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puzzles/solo.c
Simon Tatham a8980f3736 I am again gormless! When I overhauled Solo's grid generator in 
r6160, I completely failed to ensure that generated grids were _at
most_ the required difficulty. It appears to have been only random
chance that prevented a request for a Trivial puzzle from producing
Extreme. Here's a one-line fix.

[originally from svn r6298]
[r6160 == e55838bc9b0d173ca539d0cfe714495b5c12b9dd]
2005-09-12 17:09:29 +00:00

3158 lines
90 KiB
C
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/*
* solo.c: the number-placing puzzle most popularly known as `Sudoku'.
*
* TODO:
*
* - reports from users are that `Trivial'-mode puzzles are still
* rather hard compared to newspapers' easy ones, so some better
* low-end difficulty grading would be nice
* + it's possible that really easy puzzles always have
* _several_ things you can do, so don't make you hunt too
* hard for the one deduction you can currently make
* + it's also possible that easy puzzles require fewer
* cross-eliminations: perhaps there's a higher incidence of
* things you can deduce by looking only at (say) rows,
* rather than things you have to check both rows and columns
* for
* + but really, what I need to do is find some really easy
* puzzles and _play_ them, to see what's actually easy about
* them
* + while I'm revamping this area, filling in the _last_
* number in a nearly-full row or column should certainly be
* permitted even at the lowest difficulty level.
* + also Owen noticed that `Basic' grids requiring numeric
* elimination are actually very hard, so I wonder if a
* difficulty gradation between that and positional-
* elimination-only might be in order
* + but it's not good to have _too_ many difficulty levels, or
* it'll take too long to randomly generate a given level.
*
* - it might still be nice to do some prioritisation on the
* removal of numbers from the grid
* + one possibility is to try to minimise the maximum number
* of filled squares in any block, which in particular ought
* to enforce never leaving a completely filled block in the
* puzzle as presented.
*
* - alternative interface modes
* + sudoku.com's Windows program has a palette of possible
* entries; you select a palette entry first and then click
* on the square you want it to go in, thus enabling
* mouse-only play. Useful for PDAs! I don't think it's
* actually incompatible with the current highlight-then-type
* approach: you _either_ highlight a palette entry and then
* click, _or_ you highlight a square and then type. At most
* one thing is ever highlighted at a time, so there's no way
* to confuse the two.
* + then again, I don't actually like sudoku.com's interface;
* it's too much like a paint package whereas I prefer to
* think of Solo as a text editor.
* + another PDA-friendly possibility is a drag interface:
* _drag_ numbers from the palette into the grid squares.
* Thought experiments suggest I'd prefer that to the
* sudoku.com approach, but I haven't actually tried it.
*/
/*
* Solo puzzles need to be square overall (since each row and each
* column must contain one of every digit), but they need not be
* subdivided the same way internally. I am going to adopt a
* convention whereby I _always_ refer to `r' as the number of rows
* of _big_ divisions, and `c' as the number of columns of _big_
* divisions. Thus, a 2c by 3r puzzle looks something like this:
*
* 4 5 1 | 2 6 3
* 6 3 2 | 5 4 1
* ------+------ (Of course, you can't subdivide it the other way
* 1 4 5 | 6 3 2 or you'll get clashes; observe that the 4 in the
* 3 2 6 | 4 1 5 top left would conflict with the 4 in the second
* ------+------ box down on the left-hand side.)
* 5 1 4 | 3 2 6
* 2 6 3 | 1 5 4
*
* The need for a strong naming convention should now be clear:
* each small box is two rows of digits by three columns, while the
* overall puzzle has three rows of small boxes by two columns. So
* I will (hopefully) consistently use `r' to denote the number of
* rows _of small boxes_ (here 3), which is also the number of
* columns of digits in each small box; and `c' vice versa (here
* 2).
*
* I'm also going to choose arbitrarily to list c first wherever
* possible: the above is a 2x3 puzzle, not a 3x2 one.
*/
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <assert.h>
#include <ctype.h>
#include <math.h>
#ifdef STANDALONE_SOLVER
#include <stdarg.h>
int solver_show_working, 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 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_GRID,
COL_CLUE,
COL_USER,
COL_HIGHLIGHT,
COL_ERROR,
COL_PENCIL,
NCOLOURS
};
struct game_params {
int c, r, symm, diff;
};
struct game_state {
int c, r;
digit *grid;
unsigned char *pencil; /* c*r*c*r elements */
unsigned char *immutable; /* marks which digits are clues */
int completed, cheated;
};
static game_params *default_params(void)
{
game_params *ret = snew(game_params);
ret->c = ret->r = 3;
ret->symm = SYMM_ROT2; /* a plausible default */
ret->diff = DIFF_BLOCK; /* so is this */
return ret;
}
static void free_params(game_params *params)
{
sfree(params);
}
static game_params *dup_params(game_params *params)
{
game_params *ret = snew(game_params);
*ret = *params; /* structure copy */
return ret;
}
static int game_fetch_preset(int i, char **name, game_params **params)
{
static struct {
char *title;
game_params params;
} presets[] = {
{ "2x2 Trivial", { 2, 2, SYMM_ROT2, DIFF_BLOCK } },
{ "2x3 Basic", { 2, 3, SYMM_ROT2, DIFF_SIMPLE } },
{ "3x3 Trivial", { 3, 3, SYMM_ROT2, DIFF_BLOCK } },
{ "3x3 Basic", { 3, 3, SYMM_ROT2, DIFF_SIMPLE } },
{ "3x3 Intermediate", { 3, 3, SYMM_ROT2, DIFF_INTERSECT } },
{ "3x3 Advanced", { 3, 3, SYMM_ROT2, DIFF_SET } },
{ "3x3 Extreme", { 3, 3, SYMM_ROT2, DIFF_EXTREME } },
{ "3x3 Unreasonable", { 3, 3, SYMM_ROT2, DIFF_RECURSIVE } },
#ifndef SLOW_SYSTEM
{ "3x4 Basic", { 3, 4, SYMM_ROT2, DIFF_SIMPLE } },
{ "4x4 Basic", { 4, 4, SYMM_ROT2, DIFF_SIMPLE } },
#endif
};
if (i < 0 || i >= lenof(presets))
return FALSE;
*name = dupstr(presets[i].title);
*params = dup_params(&presets[i].params);
return TRUE;
}
static void decode_params(game_params *ret, char const *string)
{
ret->c = ret->r = atoi(string);
while (*string && isdigit((unsigned char)*string)) string++;
if (*string == 'x') {
string++;
ret->r = atoi(string);
while (*string && isdigit((unsigned char)*string)) string++;
}
while (*string) {
if (*string == 'r' || *string == 'm' || *string == 'a') {
int sn, sc, sd;
sc = *string++;
if (*string == 'd') {
sd = TRUE;
string++;
} else {
sd = FALSE;
}
sn = atoi(string);
while (*string && isdigit((unsigned char)*string)) string++;
if (sc == 'm' && sn == 8)
ret->symm = SYMM_REF8;
if (sc == 'm' && sn == 4)
ret->symm = sd ? SYMM_REF4D : SYMM_REF4;
if (sc == 'm' && sn == 2)
ret->symm = sd ? SYMM_REF2D : SYMM_REF2;
if (sc == 'r' && sn == 4)
ret->symm = SYMM_ROT4;
if (sc == 'r' && sn == 2)
ret->symm = SYMM_ROT2;
if (sc == 'a')
ret->symm = SYMM_NONE;
} else if (*string == 'd') {
string++;
if (*string == 't') /* trivial */
string++, ret->diff = DIFF_BLOCK;
else if (*string == 'b') /* basic */
string++, ret->diff = DIFF_SIMPLE;
else if (*string == 'i') /* intermediate */
string++, ret->diff = DIFF_INTERSECT;
else if (*string == 'a') /* advanced */
string++, ret->diff = DIFF_SET;
else if (*string == '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];
sprintf(str, "%dx%d", params->c, params->r);
if (full) {
switch (params->symm) {
case SYMM_REF8: strcat(str, "m8"); break;
case SYMM_REF4: strcat(str, "m4"); break;
case SYMM_REF4D: strcat(str, "md4"); break;
case SYMM_REF2: strcat(str, "m2"); break;
case SYMM_REF2D: strcat(str, "md2"); break;
case SYMM_ROT4: strcat(str, "r4"); break;
/* case SYMM_ROT2: strcat(str, "r2"); break; [default] */
case SYMM_NONE: strcat(str, "a"); break;
}
switch (params->diff) {
/* case DIFF_BLOCK: strcat(str, "dt"); break; [default] */
case DIFF_SIMPLE: strcat(str, "db"); break;
case DIFF_INTERSECT: strcat(str, "di"); break;
case DIFF_SET: strcat(str, "da"); break;
case DIFF_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(5, config_item);
ret[0].name = "Columns of sub-blocks";
ret[0].type = C_STRING;
sprintf(buf, "%d", params->c);
ret[0].sval = dupstr(buf);
ret[0].ival = 0;
ret[1].name = "Rows of sub-blocks";
ret[1].type = C_STRING;
sprintf(buf, "%d", params->r);
ret[1].sval = dupstr(buf);
ret[1].ival = 0;
ret[2].name = "Symmetry";
ret[2].type = C_CHOICES;
ret[2].sval = ":None:2-way rotation:4-way rotation:2-way mirror:"
"2-way diagonal mirror:4-way mirror:4-way diagonal mirror:"
"8-way mirror";
ret[2].ival = params->symm;
ret[3].name = "Difficulty";
ret[3].type = C_CHOICES;
ret[3].sval = ":Trivial:Basic:Intermediate:Advanced:Extreme:Unreasonable";
ret[3].ival = params->diff;
ret[4].name = NULL;
ret[4].type = C_END;
ret[4].sval = NULL;
ret[4].ival = 0;
return ret;
}
static game_params *custom_params(config_item *cfg)
{
game_params *ret = snew(game_params);
ret->c = atoi(cfg[0].sval);
ret->r = atoi(cfg[1].sval);
ret->symm = cfg[2].ival;
ret->diff = cfg[3].ival;
return ret;
}
static char *validate_params(game_params *params, int full)
{
if (params->c < 2 || params->r < 2)
return "Both dimensions must be at least 2";
if (params->c > ORDER_MAX || params->r > ORDER_MAX)
return "Dimensions greater than "STR(ORDER_MAX)" are not supported";
if ((params->c * params->r) > 36)
return "Unable to support more than 36 distinct symbols in a puzzle";
return NULL;
}
/* ----------------------------------------------------------------------
* 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).
*
* - Mutual neighbour elimination: find two squares A,B and a
* number N in the possible set of A, such that putting N in A
* would rule out enough possibilities from the mutual
* neighbours of A and B that there would be no possibilities
* left for B. Thereby rule out N in A.
* + The simplest case of this is if B has two possibilities
* (wlog {1,2}), and there are two mutual neighbours of A and
* B which have possibilities {1,3} and {2,3}. Thus, if A
* were to be 3, then those neighbours would contain 1 and 2,
* and hence there would be nothing left which could go in B.
* + There can be more complex cases of it too: if A and B are
* in the same column of large blocks, then they can have
* more than two mutual neighbours, some of which can also be
* neighbours of one another. Suppose, for example, that B
* has possibilities {1,2,3}; there's one square P in the
* same column as B and the same block as A, with
* possibilities {1,4}; and there are _two_ squares Q,R in
* the same column as A and the same block as B with
* possibilities {2,3,4}. Then if A contained 4, P would
* contain 1, and Q and R would have to contain 2 and 3 in
* _some_ order; therefore, once again, B would have no
* remaining possibilities.
*
* - Recursion. If all else fails, we pick one of the currently
* most constrained empty squares and take a random guess at its
* contents, then continue solving on that basis and see if we
* get any further.
*/
/*
* Within this solver, I'm going to transform all y-coordinates by
* inverting the significance of the block number and the position
* within the block. That is, we will start with the top row of
* each block in order, then the second row of each block in order,
* etc.
*
* This transformation has the enormous advantage that it means
* every row, column _and_ block is described by an arithmetic
* progression of coordinates within the cubic array, so that I can
* use the same very simple function to do blockwise, row-wise and
* column-wise elimination.
*/
#define YTRANS(y) (((y)%c)*r+(y)/c)
#define YUNTRANS(y) (((y)%r)*c+(y)/r)
struct solver_usage {
int c, r, cr;
/*
* We set up a cubic array, indexed by x, y and digit; each
* element of this array is TRUE or FALSE according to whether
* or not that digit _could_ in principle go in that position.
*
* The way to index this array is cube[(x*cr+y)*cr+n-1].
* y-coordinates in here are transformed.
*/
unsigned char *cube;
/*
* This is the grid in which we write down our final
* deductions. y-coordinates in here are _not_ transformed.
*/
digit *grid;
/*
* Now we keep track, at a slightly higher level, of what we
* have yet to work out, to prevent doing the same deduction
* many times.
*/
/* row[y*cr+n-1] TRUE if digit n has been placed in row y */
unsigned char *row;
/* col[x*cr+n-1] TRUE if digit n has been placed in row x */
unsigned char *col;
/* blk[(y*c+x)*cr+n-1] TRUE if digit n has been placed in block (x,y) */
unsigned char *blk;
};
#define cubepos(x,y,n) (((x)*usage->cr+(y))*usage->cr+(n)-1)
#define cube(x,y,n) (usage->cube[cubepos(x,y,n)])
/*
* Function called when we are certain that a particular square has
* a particular number in it. The y-coordinate passed in here is
* transformed.
*/
static void solver_place(struct solver_usage *usage, int x, int y, int n)
{
int c = usage->c, r = usage->r, cr = usage->cr;
int i, j, bx, by;
assert(cube(x,y,n));
/*
* Rule out all other numbers in this square.
*/
for (i = 1; i <= cr; i++)
if (i != n)
cube(x,y,i) = FALSE;
/*
* Rule out this number in all other positions in the row.
*/
for (i = 0; i < cr; i++)
if (i != y)
cube(x,i,n) = FALSE;
/*
* Rule out this number in all other positions in the column.
*/
for (i = 0; i < cr; i++)
if (i != x)
cube(i,y,n) = FALSE;
/*
* Rule out this number in all other positions in the block.
*/
bx = (x/r)*r;
by = y % r;
for (i = 0; i < r; i++)
for (j = 0; j < c; j++)
if (bx+i != x || by+j*r != y)
cube(bx+i,by+j*r,n) = FALSE;
/*
* Enter the number in the result grid.
*/
usage->grid[YUNTRANS(y)*cr+x] = n;
/*
* Cross out this number from the list of numbers left to place
* in its row, its column and its block.
*/
usage->row[y*cr+n-1] = usage->col[x*cr+n-1] =
usage->blk[((y%r)*c+(x/r))*cr+n-1] = TRUE;
}
static int solver_elim(struct solver_usage *usage, int start, int step
#ifdef STANDALONE_SOLVER
, char *fmt, ...
#endif
)
{
int c = usage->c, r = usage->r, cr = c*r;
int fpos, m, i;
/*
* Count the number of set bits within this section of the
* cube.
*/
m = 0;
fpos = -1;
for (i = 0; i < cr; i++)
if (usage->cube[start+i*step]) {
fpos = start+i*step;
m++;
}
if (m == 1) {
int x, y, n;
assert(fpos >= 0);
n = 1 + fpos % cr;
y = fpos / cr;
x = y / cr;
y %= cr;
if (!usage->grid[YUNTRANS(y)*cr+x]) {
#ifdef STANDALONE_SOLVER
if (solver_show_working) {
va_list ap;
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+YUNTRANS(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 start1, int step1, int start2, int step2
#ifdef STANDALONE_SOLVER
, char *fmt, ...
#endif
)
{
int c = usage->c, r = usage->r, cr = c*r;
int ret, i;
/*
* Loop over the first domain and see if there's any set bit
* not also in the second.
*/
for (i = 0; i < cr; i++) {
int p = start1+i*step1;
if (usage->cube[p] &&
!(p >= start2 && p < start2+cr*step2 &&
(p - start2) % step2 == 0))
return 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 = 0; i < cr; i++) {
int p = start2+i*step2;
if (usage->cube[p] &&
!(p >= start1 && p < start1+cr*step1 && (p - start1) % step1 == 0))
{
#ifdef STANDALONE_SOLVER
if (solver_show_working) {
int px, py, pn;
if (!ret) {
va_list ap;
printf("%*s", solver_recurse_depth*4, "");
va_start(ap, fmt);
vprintf(fmt, ap);
va_end(ap);
printf(":\n");
}
pn = 1 + p % cr;
py = p / cr;
px = py / cr;
py %= cr;
printf("%*s ruling out %d at (%d,%d)\n",
solver_recurse_depth*4, "", pn, 1+px, 1+YUNTRANS(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;
#ifdef STANDALONE_SOLVER
int *bfsprev;
#endif
};
static int solver_set(struct solver_usage *usage,
struct solver_scratch *scratch,
int start, int step1, int step2
#ifdef STANDALONE_SOLVER
, char *fmt, ...
#endif
)
{
int c = usage->c, r = usage->r, cr = c*r;
int i, j, n, count;
unsigned char *grid = scratch->grid;
unsigned char *rowidx = scratch->rowidx;
unsigned char *colidx = scratch->colidx;
unsigned char *set = scratch->set;
/*
* We are passed a cr-by-cr matrix of booleans. Our first job
* is to winnow it by finding any definite placements - i.e.
* any row with a solitary 1 - and discarding that row and the
* column containing the 1.
*/
memset(rowidx, TRUE, cr);
memset(colidx, TRUE, cr);
for (i = 0; i < cr; i++) {
int count = 0, first = -1;
for (j = 0; j < cr; j++)
if (usage->cube[start+i*step1+j*step2])
first = j, count++;
/*
* If count == 0, 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[start+rowidx[i]*step1+colidx[j]*step2];
/*
* Having done that, we now have a matrix in which every row
* has at least two 1s in. Now we search to see if we can find
* a rectangle of zeroes (in the set-theoretic sense of
* `rectangle', i.e. a subset of rows crossed with a subset of
* columns) whose width and height add up to n.
*/
memset(set, 0, n);
count = 0;
while (1) {
/*
* We have a candidate set. If its size is <=1 or >=n-1
* then we move on immediately.
*/
if (count > 1 && count < n-1) {
/*
* The number of rows we need is n-count. See if we can
* find that many rows which each have a zero in all
* the positions listed in `set'.
*/
int rows = 0;
for (i = 0; i < n; i++) {
int ok = TRUE;
for (j = 0; j < n; j++)
if (set[j] && grid[i*cr+j]) {
ok = FALSE;
break;
}
if (ok)
rows++;
}
/*
* We expect never to be able to get _more_ than
* n-count suitable rows: this would imply that (for
* example) there are four numbers which between them
* have at most three possible positions, and hence it
* indicates a faulty deduction before this point or
* even a bogus clue.
*/
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 = (start+rowidx[i]*step1+
colidx[j]*step2);
#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;
py = fpos / cr;
px = py / cr;
py %= cr;
printf("%*s ruling out %d at (%d,%d)\n",
solver_recurse_depth*4, "",
pn, 1+px, 1+YUNTRANS(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;
}
/*
* Try to find a number in the possible set of (x1,y1) which can be
* ruled out because it would leave no possibilities for (x2,y2).
*/
static int solver_mne(struct solver_usage *usage,
struct solver_scratch *scratch,
int x1, int y1, int x2, int y2)
{
int c = usage->c, r = usage->r, cr = c*r;
int *nb[2];
unsigned char *set = scratch->set;
unsigned char *numbers = scratch->rowidx;
unsigned char *numbersleft = scratch->colidx;
int nnb, count;
int i, j, n, nbi;
nb[0] = scratch->neighbours;
nb[1] = scratch->neighbours + cr;
/*
* First, work out the mutual neighbour squares of the two. We
* can assert that they're not actually in the same block,
* which leaves two possibilities: they're in different block
* rows _and_ different block columns (thus their mutual
* neighbours are precisely the other two corners of the
* rectangle), or they're in the same row (WLOG) and different
* columns, in which case their mutual neighbours are the
* column of each block aligned with the other square.
*
* We divide the mutual neighbours into two separate subsets
* nb[0] and nb[1]; squares in the same subset are not only
* adjacent to both our key squares, but are also always
* adjacent to one another.
*/
if (x1 / r != x2 / r && y1 % r != y2 % r) {
/* Corners of the rectangle. */
nnb = 1;
nb[0][0] = cubepos(x2, y1, 1);
nb[1][0] = cubepos(x1, y2, 1);
} else if (x1 / r != x2 / r) {
/* Same row of blocks; different blocks within that row. */
int x1b = x1 - (x1 % r);
int x2b = x2 - (x2 % r);
nnb = r;
for (i = 0; i < r; i++) {
nb[0][i] = cubepos(x2b+i, y1, 1);
nb[1][i] = cubepos(x1b+i, y2, 1);
}
} else {
/* Same column of blocks; different blocks within that column. */
int y1b = y1 % r;
int y2b = y2 % r;
assert(y1 % r != y2 % r);
nnb = c;
for (i = 0; i < c; i++) {
nb[0][i] = cubepos(x2, y1b+i*r, 1);
nb[1][i] = cubepos(x1, y2b+i*r, 1);
}
}
/*
* Right. Now loop over each possible number.
*/
for (n = 1; n <= cr; n++) {
if (!cube(x1, y1, n))
continue;
for (j = 0; j < cr; j++)
numbersleft[j] = cube(x2, y2, j+1);
/*
* Go over every possible subset of each neighbour list,
* and see if its union of possible numbers minus n has the
* same size as the subset. If so, add the numbers in that
* subset to the set of things which would be ruled out
* from (x2,y2) if n were placed at (x1,y1).
*/
memset(set, 0, nnb);
count = 0;
while (1) {
/*
* Binary increment: change the rightmost 0 to a 1, and
* change all 1s to the right of it to 0s.
*/
i = nnb;
while (i > 0 && set[i-1])
set[--i] = 0, count--;
if (i > 0)
set[--i] = 1, count++;
else
break; /* done */
/*
* Examine this subset of each neighbour set.
*/
for (nbi = 0; nbi < 2; nbi++) {
int *nbs = nb[nbi];
memset(numbers, 0, cr);
for (i = 0; i < nnb; i++)
if (set[i])
for (j = 0; j < cr; j++)
if (j != n-1 && usage->cube[nbs[i] + j])
numbers[j] = 1;
for (i = j = 0; j < cr; j++)
i += numbers[j];
if (i == count) {
/*
* Got one. This subset of nbs, in the absence
* of n, would definitely contain all the
* numbers listed in `numbers'. Rule them out
* of `numbersleft'.
*/
for (j = 0; j < cr; j++)
if (numbers[j])
numbersleft[j] = 0;
}
}
}
/*
* If we've got nothing left in `numbersleft', we have a
* successful mutual neighbour elimination.
*/
for (j = 0; j < cr; j++)
if (numbersleft[j])
break;
if (j == cr) {
#ifdef STANDALONE_SOLVER
if (solver_show_working) {
printf("%*smutual neighbour elimination, (%d,%d) vs (%d,%d):\n",
solver_recurse_depth*4, "",
1+x1, 1+YUNTRANS(y1), 1+x2, 1+YUNTRANS(y2));
printf("%*s ruling out %d at (%d,%d)\n",
solver_recurse_depth*4, "",
n, 1+x1, 1+YUNTRANS(y1));
}
#endif
cube(x1, y1, n) = FALSE;
return +1;
}
}
return 0; /* nothing found */
}
/*
* Look for forcing chains. A forcing chain is a path of
* pairwise-exclusive squares (i.e. each pair of adjacent squares
* in the path are in the same row, column or block) with the
* following properties:
*
* (a) Each square on the path has precisely two possible numbers.
*
* (b) Each pair of squares which are adjacent on the path share
* at least one possible number in common.
*
* (c) Each square in the middle of the path shares _both_ of its
* numbers with at least one of its neighbours (not the same
* one with both neighbours).
*
* These together imply that at least one of the possible number
* choices at one end of the path forces _all_ the rest of the
* numbers along the path. In order to make real use of this, we
* need further properties:
*
* (c) Ruling out some number N from the square at one end
* of the path forces the square at the other end to
* take number N.
*
* (d) The two end squares are both in line with some third
* square.
*
* (e) That third square currently has N as a possibility.
*
* If we can find all of that lot, we can deduce that at least one
* of the two ends of the forcing chain has number N, and that
* therefore the mutually adjacent third square does not.
*
* To find forcing chains, we're going to start a bfs at each
* suitable square, once for each of its two possible numbers.
*/
static int solver_forcing(struct solver_usage *usage,
struct solver_scratch *scratch)
{
int c = usage->c, r = usage->r, cr = c*r;
int *bfsqueue = scratch->bfsqueue;
#ifdef STANDALONE_SOLVER
int *bfsprev = scratch->bfsprev;
#endif
unsigned char *number = scratch->grid;
int *neighbours = scratch->neighbours;
int x, y;
for (y = 0; y < cr; y++)
for (x = 0; x < cr; x++) {
int count, t, n;
/*
* If this square doesn't have exactly two candidate
* numbers, don't try it.
*
* In this loop we also sum the candidate numbers,
* which is a nasty hack to allow us to quickly find
* `the other one' (since we will shortly know there
* are exactly two).
*/
for (count = t = 0, n = 1; n <= cr; n++)
if (cube(x, y, n))
count++, t += n;
if (count != 2)
continue;
/*
* Now attempt a bfs for each candidate.
*/
for (n = 1; n <= cr; n++)
if (cube(x, y, n)) {
int orign, currn, head, tail;
/*
* Begin a bfs.
*/
orign = n;
memset(number, cr+1, cr*cr);
head = tail = 0;
bfsqueue[tail++] = y*cr+x;
#ifdef STANDALONE_SOLVER
bfsprev[y*cr+x] = -1;
#endif
number[y*cr+x] = t - n;
while (head < tail) {
int xx, yy, nneighbours, xt, yt, xblk, i;
xx = bfsqueue[head++];
yy = xx / cr;
xx %= cr;
currn = number[yy*cr+xx];
/*
* Find neighbours of yy,xx.
*/
nneighbours = 0;
for (yt = 0; yt < cr; yt++)
neighbours[nneighbours++] = yt*cr+xx;
for (xt = 0; xt < cr; xt++)
neighbours[nneighbours++] = yy*cr+xt;
xblk = xx - (xx % r);
for (yt = yy % r; yt < cr; yt += r)
for (xt = xblk; xt < xblk+r; xt++)
neighbours[nneighbours++] = yt*cr+xt;
/*
* Try visiting each of those neighbours.
*/
for (i = 0; i < nneighbours; i++) {
int cc, tt, nn;
xt = neighbours[i] % cr;
yt = neighbours[i] / cr;
/*
* We need this square to not be
* already visited, and to include
* currn as a possible number.
*/
if (number[yt*cr+xt] <= cr)
continue;
if (!cube(xt, yt, currn))
continue;
/*
* Don't visit _this_ square a second
* time!
*/
if (xt == xx && yt == yy)
continue;
/*
* To continue with the bfs, we need
* this square to have exactly two
* possible numbers.
*/
for (cc = tt = 0, nn = 1; nn <= cr; nn++)
if (cube(xt, yt, nn))
cc++, tt += nn;
if (cc == 2) {
bfsqueue[tail++] = yt*cr+xt;
#ifdef STANDALONE_SOLVER
bfsprev[yt*cr+xt] = yy*cr+xx;
#endif
number[yt*cr+xt] = tt - currn;
}
/*
* One other possibility is that this
* might be the square in which we can
* make a real deduction: if it's
* adjacent to x,y, and currn is equal
* to the original number we ruled out.
*/
if (currn == orign &&
(xt == x || yt == y ||
(xt / r == x / r && yt % r == y % r))) {
#ifdef STANDALONE_SOLVER
if (solver_show_working) {
char *sep = "";
int xl, yl;
printf("%*sforcing chain, %d at ends of ",
solver_recurse_depth*4, "", orign);
xl = xx;
yl = yy;
while (1) {
printf("%s(%d,%d)", sep, 1+xl,
1+YUNTRANS(yl));
xl = bfsprev[yl*cr+xl];
if (xl < 0)
break;
yl = xl / cr;
xl %= cr;
sep = "-";
}
printf("\n%*s ruling out %d at (%d,%d)\n",
solver_recurse_depth*4, "",
orign, 1+xt, 1+YUNTRANS(yt));
}
#endif
cube(xt, yt, orign) = FALSE;
return 1;
}
}
}
}
}
return 0;
}
static struct solver_scratch *solver_new_scratch(struct solver_usage *usage)
{
struct solver_scratch *scratch = snew(struct solver_scratch);
int cr = usage->cr;
scratch->grid = snewn(cr*cr, unsigned char);
scratch->rowidx = snewn(cr, unsigned char);
scratch->colidx = snewn(cr, unsigned char);
scratch->set = snewn(cr, unsigned char);
scratch->neighbours = snewn(3*cr, int);
scratch->bfsqueue = snewn(cr*cr, int);
#ifdef STANDALONE_SOLVER
scratch->bfsprev = snewn(cr*cr, int);
#endif
return scratch;
}
static void solver_free_scratch(struct solver_scratch *scratch)
{
#ifdef STANDALONE_SOLVER
sfree(scratch->bfsprev);
#endif
sfree(scratch->bfsqueue);
sfree(scratch->neighbours);
sfree(scratch->set);
sfree(scratch->colidx);
sfree(scratch->rowidx);
sfree(scratch->grid);
sfree(scratch);
}
static int solver(int c, int r, digit *grid, int maxdiff)
{
struct solver_usage *usage;
struct solver_scratch *scratch;
int cr = c*r;
int x, y, x2, y2, n, ret;
int diff = DIFF_BLOCK;
/*
* Set up a usage structure as a clean slate (everything
* possible).
*/
usage = snew(struct solver_usage);
usage->c = c;
usage->r = r;
usage->cr = cr;
usage->cube = snewn(cr*cr*cr, unsigned char);
usage->grid = grid; /* write straight back to the input */
memset(usage->cube, TRUE, cr*cr*cr);
usage->row = snewn(cr * cr, unsigned char);
usage->col = snewn(cr * cr, unsigned char);
usage->blk = snewn(cr * cr, unsigned char);
memset(usage->row, FALSE, cr * cr);
memset(usage->col, FALSE, cr * cr);
memset(usage->blk, FALSE, cr * cr);
scratch = solver_new_scratch(usage);
/*
* Place all the clue numbers we are given.
*/
for (x = 0; x < cr; x++)
for (y = 0; y < cr; y++)
if (grid[y*cr+x])
solver_place(usage, x, YTRANS(y), grid[y*cr+x]);
/*
* Now loop over the grid repeatedly trying all permitted modes
* of reasoning. The loop terminates if we complete an
* iteration without making any progress; we then return
* failure or success depending on whether the grid is full or
* not.
*/
while (1) {
/*
* I'd like to write `continue;' inside each of the
* following loops, so that the solver returns here after
* making some progress. However, I can't specify that I
* want to continue an outer loop rather than the innermost
* one, so I'm apologetically resorting to a goto.
*/
cont:
/*
* Blockwise positional elimination.
*/
for (x = 0; x < cr; x += r)
for (y = 0; y < r; y++)
for (n = 1; n <= cr; n++)
if (!usage->blk[(y*c+(x/r))*cr+n-1]) {
ret = solver_elim(usage, cubepos(x,y,n), r*cr
#ifdef STANDALONE_SOLVER
, "positional elimination,"
" %d in block (%d,%d)", n, 1+x/r, 1+y
#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]) {
ret = solver_elim(usage, cubepos(0,y,n), cr*cr
#ifdef STANDALONE_SOLVER
, "positional elimination,"
" %d in row %d", n, 1+YUNTRANS(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]) {
ret = solver_elim(usage, cubepos(x,0,n), cr
#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;
}
}
/*
* Numeric elimination.
*/
for (x = 0; x < cr; x++)
for (y = 0; y < cr; y++)
if (!usage->grid[YUNTRANS(y)*cr+x]) {
ret = solver_elim(usage, cubepos(x,y,1), 1
#ifdef STANDALONE_SOLVER
, "numeric elimination at (%d,%d)", 1+x,
1+YUNTRANS(y)
#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 (x = 0; x < cr; x += r)
for (n = 1; n <= cr; n++)
/*
* solver_intersect() never returns -1.
*/
if (!usage->row[y*cr+n-1] &&
!usage->blk[((y%r)*c+(x/r))*cr+n-1] &&
(solver_intersect(usage, cubepos(0,y,n), cr*cr,
cubepos(x,y%r,n), r*cr
#ifdef STANDALONE_SOLVER
, "intersectional analysis,"
" %d in row %d vs block (%d,%d)",
n, 1+YUNTRANS(y), 1+x/r, 1+y%r
#endif
) ||
solver_intersect(usage, cubepos(x,y%r,n), r*cr,
cubepos(0,y,n), cr*cr
#ifdef STANDALONE_SOLVER
, "intersectional analysis,"
" %d in block (%d,%d) vs row %d",
n, 1+x/r, 1+y%r, 1+YUNTRANS(y)
#endif
))) {
diff = max(diff, DIFF_INTERSECT);
goto cont;
}
/*
* Intersectional analysis, columns vs blocks.
*/
for (x = 0; x < cr; x++)
for (y = 0; y < r; y++)
for (n = 1; n <= cr; n++)
if (!usage->col[x*cr+n-1] &&
!usage->blk[(y*c+(x/r))*cr+n-1] &&
(solver_intersect(usage, cubepos(x,0,n), cr,
cubepos((x/r)*r,y,n), r*cr
#ifdef STANDALONE_SOLVER
, "intersectional analysis,"
" %d in column %d vs block (%d,%d)",
n, 1+x, 1+x/r, 1+y
#endif
) ||
solver_intersect(usage, cubepos((x/r)*r,y,n), r*cr,
cubepos(x,0,n), cr
#ifdef STANDALONE_SOLVER
, "intersectional analysis,"
" %d in block (%d,%d) vs column %d",
n, 1+x/r, 1+y, 1+x
#endif
))) {
diff = max(diff, DIFF_INTERSECT);
goto cont;
}
if (maxdiff <= DIFF_INTERSECT)
break;
/*
* Blockwise set elimination.
*/
for (x = 0; x < cr; x += r)
for (y = 0; y < r; y++) {
ret = solver_set(usage, scratch, cubepos(x,y,1), r*cr, 1
#ifdef STANDALONE_SOLVER
, "set elimination, block (%d,%d)", 1+x/r, 1+y
#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++) {
ret = solver_set(usage, scratch, cubepos(0,y,1), cr*cr, 1
#ifdef STANDALONE_SOLVER
, "set elimination, row %d", 1+YUNTRANS(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++) {
ret = solver_set(usage, scratch, cubepos(x,0,1), cr, 1
#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;
}
}
/*
* Row-vs-column set elimination on a single number.
*/
for (n = 1; n <= cr; n++) {
ret = solver_set(usage, scratch, cubepos(0,0,n), cr*cr, cr
#ifdef STANDALONE_SOLVER
, "positional set elimination, number %d", n
#endif
);
if (ret < 0) {
diff = DIFF_IMPOSSIBLE;
goto got_result;
} else if (ret > 0) {
diff = max(diff, DIFF_EXTREME);
goto cont;
}
}
/*
* Mutual neighbour elimination.
*/
for (y = 0; y+1 < cr; y++) {
for (x = 0; x+1 < cr; x++) {
for (y2 = y+1; y2 < cr; y2++) {
for (x2 = x+1; x2 < cr; x2++) {
/*
* Can't do mutual neighbour elimination
* between elements of the same actual
* block.
*/
if (x/r == x2/r && y%r == y2%r)
continue;
/*
* Otherwise, try (x,y) vs (x2,y2) in both
* directions, and likewise (x2,y) vs
* (x,y2).
*/
if (!usage->grid[YUNTRANS(y)*cr+x] &&
!usage->grid[YUNTRANS(y2)*cr+x2] &&
(solver_mne(usage, scratch, x, y, x2, y2) ||
solver_mne(usage, scratch, x2, y2, x, y))) {
diff = max(diff, DIFF_EXTREME);
goto cont;
}
if (!usage->grid[YUNTRANS(y)*cr+x2] &&
!usage->grid[YUNTRANS(y2)*cr+x] &&
(solver_mne(usage, scratch, x2, y, x, y2) ||
solver_mne(usage, scratch, x, y2, x2, y))) {
diff = max(diff, DIFF_EXTREME);
goto cont;
}
}
}
}
}
/*
* Forcing chains.
*/
if (solver_forcing(usage, scratch)) {
diff = max(diff, DIFF_EXTREME);
goto cont;
}
/*
* If we reach here, we have made no deductions in this
* iteration, so the algorithm terminates.
*/
break;
}
/*
* Last chance: if we haven't fully solved the puzzle yet, try
* recursing based on guesses for a particular square. We pick
* one of the most constrained empty squares we can find, which
* has the effect of pruning the search tree as much as
* possible.
*/
if (maxdiff >= DIFF_RECURSIVE) {
int best, bestcount;
best = -1;
bestcount = cr+1;
for (y = 0; y < cr; y++)
for (x = 0; x < cr; x++)
if (!grid[y*cr+x]) {
int count;
/*
* An unfilled square. Count the number of
* possible digits in it.
*/
count = 0;
for (n = 1; n <= cr; n++)
if (cube(x,YTRANS(y),n))
count++;
/*
* We should have found any impossibilities
* already, so this can safely be an assert.
*/
assert(count > 1);
if (count < bestcount) {
bestcount = count;
best = y*cr+x;
}
}
if (best != -1) {
int i, j;
digit *list, *ingrid, *outgrid;
diff = DIFF_IMPOSSIBLE; /* no solution found yet */
/*
* Attempt recursion.
*/
y = best / cr;
x = best % cr;
list = snewn(cr, digit);
ingrid = snewn(cr * cr, digit);
outgrid = snewn(cr * cr, digit);
memcpy(ingrid, grid, cr * cr);
/* Make a list of the possible digits. */
for (j = 0, n = 1; n <= cr; n++)
if (cube(x,YTRANS(y),n))
list[j++] = n;
#ifdef STANDALONE_SOLVER
if (solver_show_working) {
char *sep = "";
printf("%*srecursing on (%d,%d) [",
solver_recurse_depth*4, "", x, y);
for (i = 0; i < j; i++) {
printf("%s%d", sep, list[i]);
sep = " or ";
}
printf("]\n");
}
#endif
/*
* And step along the list, recursing back into the
* main solver at every stage.
*/
for (i = 0; i < j; i++) {
int ret;
memcpy(outgrid, ingrid, cr * cr);
outgrid[y*cr+x] = list[i];
#ifdef STANDALONE_SOLVER
if (solver_show_working)
printf("%*sguessing %d at (%d,%d)\n",
solver_recurse_depth*4, "", list[i], x, y);
solver_recurse_depth++;
#endif
ret = solver(c, r, outgrid, maxdiff);
#ifdef STANDALONE_SOLVER
solver_recurse_depth--;
if (solver_show_working) {
printf("%*sretracting %d at (%d,%d)\n",
solver_recurse_depth*4, "", list[i], x, y);
}
#endif
/*
* If we have our first solution, copy it into the
* grid we will return.
*/
if (diff == DIFF_IMPOSSIBLE && ret != DIFF_IMPOSSIBLE)
memcpy(grid, outgrid, cr*cr);
if (ret == DIFF_AMBIGUOUS)
diff = DIFF_AMBIGUOUS;
else if (ret == DIFF_IMPOSSIBLE)
/* do not change our return value */;
else {
/* the recursion turned up exactly one solution */
if (diff == DIFF_IMPOSSIBLE)
diff = DIFF_RECURSIVE;
else
diff = DIFF_AMBIGUOUS;
}
/*
* As soon as we've found more than one solution,
* give up immediately.
*/
if (diff == DIFF_AMBIGUOUS)
break;
}
sfree(outgrid);
sfree(ingrid);
sfree(list);
}
} else {
/*
* We're forbidden to use recursion, so we just see whether
* our grid is fully solved, and return DIFF_IMPOSSIBLE
* otherwise.
*/
for (y = 0; y < cr; y++)
for (x = 0; x < cr; x++)
if (!grid[y*cr+x])
diff = DIFF_IMPOSSIBLE;
}
got_result:;
#ifdef STANDALONE_SOLVER
if (solver_show_working)
printf("%*s%s found\n",
solver_recurse_depth*4, "",
diff == DIFF_IMPOSSIBLE ? "no solution" :
diff == DIFF_AMBIGUOUS ? "multiple solutions" :
"one solution");
#endif
sfree(usage->cube);
sfree(usage->row);
sfree(usage->col);
sfree(usage->blk);
sfree(usage);
solver_free_scratch(scratch);
return diff;
}
/* ----------------------------------------------------------------------
* End of solver code.
*/
/* ----------------------------------------------------------------------
* Solo filled-grid generator.
*
* This grid generator works by essentially trying to solve a grid
* starting from no clues, and not worrying that there's more than
* one possible solution. Unfortunately, it isn't computationally
* feasible to do this by calling the above solver with an empty
* grid, because that one needs to allocate a lot of scratch space
* at every recursion level. Instead, I have a much simpler
* algorithm which I shamelessly copied from a Python solver
* written by Andrew Wilkinson (which is GPLed, but I've reused
* only ideas and no code). It mostly just does the obvious
* recursive thing: pick an empty square, put one of the possible
* digits in it, recurse until all squares are filled, backtrack
* and change some choices if necessary.
*
* The clever bit is that every time it chooses which square to
* fill in next, it does so by counting the number of _possible_
* numbers that can go in each square, and it prioritises so that
* it picks a square with the _lowest_ number of possibilities. The
* idea is that filling in lots of the obvious bits (particularly
* any squares with only one possibility) will cut down on the list
* of possibilities for other squares and hence reduce the enormous
* search space as much as possible as early as possible.
*/
/*
* Internal data structure used in gridgen to keep track of
* progress.
*/
struct gridgen_coord { int x, y, r; };
struct gridgen_usage {
int c, r, cr; /* cr == c*r */
/* grid is a copy of the input grid, modified as we go along */
digit *grid;
/* row[y*cr+n-1] TRUE if digit n has been placed in row y */
unsigned char *row;
/* col[x*cr+n-1] TRUE if digit n has been placed in row x */
unsigned char *col;
/* blk[(y*c+x)*cr+n-1] TRUE if digit n has been placed in block (x,y) */
unsigned char *blk;
/* This lists all the empty spaces remaining in the grid. */
struct gridgen_coord *spaces;
int nspaces;
/* If we need randomisation in the solve, this is our random state. */
random_state *rs;
};
/*
* The real recursive step in the generating function.
*/
static int gridgen_real(struct gridgen_usage *usage, digit *grid)
{
int c = usage->c, r = usage->r, cr = usage->cr;
int i, j, n, sx, sy, bestm, bestr, ret;
int *digits;
/*
* Firstly, check for completion! If there are no spaces left
* in the grid, we have a solution.
*/
if (usage->nspaces == 0) {
memcpy(grid, usage->grid, cr * cr);
return TRUE;
}
/*
* Otherwise, there must be at least one space. Find the most
* constrained space, using the `r' field as a tie-breaker.
*/
bestm = cr+1; /* so that any space will beat it */
bestr = 0;
i = sx = sy = -1;
for (j = 0; j < usage->nspaces; j++) {
int x = usage->spaces[j].x, y = usage->spaces[j].y;
int m;
/*
* Find the number of digits that could go in this space.
*/
m = 0;
for (n = 0; n < cr; n++)
if (!usage->row[y*cr+n] && !usage->col[x*cr+n] &&
!usage->blk[((y/c)*c+(x/r))*cr+n])
m++;
if (m < bestm || (m == bestm && usage->spaces[j].r < bestr)) {
bestm = m;
bestr = usage->spaces[j].r;
sx = x;
sy = y;
i = j;
}
}
/*
* Swap that square into the final place in the spaces array,
* so that decrementing nspaces will remove it from the list.
*/
if (i != usage->nspaces-1) {
struct gridgen_coord t;
t = usage->spaces[usage->nspaces-1];
usage->spaces[usage->nspaces-1] = usage->spaces[i];
usage->spaces[i] = t;
}
/*
* Now we've decided which square to start our recursion at,
* simply go through all possible values, shuffling them
* randomly first if necessary.
*/
digits = snewn(bestm, int);
j = 0;
for (n = 0; n < cr; n++)
if (!usage->row[sy*cr+n] && !usage->col[sx*cr+n] &&
!usage->blk[((sy/c)*c+(sx/r))*cr+n]) {
digits[j++] = n+1;
}
if (usage->rs)
shuffle(digits, j, sizeof(*digits), usage->rs);
/* And finally, go through the digit list and actually recurse. */
ret = FALSE;
for (i = 0; i < j; i++) {
n = digits[i];
/* Update the usage structure to reflect the placing of this digit. */
usage->row[sy*cr+n-1] = usage->col[sx*cr+n-1] =
usage->blk[((sy/c)*c+(sx/r))*cr+n-1] = TRUE;
usage->grid[sy*cr+sx] = n;
usage->nspaces--;
/* Call the solver recursively. Stop when we find a solution. */
if (gridgen_real(usage, grid))
ret = TRUE;
/* Revert the usage structure. */
usage->row[sy*cr+n-1] = usage->col[sx*cr+n-1] =
usage->blk[((sy/c)*c+(sx/r))*cr+n-1] = FALSE;
usage->grid[sy*cr+sx] = 0;
usage->nspaces++;
if (ret)
break;
}
sfree(digits);
return ret;
}
/*
* Entry point to generator. You give it dimensions and a starting
* grid, which is simply an array of cr*cr digits.
*/
static void gridgen(int c, int r, digit *grid, random_state *rs)
{
struct gridgen_usage *usage;
int x, y, cr = c*r;
/*
* Clear the grid to start with.
*/
memset(grid, 0, cr*cr);
/*
* Create a gridgen_usage structure.
*/
usage = snew(struct gridgen_usage);
usage->c = c;
usage->r = r;
usage->cr = cr;
usage->grid = snewn(cr * cr, digit);
memcpy(usage->grid, grid, cr * cr);
usage->row = snewn(cr * cr, unsigned char);
usage->col = snewn(cr * cr, unsigned char);
usage->blk = snewn(cr * cr, unsigned char);
memset(usage->row, FALSE, cr * cr);
memset(usage->col, FALSE, cr * cr);
memset(usage->blk, FALSE, cr * cr);
usage->spaces = snewn(cr * cr, struct gridgen_coord);
usage->nspaces = 0;
usage->rs = rs;
/*
* Initialise the list of grid spaces.
*/
for (y = 0; y < cr; y++) {
for (x = 0; x < cr; x++) {
usage->spaces[usage->nspaces].x = x;
usage->spaces[usage->nspaces].y = y;
usage->spaces[usage->nspaces].r = random_bits(rs, 31);
usage->nspaces++;
}
}
/*
* Run the real generator function.
*/
gridgen_real(usage, grid);
/*
* Clean up the usage structure now we have our answer.
*/
sfree(usage->spaces);
sfree(usage->blk);
sfree(usage->col);
sfree(usage->row);
sfree(usage->grid);
sfree(usage);
}
/* ----------------------------------------------------------------------
* End of grid generator code.
*/
/*
* Check whether a grid contains a valid complete puzzle.
*/
static int check_valid(int c, int r, digit *grid)
{
int cr = c*r;
unsigned char *used;
int x, y, n;
used = snewn(cr, unsigned char);
/*
* Check that each row contains precisely one of everything.
*/
for (y = 0; y < cr; y++) {
memset(used, FALSE, cr);
for (x = 0; x < cr; x++)
if (grid[y*cr+x] > 0 && grid[y*cr+x] <= cr)
used[grid[y*cr+x]-1] = TRUE;
for (n = 0; n < cr; n++)
if (!used[n]) {
sfree(used);
return FALSE;
}
}
/*
* Check that each column contains precisely one of everything.
*/
for (x = 0; x < cr; x++) {
memset(used, FALSE, cr);
for (y = 0; y < cr; y++)
if (grid[y*cr+x] > 0 && grid[y*cr+x] <= cr)
used[grid[y*cr+x]-1] = TRUE;
for (n = 0; n < cr; n++)
if (!used[n]) {
sfree(used);
return FALSE;
}
}
/*
* Check that each block contains precisely one of everything.
*/
for (x = 0; x < cr; x += r) {
for (y = 0; y < cr; y += c) {
int xx, yy;
memset(used, FALSE, cr);
for (xx = x; xx < x+r; xx++)
for (yy = 0; yy < y+c; yy++)
if (grid[yy*cr+xx] > 0 && grid[yy*cr+xx] <= cr)
used[grid[yy*cr+xx]-1] = TRUE;
for (n = 0; n < cr; n++)
if (!used[n]) {
sfree(used);
return FALSE;
}
}
}
sfree(used);
return TRUE;
}
static int symmetries(game_params *params, int x, int y, int *output, int s)
{
int c = params->c, r = params->r, cr = c*r;
int i = 0;
#define ADD(x,y) (*output++ = (x), *output++ = (y), i++)
ADD(x, y);
switch (s) {
case SYMM_NONE:
break; /* just x,y is all we need */
case SYMM_ROT2:
ADD(cr - 1 - x, cr - 1 - y);
break;
case SYMM_ROT4:
ADD(cr - 1 - y, x);
ADD(y, cr - 1 - x);
ADD(cr - 1 - x, cr - 1 - y);
break;
case SYMM_REF2:
ADD(cr - 1 - x, y);
break;
case SYMM_REF2D:
ADD(y, x);
break;
case SYMM_REF4:
ADD(cr - 1 - x, y);
ADD(x, cr - 1 - y);
ADD(cr - 1 - x, cr - 1 - y);
break;
case SYMM_REF4D:
ADD(y, x);
ADD(cr - 1 - x, cr - 1 - y);
ADD(cr - 1 - y, cr - 1 - x);
break;
case SYMM_REF8:
ADD(cr - 1 - x, y);
ADD(x, cr - 1 - y);
ADD(cr - 1 - x, cr - 1 - y);
ADD(y, x);
ADD(y, cr - 1 - x);
ADD(cr - 1 - y, x);
ADD(cr - 1 - y, cr - 1 - x);
break;
}
#undef ADD
return i;
}
static char *encode_solve_move(int cr, digit *grid)
{
int i, len;
char *ret, *p, *sep;
/*
* It's surprisingly easy to work out _exactly_ how long this
* string needs to be. To decimal-encode all the numbers from 1
* to n:
*
* - every number has a units digit; total is n.
* - all numbers above 9 have a tens digit; total is max(n-9,0).
* - all numbers above 99 have a hundreds digit; total is max(n-99,0).
* - and so on.
*/
len = 0;
for (i = 1; i <= cr; i *= 10)
len += max(cr - i + 1, 0);
len += cr; /* don't forget the commas */
len *= cr; /* there are cr rows of these */
/*
* Now len is one bigger than the total size of the
* comma-separated numbers (because we counted an
* additional leading comma). We need to have a leading S
* and a trailing NUL, so we're off by one in total.
*/
len++;
ret = snewn(len, char);
p = ret;
*p++ = 'S';
sep = "";
for (i = 0; i < cr*cr; i++) {
p += sprintf(p, "%s%d", sep, grid[i]);
sep = ",";
}
*p++ = '\0';
assert(p - ret == len);
return ret;
}
static char *new_game_desc(game_params *params, random_state *rs,
char **aux, int interactive)
{
int c = params->c, r = params->r, cr = c*r;
int area = cr*cr;
digit *grid, *grid2;
struct xy { int x, y; } *locs;
int nlocs;
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);
/*
* Loop until we get a grid of the required difficulty. This is
* nasty, but it seems to be unpleasantly hard to generate
* difficult grids otherwise.
*/
do {
/*
* Generate a random solved state.
*/
gridgen(c, r, grid, rs);
assert(check_valid(c, r, grid));
/*
* Save the solved grid in aux.
*/
{
/*
* We might already have written *aux the last time we
* went round this loop, in which case we should free
* the old aux before overwriting it with the new one.
*/
if (*aux) {
sfree(*aux);
}
*aux = encode_solve_move(cr, grid);
}
/*
* Now we have a solved grid, start removing things from it
* while preserving solubility.
*/
/*
* 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(c, r, grid2, maxdiff);
if (ret <= maxdiff) {
for (j = 0; j < ncoords; j++)
grid[coords[2*j+1]*cr+coords[2*j]] = 0;
}
}
memcpy(grid2, grid, area);
} while (solver(c, r, grid2, maxdiff) < maxdiff);
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(5 * area, char);
p = desc;
run = 0;
for (i = 0; i <= area; i++) {
int n = (i < area ? grid[i] : -1);
if (!n)
run++;
else {
if (run) {
while (run > 0) {
int c = 'a' - 1 + run;
if (run > 26)
c = 'z';
*p++ = c;
run -= c - ('a' - 1);
}
} else {
/*
* If there's a number in the very top left or
* bottom right, there's no point putting an
* unnecessary _ before or after it.
*/
if (p > desc && n > 0)
*p++ = '_';
}
if (n > 0)
p += sprintf(p, "%d", n);
run = 0;
}
}
assert(p - desc < 5 * area);
*p++ = '\0';
desc = sresize(desc, p - desc, char);
}
sfree(grid);
return desc;
}
static char *validate_desc(game_params *params, char *desc)
{
int area = params->r * params->r * params->c * params->c;
int squares = 0;
while (*desc) {
int n = *desc++;
if (n >= 'a' && n <= 'z') {
squares += n - 'a' + 1;
} else if (n == '_') {
/* do nothing */;
} else if (n > '0' && n <= '9') {
squares++;
while (*desc >= '0' && *desc <= '9')
desc++;
} else
return "Invalid character in game description";
}
if (squares < area)
return "Not enough data to fill grid";
if (squares > area)
return "Too much data to fit in grid";
return NULL;
}
static game_state *new_game(midend *me, game_params *params, char *desc)
{
game_state *state = snew(game_state);
int c = params->c, r = params->r, cr = c*r, area = cr * cr;
int i;
state->c = params->c;
state->r = params->r;
state->grid = snewn(area, digit);
state->pencil = snewn(area * cr, unsigned char);
memset(state->pencil, 0, area * cr);
state->immutable = snewn(area, unsigned char);
memset(state->immutable, FALSE, area);
state->completed = state->cheated = FALSE;
i = 0;
while (*desc) {
int n = *desc++;
if (n >= 'a' && n <= 'z') {
int run = n - 'a' + 1;
assert(i + run <= area);
while (run-- > 0)
state->grid[i++] = 0;
} else if (n == '_') {
/* do nothing */;
} else if (n > '0' && n <= '9') {
assert(i < area);
state->immutable[i] = TRUE;
state->grid[i++] = atoi(desc-1);
while (*desc >= '0' && *desc <= '9')
desc++;
} else {
assert(!"We can't get here");
}
}
assert(i == area);
return state;
}
static game_state *dup_game(game_state *state)
{
game_state *ret = snew(game_state);
int c = state->c, r = state->r, cr = c*r, area = cr * cr;
ret->c = state->c;
ret->r = state->r;
ret->grid = snewn(area, digit);
memcpy(ret->grid, state->grid, area);
ret->pencil = snewn(area * cr, unsigned char);
memcpy(ret->pencil, state->pencil, area * cr);
ret->immutable = snewn(area, unsigned char);
memcpy(ret->immutable, state->immutable, area);
ret->completed = state->completed;
ret->cheated = state->cheated;
return ret;
}
static void free_game(game_state *state)
{
sfree(state->immutable);
sfree(state->pencil);
sfree(state->grid);
sfree(state);
}
static char *solve_game(game_state *state, game_state *currstate,
char *ai, char **error)
{
int c = state->c, r = state->r, cr = c*r;
char *ret;
digit *grid;
int 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(c, r, 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 c, int r, digit *grid)
{
int cr = c*r;
int x, y;
int maxlen;
char *ret, *p;
/*
* There are cr lines of digits, plus r-1 lines of block
* separators. Each line contains cr digits, cr-1 separating
* spaces, and c-1 two-character block separators. Thus, the
* total length of a line is 2*cr+2*c-3 (not counting the
* newline), and there are cr+r-1 of them.
*/
maxlen = (cr+r-1) * (2*cr+2*c-2);
ret = snewn(maxlen+1, char);
p = ret;
for (y = 0; y < cr; y++) {
for (x = 0; x < cr; x++) {
int ch = grid[y * cr + x];
if (ch == 0)
ch = '.';
else if (ch <= 9)
ch = '0' + ch;
else
ch = 'a' + ch-10;
*p++ = ch;
if (x+1 < cr) {
*p++ = ' ';
if ((x+1) % r == 0) {
*p++ = '|';
*p++ = ' ';
}
}
}
*p++ = '\n';
if (y+1 < cr && (y+1) % c == 0) {
for (x = 0; x < cr; x++) {
*p++ = '-';
if (x+1 < cr) {
*p++ = '-';
if ((x+1) % r == 0) {
*p++ = '+';
*p++ = '-';
}
}
}
*p++ = '\n';
}
}
assert(p - ret == maxlen);
*p = '\0';
return ret;
}
static char *game_text_format(game_state *state)
{
return grid_text_format(state->c, state->r, state->grid);
}
struct game_ui {
/*
* These are the coordinates of the currently highlighted
* square on the grid, or -1,-1 if there isn't one. When there
* is, pressing a valid number or letter key or Space will
* enter that number or letter in the grid.
*/
int hx, hy;
/*
* This indicates whether the current highlight is a
* pencil-mark one or a real one.
*/
int hpencil;
};
static game_ui *new_ui(game_state *state)
{
game_ui *ui = snew(game_ui);
ui->hx = ui->hy = -1;
ui->hpencil = 0;
return ui;
}
static void free_ui(game_ui *ui)
{
sfree(ui);
}
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 c = newstate->c, r = newstate->r, cr = c*r;
/*
* We prevent pencil-mode highlighting of a filled square. So
* if the user has just filled in a square which we had a
* pencil-mode highlight in (by Undo, or by Redo, or by Solve),
* then we cancel the highlight.
*/
if (ui->hx >= 0 && ui->hy >= 0 && ui->hpencil &&
newstate->grid[ui->hy * cr + ui->hx] != 0) {
ui->hx = ui->hy = -1;
}
}
struct game_drawstate {
int started;
int c, r, cr;
int tilesize;
digit *grid;
unsigned char *pencil;
unsigned char *hl;
/* This is scratch space used within a single call to game_redraw. */
int *entered_items;
};
static char *interpret_move(game_state *state, game_ui *ui, game_drawstate *ds,
int x, int y, int button)
{
int c = state->c, r = state->r, cr = c*r;
int tx, ty;
char buf[80];
button &= ~MOD_MASK;
tx = (x + TILE_SIZE - BORDER) / TILE_SIZE - 1;
ty = (y + TILE_SIZE - BORDER) / TILE_SIZE - 1;
if (tx >= 0 && tx < cr && ty >= 0 && ty < cr) {
if (button == LEFT_BUTTON) {
if (state->immutable[ty*cr+tx]) {
ui->hx = ui->hy = -1;
} else if (tx == ui->hx && ty == ui->hy && ui->hpencil == 0) {
ui->hx = ui->hy = -1;
} else {
ui->hx = tx;
ui->hy = ty;
ui->hpencil = 0;
}
return ""; /* UI activity occurred */
}
if (button == RIGHT_BUTTON) {
/*
* Pencil-mode highlighting for non filled squares.
*/
if (state->grid[ty*cr+tx] == 0) {
if (tx == ui->hx && ty == ui->hy && ui->hpencil) {
ui->hx = ui->hy = -1;
} else {
ui->hpencil = 1;
ui->hx = tx;
ui->hy = ty;
}
} else {
ui->hx = ui->hy = -1;
}
return ""; /* UI activity occurred */
}
}
if (ui->hx != -1 && ui->hy != -1 &&
((button >= '1' && button <= '9' && button - '0' <= cr) ||
(button >= 'a' && button <= 'z' && button - 'a' + 10 <= cr) ||
(button >= 'A' && button <= 'Z' && button - 'A' + 10 <= cr) ||
button == ' ' || 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 c = from->c, r = from->r, cr = c*r;
game_state *ret;
int x, y, n;
if (move[0] == 'S') {
char *p;
ret = dup_game(from);
ret->completed = ret->cheated = TRUE;
p = move+1;
for (n = 0; n < cr*cr; n++) {
ret->grid[n] = atoi(p);
if (!*p || ret->grid[n] < 1 || ret->grid[n] > cr) {
free_game(ret);
return NULL;
}
while (*p && isdigit((unsigned char)*p)) p++;
if (*p == ',') p++;
}
return ret;
} else if ((move[0] == 'P' || move[0] == 'R') &&
sscanf(move+1, "%d,%d,%d", &x, &y, &n) == 3 &&
x >= 0 && x < cr && y >= 0 && y < cr && n >= 0 && n <= cr) {
ret = dup_game(from);
if (move[0] == 'P' && n > 0) {
int index = (y*cr+x) * cr + (n-1);
ret->pencil[index] = !ret->pencil[index];
} else {
ret->grid[y*cr+x] = n;
memset(ret->pencil + (y*cr+x)*cr, 0, cr);
/*
* We've made a real change to the grid. Check to see
* if the game has been completed.
*/
if (!ret->completed && check_valid(c, r, ret->grid)) {
ret->completed = TRUE;
}
}
return ret;
} else
return NULL; /* couldn't parse move string */
}
/* ----------------------------------------------------------------------
* Drawing routines.
*/
#define SIZE(cr) ((cr) * TILE_SIZE + 2*BORDER + 1)
#define GETTILESIZE(cr, w) ( (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, game_state *state, int *ncolours)
{
float *ret = snewn(3 * NCOLOURS, float);
frontend_default_colour(fe, &ret[COL_BACKGROUND * 3]);
ret[COL_GRID * 3 + 0] = 0.0F;
ret[COL_GRID * 3 + 1] = 0.0F;
ret[COL_GRID * 3 + 2] = 0.0F;
ret[COL_CLUE * 3 + 0] = 0.0F;
ret[COL_CLUE * 3 + 1] = 0.0F;
ret[COL_CLUE * 3 + 2] = 0.0F;
ret[COL_USER * 3 + 0] = 0.0F;
ret[COL_USER * 3 + 1] = 0.6F * ret[COL_BACKGROUND * 3 + 1];
ret[COL_USER * 3 + 2] = 0.0F;
ret[COL_HIGHLIGHT * 3 + 0] = 0.85F * ret[COL_BACKGROUND * 3 + 0];
ret[COL_HIGHLIGHT * 3 + 1] = 0.85F * ret[COL_BACKGROUND * 3 + 1];
ret[COL_HIGHLIGHT * 3 + 2] = 0.85F * ret[COL_BACKGROUND * 3 + 2];
ret[COL_ERROR * 3 + 0] = 1.0F;
ret[COL_ERROR * 3 + 1] = 0.0F;
ret[COL_ERROR * 3 + 2] = 0.0F;
ret[COL_PENCIL * 3 + 0] = 0.5F * ret[COL_BACKGROUND * 3 + 0];
ret[COL_PENCIL * 3 + 1] = 0.5F * ret[COL_BACKGROUND * 3 + 1];
ret[COL_PENCIL * 3 + 2] = ret[COL_BACKGROUND * 3 + 2];
*ncolours = NCOLOURS;
return ret;
}
static game_drawstate *game_new_drawstate(drawing *dr, game_state *state)
{
struct game_drawstate *ds = snew(struct game_drawstate);
int c = state->c, r = state->r, cr = c*r;
ds->started = FALSE;
ds->c = c;
ds->r = r;
ds->cr = cr;
ds->grid = snewn(cr*cr, digit);
memset(ds->grid, 0, cr*cr);
ds->pencil = snewn(cr*cr*cr, digit);
memset(ds->pencil, 0, cr*cr*cr);
ds->hl = snewn(cr*cr, unsigned char);
memset(ds->hl, 0, cr*cr);
ds->entered_items = snewn(cr*cr, int);
ds->tilesize = 0; /* not decided yet */
return ds;
}
static void game_free_drawstate(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 c = state->c, r = state->r, cr = c*r;
int tx, ty;
int cx, cy, cw, ch;
char str[2];
if (ds->grid[y*cr+x] == state->grid[y*cr+x] &&
ds->hl[y*cr+x] == hl &&
!memcmp(ds->pencil+(y*cr+x)*cr, state->pencil+(y*cr+x)*cr, cr))
return; /* no change required */
tx = BORDER + x * TILE_SIZE + 2;
ty = BORDER + y * TILE_SIZE + 2;
cx = tx;
cy = ty;
cw = TILE_SIZE-3;
ch = TILE_SIZE-3;
if (x % r)
cx--, cw++;
if ((x+1) % r)
cw++;
if (y % c)
cy--, ch++;
if ((y+1) % c)
ch++;
clip(dr, cx, cy, cw, ch);
/* background needs erasing */
draw_rect(dr, cx, cy, cw, ch, (hl & 15) == 1 ? COL_HIGHLIGHT : COL_BACKGROUND);
/* pencil-mode highlight */
if ((hl & 15) == 2) {
int coords[6];
coords[0] = cx;
coords[1] = cy;
coords[2] = cx+cw/2;
coords[3] = cy;
coords[4] = cx;
coords[5] = cy+ch/2;
draw_polygon(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 c = state->c, r = state->r, cr = c*r;
int x, y;
if (!ds->started) {
/*
* The initial contents of the window are not guaranteed
* and can vary with front ends. To be on the safe side,
* all games should start by drawing a big
* background-colour rectangle covering the whole window.
*/
draw_rect(dr, 0, 0, SIZE(cr), SIZE(cr), COL_BACKGROUND);
/*
* Draw the grid.
*/
for (x = 0; x <= cr; x++) {
int thick = (x % r ? 0 : 1);
draw_rect(dr, BORDER + x*TILE_SIZE - thick, BORDER-1,
1+2*thick, cr*TILE_SIZE+3, COL_GRID);
}
for (y = 0; y <= cr; y++) {
int thick = (y % c ? 0 : 1);
draw_rect(dr, BORDER-1, BORDER + y*TILE_SIZE - thick,
cr*TILE_SIZE+3, 1+2*thick, COL_GRID);
}
}
/*
* This array is used to keep track of rows, columns and boxes
* which contain a number more than once.
*/
for (x = 0; x < cr * cr; x++)
ds->entered_items[x] = 0;
for (x = 0; x < cr; x++)
for (y = 0; y < cr; y++) {
digit d = state->grid[y*cr+x];
if (d) {
int box = (x/r)+(y/c)*c;
ds->entered_items[x*cr+d-1] |= ((ds->entered_items[x*cr+d-1] & 1) << 1) | 1;
ds->entered_items[y*cr+d-1] |= ((ds->entered_items[y*cr+d-1] & 4) << 1) | 4;
ds->entered_items[box*cr+d-1] |= ((ds->entered_items[box*cr+d-1] & 16) << 1) | 16;
}
}
/*
* Draw any numbers which need redrawing.
*/
for (x = 0; x < cr; x++) {
for (y = 0; y < cr; y++) {
int highlight = 0;
digit d = state->grid[y*cr+x];
if (flashtime > 0 &&
(flashtime <= FLASH_TIME/3 ||
flashtime >= FLASH_TIME*2/3))
highlight = 1;
/* Highlight active input areas. */
if (x == ui->hx && y == ui->hy)
highlight = ui->hpencil ? 2 : 1;
/* Mark obvious errors (ie, numbers which occur more than once
* in a single row, column, or box). */
if (d && ((ds->entered_items[x*cr+d-1] & 2) ||
(ds->entered_items[y*cr+d-1] & 8) ||
(ds->entered_items[((x/r)+(y/c)*c)*cr+d-1] & 32)))
highlight |= 16;
draw_number(dr, ds, state, x, y, highlight);
}
}
/*
* Update the _entire_ grid if necessary.
*/
if (!ds->started) {
draw_update(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_wants_statusbar(void)
{
return FALSE;
}
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 c = state->c, r = state->r, cr = c*r;
int ink = print_mono_colour(dr, 0);
int x, y;
/* Ick: fake up `ds->tilesize' for macro expansion purposes */
game_drawstate ads, *ds = &ads;
ads.tilesize = tilesize;
/*
* Border.
*/
print_line_width(dr, 3 * TILE_SIZE / 40);
draw_rect_outline(dr, BORDER, BORDER, cr*TILE_SIZE, cr*TILE_SIZE, ink);
/*
* Grid.
*/
for (x = 1; x < cr; x++) {
print_line_width(dr, (x % r ? 1 : 3) * TILE_SIZE / 40);
draw_line(dr, BORDER+x*TILE_SIZE, BORDER,
BORDER+x*TILE_SIZE, BORDER+cr*TILE_SIZE, ink);
}
for (y = 1; y < cr; y++) {
print_line_width(dr, (y % c ? 1 : 3) * TILE_SIZE / 40);
draw_line(dr, BORDER, BORDER+y*TILE_SIZE,
BORDER+cr*TILE_SIZE, BORDER+y*TILE_SIZE, ink);
}
/*
* Numbers.
*/
for (y = 0; y < cr; y++)
for (x = 0; x < cr; x++)
if (state->grid[y*cr+x]) {
char str[2];
str[1] = '\0';
str[0] = state->grid[y*cr+x] + '0';
if (str[0] > '9')
str[0] += 'a' - ('9'+1);
draw_text(dr, BORDER + x*TILE_SIZE + TILE_SIZE/2,
BORDER + y*TILE_SIZE + TILE_SIZE/2,
FONT_VARIABLE, TILE_SIZE/2,
ALIGN_VCENTRE | ALIGN_HCENTRE, ink, str);
}
}
#ifdef COMBINED
#define thegame solo
#endif
const struct game thegame = {
"Solo", "games.solo",
default_params,
game_fetch_preset,
decode_params,
encode_params,
free_params,
dup_params,
TRUE, game_configure, custom_params,
validate_params,
new_game_desc,
validate_desc,
new_game,
dup_game,
free_game,
TRUE, solve_game,
TRUE, game_text_format,
new_ui,
free_ui,
encode_ui,
decode_ui,
game_changed_state,
interpret_move,
execute_move,
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,
game_wants_statusbar,
FALSE, game_timing_state,
0, /* mouse_priorities */
};
#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(p->c, p->r, s->grid, DIFF_RECURSIVE);
if (grade) {
printf("Difficulty rating: %s\n",
ret==DIFF_BLOCK ? "Trivial (blockwise positional elimination only)":
ret==DIFF_SIMPLE ? "Basic (row/column/number elimination required)":
ret==DIFF_INTERSECT ? "Intermediate (intersectional analysis required)":
ret==DIFF_SET ? "Advanced (set elimination required)":
ret==DIFF_EXTREME ? "Extreme (complex non-recursive techniques required)":
ret==DIFF_RECURSIVE ? "Unreasonable (guesswork and backtracking required)":
ret==DIFF_AMBIGUOUS ? "Ambiguous (multiple solutions exist)":
ret==DIFF_IMPOSSIBLE ? "Impossible (no solution exists)":
"INTERNAL ERROR: unrecognised difficulty code");
} else {
printf("%s\n", grid_text_format(p->c, p->r, s->grid));
}
return 0;
}
#endif
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