ref: f01b1674bd862e1792b5216bd5d832cd52a170ae
dir: /laydomino.c/
/* * laydomino.c: code for performing a domino (2x1 tile) layout of * a given area of code. */ #include <stdio.h> #include <stdlib.h> #include <assert.h> #include "puzzles.h" /* * This function returns an array size w x h representing a grid: * each grid[i] = j, where j is the other end of a 2x1 domino. * If w*h is odd, one square will remain referring to itself. */ int *domino_layout(int w, int h, random_state *rs) { int *grid, *grid2, *list; int wh = w*h; /* * Allocate space in which to lay the grid out. */ grid = snewn(wh, int); grid2 = snewn(wh, int); list = snewn(2*wh, int); domino_layout_prealloc(w, h, rs, grid, grid2, list); sfree(grid2); sfree(list); return grid; } /* * As for domino_layout, but with preallocated buffers. * grid and grid2 should be size w*h, and list size 2*w*h. */ void domino_layout_prealloc(int w, int h, random_state *rs, int *grid, int *grid2, int *list) { int i, j, k, m, wh = w*h, todo, done; /* * To begin with, set grid[i] = i for all i to indicate * that all squares are currently singletons. Later we'll * set grid[i] to be the index of the other end of the * domino on i. */ for (i = 0; i < wh; i++) grid[i] = i; /* * Now prepare a list of the possible domino locations. There * are w*(h-1) possible vertical locations, and (w-1)*h * horizontal ones, for a total of 2*wh - h - w. * * I'm going to denote the vertical domino placement with * its top in square i as 2*i, and the horizontal one with * its left half in square i as 2*i+1. */ k = 0; for (j = 0; j < h-1; j++) for (i = 0; i < w; i++) list[k++] = 2 * (j*w+i); /* vertical positions */ for (j = 0; j < h; j++) for (i = 0; i < w-1; i++) list[k++] = 2 * (j*w+i) + 1; /* horizontal positions */ assert(k == 2*wh - h - w); /* * Shuffle the list. */ shuffle(list, k, sizeof(*list), rs); /* * Work down the shuffled list, placing a domino everywhere * we can. */ for (i = 0; i < k; i++) { int horiz, xy, xy2; horiz = list[i] % 2; xy = list[i] / 2; xy2 = xy + (horiz ? 1 : w); if (grid[xy] == xy && grid[xy2] == xy2) { /* * We can place this domino. Do so. */ grid[xy] = xy2; grid[xy2] = xy; } } #ifdef GENERATION_DIAGNOSTICS printf("generated initial layout\n"); #endif /* * Now we've placed as many dominoes as we can immediately * manage. There will be squares remaining, but they'll be * singletons. So loop round and deal with the singletons * two by two. */ while (1) { #ifdef GENERATION_DIAGNOSTICS for (j = 0; j < h; j++) { for (i = 0; i < w; i++) { int xy = j*w+i; int v = grid[xy]; int c = (v == xy+1 ? '[' : v == xy-1 ? ']' : v == xy+w ? 'n' : v == xy-w ? 'U' : '.'); putchar(c); } putchar('\n'); } putchar('\n'); #endif /* * Our strategy is: * * First find a singleton square. * * Then breadth-first search out from the starting * square. From that square (and any others we reach on * the way), examine all four neighbours of the square. * If one is an end of a domino, we move to the _other_ * end of that domino before looking at neighbours * again. When we encounter another singleton on this * search, stop. * * This will give us a path of adjacent squares such * that all but the two ends are covered in dominoes. * So we can now shuffle every domino on the path up by * one. * * (Chessboard colours are mathematically important * here: we always end up pairing each singleton with a * singleton of the other colour. However, we never * have to track this manually, since it's * automatically taken care of by the fact that we * always make an even number of orthogonal moves.) */ k = 0; for (j = 0; j < wh; j++) { if (grid[j] == j) { k++; i = j; /* start BFS here. */ } } if (k == (wh % 2)) break; /* if area is even, we have no more singletons; if area is odd, we have one singleton. either way, we're done. */ #ifdef GENERATION_DIAGNOSTICS printf("starting b.f.s. at singleton %d\n", i); #endif /* * Set grid2 to -1 everywhere. It will hold our * distance-from-start values, and also our * backtracking data, during the b.f.s. */ for (j = 0; j < wh; j++) grid2[j] = -1; grid2[i] = 0; /* starting square has distance zero */ /* * Start our to-do list of squares. It'll live in * `list'; since the b.f.s can cover every square at * most once there is no need for it to be circular. * We'll just have two counters tracking the end of the * list and the squares we've already dealt with. */ done = 0; todo = 1; list[0] = i; /* * Now begin the b.f.s. loop. */ while (done < todo) { int d[4], nd, x, y; i = list[done++]; #ifdef GENERATION_DIAGNOSTICS printf("b.f.s. iteration from %d\n", i); #endif x = i % w; y = i / w; nd = 0; if (x > 0) d[nd++] = i - 1; if (x+1 < w) d[nd++] = i + 1; if (y > 0) d[nd++] = i - w; if (y+1 < h) d[nd++] = i + w; /* * To avoid directional bias, process the * neighbours of this square in a random order. */ shuffle(d, nd, sizeof(*d), rs); for (j = 0; j < nd; j++) { k = d[j]; if (grid[k] == k) { #ifdef GENERATION_DIAGNOSTICS printf("found neighbouring singleton %d\n", k); #endif grid2[k] = i; break; /* found a target singleton! */ } /* * We're moving through a domino here, so we * have two entries in grid2 to fill with * useful data. In grid[k] - the square * adjacent to where we came from - I'm going * to put the address _of_ the square we came * from. In the other end of the domino - the * square from which we will continue the * search - I'm going to put the distance. */ m = grid[k]; if (grid2[m] < 0 || grid2[m] > grid2[i]+1) { #ifdef GENERATION_DIAGNOSTICS printf("found neighbouring domino %d/%d\n", k, m); #endif grid2[m] = grid2[i]+1; grid2[k] = i; /* * And since we've now visited a new * domino, add m to the to-do list. */ assert(todo < wh); list[todo++] = m; } } if (j < nd) { i = k; #ifdef GENERATION_DIAGNOSTICS printf("terminating b.f.s. loop, i = %d\n", i); #endif break; } i = -1; /* just in case the loop terminates */ } /* * We expect this b.f.s. to have found us a target * square. */ assert(i >= 0); /* * Now we can follow the trail back to our starting * singleton, re-laying dominoes as we go. */ while (1) { j = grid2[i]; assert(j >= 0 && j < wh); k = grid[j]; grid[i] = j; grid[j] = i; #ifdef GENERATION_DIAGNOSTICS printf("filling in domino %d/%d (next %d)\n", i, j, k); #endif if (j == k) break; /* we've reached the other singleton */ i = k; } #ifdef GENERATION_DIAGNOSTICS printf("fixup path completed\n"); #endif } } /* vim: set shiftwidth=4 :set textwidth=80: */