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Enhance Filling's solver to handle large ghost regions.

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The previous solver could cope with inferring a '1' in an empty
square, but had no deductions that would enable it to infer the
existence of a '4'-sized region in 5x3:52d5b1a5b3. The new solver can
handle that, and I've made a companion change to the clue-stripping
code so that it aims to erase whole regions where possible so as to
actually present this situation to the player.

Current testing suggests that at the smallest preset a nontrivial
ghost region comes up in about 1/3 of games, and at the largest, more
like 1/2 of games. I may yet decide to introduce a difficulty level at
which it's skewed to happen more often still and one at which it
doesn't happen at all; but for the moment, this at least gets the
basic functionality into the code.
This commit is contained in:
Simon Tatham
2015-10-20 20:33:53 +01:00
parent 3c0b01114c
commit 90af15b43e
1 changed files with 341 additions and 8 deletions
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349
filling.c
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@ -11,13 +11,6 @@
* - the type should be somewhat big: board[i] = i
* - Using shorts gives us 181x181 puzzles as upper bound.
*
* - make a somewhat more clever solver
* + enable "ghost regions" of size > 1
* - one can put an upper bound on the size of a ghost region
* by considering the board size and summing present hints.
* + for each square, for i=1..n, what is the distance to a region
* containing i? How full is the region? How is this useful?
*
* - 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
@ -304,6 +297,10 @@ struct solver_state
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, *bmdsf, *bmminsize;
};
static void print_board(int *board, int w, int h) {
@ -817,6 +814,262 @@ static int learn_critical_square(struct solver_state *s, int w, int h) {
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 int learn_bitmap_deductions(struct solver_state *s, int w, int h)
{
const int sz = w * h;
int *bm = s->bm;
int *dsf = s->bmdsf;
int *minsize = s->bmminsize;
int x, y, i, j, n;
int 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_init(dsf, sz);
/* 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;
}
}
for (j = 1; j < n; j++) {
/*
* Find neighbours of cells scoring j, and set their score
* to at most j+1.
*
* Doing the BFS this way means we need n passes over the
* grid, which isn't entirely optimal but it seems to be
* fast enough for the moment. This could probably be
* improved by keeping a linked-list queue of cells in
* some way, but I think you'd have to be a bit careful to
* insert things into the right place in the queue; this
* way is easier not to get wrong.
*/
for (y = 0; y < h; y++) {
for (x = 0; x < w; x++) {
i = y*w+x;
if (minsize[i] == j) {
if (x > 0 && minsize[i-1] > j+1)
minsize[i-1] = j+1;
if (x+1 < w && minsize[i+1] > j+1)
minsize[i+1] = j+1;
if (y > 0 && minsize[i-w] > j+1)
minsize[i-w] = j+1;
if (y+1 < h && minsize[i+w] > j+1)
minsize[i+w] = j+1;
}
}
}
}
/*
* Now, every cell scoring at most n should have its 1<<n bit
* in the bitmap reinstated, because we've found that it's
* potentially reachable by extending an existing CC.
*/
for (i = 0; i <= sz; i++)
if (minsize[i] <= n)
bm[i] |= 1<<n;
}
#if 0
printv("bitmap after bfs:\n");
print_bitmap(bm, w, h);
#endif
/*
* Now our bitmap is complete. Look for entries with only one bit
* set; those are squares with only one possible number, in which
* case we can fill that number in.
*/
for (i = 0; i < sz; i++) {
if (bm[i] && !(bm[i] & (bm[i]-1))) { /* is bm[i] a power of two? */
int val = bm[i];
/* Integer log2, by simple binary search. */
n = 0;
if (val >> 8) { val >>= 8; n += 8; }
if (val >> 4) { val >>= 4; n += 4; }
if (val >> 2) { val >>= 2; n += 2; }
if (val >> 1) { val >>= 1; n += 1; }
/* Double-check that we ended up with a sensible
* answer. */
assert(1 <= n);
assert(n <= 9);
assert(bm[i] == (1 << n));
if (s->board[i] == EMPTY) {
printv("learn: %d is only possibility at (%d, %d)\n",
n, i % w, i / w);
s->board[i] = n;
filled_square(s, w, h, i);
assert(s->nempty);
--s->nempty;
learn = TRUE;
}
}
}
return learn;
}
static int solver(const int *orig, int w, int h, char **solution) {
const int sz = w * h;
@ -826,6 +1079,9 @@ static int solver(const int *orig, int w, int h, char **solution) {
ss.connected = snewn(sz, int); /* connected[n] := n.next; */
/* cyclic disjoint singly linked lists, same partitioning as dsf.
* The lists lets you iterate over a partition given any member */
ss.bm = snewn(sz, int);
ss.bmdsf = snew_dsf(sz);
ss.bmminsize = snewn(sz, int);
printv("trying to solve this:\n");
print_board(ss.board, w, h);
@ -835,6 +1091,7 @@ static int solver(const int *orig, int w, int h, char **solution) {
if (learn_blocked_expansion(&ss, w, h)) continue;
if (learn_expand_or_one(&ss, w, h)) continue;
if (learn_critical_square(&ss, w, h)) continue;
if (learn_bitmap_deductions(&ss, w, h)) continue;
break;
} while (ss.nempty);
@ -854,6 +1111,9 @@ static int solver(const int *orig, int w, int h, char **solution) {
sfree(ss.dsf);
sfree(ss.board);
sfree(ss.connected);
sfree(ss.bm);
sfree(ss.bmdsf);
sfree(ss.bmminsize);
return !ss.nempty;
}
@ -884,11 +1144,84 @@ static void minimize_clue_set(int *board, int w, int h, random_state *rs)
{
const int sz = w * h;
int *shuf = snewn(sz, int), i;
int *dsf, *next;
for (i = 0; i < sz; ++i) shuf[i] = i;
shuffle(shuf, sz, sizeof (int), rs);
/* the solver is monotone, so a second pass is superfluous. */
/*
* First, try to eliminate an entire region at a time if possible,
* because inferring the existence of a completely unclued region
* is a particularly good aspect of this puzzle type and we want
* to encourage it to happen.
*
* Begin by identifying the regions as linked lists of cells using
* the 'next' array.
*/
dsf = make_dsf(NULL, board, w, h);
next = snewn(sz, int);
for (i = 0; i < sz; ++i) {
int j = dsf_canonify(dsf, i);
if (i == j) {
/* First cell of a region; set next[i] = -1 to indicate
* end-of-list. */
next[i] = -1;
} else {
/* Add this cell to a region which already has a
* linked-list head, by pointing the canonical element j
* at this one, and pointing this one in turn at wherever
* j previously pointed. (This should end up with the
* elements linked in the order 1,n,n-1,n-2,...,2, which
* is a bit weird-looking, but any order is fine.)
*/
assert(j < i);
next[i] = next[j];
next[j] = i;
}
}
/*
* Now loop over the grid cells in our shuffled order, and each
* time we encounter a region for the first time, try to remove it
* all. Then we set next[canonical index] to -2 rather than -1, to
* mark it as already tried.
*
* Doing this in a loop over _cells_, rather than extracting and
* shuffling a list of _regions_, is intended to skew the
* probabilities towards trying to remove larger regions first
* (but without anything as crudely predictable as enforcing that
* we _always_ process regions in descending size order). Region
* removals might well be mutually exclusive, and larger ghost
* regions are more interesting, so we want to bias towards them
* if we can.
*/
for (i = 0; i < sz; ++i) {
int j = dsf_canonify(dsf, shuf[i]);
if (next[j] != -2) {
int tmp = board[j];
int k;
/* Blank out the whole thing. */
for (k = j; k >= 0; k = next[k])
board[k] = EMPTY;
if (!solver(board, w, h, NULL)) {
/* Wasn't still solvable; reinstate it all */
for (k = j; k >= 0; k = next[k])
board[k] = tmp;
}
/* Either way, don't try this region again. */
next[j] = -2;
}
}
sfree(next);
sfree(dsf);
/*
* Now go through individual cells, in the same shuffled order,
* and try to remove each one by itself.
*/
for (i = 0; i < sz; ++i) {
int tmp = board[shuf[i]];
board[shuf[i]] = EMPTY;