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A piece of library code which constructs a random division of a
rectangle into equally sized ominoes. I have a couple of potential applications for this, but none I've actually implemented yet, so for the moment it's living in `unfinished'. [originally from svn r7690]
This commit is contained in:
Simon Tatham
533
unfinished/divvy.c
Normal file
533
unfinished/divvy.c
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@ -0,0 +1,533 @@
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/*
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* Library code to divide up a rectangle into a number of equally
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* sized ominoes, in a random fashion.
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*
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* Could use this for generating solved grids of
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* http://www.nikoli.co.jp/ja/puzzles/block_puzzle/
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* or for generating the playfield for Jigsaw Sudoku.
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*/
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#include <assert.h>
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#include <stdio.h>
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#include <stdlib.h>
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#include <stddef.h>
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#include "puzzles.h"
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/*
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* Subroutine which implements a function used in computing both
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* whether a square can safely be added to an omino, and whether
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* it can safely be removed.
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*
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* We enumerate the eight squares 8-adjacent to this one, in
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* cyclic order. We go round that loop and count the number of
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* times we find a square owned by the target omino next to one
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* not owned by it. We then return success iff that count is 2.
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*
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* When adding a square to an omino, this is precisely the
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* criterion which tells us that adding the square won't leave a
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* hole in the middle of the omino. (There's no explicit
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* requirement in the statement of our problem that the ominoes be
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* simply connected, but we do know they must be all of equal size
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* and so it's clear that we must avoid leaving holes, since a
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* hole would necessarily be smaller than the maximum omino size.)
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*
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* When removing a square from an omino, the _same_ criterion
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* tells us that removing the square won't disconnect the omino.
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*/
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static int addremcommon(int w, int h, int x, int y, int *own, int val)
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{
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int neighbours[8];
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int dir, count;
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for (dir = 0; dir < 8; dir++) {
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int dx = ((dir & 3) == 2 ? 0 : dir > 2 && dir < 6 ? +1 : -1);
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int dy = ((dir & 3) == 0 ? 0 : dir < 4 ? -1 : +1);
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int sx = x+dx, sy = y+dy;
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if (sx < 0 || sx >= w || sy < 0 || sy >= h)
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neighbours[dir] = -1; /* outside the grid */
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else
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neighbours[dir] = own[sy*w+sx];
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}
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/*
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* To begin with, check 4-adjacency.
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*/
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if (neighbours[0] != val && neighbours[2] != val &&
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neighbours[4] != val && neighbours[6] != val)
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return FALSE;
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count = 0;
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for (dir = 0; dir < 8; dir++) {
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int next = (dir + 1) & 7;
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int gotthis = (neighbours[dir] == val);
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int gotnext = (neighbours[next] == val);
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if (gotthis != gotnext)
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count++;
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}
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return (count == 2);
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}
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/*
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* w and h are the dimensions of the rectangle.
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*
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* k is the size of the required ominoes. (So k must divide w*h,
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* of course.)
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*
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* The returned result is a w*h-sized dsf.
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*
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* In both of the above suggested use cases, the user would
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* probably want w==h==k, but that isn't a requirement.
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*/
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int *divvy_rectangle(int w, int h, int k, random_state *rs)
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{
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int *order, *queue, *tmp, *own, *sizes, *addable, *removable, *retdsf;
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int wh = w*h;
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int i, j, n, x, y, qhead, qtail;
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n = wh / k;
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assert(wh == k*n);
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order = snewn(wh, int);
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tmp = snewn(wh, int);
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own = snewn(wh, int);
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sizes = snewn(n, int);
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queue = snewn(n, int);
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addable = snewn(wh*4, int);
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removable = snewn(wh, int);
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/*
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* Permute the grid squares into a random order, which will be
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* used for iterating over the grid whenever we need to search
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* for something. This prevents directional bias and arranges
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* for the answer to be non-deterministic.
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*/
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for (i = 0; i < wh; i++)
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order[i] = i;
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shuffle(order, wh, sizeof(*order), rs);
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/*
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* Begin by choosing a starting square at random for each
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* omino.
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*/
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for (i = 0; i < wh; i++) {
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own[i] = -1;
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}
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for (i = 0; i < n; i++) {
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own[order[i]] = i;
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sizes[i] = 1;
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}
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/*
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* Now repeatedly pick a random omino which isn't already at
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* the target size, and find a way to expand it by one. This
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* may involve stealing a square from another omino, in which
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* case we then re-expand that omino, forming a chain of
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* square-stealing which terminates in an as yet unclaimed
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* square. Hence every successful iteration around this loop
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* causes the number of unclaimed squares to drop by one, and
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* so the process is bounded in duration.
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*/
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while (1) {
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#ifdef DIVVY_DIAGNOSTICS
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{
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int x, y;
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printf("Top of loop. Current grid:\n");
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for (y = 0; y < h; y++) {
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for (x = 0; x < w; x++)
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printf("%3d", own[y*w+x]);
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printf("\n");
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}
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}
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#endif
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/*
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* Go over the grid and figure out which squares can
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* safely be added to, or removed from, each omino. We
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* don't take account of other ominoes in this process, so
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* we will often end up knowing that a square can be
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* poached from one omino by another.
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*
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* For each square, there may be up to four ominoes to
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* which it can be added (those to which it is
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* 4-adjacent).
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*/
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for (y = 0; y < h; y++) {
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for (x = 0; x < w; x++) {
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int yx = y*w+x;
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int curr = own[yx];
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int dir;
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if (curr < 0) {
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removable[yx] = 0; /* can't remove if it's not owned! */
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} else {
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/*
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* See if this square can be removed from its
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* omino without disconnecting it.
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*/
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removable[yx] = addremcommon(w, h, x, y, own, curr);
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}
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for (dir = 0; dir < 4; dir++) {
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int dx = (dir == 0 ? -1 : dir == 1 ? +1 : 0);
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int dy = (dir == 2 ? -1 : dir == 3 ? +1 : 0);
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int sx = x + dx, sy = y + dy;
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int syx = sy*w+sx;
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addable[yx*4+dir] = -1;
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if (sx < 0 || sx >= w || sy < 0 || sy >= h)
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continue; /* no omino here! */
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if (own[syx] < 0)
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continue; /* also no omino here */
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if (own[syx] == own[yx])
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continue; /* we already got one */
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if (!addremcommon(w, h, x, y, own, own[syx]))
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continue; /* would non-simply connect the omino */
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addable[yx*4+dir] = own[syx];
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}
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}
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}
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for (i = j = 0; i < n; i++)
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if (sizes[i] < k)
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tmp[j++] = i;
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if (j == 0)
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break; /* all ominoes are complete! */
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j = tmp[random_upto(rs, j)];
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/*
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* So we're trying to expand omino j. We breadth-first
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* search out from j across the space of ominoes.
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*
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* For bfs purposes, we use two elements of tmp per omino:
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* tmp[2*i+0] tells us which omino we got to i from, and
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* tmp[2*i+1] numbers the grid square that omino stole
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* from us.
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*
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* This requires that wh (the size of tmp) is at least 2n,
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* i.e. k is at least 2. There would have been nothing to
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* stop a user calling this function with k=1, but if they
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* did then we wouldn't have got to _here_ in the code -
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* we would have noticed above that all ominoes were
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* already at their target sizes, and terminated :-)
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*/
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assert(wh >= 2*n);
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for (i = 0; i < n; i++)
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tmp[2*i] = tmp[2*i+1] = -1;
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qhead = qtail = 0;
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queue[qtail++] = j;
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tmp[2*j] = tmp[2*j+1] = -2; /* special value: `starting point' */
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while (qhead < qtail) {
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int tmpsq;
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j = queue[qhead];
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/*
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* We wish to expand omino j. However, we might have
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* got here by omino j having a square stolen from it,
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* so first of all we must temporarily mark that
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* square as not belonging to j, so that our adjacency
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* calculations don't assume j _does_ belong to us.
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*/
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tmpsq = tmp[2*j+1];
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if (tmpsq >= 0) {
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assert(own[tmpsq] == j);
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own[tmpsq] = -1;
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}
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/*
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* OK. Now begin by seeing if we can find any
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* unclaimed square into which we can expand omino j.
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* If we find one, the entire bfs terminates.
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*/
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for (i = 0; i < wh; i++) {
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int dir;
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if (own[order[i]] >= 0)
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continue; /* this square is claimed */
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for (dir = 0; dir < 4; dir++)
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if (addable[order[i]*4+dir] == j) {
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/*
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* We know this square is addable to this
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* omino with the grid in the state it had
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* at the top of the loop. However, we
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* must now check that it's _still_
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* addable to this omino when the omino is
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* missing a square. To do this it's only
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* necessary to re-check addremcommon.
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~|~ */
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if (!addremcommon(w, h, order[i]%w, order[i]/w,
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own, j))
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continue;
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break;
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}
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if (dir == 4)
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continue; /* we can't add this square to j */
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break; /* got one! */
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}
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if (i < wh) {
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i = order[i];
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/*
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* We are done. We can add square i to omino j,
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* and then backtrack along the trail in tmp
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* moving squares between ominoes, ending up
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* expanding our starting omino by one.
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*/
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while (1) {
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own[i] = j;
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if (tmp[2*j] == -2)
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break;
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i = tmp[2*j+1];
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j = tmp[2*j];
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}
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/*
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* Increment the size of the starting omino.
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*/
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sizes[j]++;
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/*
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* Terminate the bfs loop.
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*/
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break;
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}
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/*
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* If we get here, we haven't been able to expand
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* omino j into an unclaimed square. So now we begin
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* to investigate expanding it into squares which are
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* claimed by ominoes the bfs has not yet visited.
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*/
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for (i = 0; i < wh; i++) {
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int dir, nj;
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nj = own[order[i]];
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if (nj < 0 || tmp[2*nj] != -1)
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continue; /* unclaimed, or owned by wrong omino */
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if (!removable[order[i]])
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continue; /* its omino won't let it go */
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for (dir = 0; dir < 4; dir++)
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if (addable[order[i]*4+dir] == j) {
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/*
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* As above, re-check addremcommon.
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*/
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if (!addremcommon(w, h, order[i]%w, order[i]/w,
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own, j))
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continue;
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/*
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* We have found a square we can use to
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* expand omino j, at the expense of the
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* as-yet unvisited omino nj. So add this
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* to the bfs queue.
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*/
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assert(qtail < n);
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queue[qtail++] = nj;
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tmp[2*nj] = j;
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tmp[2*nj+1] = order[i];
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/*
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* Now terminate the loop over dir, to
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* ensure we don't accidentally add the
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* same omino twice to the queue.
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*/
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break;
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}
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}
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/*
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* Restore the temporarily removed square.
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*/
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if (tmpsq >= 0)
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own[tmpsq] = j;
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/*
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* Advance the queue head.
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*/
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qhead++;
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}
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if (qhead == qtail) {
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/*
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* We have finished the bfs and not found any way to
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* expand omino j. Panic, and return failure.
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*
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* FIXME: or should we loop over all ominoes before we
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* give up?
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*/
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retdsf = NULL;
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goto cleanup;
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}
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}
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/*
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* Construct the output dsf.
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*/
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for (i = 0; i < wh; i++) {
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assert(own[i] >= 0 && own[i] < n);
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tmp[own[i]] = i;
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}
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retdsf = snew_dsf(wh);
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for (i = 0; i < wh; i++) {
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dsf_merge(retdsf, i, tmp[own[i]]);
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}
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/*
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* Construct the output dsf a different way, to verify that
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* the ominoes really are k-ominoes and we haven't
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* accidentally split one into two disconnected pieces.
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*/
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dsf_init(tmp, wh);
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for (y = 0; y < h; y++)
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for (x = 0; x+1 < w; x++)
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if (own[y*w+x] == own[y*w+(x+1)])
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dsf_merge(tmp, y*w+x, y*w+(x+1));
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for (x = 0; x < w; x++)
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for (y = 0; y+1 < h; y++)
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if (own[y*w+x] == own[(y+1)*w+x])
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dsf_merge(tmp, y*w+x, (y+1)*w+x);
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for (i = 0; i < wh; i++) {
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j = dsf_canonify(retdsf, i);
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assert(dsf_canonify(tmp, j) == dsf_canonify(tmp, i));
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}
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cleanup:
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/*
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* Free our temporary working space.
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*/
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sfree(order);
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sfree(tmp);
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sfree(own);
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sfree(sizes);
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sfree(queue);
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sfree(addable);
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sfree(removable);
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/*
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* And we're done.
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*/
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return retdsf;
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}
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#ifdef TESTMODE
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/*
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* gcc -g -O0 -DTESTMODE -I.. -o divvy divvy.c ../random.c ../malloc.c ../dsf.c ../misc.c ../nullfe.c
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*
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* or to debug
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*
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||||
* gcc -g -O0 -DDIVVY_DIAGNOSTICS -DTESTMODE -I.. -o divvy divvy.c ../random.c ../malloc.c ../dsf.c ../misc.c ../nullfe.c
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*/
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int main(int argc, char **argv)
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{
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int *dsf;
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int i, successes;
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int w = 9, h = 4, k = 6, tries = 100;
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random_state *rs;
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rs = random_new("123456", 6);
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if (argc > 1)
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w = atoi(argv[1]);
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if (argc > 2)
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h = atoi(argv[2]);
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||||
if (argc > 3)
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k = atoi(argv[3]);
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||||
if (argc > 4)
|
||||
tries = atoi(argv[4]);
|
||||
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||||
successes = 0;
|
||||
for (i = 0; i < tries; i++) {
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||||
dsf = divvy_rectangle(w, h, k, rs);
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||||
if (dsf) {
|
||||
int x, y;
|
||||
|
||||
successes++;
|
||||
|
||||
for (y = 0; y <= 2*h; y++) {
|
||||
for (x = 0; x <= 2*w; x++) {
|
||||
int miny = y/2 - 1, maxy = y/2;
|
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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 successes out of %d tries\n", successes, tries);
|
||||
|
||||
return 0;
|
||||
}
|
||||
|
||||
#endif
|
||||
Reference in New Issue
Block a user