puzzles: refactor and resync with upstream

This brings puzzles up-to-date with upstream revision
2d333750272c3967cfd5cd3677572cddeaad5932, though certain changes made
by me, including cursor-only Untangle and some compilation fixes
remain. Upstream code has been moved to its separate subdirectory and
future syncs can be done by simply copying over the new sources.

Change-Id: Ia6506ca5f78c3627165ea6791d38db414ace0804

1 files changed, 781 insertions, 0 deletions

diff --git a/apps/plugins/puzzles/src/divvy.c b/apps/plugins/puzzles/src/divvy.c
new file mode 100644
index 0000000..517e3dd
--- /dev/null
+++ b/apps/plugins/puzzles/src/divvy.c
@@ -0,0 +1,781 @@
+/*
+ * 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 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