I've just realised that my deliberate avoidance of non-simply

[sgt/puzzles] / 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.
 */

/*
 * 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 believe the smallest k which can manage this is 23, which I
 * derive by considering the largest _rectangle_ a k-omino can
 * surround. Consider: 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 n-sides, and fill in three of the corners. So
 * m+n must be at most (k-3)/2. Hence, to find the largest area of
 * such a rectangle, we must find m so as to maximise the product
 * m((k-3)/2-m). This is achieved when m is as close as possible
 * to half of (k-3)/2; so the largest rectangle surroundable by a
 * k-omino is equal to floor(k'/2)*ceil(k'/2), with k'=(k-3)/2.
 * The smallest k for which this is at least k is 23: a 23-omino
 * can surround a 5x5 rectangle, whereas the best a 22-omino can
 * manage is a 5x4.
 * 
 * (I don't have a definite argument to show that a k-omino cannot
 * surround a larger area in non-rectangular than rectangular
 * form, but it seems intuitively obvious to me that it can't. I
 * may update this with a more rigorous proof if I think of one.)
 * 
 * Anyway: 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. (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