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Allow a 1-omino to be completely destroyed and recreated in an

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arbitrary unclaimed square. This cures the most common cause of
generation failures (covering a large area in dominoes was the most
difficult case, and would fail even if the large area was 1xn!); the
failure rate is now sufficiently low under all circumstances I've
found that I'm willing to just loop until I get a success.

[originally from svn r7693]
This commit is contained in:
Simon Tatham
2007-08-18 13:30:13 +00:00
parent 333d57bf6e
commit 4c1e3ca7cb
1 changed files with 128 additions and 72 deletions
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200
unfinished/divvy.c
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@ -7,6 +7,21 @@
* or for generating the playfield for Jigsaw Sudoku.
*/
/*
* Possible improvements which might cut the fail rate:
*
* - instead of picking one omino to extend in an iteration, try
* them all in succession (in a randomised order)
*
* - (for real rigour) instead of bfsing over ominoes, bfs over
* the space of possible _removed squares_. That way we aren't
* limited to randomly choosing a single square to remove from
* an omino and failing if that particular square doesn't
* happen to work.
*
* However, I don't currently think it's neecss~|~
*/
#include <assert.h>
#include <stdio.h>
#include <stdlib.h>
@ -83,7 +98,7 @@ static int addremcommon(int w, int h, int x, int y, int *own, int val)
* In both of the above suggested use cases, the user would
* probably want w==h==k, but that isn't a requirement.
*/
int *divvy_rectangle(int w, int h, int k, random_state *rs)
static int *divvy_internal(int w, int h, int k, random_state *rs)
{
int *order, *queue, *tmp, *own, *sizes, *addable, *removable, *retdsf;
int wh = w*h;
@ -164,7 +179,9 @@ int *divvy_rectangle(int w, int h, int k, random_state *rs)
int dir;
if (curr < 0) {
removable[yx] = 0; /* can't remove if it's not owned! */
removable[yx] = FALSE; /* can't remove if not owned! */
} else if (sizes[curr] == 1) {
removable[yx] = TRUE; /* can always remove a singleton */
} else {
/*
* See if this square can be removed from its
@ -243,7 +260,7 @@ int *divvy_rectangle(int w, int h, int k, random_state *rs)
tmpsq = tmp[2*j+1];
if (tmpsq >= 0) {
assert(own[tmpsq] == j);
own[tmpsq] = -1;
own[tmpsq] = -3;
}
/*
@ -254,8 +271,22 @@ int *divvy_rectangle(int w, int h, int k, random_state *rs)
for (i = 0; i < wh; i++) {
int dir;
if (own[order[i]] >= 0)
if (own[order[i]] != -1)
continue; /* this square is claimed */
/*
* Special case: if our current omino was size 1
* and then had a square stolen from it, it's now
* size zero, which means it's valid to `expand'
* it into _any_ unclaimed square.
*/
if (sizes[j] == 1 && tmpsq >= 0)
break; /* got one */
/*
* Failing that, we must do the full test for
* addability.
*/
for (dir = 0; dir < 4; dir++)
if (addable[order[i]*4+dir] == j) {
/*
@ -274,11 +305,19 @@ int *divvy_rectangle(int w, int h, int k, random_state *rs)
}
if (dir == 4)
continue; /* we can't add this square to j */
break; /* got one! */
}
if (i < wh) {
i = order[i];
/*
* Restore the temporarily removed square _before_
* we start shifting ownerships about.
*/
if (tmpsq >= 0)
own[tmpsq] = j;
/*
* We are done. We can add square i to omino j,
* and then backtrack along the trail in tmp
@ -450,6 +489,27 @@ int *divvy_rectangle(int w, int h, int k, random_state *rs)
return retdsf;
}
#ifdef TESTMODE
static int fail_counter = 0;
#endif
int *divvy_rectangle(int w, int h, int k, random_state *rs)
{
int *ret;
do {
ret = divvy_internal(w, h, k, rs);
#ifdef TESTMODE
if (!ret)
fail_counter++;
#endif
} while (!ret);
return ret;
}
#ifdef TESTMODE
/*
@ -463,7 +523,7 @@ int *divvy_rectangle(int w, int h, int k, random_state *rs)
int main(int argc, char **argv)
{
int *dsf;
int i, successes;
int i;
int w = 9, h = 4, k = 6, tries = 100;
random_state *rs;
@ -478,84 +538,80 @@ int main(int argc, char **argv)
if (argc > 4)
tries = atoi(argv[4]);
successes = 0;
for (i = 0; i < tries; i++) {
int x, y;
dsf = divvy_rectangle(w, h, k, rs);
if (dsf) {
int x, y;
assert(dsf);
successes++;
for (y = 0; y <= 2*h; y++) {
for (x = 0; x <= 2*w; x++) {
int miny = y/2 - 1, maxy = y/2;
int minx = x/2 - 1, maxx = x/2;
int classes[4], tx, ty;
for (ty = 0; ty < 2; ty++)
for (tx = 0; tx < 2; tx++) {
int cx = minx+tx, cy = miny+ty;
if (cx < 0 || cx >= w || cy < 0 || cy >= h)
classes[ty*2+tx] = -1;
else
classes[ty*2+tx] = dsf_canonify(dsf, cy*w+cx);
}
switch (y%2 * 2 + x%2) {
case 0: /* corner */
/*
* Cases for the corner:
*
* - if all four surrounding squares
* belong to the same omino, we print a
* space.
*
* - if the top two are the same and the
* bottom two are the same, we print a
* horizontal line.
*
* - if the left two are the same and the
* right two are the same, we print a
* vertical line.
*
* - otherwise, we print a cross.
*/
if (classes[0] == classes[1] &&
classes[1] == classes[2] &&
classes[2] == classes[3])
printf(" ");
else if (classes[0] == classes[1] &&
classes[2] == classes[3])
printf("-");
else if (classes[0] == classes[2] &&
classes[1] == classes[3])
printf("|");
for (y = 0; y <= 2*h; y++) {
for (x = 0; x <= 2*w; x++) {
int miny = y/2 - 1, maxy = y/2;
int minx = x/2 - 1, maxx = x/2;
int classes[4], tx, ty;
for (ty = 0; ty < 2; ty++)
for (tx = 0; tx < 2; tx++) {
int cx = minx+tx, cy = miny+ty;
if (cx < 0 || cx >= w || cy < 0 || cy >= h)
classes[ty*2+tx] = -1;
else
printf("+");
break;
case 1: /* horiz edge */
if (classes[1] == classes[3])
printf(" ");
else
printf("--");
break;
case 2: /* vert edge */
if (classes[2] == classes[3])
printf(" ");
else
printf("|");
break;
case 3: /* square centre */
printf(" ");
break;
classes[ty*2+tx] = dsf_canonify(dsf, cy*w+cx);
}
switch (y%2 * 2 + x%2) {
case 0: /* corner */
/*
* Cases for the corner:
*
* - if all four surrounding squares belong
* to the same omino, we print a space.
*
* - if the top two are the same and the
* bottom two are the same, we print a
* horizontal line.
*
* - if the left two are the same and the
* right two are the same, we print a
* vertical line.
*
* - otherwise, we print a cross.
*/
if (classes[0] == classes[1] &&
classes[1] == classes[2] &&
classes[2] == classes[3])
printf(" ");
else if (classes[0] == classes[1] &&
classes[2] == classes[3])
printf("-");
else if (classes[0] == classes[2] &&
classes[1] == classes[3])
printf("|");
else
printf("+");
break;
case 1: /* horiz edge */
if (classes[1] == classes[3])
printf(" ");
else
printf("--");
break;
case 2: /* vert edge */
if (classes[2] == classes[3])
printf(" ");
else
printf("|");
break;
case 3: /* square centre */
printf(" ");
break;
}
printf("\n");
}
printf("\n");
sfree(dsf);
}
printf("\n");
sfree(dsf);
}
printf("%d successes out of %d tries\n", successes, tries);
printf("%d retries needed for %d successes\n", fail_counter, tries);
return 0;
}