ref: 4ff90bd47f4bd3b11c4fb8b1ea25afb7b95f76da
dir: /unfinished/divvy.c/
/*
* Library code to divide up a rectangle into a number of equally
* sized ominoes, in a random fashion.
*
* Could use this for generating solved grids of
* http://www.nikoli.co.jp/ja/puzzles/block_puzzle/
* 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 necessary to do either of
* these, because the failure rate is already low enough to be
* easily tolerable, under all circumstances I've been able to
* think of.
*/
#include <assert.h>
#include <stdio.h>
#include <stdlib.h>
#include <stddef.h>
#include "puzzles.h"
/*
* Subroutine which implements a function used in computing both
* whether a square can safely be added to an omino, and whether
* it can safely be removed.
*
* We enumerate the eight squares 8-adjacent to this one, in
* cyclic order. We go round that loop and count the number of
* times we find a square owned by the target omino next to one
* not owned by it. We then return success iff that count is 2.
*
* When adding a square to an omino, this is precisely the
* criterion which tells us that adding the square won't leave a
* hole in the middle of the omino. (There's no explicit
* requirement in the statement of our problem that the ominoes be
* simply connected, but we do know they must be all of equal size
* and so it's clear that we must avoid leaving holes, since a
* hole would necessarily be smaller than the maximum omino size.)
*
* When removing a square from an omino, the _same_ criterion
* tells us that removing the square won't disconnect the omino.
*/
static int addremcommon(int w, int h, int x, int y, int *own, int val)
{
int neighbours[8];
int dir, count;
for (dir = 0; dir < 8; dir++) {
int dx = ((dir & 3) == 2 ? 0 : dir > 2 && dir < 6 ? +1 : -1);
int dy = ((dir & 3) == 0 ? 0 : dir < 4 ? -1 : +1);
int sx = x+dx, sy = y+dy;
if (sx < 0 || sx >= w || sy < 0 || sy >= h)
neighbours[dir] = -1; /* outside the grid */
else
neighbours[dir] = own[sy*w+sx];
}
/*
* To begin with, check 4-adjacency.
*/
if (neighbours[0] != val && neighbours[2] != val &&
neighbours[4] != val && neighbours[6] != val)
return FALSE;
count = 0;
for (dir = 0; dir < 8; dir++) {
int next = (dir + 1) & 7;
int gotthis = (neighbours[dir] == val);
int gotnext = (neighbours[next] == val);
if (gotthis != gotnext)
count++;
}
return (count == 2);
}
/*
* w and h are the dimensions of the rectangle.
*
* k is the size of the required ominoes. (So k must divide w*h,
* of course.)
*
* The returned result is a w*h-sized dsf.
*
* In both of the above suggested use cases, the user would
* probably want w==h==k, but that isn't a requirement.
*/
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;
int i, j, n, x, y, qhead, qtail;
n = wh / k;
assert(wh == k*n);
order = snewn(wh, int);
tmp = snewn(wh, int);
own = snewn(wh, int);
sizes = snewn(n, int);
queue = snewn(n, int);
addable = snewn(wh*4, int);
removable = snewn(wh, int);
/*
* Permute the grid squares into a random order, which will be
* used for iterating over the grid whenever we need to search
* for something. This prevents directional bias and arranges
* for the answer to be non-deterministic.
*/
for (i = 0; i < wh; i++)
order[i] = i;
shuffle(order, wh, sizeof(*order), rs);
/*
* Begin by choosing a starting square at random for each
* omino.
*/
for (i = 0; i < wh; i++) {
own[i] = -1;
}
for (i = 0; i < n; i++) {
own[order[i]] = i;
sizes[i] = 1;
}
/*
* Now repeatedly pick a random omino which isn't already at
* the target size, and find a way to expand it by one. This
* may involve stealing a square from another omino, in which
* case we then re-expand that omino, forming a chain of
* square-stealing which terminates in an as yet unclaimed
* square. Hence every successful iteration around this loop
* causes the number of unclaimed squares to drop by one, and
* so the process is bounded in duration.
*/
while (1) {
#ifdef DIVVY_DIAGNOSTICS
{
int x, y;
printf("Top of loop. Current grid:\n");
for (y = 0; y < h; y++) {
for (x = 0; x < w; x++)
printf("%3d", own[y*w+x]);
printf("\n");
}
}
#endif
/*
* Go over the grid and figure out which squares can
* safely be added to, or removed from, each omino. We
* don't take account of other ominoes in this process, so
* we will often end up knowing that a square can be
* poached from one omino by another.
*
* For each square, there may be up to four ominoes to
* which it can be added (those to which it is
* 4-adjacent).
*/
for (y = 0; y < h; y++) {
for (x = 0; x < w; x++) {
int yx = y*w+x;
int curr = own[yx];
int dir;
if (curr < 0) {
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
* omino without disconnecting it.
*/
removable[yx] = addremcommon(w, h, x, y, own, curr);
}
for (dir = 0; dir < 4; dir++) {
int dx = (dir == 0 ? -1 : dir == 1 ? +1 : 0);
int dy = (dir == 2 ? -1 : dir == 3 ? +1 : 0);
int sx = x + dx, sy = y + dy;
int syx = sy*w+sx;
addable[yx*4+dir] = -1;
if (sx < 0 || sx >= w || sy < 0 || sy >= h)
continue; /* no omino here! */
if (own[syx] < 0)
continue; /* also no omino here */
if (own[syx] == own[yx])
continue; /* we already got one */
if (!addremcommon(w, h, x, y, own, own[syx]))
continue; /* would non-simply connect the omino */
addable[yx*4+dir] = own[syx];
}
}
}
for (i = j = 0; i < n; i++)
if (sizes[i] < k)
tmp[j++] = i;
if (j == 0)
break; /* all ominoes are complete! */
j = tmp[random_upto(rs, j)];
#ifdef DIVVY_DIAGNOSTICS
printf("Trying to extend %d\n", j);
#endif
/*
* So we're trying to expand omino j. We breadth-first
* search out from j across the space of ominoes.
*
* For bfs purposes, we use two elements of tmp per omino:
* tmp[2*i+0] tells us which omino we got to i from, and
* tmp[2*i+1] numbers the grid square that omino stole
* from us.
*
* This requires that wh (the size of tmp) is at least 2n,
* i.e. k is at least 2. There would have been nothing to
* stop a user calling this function with k=1, but if they
* did then we wouldn't have got to _here_ in the code -
* we would have noticed above that all ominoes were
* already at their target sizes, and terminated :-)
*/
assert(wh >= 2*n);
for (i = 0; i < n; i++)
tmp[2*i] = tmp[2*i+1] = -1;
qhead = qtail = 0;
queue[qtail++] = j;
tmp[2*j] = tmp[2*j+1] = -2; /* special value: `starting point' */
while (qhead < qtail) {
int tmpsq;
j = queue[qhead];
/*
* We wish to expand omino j. However, we might have
* got here by omino j having a square stolen from it,
* so first of all we must temporarily mark that
* square as not belonging to j, so that our adjacency
* calculations don't assume j _does_ belong to us.
*/
tmpsq = tmp[2*j+1];
if (tmpsq >= 0) {
assert(own[tmpsq] == j);
own[tmpsq] = -3;
}
/*
* OK. Now begin by seeing if we can find any
* unclaimed square into which we can expand omino j.
* If we find one, the entire bfs terminates.
*/
for (i = 0; i < wh; i++) {
int dir;
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) {
/*
* We know this square is addable to this
* omino with the grid in the state it had
* at the top of the loop. However, we
* must now check that it's _still_
* addable to this omino when the omino is
* missing a square. To do this it's only
* necessary to re-check addremcommon.
*/
if (!addremcommon(w, h, order[i]%w, order[i]/w,
own, j))
continue;
break;
}
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
* moving squares between ominoes, ending up
* expanding our starting omino by one.
*/
#ifdef DIVVY_DIAGNOSTICS
printf("(%d,%d)", i%w, i/w);
#endif
while (1) {
own[i] = j;
#ifdef DIVVY_DIAGNOSTICS
printf(" -> %d", j);
#endif
if (tmp[2*j] == -2)
break;
i = tmp[2*j+1];
j = tmp[2*j];
#ifdef DIVVY_DIAGNOSTICS
printf("; (%d,%d)", i%w, i/w);
#endif
}
#ifdef DIVVY_DIAGNOSTICS
printf("\n");
#endif
/*
* Increment the size of the starting omino.
*/
sizes[j]++;
/*
* Terminate the bfs loop.
*/
break;
}
/*
* If we get here, we haven't been able to expand
* omino j into an unclaimed square. So now we begin
* to investigate expanding it into squares which are
* claimed by ominoes the bfs has not yet visited.
*/
for (i = 0; i < wh; i++) {
int dir, nj;
nj = own[order[i]];
if (nj < 0 || tmp[2*nj] != -1)
continue; /* unclaimed, or owned by wrong omino */
if (!removable[order[i]])
continue; /* its omino won't let it go */
for (dir = 0; dir < 4; dir++)
if (addable[order[i]*4+dir] == j) {
/*
* As above, re-check addremcommon.
*/
if (!addremcommon(w, h, order[i]%w, order[i]/w,
own, j))
continue;
/*
* We have found a square we can use to
* expand omino j, at the expense of the
* as-yet unvisited omino nj. So add this
* to the bfs queue.
*/
assert(qtail < n);
queue[qtail++] = nj;
tmp[2*nj] = j;
tmp[2*nj+1] = order[i];
/*
* Now terminate the loop over dir, to
* ensure we don't accidentally add the
* same omino twice to the queue.
*/
break;
}
}
/*
* Restore the temporarily removed square.
*/
if (tmpsq >= 0)
own[tmpsq] = j;
/*
* Advance the queue head.
*/
qhead++;
}
if (qhead == qtail) {
/*
* We have finished the bfs and not found any way to
* expand omino j. Panic, and return failure.
*
* FIXME: or should we loop over all ominoes before we
* give up?
*/
#ifdef DIVVY_DIAGNOSTICS
printf("FAIL!\n");
#endif
retdsf = NULL;
goto cleanup;
}
}
#ifdef DIVVY_DIAGNOSTICS
{
int x, y;
printf("SUCCESS! Final grid:\n");
for (y = 0; y < h; y++) {
for (x = 0; x < w; x++)
printf("%3d", own[y*w+x]);
printf("\n");
}
}
#endif
/*
* Construct the output dsf.
*/
for (i = 0; i < wh; i++) {
assert(own[i] >= 0 && own[i] < n);
tmp[own[i]] = i;
}
retdsf = snew_dsf(wh);
for (i = 0; i < wh; i++) {
dsf_merge(retdsf, i, tmp[own[i]]);
}
/*
* Construct the output dsf a different way, to verify that
* the ominoes really are k-ominoes and we haven't
* accidentally split one into two disconnected pieces.
*/
dsf_init(tmp, wh);
for (y = 0; y < h; y++)
for (x = 0; x+1 < w; x++)
if (own[y*w+x] == own[y*w+(x+1)])
dsf_merge(tmp, y*w+x, y*w+(x+1));
for (x = 0; x < w; x++)
for (y = 0; y+1 < h; y++)
if (own[y*w+x] == own[(y+1)*w+x])
dsf_merge(tmp, y*w+x, (y+1)*w+x);
for (i = 0; i < wh; i++) {
j = dsf_canonify(retdsf, i);
assert(dsf_canonify(tmp, j) == dsf_canonify(tmp, i));
}
cleanup:
/*
* Free our temporary working space.
*/
sfree(order);
sfree(tmp);
sfree(own);
sfree(sizes);
sfree(queue);
sfree(addable);
sfree(removable);
/*
* And we're done.
*/
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
/*
* gcc -g -O0 -DTESTMODE -I.. -o divvy divvy.c ../random.c ../malloc.c ../dsf.c ../misc.c ../nullfe.c
*
* or to debug
*
* gcc -g -O0 -DDIVVY_DIAGNOSTICS -DTESTMODE -I.. -o divvy divvy.c ../random.c ../malloc.c ../dsf.c ../misc.c ../nullfe.c
*/
int main(int argc, char **argv)
{
int *dsf;
int i;
int w = 9, h = 4, k = 6, tries = 100;
random_state *rs;
rs = random_new("123456", 6);
if (argc > 1)
w = atoi(argv[1]);
if (argc > 2)
h = atoi(argv[2]);
if (argc > 3)
k = atoi(argv[3]);
if (argc > 4)
tries = atoi(argv[4]);
for (i = 0; i < tries; i++) {
int x, y;
dsf = divvy_rectangle(w, h, k, rs);
assert(dsf);
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("|");
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("%d retries needed for %d successes\n", fail_counter, tries);
return 0;
}
#endif