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puzzles/solo.c
Simon Tatham c8a843ee62 Improvements to filled-grid generation. Introduced a cunning idea 
suggested by IWJ last night: grid generation can immediately choose
an entire grid row randomly, since all that's doing is nailing down
the names of the numbers, and that gets the whole thing started more
efficiently. But the main difference is that now grid generation is
given only area^2 steps to come up with a filled grid, and then cut
off unceremoniously, causing grid generation to fail and be retried
from scratch. This seems to prevent hangups on jigsaw layouts that
admit few useful solutions, by changing layout constantly. 9j
puzzles now generate at a sensible rate, and as an added bonus so do
5x5 normal puzzles, which they never used to.

[originally from svn r7978]
2008-04-08 09:36:33 +00:00

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/*
* solo.c: the number-placing puzzle most popularly known as `Sudoku'.
*
* TODO:
*
* - 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.
*
* - 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, solver_recurse_depth;
#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 GRIDEXTRA (TILE_SIZE / 32)
#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_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 {
/*
* 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 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, 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 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, 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)
{
int seen_r = FALSE;
ret->c = ret->r = atoi(string);
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++;
}
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 */
}
}
static char *encode_params(game_params *params, int full)
{
char str[80];
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);
}
static config_item *game_configure(game_params *params)
{
config_item *ret;
char buf[80];
ret = snewn(6, 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 = "\"X\" (require every number in each main diagonal)";
ret[2].type = C_BOOLEAN;
ret[2].sval = NULL;
ret[2].ival = params->xtype;
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[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;
}
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->xtype = cfg[2].ival;
ret->symm = cfg[3].ival;
ret->diff = cfg[4].ival;
return ret;
}
static char *validate_params(game_params *params, int full)
{
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;
}
/* ----------------------------------------------------------------------
* 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.
*/
/*
* 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).
*
* - Forcing chains (see comment for solver_forcing().)
*
* - 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.
*/
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[(y*cr+x)*cr+n-1]; there
* are macros below to help with this.
*/
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[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 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 solver_place(struct solver_usage *usage, int x, int y, int n)
{
int cr = usage->cr;
int sqindex = y*cr+x;
int i, bi;
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.
*/
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[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[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 solver_elim(struct solver_usage *usage, int *indices
#ifdef STANDALONE_SOLVER
, char *fmt, ...
#endif
)
{
int cr = usage->cr;
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[indices[i]]) {
fpos = indices[i];
m++;
}
if (m == 1) {
int x, y, n;
assert(fpos >= 0);
n = 1 + fpos % cr;
x = fpos / cr;
y = x / cr;
x %= cr;
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;
}
} 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 solver_intersect(struct solver_usage *usage,
int *indices1, int *indices2
#ifdef STANDALONE_SOLVER
, char *fmt, ...
#endif
)
{
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 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);
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(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;
};
static void gridgen_place(struct gridgen_usage *usage, int x, int y, digit n,
int placing)
{
int cr = usage->cr;
usage->row[y*cr+n-1] = usage->col[x*cr+n-1] =
usage->blk[usage->blocks->whichblock[y*cr+x]*cr+n-1] = placing;
if (usage->diag) {
if (ondiag0(y*cr+x))
usage->diag[n-1] = placing;
if (ondiag1(y*cr+x))
usage->diag[cr+n-1] = placing;
}
usage->grid[y*cr+x] = placing ? n : 0;
}
/*
* 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 *steps)
{
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)
return TRUE;
/*
* Next, abandon generation if we went over our steps limit.
*/
if (*steps <= 0)
return FALSE;
(*steps)--;
/*
* 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. */
gridgen_place(usage, sx, sy, n, TRUE);
usage->nspaces--;
/* Call the solver recursively. Stop when we find a solution. */
if (gridgen_real(usage, grid, steps)) {
ret = TRUE;
break;
}
/* Revert the usage structure. */
gridgen_place(usage, sx, sy, n, FALSE);
usage->nspaces++;
}
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, int maxsteps)
{
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 = grid;
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;
}
/*
* Begin by filling in the whole top row with randomly chosen
* numbers. This cannot introduce any bias or restriction on
* the available grids, since we already know those numbers
* are all distinct so all we're doing is choosing their
* labels.
*/
for (x = 0; x < cr; x++)
grid[x] = x+1;
shuffle(grid, cr, sizeof(*grid), rs);
for (x = 0; x < cr; x++)
gridgen_place(usage, x, 0, grid[x], TRUE);
usage->spaces = snewn(cr * cr, struct gridgen_coord);
usage->nspaces = 0;
usage->rs = rs;
/*
* Initialise the list of grid spaces, taking care to leave
* out the row I've already filled in above.
*/
for (y = 1; 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++;
}
}
/*
* Run the real generator function.
*/
ret = gridgen_real(usage, grid, &maxsteps);
/*
* Clean up the usage structure now we have our answer.
*/
sfree(usage->spaces);
sfree(usage->blk);
sfree(usage->col);
sfree(usage->row);
sfree(usage);
return ret;
}
/* ----------------------------------------------------------------------
* End of grid generator code.
*/
/*
* Check whether a grid contains a valid complete puzzle.
*/
static int check_valid(int cr, struct block_structure *blocks, int xtype,
digit *grid)
{
unsigned char *used;
int x, y, i, j, 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 (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 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;
struct block_structure *blocks;
digit *grid, *grid2;
struct xy { int x, y; } *locs;
int nlocs;
char *desc;
int coords[16], ncoords;
int maxdiff;
int x, y, i, j;
/*
* 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);
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
/*
* 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.
*/
while (1) {
/*
* 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, area*area))
continue;
assert(check_valid(cr, blocks, params->xtype, 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.
*/
/*
* 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);
/*
* 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(7 * 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;
}
}
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';
desc = sresize(desc, p - desc, char);
}
sfree(grid);
return desc;
}
static char *validate_desc(game_params *params, char *desc)
{
int cr = params->c * params->r, area = cr*cr;
int squares = 0;
int *dsf;
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 (*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";
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(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->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->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 (*desc && *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);
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 cr = state->cr, area = cr * cr;
ret->cr = state->cr;
ret->xtype = state->xtype;
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
* 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);
}
static char *encode_ui(game_ui *ui)
{
return NULL;
}
static void decode_ui(game_ui *ui, char *encoding)
{
}
static void game_changed_state(game_ui *ui, game_state *oldstate,
game_state *newstate)
{
int 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;
}
}
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 == '\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 == ' ' || button == '\010' || button == '\177')
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",
(char)(ui->hpencil && n > 0 ? 'P' : 'R'), ui->hx, ui->hy, n);
ui->hx = ui->hy = -1;
return dupstr(buf);
}
return NULL;
}
static game_state *execute_move(game_state *from, char *move)
{
int 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.
*/
#define SIZE(cr) ((cr) * TILE_SIZE + 2*BORDER + 1)
#define GETTILESIZE(cr, w) ( (double)(w-1) / (double)(cr+1) )
static void game_compute_size(game_params *params, int tilesize,
int *x, int *y)
{
/* Ick: fake up `ds->tilesize' for macro expansion purposes */
struct { int tilesize; } ads, *ds = &ads;
ads.tilesize = tilesize;
*x = SIZE(params->c * params->r);
*y = SIZE(params->c * params->r);
}
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_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] = 0.78F * 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(drawing *dr, game_state *state)
{
struct game_drawstate *ds = snew(struct game_drawstate);
int cr = state->cr;
ds->started = FALSE;
ds->cr = cr;
ds->xtype = state->xtype;
ds->grid = snewn(cr*cr, digit);
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(drawing *dr, game_drawstate *ds)
{
sfree(ds->hl);
sfree(ds->pencil);
sfree(ds->grid);
sfree(ds->entered_items);
sfree(ds);
}
static void draw_number(drawing *dr, game_drawstate *ds, game_state *state,
int x, int y, int hl)
{
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 &&
!memcmp(ds->pencil+(y*cr+x)*cr, state->pencil+(y*cr+x)*cr, cr))
return; /* no change required */
tx = BORDER + x * TILE_SIZE + 1 + GRIDEXTRA;
ty = BORDER + y * TILE_SIZE + 1 + GRIDEXTRA;
cx = tx;
cy = ty;
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));
/*
* 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[1] = '\0';
str[0] = state->grid[y*cr+x] + '0';
if (str[0] > '9')
str[0] += 'a' - ('9'+1);
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 : (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(dr);
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(drawing *dr, game_drawstate *ds, game_state *oldstate,
game_state *state, int dir, game_ui *ui,
float animtime, float flashtime)
{
int cr = state->cr;
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(dr, 0, 0, SIZE(cr), SIZE(cr), COL_BACKGROUND);
/*
* 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.
*/
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++) {
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(dr, 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_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)
{
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
#define thegame solo
#endif
const struct game thegame = {
"Solo", "games.solo", "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,
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,
TRUE, FALSE, game_print_size, game_print,
FALSE, /* wants_statusbar */
FALSE, game_timing_state,
REQUIRE_RBUTTON | REQUIRE_NUMPAD, /* flags */
};
#ifdef STANDALONE_SOLVER
int main(int argc, char **argv)
{
game_params *p;
game_state *s;
char *id = NULL, *desc, *err;
int grade = FALSE;
int ret;
while (--argc > 0) {
char *p = *++argv;
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], p);
return 1;
} else {
id = p;
}
}
if (!id) {
fprintf(stderr, "usage: %s [-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);
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 {
printf("%s\n", grid_text_format(s->cr, s->blocks, s->xtype, s->grid));
}
return 0;
}
#endif
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