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706e27de8d
randomly generates a tiling of a rectangle with dominoes, since he wants to reuse that function in another puzzle. [originally from svn r8488]
292 lines
8.5 KiB
C
292 lines
8.5 KiB
C
/*
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* laydomino.c: code for performing a domino (2x1 tile) layout of
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* a given area of code.
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*/
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#include <stdio.h>
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#include <stdlib.h>
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#include <assert.h>
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#include "puzzles.h"
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/*
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* This function returns an array size w x h representing a grid:
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* each grid[i] = j, where j is the other end of a 2x1 domino.
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* If w*h is odd, one square will remain referring to itself.
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*/
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int *domino_layout(int w, int h, random_state *rs)
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{
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int *grid, *grid2, *list;
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int wh = w*h;
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/*
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* Allocate space in which to lay the grid out.
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*/
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grid = snewn(wh, int);
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grid2 = snewn(wh, int);
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list = snewn(2*wh, int);
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domino_layout_prealloc(w, h, rs, grid, grid2, list);
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sfree(grid2);
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sfree(list);
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return grid;
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}
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/*
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* As for domino_layout, but with preallocated buffers.
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* grid and grid2 should be size w*h, and list size 2*w*h.
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*/
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void domino_layout_prealloc(int w, int h, random_state *rs,
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int *grid, int *grid2, int *list)
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{
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int i, j, k, m, wh = w*h, todo, done;
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/*
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* To begin with, set grid[i] = i for all i to indicate
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* that all squares are currently singletons. Later we'll
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* set grid[i] to be the index of the other end of the
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* domino on i.
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*/
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for (i = 0; i < wh; i++)
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grid[i] = i;
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/*
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* Now prepare a list of the possible domino locations. There
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* are w*(h-1) possible vertical locations, and (w-1)*h
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* horizontal ones, for a total of 2*wh - h - w.
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*
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* I'm going to denote the vertical domino placement with
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* its top in square i as 2*i, and the horizontal one with
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* its left half in square i as 2*i+1.
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*/
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k = 0;
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for (j = 0; j < h-1; j++)
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for (i = 0; i < w; i++)
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list[k++] = 2 * (j*w+i); /* vertical positions */
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for (j = 0; j < h; j++)
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for (i = 0; i < w-1; i++)
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list[k++] = 2 * (j*w+i) + 1; /* horizontal positions */
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assert(k == 2*wh - h - w);
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/*
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* Shuffle the list.
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*/
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shuffle(list, k, sizeof(*list), rs);
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/*
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* Work down the shuffled list, placing a domino everywhere
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* we can.
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*/
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for (i = 0; i < k; i++) {
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int horiz, xy, xy2;
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horiz = list[i] % 2;
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xy = list[i] / 2;
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xy2 = xy + (horiz ? 1 : w);
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if (grid[xy] == xy && grid[xy2] == xy2) {
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/*
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* We can place this domino. Do so.
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*/
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grid[xy] = xy2;
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grid[xy2] = xy;
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}
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}
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#ifdef GENERATION_DIAGNOSTICS
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printf("generated initial layout\n");
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#endif
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/*
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* Now we've placed as many dominoes as we can immediately
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* manage. There will be squares remaining, but they'll be
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* singletons. So loop round and deal with the singletons
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* two by two.
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*/
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while (1) {
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#ifdef GENERATION_DIAGNOSTICS
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for (j = 0; j < h; j++) {
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for (i = 0; i < w; i++) {
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int xy = j*w+i;
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int v = grid[xy];
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int c = (v == xy+1 ? '[' : v == xy-1 ? ']' :
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v == xy+w ? 'n' : v == xy-w ? 'U' : '.');
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putchar(c);
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}
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putchar('\n');
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}
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putchar('\n');
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#endif
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/*
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* Our strategy is:
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*
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* First find a singleton square.
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*
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* Then breadth-first search out from the starting
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* square. From that square (and any others we reach on
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* the way), examine all four neighbours of the square.
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* If one is an end of a domino, we move to the _other_
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* end of that domino before looking at neighbours
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* again. When we encounter another singleton on this
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* search, stop.
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*
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* This will give us a path of adjacent squares such
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* that all but the two ends are covered in dominoes.
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* So we can now shuffle every domino on the path up by
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* one.
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*
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* (Chessboard colours are mathematically important
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* here: we always end up pairing each singleton with a
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* singleton of the other colour. However, we never
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* have to track this manually, since it's
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* automatically taken care of by the fact that we
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* always make an even number of orthogonal moves.)
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*/
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k = 0;
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for (j = 0; j < wh; j++) {
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if (grid[j] == j) {
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k++;
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i = j; /* start BFS here. */
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}
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}
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if (k == (wh % 2))
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break; /* if area is even, we have no more singletons;
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if area is odd, we have one singleton.
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either way, we're done. */
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#ifdef GENERATION_DIAGNOSTICS
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printf("starting b.f.s. at singleton %d\n", i);
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#endif
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/*
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* Set grid2 to -1 everywhere. It will hold our
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* distance-from-start values, and also our
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* backtracking data, during the b.f.s.
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*/
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for (j = 0; j < wh; j++)
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grid2[j] = -1;
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grid2[i] = 0; /* starting square has distance zero */
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/*
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* Start our to-do list of squares. It'll live in
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* `list'; since the b.f.s can cover every square at
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* most once there is no need for it to be circular.
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* We'll just have two counters tracking the end of the
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* list and the squares we've already dealt with.
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*/
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done = 0;
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todo = 1;
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list[0] = i;
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/*
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* Now begin the b.f.s. loop.
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*/
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while (done < todo) {
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int d[4], nd, x, y;
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i = list[done++];
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#ifdef GENERATION_DIAGNOSTICS
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printf("b.f.s. iteration from %d\n", i);
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#endif
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x = i % w;
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y = i / w;
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nd = 0;
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if (x > 0)
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d[nd++] = i - 1;
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if (x+1 < w)
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d[nd++] = i + 1;
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if (y > 0)
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d[nd++] = i - w;
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if (y+1 < h)
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d[nd++] = i + w;
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/*
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* To avoid directional bias, process the
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* neighbours of this square in a random order.
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*/
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shuffle(d, nd, sizeof(*d), rs);
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for (j = 0; j < nd; j++) {
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k = d[j];
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if (grid[k] == k) {
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#ifdef GENERATION_DIAGNOSTICS
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printf("found neighbouring singleton %d\n", k);
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#endif
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grid2[k] = i;
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break; /* found a target singleton! */
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}
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/*
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* We're moving through a domino here, so we
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* have two entries in grid2 to fill with
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* useful data. In grid[k] - the square
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* adjacent to where we came from - I'm going
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* to put the address _of_ the square we came
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* from. In the other end of the domino - the
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* square from which we will continue the
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* search - I'm going to put the distance.
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*/
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m = grid[k];
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if (grid2[m] < 0 || grid2[m] > grid2[i]+1) {
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#ifdef GENERATION_DIAGNOSTICS
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printf("found neighbouring domino %d/%d\n", k, m);
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#endif
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grid2[m] = grid2[i]+1;
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grid2[k] = i;
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/*
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* And since we've now visited a new
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* domino, add m to the to-do list.
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*/
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assert(todo < wh);
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list[todo++] = m;
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}
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}
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if (j < nd) {
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i = k;
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#ifdef GENERATION_DIAGNOSTICS
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printf("terminating b.f.s. loop, i = %d\n", i);
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#endif
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break;
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}
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i = -1; /* just in case the loop terminates */
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}
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/*
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* We expect this b.f.s. to have found us a target
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* square.
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*/
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assert(i >= 0);
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/*
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* Now we can follow the trail back to our starting
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* singleton, re-laying dominoes as we go.
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*/
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while (1) {
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j = grid2[i];
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assert(j >= 0 && j < wh);
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k = grid[j];
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grid[i] = j;
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grid[j] = i;
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#ifdef GENERATION_DIAGNOSTICS
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printf("filling in domino %d/%d (next %d)\n", i, j, k);
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#endif
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if (j == k)
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break; /* we've reached the other singleton */
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i = k;
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}
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#ifdef GENERATION_DIAGNOSTICS
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printf("fixup path completed\n");
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#endif
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}
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}
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/* vim: set shiftwidth=4 :set textwidth=80: */
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