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filling.c
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filling.c
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/*
* filling.c: An implementation of the Nikoli game fillomino.
* Copyright (C) 2007 Jonas Kölker. See LICENSE for the license.
*/
/* TODO:
*
* - use a typedef instead of int for numbers on the board
* + replace int with something else (signed short?)
* - the type should be signed (for -board[i] and -SENTINEL)
* - the type should be somewhat big: board[i] = i
* - Using shorts gives us 181x181 puzzles as upper bound.
*
* - in board generation, after having merged regions such that no
* more merges are necessary, try splitting (big) regions.
* + it seems that smaller regions make for better puzzles; see
* for instance the 7x7 puzzle in this file (grep for 7x7:).
*
* - symmetric hints (solo-style)
* + right now that means including _many_ hints, and the puzzles
* won't look any nicer. Not worth it (at the moment).
*
* - make the solver do recursion/backtracking.
* + This is for user-submitted puzzles, not for puzzle
* generation (on the other hand, never say never).
*
* - prove that only w=h=2 needs a special case
*
* - solo-like pencil marks?
*
* - a user says that the difficulty is unevenly distributed.
* + partition into levels? Will they be non-crap?
*
* - Allow square contents > 9?
* + I could use letters for digits (solo does this), but
* letters don't have numeric significance (normal people hate
* base36), which is relevant here (much more than in solo).
* + [click, 1, 0, enter] => [10 in clicked square]?
* + How much information is needed to solve? Does one need to
* know the algorithm by which the largest number is set?
*
* - eliminate puzzle instances with done chunks (1's in particular)?
* + that's what the qsort call is all about.
* + the 1's don't bother me that much.
* + but this takes a LONG time (not always possible)?
* - this may be affected by solver (lack of) quality.
* - weed them out by construction instead of post-cons check
* + but that interleaves make_board and new_game_desc: you
* have to alternate between changing the board and
* changing the hint set (instead of just creating the
* board once, then changing the hint set once -> done).
*
* - use binary search when discovering the minimal sovable point
* + profile to show a need (but when the solver gets slower...)
* + 7x9 @ .011s, 9x13 @ .075s, 17x13 @ .661s (all avg with n=100)
* + but the hints are independent, not linear, so... what?
*/
#include <assert.h>
#include <ctype.h>
#ifdef NO_TGMATH_H
# include <math.h>
#else
# include <tgmath.h>
#endif
#include <stdarg.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include "puzzles.h"
static bool verbose;
#ifdef STANDALONE_SOLVER
#define printv if (!verbose); else printf
#else
#define printv(...)
#endif
/*****************************************************************************
* GAME CONFIGURATION AND PARAMETERS *
*****************************************************************************/
struct game_params {
int w, h;
};
struct shared_state {
struct game_params params;
int *clues;
int refcnt;
};
struct game_state {
int *board;
struct shared_state *shared;
bool completed, cheated;
};
static const struct game_params filling_defaults[3] = {
{9, 7}, {13, 9}, {17, 13}
};
static game_params *default_params(void)
{
game_params *ret = snew(game_params);
*ret = filling_defaults[1]; /* struct copy */
return ret;
}
static bool game_fetch_preset(int i, char **name, game_params **params)
{
char buf[64];
if (i < 0 || i >= lenof(filling_defaults)) return false;
*params = snew(game_params);
**params = filling_defaults[i]; /* struct copy */
sprintf(buf, "%dx%d", filling_defaults[i].w, filling_defaults[i].h);
*name = dupstr(buf);
return true;
}
static void free_params(game_params *params)
{
sfree(params);
}
static game_params *dup_params(const game_params *params)
{
game_params *ret = snew(game_params);
*ret = *params; /* struct copy */
return ret;
}
static void decode_params(game_params *ret, char const *string)
{
ret->w = ret->h = atoi(string);
while (*string && isdigit((unsigned char) *string)) ++string;
if (*string == 'x') ret->h = atoi(++string);
}
static char *encode_params(const game_params *params, bool full)
{
char buf[64];
sprintf(buf, "%dx%d", params->w, params->h);
return dupstr(buf);
}
static config_item *game_configure(const game_params *params)
{
config_item *ret;
char buf[64];
ret = snewn(3, config_item);
ret[0].name = "Width";
ret[0].type = C_STRING;
sprintf(buf, "%d", params->w);
ret[0].u.string.sval = dupstr(buf);
ret[1].name = "Height";
ret[1].type = C_STRING;
sprintf(buf, "%d", params->h);
ret[1].u.string.sval = dupstr(buf);
ret[2].name = NULL;
ret[2].type = C_END;
return ret;
}
static game_params *custom_params(const config_item *cfg)
{
game_params *ret = snew(game_params);
ret->w = atoi(cfg[0].u.string.sval);
ret->h = atoi(cfg[1].u.string.sval);
return ret;
}
static const char *validate_params(const game_params *params, bool full)
{
if (params->w < 1) return "Width must be at least one";
if (params->h < 1) return "Height must be at least one";
if (params->w > INT_MAX / params->h)
return "Width times height must not be unreasonably large";
return NULL;
}
/*****************************************************************************
* STRINGIFICATION OF GAME STATE *
*****************************************************************************/
#define EMPTY 0
/* Example of plaintext rendering:
* +---+---+---+---+---+---+---+
* | 6 | | | 2 | | | 2 |
* +---+---+---+---+---+---+---+
* | | 3 | | 6 | | 3 | |
* +---+---+---+---+---+---+---+
* | 3 | | | | | | 1 |
* +---+---+---+---+---+---+---+
* | | 2 | 3 | | 4 | 2 | |
* +---+---+---+---+---+---+---+
* | 2 | | | | | | 3 |
* +---+---+---+---+---+---+---+
* | | 5 | | 1 | | 4 | |
* +---+---+---+---+---+---+---+
* | 4 | | | 3 | | | 3 |
* +---+---+---+---+---+---+---+
*
* This puzzle instance is taken from the nikoli website
* Encoded (unsolved and solved), the strings are these:
* 7x7:6002002030603030000010230420200000305010404003003
* 7x7:6662232336663232331311235422255544325413434443313
*/
static char *board_to_string(int *board, int w, int h) {
const int sz = w * h;
const int chw = (4*w + 2); /* +2 for trailing '+' and '\n' */
const int chh = (2*h + 1); /* +1: n fence segments, n+1 posts */
const int chlen = chw * chh;
char *repr = snewn(chlen + 1, char);
int i;
assert(board);
/* build the first line ("^(\+---){n}\+$") */
for (i = 0; i < w; ++i) {
repr[4*i + 0] = '+';
repr[4*i + 1] = '-';
repr[4*i + 2] = '-';
repr[4*i + 3] = '-';
}
repr[4*i + 0] = '+';
repr[4*i + 1] = '\n';
/* ... and copy it onto the odd-numbered lines */
for (i = 0; i < h; ++i) memcpy(repr + (2*i + 2) * chw, repr, chw);
/* build the second line ("^(\|\t){n}\|$") */
for (i = 0; i < w; ++i) {
repr[chw + 4*i + 0] = '|';
repr[chw + 4*i + 1] = ' ';
repr[chw + 4*i + 2] = ' ';
repr[chw + 4*i + 3] = ' ';
}
repr[chw + 4*i + 0] = '|';
repr[chw + 4*i + 1] = '\n';
/* ... and copy it onto the even-numbered lines */
for (i = 1; i < h; ++i) memcpy(repr + (2*i + 1) * chw, repr + chw, chw);
/* fill in the numbers */
for (i = 0; i < sz; ++i) {
const int x = i % w;
const int y = i / w;
if (board[i] == EMPTY) continue;
repr[chw*(2*y + 1) + (4*x + 2)] = board[i] + '0';
}
repr[chlen] = '\0';
return repr;
}
static bool game_can_format_as_text_now(const game_params *params)
{
return true;
}
static char *game_text_format(const game_state *state)
{
const int w = state->shared->params.w;
const int h = state->shared->params.h;
return board_to_string(state->board, w, h);
}
/*****************************************************************************
* GAME GENERATION AND SOLVER *
*****************************************************************************/
static const int dx[4] = {-1, 1, 0, 0};
static const int dy[4] = {0, 0, -1, 1};
struct solver_state
{
DSF *dsf;
int *board;
int *connected;
int nempty;
/* Used internally by learn_bitmap_deductions; kept here to avoid
* mallocing/freeing them every time that function is called. */
int *bm, *bmminsize;
DSF *bmdsf;
};
static void print_board(int *board, int w, int h) {
if (verbose) {
char *repr = board_to_string(board, w, h);
printv("%s\n", repr);
free(repr);
}
}
static game_state *new_game(midend *, const game_params *, const char *);
static void free_game(game_state *);
#define SENTINEL (sz+1)
static bool mark_region(int *board, int w, int h, int i, int n, int m) {
int j;
board[i] = -1;
for (j = 0; j < 4; ++j) {
const int x = (i % w) + dx[j], y = (i / w) + dy[j], ii = w*y + x;
if (x < 0 || x >= w || y < 0 || y >= h) continue;
if (board[ii] == m) return false;
if (board[ii] != n) continue;
if (!mark_region(board, w, h, ii, n, m)) return false;
}
return true;
}
static int region_size(int *board, int w, int h, int i) {
const int sz = w * h;
int j, size, copy;
if (board[i] == 0) return 0;
copy = board[i];
mark_region(board, w, h, i, board[i], SENTINEL);
for (size = j = 0; j < sz; ++j) {
if (board[j] != -1) continue;
++size;
board[j] = copy;
}
return size;
}
static void merge_ones(int *board, int w, int h)
{
const int sz = w * h;
const int maxsize = min(max(max(w, h), 3), 9);
int i, j, k;
bool change;
do {
change = false;
for (i = 0; i < sz; ++i) {
if (board[i] != 1) continue;
for (j = 0; j < 4; ++j, board[i] = 1) {
const int x = (i % w) + dx[j], y = (i / w) + dy[j];
int oldsize, newsize, ii = w*y + x;
bool ok;
if (x < 0 || x >= w || y < 0 || y >= h) continue;
if (board[ii] == maxsize) continue;
oldsize = board[ii];
board[i] = oldsize;
newsize = region_size(board, w, h, i);
if (newsize > maxsize) continue;
ok = mark_region(board, w, h, i, oldsize, newsize);
for (k = 0; k < sz; ++k)
if (board[k] == -1)
board[k] = ok ? newsize : oldsize;
if (ok) break;
}
if (j < 4) change = true;
}
} while (change);
}
/* generate a random valid board; uses validate_board. */
static void make_board(int *board, int w, int h, random_state *rs) {
const int sz = w * h;
/* w=h=2 is a special case which requires a number > max(w, h) */
/* TODO prove that this is the case ONLY for w=h=2. */
const int maxsize = min(max(max(w, h), 3), 9);
/* Note that if 1 in {w, h} then it's impossible to have a region
* of size > w*h, so the special case only affects w=h=2. */
int i;
DSF *dsf;
bool change;
assert(w >= 1);
assert(h >= 1);
assert(board);
/* I abuse the board variable: when generating the puzzle, it
* contains a shuffled list of numbers {0, ..., sz-1}. */
for (i = 0; i < sz; ++i) board[i] = i;
dsf = dsf_new(sz);
retry:
dsf_reinit(dsf);
shuffle(board, sz, sizeof (int), rs);
do {
change = false; /* as long as the board potentially has errors */
for (i = 0; i < sz; ++i) {
const int square = dsf_canonify(dsf, board[i]);
const int size = dsf_size(dsf, square);
int merge = SENTINEL, min = maxsize - size + 1;
bool error = false;
int neighbour, neighbour_size, j;
int directions[4];
for (j = 0; j < 4; ++j)
directions[j] = j;
shuffle(directions, 4, sizeof(int), rs);
for (j = 0; j < 4; ++j) {
const int x = (board[i] % w) + dx[directions[j]];
const int y = (board[i] / w) + dy[directions[j]];
if (x < 0 || x >= w || y < 0 || y >= h) continue;
neighbour = dsf_canonify(dsf, w*y + x);
if (square == neighbour) continue;
neighbour_size = dsf_size(dsf, neighbour);
if (size == neighbour_size) error = true;
/* find the smallest neighbour to merge with, which
* wouldn't make the region too large. (This is
* guaranteed by the initial value of `min'.) */
if (neighbour_size < min && random_upto(rs, 10)) {
min = neighbour_size;
merge = neighbour;
}
}
/* if this square is not in error, leave it be */
if (!error) continue;
/* if it is, but we can't fix it, retry the whole board.
* Maybe we could fix it by merging the conflicting
* neighbouring region(s) into some of their neighbours,
* but just restarting works out fine. */
if (merge == SENTINEL) goto retry;
/* merge with the smallest neighbouring workable region. */
dsf_merge(dsf, square, merge);
change = true;
}
} while (change);
for (i = 0; i < sz; ++i) board[i] = dsf_size(dsf, i);
merge_ones(board, w, h);
dsf_free(dsf);
}
static void merge(DSF *dsf, int *connected, int a, int b) {
int c;
assert(dsf);
assert(connected);
a = dsf_canonify(dsf, a);
b = dsf_canonify(dsf, b);
if (a == b) return;
dsf_merge(dsf, a, b);
c = connected[a];
connected[a] = connected[b];
connected[b] = c;
}
static void *memdup(const void *ptr, size_t len, size_t esz) {
void *dup = smalloc(len * esz);
assert(ptr);
memcpy(dup, ptr, len * esz);
return dup;
}
static void expand(struct solver_state *s, int w, int h, int t, int f) {
int j;
assert(s);
assert(s->board[t] == EMPTY); /* expand to empty square */
assert(s->board[f] != EMPTY); /* expand from non-empty square */
printv(
"learn: expanding %d from (%d, %d) into (%d, %d)\n",
s->board[f], f % w, f / w, t % w, t / w);
s->board[t] = s->board[f];
for (j = 0; j < 4; ++j) {
const int x = (t % w) + dx[j];
const int y = (t / w) + dy[j];
const int idx = w*y + x;
if (x < 0 || x >= w || y < 0 || y >= h) continue;
if (s->board[idx] != s->board[t]) continue;
merge(s->dsf, s->connected, t, idx);
}
--s->nempty;
}
static void clear_count(int *board, int sz) {
int i;
for (i = 0; i < sz; ++i) {
if (board[i] >= 0) continue;
else if (board[i] == -SENTINEL) board[i] = EMPTY;
else board[i] = -board[i];
}
}
static void flood_count(int *board, int w, int h, int i, int n, int *c) {
const int sz = w * h;
int k;
if (board[i] == EMPTY) board[i] = -SENTINEL;
else if (board[i] == n) board[i] = -board[i];
else return;
if (--*c == 0) return;
for (k = 0; k < 4; ++k) {
const int x = (i % w) + dx[k];
const int y = (i / w) + dy[k];
const int idx = w*y + x;
if (x < 0 || x >= w || y < 0 || y >= h) continue;
flood_count(board, w, h, idx, n, c);
if (*c == 0) return;
}
}
static bool check_capacity(int *board, int w, int h, int i) {
int n = board[i];
flood_count(board, w, h, i, board[i], &n);
clear_count(board, w * h);
return n == 0;
}
static int expandsize(const int *board, DSF *dsf, int w, int h, int i, int n) {
int j;
int nhits = 0;
int hits[4];
int size = 1;
for (j = 0; j < 4; ++j) {
const int x = (i % w) + dx[j];
const int y = (i / w) + dy[j];
const int idx = w*y + x;
int root;
int m;
if (x < 0 || x >= w || y < 0 || y >= h) continue;
if (board[idx] != n) continue;
root = dsf_canonify(dsf, idx);
for (m = 0; m < nhits && root != hits[m]; ++m);
if (m < nhits) continue;
printv("\t (%d, %d) contrib %d to size\n", x, y, dsf_size(dsf, root));
size += dsf_size(dsf, root);
assert(dsf_size(dsf, root) >= 1);
hits[nhits++] = root;
}
return size;
}
/*
* +---+---+---+---+---+---+---+
* | 6 | | | 2 | | | 2 |
* +---+---+---+---+---+---+---+
* | | 3 | | 6 | | 3 | |
* +---+---+---+---+---+---+---+
* | 3 | | | | | | 1 |
* +---+---+---+---+---+---+---+
* | | 2 | 3 | | 4 | 2 | |
* +---+---+---+---+---+---+---+
* | 2 | | | | | | 3 |
* +---+---+---+---+---+---+---+
* | | 5 | | 1 | | 4 | |
* +---+---+---+---+---+---+---+
* | 4 | | | 3 | | | 3 |
* +---+---+---+---+---+---+---+
*/
/* Solving techniques:
*
* CONNECTED COMPONENT FORCED EXPANSION (too big):
* When a CC can only be expanded in one direction, because all the
* other ones would make the CC too big.
* +---+---+---+---+---+
* | 2 | 2 | | 2 | _ |
* +---+---+---+---+---+
*
* CONNECTED COMPONENT FORCED EXPANSION (too small):
* When a CC must include a particular square, because otherwise there
* would not be enough room to complete it. This includes squares not
* adjacent to the CC through learn_critical_square.
* +---+---+
* | 2 | _ |
* +---+---+
*
* DROPPING IN A ONE:
* When an empty square has no neighbouring empty squares and only a 1
* will go into the square (or other CCs would be too big).
* +---+---+---+
* | 2 | 2 | _ |
* +---+---+---+
*
* TODO: generalise DROPPING IN A ONE: find the size of the CC of
* empty squares and a list of all adjacent numbers. See if only one
* number in {1, ..., size} u {all adjacent numbers} is possible.
* Probably this is only effective for a CC size < n for some n (4?)
*
* TODO: backtracking.
*/
static void filled_square(struct solver_state *s, int w, int h, int i) {
int j;
for (j = 0; j < 4; ++j) {
const int x = (i % w) + dx[j];
const int y = (i / w) + dy[j];
const int idx = w*y + x;
if (x < 0 || x >= w || y < 0 || y >= h) continue;
if (s->board[i] == s->board[idx])
merge(s->dsf, s->connected, i, idx);
}
}
static void init_solver_state(struct solver_state *s, int w, int h) {
const int sz = w * h;
int i;
assert(s);
s->nempty = 0;
for (i = 0; i < sz; ++i) s->connected[i] = i;
for (i = 0; i < sz; ++i)
if (s->board[i] == EMPTY) ++s->nempty;
else filled_square(s, w, h, i);
}
static bool learn_expand_or_one(struct solver_state *s, int w, int h) {
const int sz = w * h;
int i;
bool learn = false;
assert(s);
for (i = 0; i < sz; ++i) {
int j;
bool one = true;
if (s->board[i] != EMPTY) continue;
for (j = 0; j < 4; ++j) {
const int x = (i % w) + dx[j];
const int y = (i / w) + dy[j];
const int idx = w*y + x;
if (x < 0 || x >= w || y < 0 || y >= h) continue;
if (s->board[idx] == EMPTY) {
one = false;
continue;
}
if (one &&
(s->board[idx] == 1 ||
(s->board[idx] >= expandsize(s->board, s->dsf, w, h,
i, s->board[idx]))))
one = false;
if (dsf_size(s->dsf, idx) == s->board[idx]) continue;
assert(s->board[i] == EMPTY);
s->board[i] = -SENTINEL;
if (check_capacity(s->board, w, h, idx)) continue;
assert(s->board[i] == EMPTY);
printv("learn: expanding in one\n");
expand(s, w, h, i, idx);
learn = true;
break;
}
if (j == 4 && one) {
printv("learn: one at (%d, %d)\n", i % w, i / w);
assert(s->board[i] == EMPTY);
s->board[i] = 1;
assert(s->nempty);
--s->nempty;
learn = true;
}
}
return learn;
}
static bool learn_blocked_expansion(struct solver_state *s, int w, int h) {
const int sz = w * h;
int i;
bool learn = false;
assert(s);
/* for every connected component */
for (i = 0; i < sz; ++i) {
int exp = SENTINEL;
int j;
if (s->board[i] == EMPTY) continue;
j = dsf_canonify(s->dsf, i);
/* (but only for each connected component) */
if (i != j) continue;
/* (and not if it's already complete) */
if (dsf_size(s->dsf, j) == s->board[j]) continue;
/* for each square j _in_ the connected component */
do {
int k;
printv(" looking at (%d, %d)\n", j % w, j / w);
/* for each neighbouring square (idx) */
for (k = 0; k < 4; ++k) {
const int x = (j % w) + dx[k];
const int y = (j / w) + dy[k];
const int idx = w*y + x;
int size;
/* int l;
int nhits = 0;
int hits[4]; */
if (x < 0 || x >= w || y < 0 || y >= h) continue;
if (s->board[idx] != EMPTY) continue;
if (exp == idx) continue;
printv("\ttrying to expand onto (%d, %d)\n", x, y);
/* find out the would-be size of the new connected
* component if we actually expanded into idx */
/*
size = 1;
for (l = 0; l < 4; ++l) {
const int lx = x + dx[l];
const int ly = y + dy[l];
const int idxl = w*ly + lx;
int root;
int m;
if (lx < 0 || lx >= w || ly < 0 || ly >= h) continue;
if (board[idxl] != board[j]) continue;
root = dsf_canonify(dsf, idxl);
for (m = 0; m < nhits && root != hits[m]; ++m);
if (m != nhits) continue;
// printv("\t (%d, %d) contributed %d to size\n", lx, ly, dsf[root] >> 2);
size += dsf_size(dsf, root);
assert(dsf_size(dsf, root) >= 1);
hits[nhits++] = root;
}
*/
size = expandsize(s->board, s->dsf, w, h, idx, s->board[j]);
/* ... and see if that size is too big, or if we
* have other expansion candidates. Otherwise
* remember the (so far) only candidate. */
printv("\tthat would give a size of %d\n", size);
if (size > s->board[j]) continue;
/* printv("\tnow knowing %d expansions\n", nexpand + 1); */
if (exp != SENTINEL) goto next_i;
assert(exp != idx);
exp = idx;
}
j = s->connected[j]; /* next square in the same CC */
assert(s->board[i] == s->board[j]);
} while (j != i);
/* end: for each square j _in_ the connected component */
if (exp == SENTINEL) continue;
printv("learning to expand\n");
expand(s, w, h, exp, i);
learn = true;
next_i:
;
}
/* end: for each connected component */
return learn;
}
static bool learn_critical_square(struct solver_state *s, int w, int h) {
const int sz = w * h;
int i;
bool learn = false;
assert(s);
/* for each connected component */
for (i = 0; i < sz; ++i) {
int j, slack;
if (s->board[i] == EMPTY) continue;
if (i != dsf_canonify(s->dsf, i)) continue;
slack = s->board[i] - dsf_size(s->dsf, i);
if (slack == 0) continue;
assert(s->board[i] != 1);
/* for each empty square */
for (j = 0; j < sz; ++j) {
if (s->board[j] == EMPTY) {
/* if it's too far away from the CC, don't bother */
int k = i, jx = j % w, jy = j / w;
do {
int kx = k % w, ky = k / w;
if (abs(kx - jx) + abs(ky - jy) <= slack) break;
k = s->connected[k];
} while (i != k);
if (i == k) continue; /* not within range */
} else continue;
s->board[j] = -SENTINEL;
if (check_capacity(s->board, w, h, i)) continue;
/* if not expanding s->board[i] to s->board[j] implies
* that s->board[i] can't reach its full size, ... */
assert(s->nempty);
printv(
"learn: ds %d at (%d, %d) blocking (%d, %d)\n",
s->board[i], j % w, j / w, i % w, i / w);
--s->nempty;
s->board[j] = s->board[i];
filled_square(s, w, h, j);
learn = true;
}
}
return learn;
}
#if 0
static void print_bitmap(int *bitmap, int w, int h) {
if (verbose) {
int x, y;
for (y = 0; y < h; y++) {
for (x = 0; x < w; x++) {
printv(" %03x", bm[y*w+x]);
}
printv("\n");
}
}
}
#endif
static bool learn_bitmap_deductions(struct solver_state *s, int w, int h)
{
const int sz = w * h;
int *bm = s->bm;
DSF *dsf = s->bmdsf;
int *minsize = s->bmminsize;
int x, y, i, j, n;
bool learn = false;
/*
* This function does deductions based on building up a bitmap
* which indicates the possible numbers that can appear in each
* grid square. If we can rule out all but one possibility for a
* particular square, then we've found out the value of that
* square. In particular, this is one of the few forms of
* deduction capable of inferring the existence of a 'ghost
* region', i.e. a region which has none of its squares filled in
* at all.
*
* The reasoning goes like this. A currently unfilled square S can
* turn out to contain digit n in exactly two ways: either S is
* part of an n-region which also includes some currently known
* connected component of squares with n in, or S is part of an
* n-region separate from _all_ currently known connected
* components. If we can rule out both possibilities, then square
* S can't contain digit n at all.
*
* The former possibility: if there's a region of size n
* containing both S and some existing component C, then that
* means the distance from S to C must be small enough that C
* could be extended to include S without becoming too big. So we
* can do a breadth-first search out from all existing components
* with n in them, to identify all the squares which could be
* joined to any of them.
*
* The latter possibility: if there's a region of size n that
* doesn't contain _any_ existing component, then it also can't
* contain any square adjacent to an existing component either. So
* we can identify all the EMPTY squares not adjacent to any
* existing square with n in, and group them into connected
* components; then any component of size less than n is ruled
* out, because there wouldn't be room to create a completely new
* n-region in it.
*
* In fact we process these possibilities in the other order.
* First we find all the squares not adjacent to an existing
* square with n in; then we winnow those by removing too-small
* connected components, to get the set of squares which could
* possibly be part of a brand new n-region; and finally we do the
* breadth-first search to add in the set of squares which could
* possibly be added to some existing n-region.
*/
/*
* Start by initialising our bitmap to 'all numbers possible in
* all squares'.
*/
for (y = 0; y < h; y++)
for (x = 0; x < w; x++)
bm[y*w+x] = (1 << 10) - (1 << 1); /* bits 1,2,...,9 now set */
#if 0
printv("initial bitmap:\n");
print_bitmap(bm, w, h);
#endif
/*
* Now completely zero out the bitmap for squares that are already
* filled in (we aren't interested in those anyway). Also, for any
* filled square, eliminate its number from all its neighbours
* (because, as discussed above, the neighbours couldn't be part
* of a _new_ region with that number in it, and that's the case
* we consider first).
*/
for (y = 0; y < h; y++) {
for (x = 0; x < w; x++) {
i = y*w+x;
n = s->board[i];
if (n != EMPTY) {
bm[i] = 0;
if (x > 0)
bm[i-1] &= ~(1 << n);
if (x+1 < w)
bm[i+1] &= ~(1 << n);
if (y > 0)
bm[i-w] &= ~(1 << n);
if (y+1 < h)
bm[i+w] &= ~(1 << n);
}
}
}
#if 0
printv("bitmap after filled squares:\n");
print_bitmap(bm, w, h);
#endif
/*
* Now, for each n, we separately find the connected components of
* squares for which n is still a possibility. Then discard any
* component of size < n, because that component is too small to
* have a completely new n-region in it.
*/
for (n = 1; n <= 9; n++) {
dsf_reinit(dsf);
/* Build the dsf */
for (y = 0; y < h; y++)
for (x = 0; x+1 < w; x++)
if (bm[y*w+x] & bm[y*w+(x+1)] & (1 << n))
dsf_merge(dsf, y*w+x, y*w+(x+1));
for (y = 0; y+1 < h; y++)
for (x = 0; x < w; x++)
if (bm[y*w+x] & bm[(y+1)*w+x] & (1 << n))
dsf_merge(dsf, y*w+x, (y+1)*w+x);
/* Query the dsf */
for (i = 0; i < sz; i++)
if ((bm[i] & (1 << n)) && dsf_size(dsf, i) < n)
bm[i] &= ~(1 << n);
}
#if 0
printv("bitmap after winnowing small components:\n");
print_bitmap(bm, w, h);
#endif
/*
* Now our bitmap includes every square which could be part of a
* completely new region, of any size. Extend it to include
* squares which could be part of an existing region.
*/
for (n = 1; n <= 9; n++) {
/*
* We're going to do a breadth-first search starting from
* existing connected components with cell value n, to find
* all cells they might possibly extend into.
*
* The quantity we compute, for each square, is 'minimum size
* that any existing CC would have to have if extended to
* include this square'. So squares already _in_ an existing
* CC are initialised to the size of that CC; then we search
* outwards using the rule that if a square's score is j, then
* its neighbours can't score more than j+1.
*
* Scores are capped at n+1, because if a square scores more
* than n then that's enough to know it can't possibly be
* reached by extending an existing region - we don't need to
* know exactly _how far_ out of reach it is.
*/
for (i = 0; i < sz; i++) {
if (s->board[i] == n) {
/* Square is part of an existing CC. */
minsize[i] = dsf_size(s->dsf, i);
} else {
/* Otherwise, initialise to the maximum score n+1;
* we'll reduce this later if we find a neighbouring
* square with a lower score. */
minsize[i] = n+1;
}
}