ref: f279c5eba035d018a8805a65a8a245b8b8d2ea2e
dir: /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.
*/
/*
* This code is restricted to simply connected solutions: that is,
* no single polyomino may completely surround another (not even
* with a corner visible to the outside world, in the sense that a
* 7-omino can `surround' a single square).
*
* It's tempting to think that this is a natural consequence of
* all the ominoes being the same size - after all, a division of
* anything into 7-ominoes must necessarily have all of them
* simply connected, because if one was not then the 1-square
* space in the middle could not be part of any 7-omino - but in
* fact, for sufficiently large k, it is perfectly possible for a
* k-omino to completely surround another k-omino. A simple
* example is this one with two 25-ominoes:
*
* +--+--+--+--+--+--+--+
* | |
* + +--+--+--+--+--+ +
* | | | |
* + + + +
* | | | |
* + + + +--+
* | | | |
* + + + +--+
* | | | |
* + + + +
* | | | |
* + +--+--+--+--+--+ +
* | |
* +--+--+--+--+--+--+--+
*
* I claim the smallest k which can manage this is 23. More
* formally:
*
* If a k-omino P is completely surrounded by another k-omino Q,
* such that every edge of P borders on Q, then k >= 23.
*
* Proof:
*
* It's relatively simple to find the largest _rectangle_ a
* k-omino can enclose. So I'll construct my proof in two parts:
* firstly, show that no 22-omino or smaller can enclose a
* rectangle as large as itself, and secondly, show that no
* polyomino can enclose a larger non-rectangle than a rectangle.
*
* The first of those claims:
*
* To surround an m x n rectangle, a polyomino must have 2m
* squares along the two m-sides of the rectangle, 2n squares
* along the two n-sides, and must fill in at least three of the
* corners in order to be connected. Thus, 2(m+n)+3 <= k. We wish
* to find the largest value of mn subject to that constraint, and
* it's clear that this is achieved when m and n are as close to
* equal as possible. (If they aren't, WLOG suppose m < n; then
* (m+1)(n-1) = mn + n - m - 1 >= mn, with equality only when
* m=n-1.)
*
* So the area of the largest rectangle which can be enclosed by a
* k-omino is given by floor(k'/2) * ceil(k'/2), where k' =
* (k-3)/2. This is a monotonic function in k, so there will be a
* unique point at which it goes from being smaller than k to
* being larger than k. That point is between 22 (maximum area 20)
* and 23 (maximum area 25).
*
* The second claim:
*
* Suppose we have an inner polyomino P surrounded by an outer
* polyomino Q. I seek to show that if P is non-rectangular, then
* P is also non-maximal, in the sense that we can transform P and
* Q into a new pair of polyominoes in which P is larger and Q is
* at most the same size.
*
* Consider walking along the boundary of P in a clockwise
* direction. (We may assume, of course, that there is only _one_
* boundary of P, i.e. P has no hole in the middle. If it does
* have a hole in the middle, it's _trivially_ non-maximal because
* we can just fill the hole in!) Our walk will take us along many
* edges between squares; sometimes we might turn left, and
* certainly sometimes we will turn right. Always there will be a
* square of P on our right, and a square of Q on our left.
*
* The net angle through which we turn during the entire walk must
* add up to 360 degrees rightwards. So if there are no left
* turns, then we must turn right exactly four times, meaning we
* have described a rectangle. Hence, if P is _not_ rectangular,
* then there must have been a left turn at some point. A left
* turn must mean we walk along two edges of the same square of Q.
*
* Thus, there is some square X in Q which is adjacent to two
* diagonally separated squares in P. Let us call those two
* squares N and E; let us refer to the other two neighbours of X
* as S and W; let us refer to the other mutual neighbour of S and
* W as D; and let us refer to the other mutual neighbour of S and
* E as Y. In other words, we have named seven squares, arranged
* thus:
*
* N
* W X E
* D S Y
*
* where N and E are in P, and X is in Q.
*
* Clearly at least one of W and S must be in Q (because otherwise
* X would not be connected to any other square in Q, and would
* hence have to be the whole of Q; and evidently if Q were a
* 1-omino it could not enclose _anything_). So we divide into
* cases:
*
* If both W and S are in Q, then we take X out of Q and put it in
* P, which does not expose any edge of P. If this disconnects Q,
* then we can reconnect it by adding D to Q.
*
* If only one of W and S is in Q, then wlog let it be W. If S is
* in _P_, then we have a particularly easy case: we can simply
* take X out of Q and add it to P, and this cannot disconnect X
* since X was a leaf square of Q.
*
* Our remaining case is that W is in Q and S is in neither P nor
* Q. Again we take X out of Q and put it in P; we also add S to
* Q. This ensures we do not expose an edge of P, but we must now
* prove that S is adjacent to some other existing square of Q so
* that we haven't disconnected Q by adding it.
*
* To do this, we recall that we walked along the edge XE, and
* then turned left to walk along XN. So just before doing all
* that, we must have reached the corner XSE, and we must have
* done it by walking along one of the three edges meeting at that
* corner which are _not_ XE. It can't have been SY, since S would
* then have been on our left and it isn't in Q; and it can't have
* been XS, since S would then have been on our right and it isn't
* in P. So it must have been YE, in which case Y was on our left,
* and hence is in Q.
*
* So in all cases we have shown that we can take X out of Q and
* add it to P, and add at most one square to Q to restore the
* containment and connectedness properties. Hence, we can keep
* doing this until we run out of left turns and P becomes
* rectangular. []
*
* ------------
*
* Anyway, that entire proof was a bit of a sidetrack. The point
* is, although constructions of this type are possible for
* sufficiently large k, divvy_rectangle() will never generate
* them. This could be considered a weakness for some purposes, in
* the sense that we can't generate all possible divisions.
* However, there are many divisions which we are highly unlikely
* to generate anyway, so in practice it probably isn't _too_ bad.
*
* If I wanted to fix this issue, I would have to make the rules
* more complicated for determining when a square can safely be
* _removed_ from a polyomino. Adding one becomes easier (a square
* may be added to a polyomino iff it is 4-adjacent to any square
* currently part of the polyomino, and the current test for loop
* formation may be dispensed with), but to determine which
* squares may be removed we must now resort to analysis of the
* overall structure of the polyomino rather than the simple local
* properties we can currently get away with measuring.
*/
/*
* 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. (If it did, then things get
* more complicated; see above.)
*
* When removing a square from an omino, the _same_ criterion
* tells us that removing the square won't disconnect the omino.
* (This only works _because_ we've ensured the omino is simply
* connected.)
*/
static bool 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;
bool gotthis = (neighbours[dir] == val);
bool 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.
*/
DSF *divvy_rectangle_attempt(int w, int h, int k, random_state *rs)
{
int *order, *queue, *tmp, *own, *sizes, *addable;
DSF *retdsf, *tmpdsf;
bool *removable;
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, bool);
retdsf = tmpdsf = NULL;
/*
* 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 = dsf_new(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.
*/
tmpdsf = dsf_new(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(tmpdsf, 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(tmpdsf, y*w+x, (y+1)*w+x);
for (i = 0; i < wh; i++) {
j = dsf_canonify(retdsf, i);
assert(dsf_equivalent(tmpdsf, j, i));
}
cleanup:
/*
* Free our temporary working space.
*/
sfree(order);
sfree(tmp);
dsf_free(tmpdsf);
sfree(own);
sfree(sizes);
sfree(queue);
sfree(addable);
sfree(removable);
/*
* And we're done.
*/
return retdsf;
}
DSF *divvy_rectangle(int w, int h, int k, random_state *rs)
{
DSF *ret;
do {
ret = divvy_rectangle_attempt(w, h, k, rs);
} while (!ret);
return ret;
}