
|
sfree(grid);
|
|
|
|
return ret;
|
|
}
|
|
|
|
static char *grid_text_format(int c, int r, digit *grid)
|
|
{
|
|
int cr = c*r;
|
|
int x, y;
|
|
int maxlen;
|
|
char *ret, *p;
|
|
|
|
/*
|
|
* There are cr lines of digits, plus r-1 lines of block
|
|
* separators. Each line contains cr digits, cr-1 separating
|
|
* spaces, and c-1 two-character block separators. Thus, the
|
|
* total length of a line is 2*cr+2*c-3 (not counting the
|
|
* newline), and there are cr+r-1 of them.
|
|
*/
|
|
maxlen = (cr+r-1) * (2*cr+2*c-2);
|
|
ret = snewn(maxlen+1, char);
|
|
p = ret;
|
|
|
|
for (y = 0; y < cr; y++) {
|
|
for (x = 0; x < cr; x++) {
|
|
int ch = grid[y * cr + x];
|
|
if (ch == 0)
|
|
ch = ' ';
|
|
else if (ch <= 9)
|
|
ch = '0' + ch;
|
|
else
|
|
ch = 'a' + ch-10;
|
|
*p++ = ch;
|
|
if (x+1 < cr) {
|
|
*p++ = ' ';
|
|
if ((x+1) % r == 0) {
|
|
*p++ = '|';
|
|
*p++ = ' ';
|
|
}
|
|
}
|
|
}
|
|
*p++ = '\n';
|
|
if (y+1 < cr && (y+1) % c == 0) {
|
|
for (x = 0; x < cr; x++) {
|
|
*p++ = '-';
|
|
if (x+1 < cr) {
|
|
*p++ = '-';
|
|
if ((x+1) % r == 0) {
|
|
*p++ = '+';
|
|
*p++ = '-';
|
|
}
|
|
}
|
|
}
|
|
*p++ = '\n';
|
|
}
|
|
}
|
|
|
|
assert(p - ret == maxlen);
|
|
*p = '\0';
|
|
return ret;
|
|
}
|
|
|
|
static char *game_text_format(game_state *state)
|
|
{
|
|
return grid_text_format(state->c, state->r, state->grid);
|
|
}
|
|
|
|
struct game_ui {
|
|
/*
|
|
* These are the coordinates of the currently highlighted
|
|
* square on the grid, or -1,-1 if there isn't one. When there
|
|
* is, pressing a valid number or letter key or Space will
|
|
* 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;
|
|
}
|
|
|
|
static void free_ui(game_ui *ui)
|
|
{
|
|
sfree(ui);
|
|
}
|
|
|
|
char *encode_ui(game_ui *ui)
|
|
{
|
|
return NULL;
|
|
}
|
|
|
|
void decode_ui(game_ui *ui, char *encoding)
|
|
{
|
|
}
|
|
|
|
static void game_changed_state(game_ui *ui, game_state *oldstate,
|
|
game_state *newstate)
|
|
{
|
|
int c = newstate->c, r = newstate->r, cr = c*r;
|
|
/*
|
|
* We prevent pencil-mode highlighting of a filled square. So
|
|
* if the user has just filled in a square which we had a
|
|
* pencil-mode highlight in (by Undo, or by Redo, or by Solve),
|
|
* then we cancel the highlight.
|
|
*/
|
|
if (ui->hx >= 0 && ui->hy >= 0 && ui->hpencil &&
|
|
newstate->grid[ui->hy * cr + ui->hx] != 0) {
|
|
ui->hx = ui->hy = -1;
|
|
}
|
|
}
|
|
|
|
struct game_drawstate {
|
|
int started;
|
|
int c, r, cr;
|
|
int tilesize;
|
|
digit *grid;
|
|
unsigned char *pencil;
|
|
unsigned char *hl;
|
|
/* This is scratch space used within a single call to game_redraw. */
|
|
int *entered_items;
|
|
};
|
|
|
|
static char *interpret_move(game_state *state, game_ui *ui, game_drawstate *ds,
|
|
int x, int y, int button)
|
|
{
|
|
int c = state->c, r = state->r, cr = c*r;
|
|
int tx, ty;
|
|
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 == ' ')) {
|
|
int n = button - '0';
|
|
if (button >= 'A' && button <= 'Z')
|
|
n = button - 'A' + 10;
|
|
if (button >= 'a' && button <= 'z')
|
|
n = button - 'a' + 10;
|
|
if (button == ' ')
|
|
n = 0;
|
|
|
|
/*
|
|
* 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",
|
|
ui->hpencil && n > 0 ? 'P' : 'R', ui->hx, ui->hy, n);
|
|
|
|
ui->hx = ui->hy = -1;
|
|
|
|
return dupstr(buf);
|
|
}
|
|
|
|
return NULL;
|
|
}
|
|
|
|
static game_state *execute_move(game_state *from, char *move)
|
|
{
|
|
int c = from->c, r = from->r, cr = c*r;
|
|
game_state *ret;
|
|
int x, y, n;
|
|
|
|
if (move[0] == 'S') {
|
|
char *p;
|
|
|
|
ret = dup_game(from);
|
|
ret->completed = ret->cheated = TRUE;
|
|
|
|
p = move+1;
|
|
for (n = 0; n < cr*cr; n++) {
|
|
ret->grid[n] = atoi(p);
|
|
|
|
if (!*p || ret->grid[n] < 1 || ret->grid[n] > cr) {
|
|
free_game(ret);
|
|
return NULL;
|
|
}
|
|
|
|
while (*p && isdigit((unsigned char)*p)) p++;
|
|
if (*p == ',') p++;
|
|
}
|
|
|
|
return ret;
|
|
} else if ((move[0] == 'P' || move[0] == 'R') &&
|
|
sscanf(move+1, "%d,%d,%d", &x, &y, &n) == 3 &&
|
|
x >= 0 && x < cr && y >= 0 && y < cr && n >= 0 && n <= cr) {
|
|
|
|
ret = dup_game(from);
|
|
if (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(c, r, ret->grid)) {
|
|
ret->completed = TRUE;
|
|
}
|
|
}
|
|
return ret;
|
|
} else
|
|
return NULL; /* couldn't parse move string */
|
|
}
|
|
|
|
/* ----------------------------------------------------------------------
|
|
* Drawing routines.
|
|
*/
|
|
|
|
#define SIZE(cr) ((cr) * TILE_SIZE + 2*BORDER + 1)
|
|
#define GETTILESIZE(cr, w) ( (w-1) / (cr+1) )
|
|
|
|
static void game_size(game_params *params, game_drawstate *ds,
|
|
int *x, int *y, int expand)
|
|
{
|
|
int c = params->c, r = params->r, cr = c*r;
|
|
int ts;
|
|
|
|
ts = min(GETTILESIZE(cr, *x), GETTILESIZE(cr, *y));
|
|
if (expand)
|
|
ds->tilesize = ts;
|
|
else
|
|
ds->tilesize = min(ts, PREFERRED_TILE_SIZE);
|
|
|
|
*x = SIZE(cr);
|
|
*y = SIZE(cr);
|
|
}
|
|
|
|
static float *game_colours(frontend *fe, game_state *state, int *ncolours)
|
|
{
|
|
float *ret = snewn(3 * NCOLOURS, float);
|
|
|
|
frontend_default_colour(fe, &ret[COL_BACKGROUND * 3]);
|
|
|
|
ret[COL_GRID * 3 + 0] = 0.0F;
|
|
ret[COL_GRID * 3 + 1] = 0.0F;
|
|
ret[COL_GRID * 3 + 2] = 0.0F;
|
|
|
|
ret[COL_CLUE * 3 + 0] = 0.0F;
|
|
ret[COL_CLUE * 3 + 1] = 0.0F;
|
|
ret[COL_CLUE * 3 + 2] = 0.0F;
|
|
|
|
ret[COL_USER * 3 + 0] = 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_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)
|
|
{
|
|
struct game_drawstate *ds = snew(struct game_drawstate);
|
|
int c = state->c, r = state->r, cr = c*r;
|
|
|
|
ds->started = FALSE;
|
|
ds->c = c;
|
|
ds->r = r;
|
|
ds->cr = cr;
|
|
ds->grid = snewn(cr*cr, digit);
|
|
memset(ds->grid, 0, 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)
|
|
{
|
|
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,
|
|
int x, int y, int hl)
|
|
{
|
|
int c = state->c, r = state->r, cr = c*r;
|
|
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 &&
|
|
!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;
|
|
|
|
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++;
|
|
|
|
clip(fe, cx, cy, cw, ch);
|
|
|
|
/* background needs erasing */
|
|
draw_rect(fe, cx, cy, cw, ch, (hl & 15) == 1 ? COL_HIGHLIGHT : COL_BACKGROUND);
|
|
|
|
/* 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(fe, coords, 3, TRUE, COL_HIGHLIGHT);
|
|
}
|
|
|
|
/* new number needs drawing? */
|
|
if (state->grid[y*cr+x]) {
|
|
str[1] = '\0';
|
|
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,
|
|
FONT_VARIABLE, TILE_SIZE/2, ALIGN_VCENTRE | ALIGN_HCENTRE,
|
|
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(fe, 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);
|
|
|
|
draw_update(fe, 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,
|
|
game_state *state, int dir, game_ui *ui,
|
|
float animtime, float flashtime)
|
|
{
|
|
int c = state->c, r = state->r, cr = c*r;
|
|
int x, y;
|
|
|
|
if (!ds->started) {
|
|
/*
|
|
* The initial contents of the window are not guaranteed
|
|
* and can vary with front ends. To be on the safe side,
|
|
* all games should start by drawing a big
|
|
* background-colour rectangle covering the whole window.
|
|
*/
|
|
draw_rect(fe, 0, 0, SIZE(cr), SIZE(cr), COL_BACKGROUND);
|
|
|
|
/*
|
|
* Draw the grid.
|
|
*/
|
|
for (x = 0; x <= cr; x++) {
|
|
int thick = (x % r ? 0 : 1);
|
|
draw_rect(fe, BORDER + x*TILE_SIZE - thick, BORDER-1,
|
|
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);
|
|
}
|
|
}
|
|
|
|
/*
|
|
* This array is used to keep track of rows, columns and boxes
|
|
* which contain a number more than once.
|
|
*/
|
|
for (x = 0; x < cr * cr; x++)
|
|
ds->entered_items[x] = 0;
|
|
for (x = 0; x < cr; x++)
|
|
for (y = 0; y < cr; y++) {
|
|
digit d = state->grid[y*cr+x];
|
|
if (d) {
|
|
int box = (x/r)+(y/c)*c;
|
|
ds->entered_items[x*cr+d-1] |= ((ds->entered_items[x*cr+d-1] & 1) << 1) | 1;
|
|
ds->entered_items[y*cr+d-1] |= ((ds->entered_items[y*cr+d-1] & 4) << 1) | 4;
|
|
ds->entered_items[box*cr+d-1] |= ((ds->entered_items[box*cr+d-1] & 16) << 1) | 16;
|
|
}
|
|
}
|
|
|
|
/*
|
|
* Draw any numbers which need redrawing.
|
|
*/
|
|
for (x = 0; x < cr; x++) {
|
|
for (y = 0; y < cr; y++) {
|
|
int highlight = 0;
|
|
digit d = state->grid[y*cr+x];
|
|
|
|
if (flashtime > 0 &&
|
|
(flashtime <= FLASH_TIME/3 ||
|
|
flashtime >= FLASH_TIME*2/3))
|
|
highlight = 1;
|
|
|
|
/* Highlight active input areas. */
|
|
if (x == ui->hx && y == ui->hy)
|
|
highlight = ui->hpencil ? 2 : 1;
|
|
|
|
/* Mark obvious errors (ie, numbers which occur more than once
|
|
* in a single row, column, or box). */
|
|
if (d && ((ds->entered_items[x*cr+d-1] & 2) ||
|
|
(ds->entered_items[y*cr+d-1] & 8) ||
|
|
(ds->entered_items[((x/r)+(y/c)*c)*cr+d-1] & 32)))
|
|
highlight |= 16;
|
|
|
|
draw_number(fe, ds, state, x, y, highlight);
|
|
}
|
|
}
|
|
|
|
/*
|
|
* Update the _entire_ grid if necessary.
|
|
*/
|
|
if (!ds->started) {
|
|
draw_update(fe, 0, 0, SIZE(cr), SIZE(cr));
|
|
ds->started = TRUE;
|
|
}
|
|
}
|
|
|
|
static float game_anim_length(game_state *oldstate, game_state *newstate,
|
|
int dir, game_ui *ui)
|
|
{
|
|
return 0.0F;
|
|
}
|
|
|
|
static float game_flash_length(game_state *oldstate, game_state *newstate,
|
|
int dir, game_ui *ui)
|
|
{
|
|
if (!oldstate->completed && newstate->completed &&
|
|
!oldstate->cheated && !newstate->cheated)
|
|
return FLASH_TIME;
|
|
return 0.0F;
|
|
}
|
|
|
|
static int game_wants_statusbar(void)
|
|
{
|
|
return FALSE;
|
|
}
|
|
|
|
static int game_timing_state(game_state *state)
|
|
{
|
|
return TRUE;
|
|
}
|
|
|
|
#ifdef COMBINED
|
|
#define thegame solo
|
|
#endif
|
|
|
|
const struct game thegame = {
|
|
"Solo", "games.solo",
|
|
default_params,
|
|
game_fetch_preset,
|
|
decode_params,
|
|
encode_params,
|
|
free_params,
|
|
dup_params,
|
|
TRUE, game_configure, custom_params,
|
|
validate_params,
|
|
new_game_desc,
|
|
validate_desc,
|
|
new_game,
|
|
dup_game,
|
|
free_game,
|
|
TRUE, solve_game,
|
|
TRUE, game_text_format,
|
|
new_ui,
|
|
free_ui,
|
|
encode_ui,
|
|
decode_ui,
|
|
game_changed_state,
|
|
interpret_move,
|
|
execute_move,
|
|
game_size,
|
|
game_colours,
|
|
game_new_drawstate,
|
|
game_free_drawstate,
|
|
game_redraw,
|
|
game_anim_length,
|
|
game_flash_length,
|
|
game_wants_statusbar,
|
|
FALSE, game_timing_state,
|
|
0, /* mouse_priorities */
|
|
};
|
|
|
|
#ifdef STANDALONE_SOLVER
|
|
|
|
/*
|
|
* gcc -DSTANDALONE_SOLVER -o solosolver solo.c malloc.c
|
|
*/
|
|
|
|
void frontend_default_colour(frontend *fe, float *output) {}
|
|
void draw_text(frontend *fe, int x, int y, int fonttype, int fontsize,
|
|
int align, int colour, char *text) {}
|
|
void draw_rect(frontend *fe, int x, int y, int w, int h, int colour) {}
|
|
void draw_line(frontend *fe, int x1, int y1, int x2, int y2, int colour) {}
|
|
void draw_polygon(frontend *fe, int *coords, int npoints,
|
|
int fill, int colour) {}
|
|
void clip(frontend *fe, int x, int y, int w, int h) {}
|
|
void unclip(frontend *fe) {}
|
|
void start_draw(frontend *fe) {}
|
|
void draw_update(frontend *fe, int x, int y, int w, int h) {}
|
|
void end_draw(frontend *fe) {}
|
|
unsigned long random_bits(random_state *state, int bits)
|
|
{ assert(!"Shouldn't get randomness"); return 0; }
|
|
unsigned long random_upto(random_state *state, unsigned long limit)
|
|
{ assert(!"Shouldn't get randomness"); return 0; }
|
|
|
|
void fatal(char *fmt, ...)
|
|
{
|
|
va_list ap;
|
|
|
|
fprintf(stderr, "fatal error: ");
|
|
|
|
va_start(ap, fmt);
|
|
vfprintf(stderr, fmt, ap);
|
|
va_end(ap);
|
|
|
|
fprintf(stderr, "\n");
|
|
exit(1);
|
|
}
|
|
|
|
int main(int argc, char **argv)
|
|
{
|
|
game_params *p;
|
|
game_state *s;
|
|
int recurse = TRUE;
|
|
char *id = NULL, *desc, *err;
|
|
int y, x;
|
|
int grade = FALSE;
|
|
|
|
while (--argc > 0) {
|
|
char *p = *++argv;
|
|
if (!strcmp(p, "-r")) {
|
|
recurse = TRUE;
|
|
} else if (!strcmp(p, "-n")) {
|
|
recurse = FALSE;
|
|
} else if (!strcmp(p, "-v")) {
|
|
solver_show_working = TRUE;
|
|
recurse = FALSE;
|
|
} else if (!strcmp(p, "-g")) {
|
|
grade = TRUE;
|
|
recurse = FALSE;
|
|
} else if (*p == '-') {
|
|
fprintf(stderr, "%s: unrecognised option `%s'\n", argv[0]);
|
|
return 1;
|
|
} else {
|
|
id = p;
|
|
}
|
|
}
|
|
|
|
if (!id) {
|
|
fprintf(stderr, "usage: %s [-n | -r | -g | -v] <game_id>\n", argv[0]);
|
|
return 1;
|
|
}
|
|
|
|
desc = strchr(id, ':');
|
|
if (!desc) {
|
|
fprintf(stderr, "%s: game id expects a colon in it\n", argv[0]);
|
|
return 1;
|
|
}
|
|
*desc++ = '\0';
|
|
|
|
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(NULL, p, desc);
|
|
|
|
if (recurse) {
|
|
int ret = rsolve(p->c, p->r, s->grid, NULL, 2);
|
|
if (ret > 1) {
|
|
fprintf(stderr, "%s: rsolve: multiple solutions detected\n",
|
|
argv[0]);
|
|
}
|
|
} else {
|
|
int ret = nsolve(p->c, p->r, s->grid);
|
|
if (grade) {
|
|
if (ret == DIFF_IMPOSSIBLE) {
|
|
/*
|
|
* Now resort to rsolve to determine whether it's
|
|
* really soluble.
|
|
*/
|
|
ret = rsolve(p->c, p->r, s->grid, NULL, 2);
|
|
if (ret == 0)
|
|
ret = DIFF_IMPOSSIBLE;
|
|
else if (ret == 1)
|
|
ret = DIFF_RECURSIVE;
|
|
else
|
|
ret = DIFF_AMBIGUOUS;
|
|
}
|
|
printf("Difficulty rating: %s\n",
|
|
ret==DIFF_BLOCK ? "Trivial (blockwise positional elimination only)":
|
|
ret==DIFF_SIMPLE ? "Basic (row/column/number elimination required)":
|
|
ret==DIFF_INTERSECT ? "Intermediate (intersectional analysis required)":
|
|
ret==DIFF_SET ? "Advanced (set elimination required)":
|
|
ret==DIFF_RECURSIVE ? "Unreasonable (guesswork and backtracking required)":
|
|
ret==DIFF_AMBIGUOUS ? "Ambiguous (multiple solutions exist)":
|
|
ret==DIFF_IMPOSSIBLE ? "Impossible (no solution exists)":
|
|
"INTERNAL ERROR: unrecognised difficulty code");
|
|
}
|
|
}
|
|
|
|
printf("%s\n", grid_text_format(p->c, p->r, s->grid));
|
|
|
|
return 0;
|
|
}
|
|
|
|
#endif
|