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puzzles/unfinished/path.c
Simon Tatham 72b28b5e71 Fix bit rot in the 'unfinished' subdir. 
Several of the source files here won't quite compile any more, because
of minor things like const-correctness and the UI_UPDATE change. Now
they should all build again (without prejudice to how useful they are
once they have built).

The biggest change was to remove the fatal() implementation from the
standalone path.c, because my new plan is that basically everything
that's not linked against a true puzzle frontend will be linked
against nullfe.c, which provides that function anyway.
2021-03-29 18:22:20 +01:00

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/*
* Experimental grid generator for Nikoli's `Number Link' puzzle.
*/
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <assert.h>
#include "puzzles.h"
/*
* 2005-07-08: This is currently a Path grid generator which will
* construct valid grids at a plausible speed. However, the grids
* are not of suitable quality to be used directly as puzzles.
*
* The basic strategy is to start with an empty grid, and
* repeatedly either (a) add a new path to it, or (b) extend one
* end of a path by one square in some direction and push other
* paths into new shapes in the process. The effect of this is that
* we are able to construct a set of paths which between them fill
* the entire grid.
*
* Quality issues: if we set the main loop to do (a) where possible
* and (b) only where necessary, we end up with a grid containing a
* few too many small paths, which therefore doesn't make for an
* interesting puzzle. If we reverse the priority so that we do (b)
* where possible and (a) only where necessary, we end up with some
* staggeringly interwoven grids with very very few separate paths,
* but the result of this is that there's invariably a solution
* other than the intended one which leaves many grid squares
* unfilled. There's also a separate problem which is that many
* grids have really boring and obvious paths in them, such as the
* entire bottom row of the grid being taken up by a single path.
*
* It's not impossible that a few tweaks might eliminate or reduce
* the incidence of boring paths, and might also find a happy
* medium between too many and too few. There remains the question
* of unique solutions, however. I fear there is no alternative but
* to write - somehow! - a solver.
*
* While I'm here, some notes on UI strategy for the parts of the
* puzzle implementation that _aren't_ the generator:
*
* - data model is to track connections between adjacent squares,
* so that you aren't limited to extending a path out from each
* number but can also mark sections of path which you know
* _will_ come in handy later.
*
* - user interface is to click in one square and drag to an
* adjacent one, thus creating a link between them. We can
* probably tolerate rapid mouse motion causing a drag directly
* to a square which is a rook move away, but any other rapid
* motion is ambiguous and probably the best option is to wait
* until the mouse returns to a square we know how to reach.
*
* - a drag causing the current path to backtrack has the effect
* of removing bits of it.
*
* - the UI should enforce at all times the constraint that at
* most two links can come into any square.
*
* - my Cunning Plan for actually implementing this: the game_ui
* contains a grid-sized array, which is copied from the current
* game_state on starting a drag. While a drag is active, the
* contents of the game_ui is adjusted with every mouse motion,
* and is displayed _in place_ of the game_state itself. On
* termination of a drag, the game_ui array is copied back into
* the new game_state (or rather, a string move is encoded which
* has precisely the set of link changes to cause that effect).
*/
/*
* 2020-05-11: some thoughts on a solver.
*
* Consider this example puzzle, from Wikipedia:
*
* ---4---
* -3--25-
* ---31--
* ---5---
* -------
* --1----
* 2---4--
*
* The kind of deduction that a human wants to make here is: which way
* does the path between the 4s go? In particular, does it go round
* the left of the W-shaped cluster of endpoints, or round the right
* of it? It's clear at a glance that it must go to the right, because
* _any_ path between the 4s that goes to the left of that cluster, no
* matter what detailed direction it takes, will disconnect the
* remaining grid squares into two components, with the two 2s not in
* the same component. So we immediately know that the path between
* the 4s _must_ go round the right-hand side of the grid.
*
* How do you model that global and topological reasoning in a
* computer?
*
* The most plausible idea I've seen so far is to use fundamental
* groups. The fundamental group of loops based at a given point in a
* space is a free group, under loop concatenation and up to homotopy,
* generated by the loops that go in each direction around each hole
* in the space. In this case, the 'holes' are clues, or connected
* groups of clues.
*
* So you might be able to enumerate all the homotopy classes of paths
* between (say) the two 4s as follows. Start with any old path
* between them (say, find the first one that breadth-first search
* will give you). Choose one of the 4s to regard as the base point
* (arbitrarily). Then breadth-first search among the space of _paths_
* by the following procedure. Given a candidate path, append to it
* each of the possible loops that starts from the base point,
* circumnavigates one clue cluster, and returns to the base point.
* The result will typically be a path that retraces its steps and
* self-intersects. Now adjust it homotopically so that it doesn't. If
* that can't be done, then we haven't generated a fresh candidate
* path; if it can, then we've got a new path that is not homotopic to
* any path we already had, so add it to our list and queue it up to
* become the starting point of this search later.
*
* The idea is that this should exhaustively enumerate, up to
* homotopy, the different ways in which the two 4s can connect to
* each other within the constraint that you have to actually fit the
* path non-self-intersectingly into this grid. Then you can keep a
* list of those homotopy classes in mind, and start ruling them out
* by techniques like the connectivity approach described above.
* Hopefully you end up narrowing down to few enough homotopy classes
* that you can deduce something concrete about actual squares of the
* grid - for example, here, that if the path between 4s has to go
* round the right, then we know some specific squares it must go
* through, so we can fill those in. And then, having filled in a
* piece of the middle of a path, you can now regard connecting the
* ultimate endpoints to that mid-section as two separate subproblems,
* so you've reduced to a simpler instance of the same puzzle.
*
* But I don't know whether all of this actually works. I more or less
* believe the process for enumerating elements of the free group; but
* I'm not as confident that when you find a group element that won't
* fit in the grid, you'll never have to consider its descendants in
* the BFS either. And I'm assuming that 'unwind the self-intersection
* homotopically' is a thing that can actually be turned into a
* sensible algorithm.
*
* --------
*
* Another thing that might be needed is to characterise _which_
* homotopy class a given path is in.
*
* For this I think it's sufficient to choose a collection of paths
* along the _edges_ of the square grid, each of which connects two of
* the holes in the grid (including the grid exterior, which counts as
* a huge hole), such that they form a spanning tree between the
* holes. Then assign each of those paths an orientation, so that
* crossing it in one direction counts as 'positive' and the other
* 'negative'. Now analyse a candidate path from one square to another
* by following it and noting down which of those paths it crosses in
* which direction, then simplifying the result like a free group word
* (i.e. adjacent + and - crossings of the same path cancel out).
*
* --------
*
* If we choose those paths to be of minimal length, then we can get
* an upper bound on the number of homotopy classes by observing that
* you can't traverse any of those barriers more times than will fit
* non-self-intersectingly in the grid. That might be an alternative
* method of bounding the search through the fundamental group to only
* finitely many possibilities.
*/
/*
* Standard notation for directions.
*/
#define L 0
#define U 1
#define R 2
#define D 3
#define DX(dir) ( (dir)==L ? -1 : (dir)==R ? +1 : 0)
#define DY(dir) ( (dir)==U ? -1 : (dir)==D ? +1 : 0)
/*
* Perform a breadth-first search over a grid of squares with the
* colour of square (X,Y) given by grid[Y*w+X]. The search begins
* at (x,y), and finds all squares which are the same colour as
* (x,y) and reachable from it by orthogonal moves. On return:
* - dist[Y*w+X] gives the distance of (X,Y) from (x,y), or -1 if
* unreachable or a different colour
* - the returned value is the number of reachable squares,
* including (x,y) itself
* - list[0] up to list[returned value - 1] list those squares, in
* increasing order of distance from (x,y) (and in arbitrary
* order within that).
*/
static int bfs(int w, int h, int *grid, int x, int y, int *dist, int *list)
{
int i, j, c, listsize, listdone;
/*
* Start by clearing the output arrays.
*/
for (i = 0; i < w*h; i++)
dist[i] = list[i] = -1;
/*
* Set up the initial list.
*/
listsize = 1;
listdone = 0;
list[0] = y*w+x;
dist[y*w+x] = 0;
c = grid[y*w+x];
/*
* Repeatedly process a square and add any extra squares to the
* end of list.
*/
while (listdone < listsize) {
i = list[listdone++];
y = i / w;
x = i % w;
for (j = 0; j < 4; j++) {
int xx, yy, ii;
xx = x + DX(j);
yy = y + DY(j);
ii = yy*w+xx;
if (xx >= 0 && xx < w && yy >= 0 && yy < h &&
grid[ii] == c && dist[ii] == -1) {
dist[ii] = dist[i] + 1;
assert(listsize < w*h);
list[listsize++] = ii;
}
}
}
return listsize;
}
struct genctx {
int w, h;
int *grid, *sparegrid, *sparegrid2, *sparegrid3;
int *dist, *list;
int npaths, pathsize;
int *pathends, *sparepathends; /* 2*npaths entries */
int *pathspare; /* npaths entries */
int *extends; /* 8*npaths entries */
};
static struct genctx *new_genctx(int w, int h)
{
struct genctx *ctx = snew(struct genctx);
ctx->w = w;
ctx->h = h;
ctx->grid = snewn(w * h, int);
ctx->sparegrid = snewn(w * h, int);
ctx->sparegrid2 = snewn(w * h, int);
ctx->sparegrid3 = snewn(w * h, int);
ctx->dist = snewn(w * h, int);
ctx->list = snewn(w * h, int);
ctx->npaths = ctx->pathsize = 0;
ctx->pathends = ctx->sparepathends = ctx->pathspare = ctx->extends = NULL;
return ctx;
}
static void free_genctx(struct genctx *ctx)
{
sfree(ctx->grid);
sfree(ctx->sparegrid);
sfree(ctx->sparegrid2);
sfree(ctx->sparegrid3);
sfree(ctx->dist);
sfree(ctx->list);
sfree(ctx->pathends);
sfree(ctx->sparepathends);
sfree(ctx->pathspare);
sfree(ctx->extends);
}
static int newpath(struct genctx *ctx)
{
int n;
n = ctx->npaths++;
if (ctx->npaths > ctx->pathsize) {
ctx->pathsize += 16;
ctx->pathends = sresize(ctx->pathends, ctx->pathsize*2, int);
ctx->sparepathends = sresize(ctx->sparepathends, ctx->pathsize*2, int);
ctx->pathspare = sresize(ctx->pathspare, ctx->pathsize, int);
ctx->extends = sresize(ctx->extends, ctx->pathsize*8, int);
}
return n;
}
static int is_endpoint(struct genctx *ctx, int x, int y)
{
int w = ctx->w, h = ctx->h, c;
assert(x >= 0 && x < w && y >= 0 && y < h);
c = ctx->grid[y*w+x];
if (c < 0)
return false; /* empty square is not an endpoint! */
assert(c >= 0 && c < ctx->npaths);
if (ctx->pathends[c*2] == y*w+x || ctx->pathends[c*2+1] == y*w+x)
return true;
return false;
}
/*
* Tries to extend a path by one square in the given direction,
* pushing other paths around if necessary. Returns true on success
* or false on failure.
*/
static int extend_path(struct genctx *ctx, int path, int end, int direction)
{
int w = ctx->w, h = ctx->h;
int x, y, xe, ye, cut;
int i, j, jp, n, first, last;
assert(path >= 0 && path < ctx->npaths);
assert(end == 0 || end == 1);
/*
* Find the endpoint of the path and the point we plan to
* extend it into.
*/
y = ctx->pathends[path * 2 + end] / w;
x = ctx->pathends[path * 2 + end] % w;
assert(x >= 0 && x < w && y >= 0 && y < h);
xe = x + DX(direction);
ye = y + DY(direction);
if (xe < 0 || xe >= w || ye < 0 || ye >= h)
return false; /* could not extend in this direction */
/*
* We don't extend paths _directly_ into endpoints of other
* paths, although we don't mind too much if a knock-on effect
* of an extension is to push part of another path into a third
* path's endpoint.
*/
if (is_endpoint(ctx, xe, ye))
return false;
/*
* We can't extend a path back the way it came.
*/
if (ctx->grid[ye*w+xe] == path)
return false;
/*
* Paths may not double back on themselves. Check if the new
* point is adjacent to any point of this path other than (x,y).
*/
for (j = 0; j < 4; j++) {
int xf, yf;
xf = xe + DX(j);
yf = ye + DY(j);
if (xf >= 0 && xf < w && yf >= 0 && yf < h &&
(xf != x || yf != y) && ctx->grid[yf*w+xf] == path)
return false;
}
/*
* Now we're convinced it's valid to _attempt_ the extension.
* It may still fail if we run out of space to push other paths
* into.
*
* So now we can set up our temporary data structures. We will
* need:
*
* - a spare copy of the grid on which to gradually move paths
* around (sparegrid)
*
* - a second spare copy with which to remember how paths
* looked just before being cut (sparegrid2). FIXME: is
* sparegrid2 necessary? right now it's never different from
* grid itself
*
* - a third spare copy with which to do the internal
* calculations involved in reconstituting a cut path
* (sparegrid3)
*
* - something to track which paths currently need
* reconstituting after being cut, and which have already
* been cut (pathspare)
*
* - a spare copy of pathends to store the altered states in
* (sparepathends)
*/
memcpy(ctx->sparegrid, ctx->grid, w*h*sizeof(int));
memcpy(ctx->sparegrid2, ctx->grid, w*h*sizeof(int));
memcpy(ctx->sparepathends, ctx->pathends, ctx->npaths*2*sizeof(int));
for (i = 0; i < ctx->npaths; i++)
ctx->pathspare[i] = 0; /* 0=untouched, 1=broken, 2=fixed */
/*
* Working in sparegrid, actually extend the path. If it cuts
* another, begin a loop in which we restore any cut path by
* moving it out of the way.
*/
cut = ctx->sparegrid[ye*w+xe];
ctx->sparegrid[ye*w+xe] = path;
ctx->sparepathends[path*2+end] = ye*w+xe;
ctx->pathspare[path] = 2; /* this one is sacrosanct */
if (cut >= 0) {
assert(cut >= 0 && cut < ctx->npaths);
ctx->pathspare[cut] = 1; /* broken */
while (1) {
for (i = 0; i < ctx->npaths; i++)
if (ctx->pathspare[i] == 1)
break;
if (i == ctx->npaths)
break; /* we're done */
/*
* Path i needs restoring. So walk along its original
* track (as given in sparegrid2) and see where it's
* been cut. Where it has, surround the cut points in
* the same colour, without overwriting already-fixed
* paths.
*/
memcpy(ctx->sparegrid3, ctx->sparegrid, w*h*sizeof(int));
n = bfs(w, h, ctx->sparegrid2,
ctx->pathends[i*2] % w, ctx->pathends[i*2] / w,
ctx->dist, ctx->list);
first = last = -1;
if (ctx->sparegrid3[ctx->pathends[i*2]] != i ||
ctx->sparegrid3[ctx->pathends[i*2+1]] != i) return false;/* FIXME */
for (j = 0; j < n; j++) {
jp = ctx->list[j];
assert(ctx->dist[jp] == j);
assert(ctx->sparegrid2[jp] == i);
/*
* Wipe out the original path in sparegrid.
*/
if (ctx->sparegrid[jp] == i)
ctx->sparegrid[jp] = -1;
/*
* Be prepared to shorten the path at either end if
* the endpoints have been stomped on.
*/
if (ctx->sparegrid3[jp] == i) {
if (first < 0)
first = jp;
last = jp;
}
if (ctx->sparegrid3[jp] != i) {
int jx = jp % w, jy = jp / w;
int dx, dy;
for (dy = -1; dy <= +1; dy++)
for (dx = -1; dx <= +1; dx++) {
int newp, newv;
if (!dy && !dx)
continue; /* central square */
if (jx+dx < 0 || jx+dx >= w ||
jy+dy < 0 || jy+dy >= h)
continue; /* out of range */
newp = (jy+dy)*w+(jx+dx);
newv = ctx->sparegrid3[newp];
if (newv >= 0 && (newv == i ||
ctx->pathspare[newv] == 2))
continue; /* can't use this square */
ctx->sparegrid3[newp] = i;
}
}
}
if (first < 0 || last < 0)
return false; /* path is completely wiped out! */
/*
* Now we've covered sparegrid3 in possible squares for
* the new layout of path i. Find the actual layout
* we're going to use by bfs: we want the shortest path
* from one endpoint to the other.
*/
n = bfs(w, h, ctx->sparegrid3, first % w, first / w,
ctx->dist, ctx->list);
if (ctx->dist[last] < 2) {
/*
* Either there is no way to get between the path's
* endpoints, or the remaining endpoints simply
* aren't far enough apart to make the path viable
* any more. This means the entire push operation
* has failed.
*/
return false;
}
/*
* Write the new path into sparegrid. Also save the new
* endpoint locations, in case they've changed.
*/
jp = last;
j = ctx->dist[jp];
while (1) {
int d;
if (ctx->sparegrid[jp] >= 0) {
if (ctx->pathspare[ctx->sparegrid[jp]] == 2)
return false; /* somehow we've hit a fixed path */
ctx->pathspare[ctx->sparegrid[jp]] = 1; /* broken */
}
ctx->sparegrid[jp] = i;
if (j == 0)
break;
/*
* Now look at the neighbours of jp to find one
* which has dist[] one less.
*/
for (d = 0; d < 4; d++) {
int jx = (jp % w) + DX(d), jy = (jp / w) + DY(d);
if (jx >= 0 && jx < w && jy >= 0 && jy < w &&
ctx->dist[jy*w+jx] == j-1) {
jp = jy*w+jx;
j--;
break;
}
}
assert(d < 4);
}
ctx->sparepathends[i*2] = first;
ctx->sparepathends[i*2+1] = last;
/* printf("new ends of path %d: %d,%d\n", i, first, last); */
ctx->pathspare[i] = 2; /* fixed */
}
}
/*
* If we got here, the extension was successful!
*/
memcpy(ctx->grid, ctx->sparegrid, w*h*sizeof(int));
memcpy(ctx->pathends, ctx->sparepathends, ctx->npaths*2*sizeof(int));
return true;
}
/*
* Tries to add a new path to the grid.
*/
static int add_path(struct genctx *ctx, random_state *rs)
{
int w = ctx->w, h = ctx->h;
int i, ii, n;
/*
* Our strategy is:
* - randomly choose an empty square in the grid
* - do a BFS from that point to find a long path starting
* from it
* - if we run out of viable empty squares, return failure.
*/
/*
* Use `sparegrid' to collect a list of empty squares.
*/
n = 0;
for (i = 0; i < w*h; i++)
if (ctx->grid[i] == -1)
ctx->sparegrid[n++] = i;
/*
* Shuffle the grid.
*/
for (i = n; i-- > 1 ;) {
int k = random_upto(rs, i+1);
if (k != i) {
int t = ctx->sparegrid[i];
ctx->sparegrid[i] = ctx->sparegrid[k];
ctx->sparegrid[k] = t;
}
}
/*
* Loop over it trying to add paths. This looks like a
* horrifying N^4 algorithm (that is, (w*h)^2), but I predict
* that in fact the worst case will very rarely arise because
* when there's lots of grid space an attempt will succeed very
* quickly.
*/
for (ii = 0; ii < n; ii++) {
int i = ctx->sparegrid[ii];
int y = i / w, x = i % w, nsq;
int r, c, j;
/*
* BFS from here to find long paths.
*/
nsq = bfs(w, h, ctx->grid, x, y, ctx->dist, ctx->list);
/*
* If there aren't any long enough, give up immediately.
*/
assert(nsq > 0); /* must be the start square at least! */
if (ctx->dist[ctx->list[nsq-1]] < 3)
continue;
/*
* Find the first viable endpoint in ctx->list (i.e. the
* first point with distance at least three). I could
* binary-search for this, but that would be O(log N)
* whereas in fact I can get a constant time bound by just
* searching up from the start - after all, there can be at
* most 13 points at _less_ than distance 3 from the
* starting one!
*/
for (j = 0; j < nsq; j++)
if (ctx->dist[ctx->list[j]] >= 3)
break;
assert(j < nsq); /* we tested above that there was one */
/*
* Now we know that any element of `list' between j and nsq
* would be valid in principle. However, we want a few long
* paths rather than many small ones, so select only those
* elements which are either the maximum length or one
* below it.
*/
while (ctx->dist[ctx->list[j]] + 1 < ctx->dist[ctx->list[nsq-1]])
j++;
r = j + random_upto(rs, nsq - j);
j = ctx->list[r];
/*
* And that's our endpoint. Mark the new path on the grid.
*/
c = newpath(ctx);
ctx->pathends[c*2] = i;
ctx->pathends[c*2+1] = j;
ctx->grid[j] = c;
while (j != i) {
int d, np, index, pts[4];
np = 0;
for (d = 0; d < 4; d++) {
int xn = (j % w) + DX(d), yn = (j / w) + DY(d);
if (xn >= 0 && xn < w && yn >= 0 && yn < w &&
ctx->dist[yn*w+xn] == ctx->dist[j] - 1)
pts[np++] = yn*w+xn;
}
if (np > 1)
index = random_upto(rs, np);
else
index = 0;
j = pts[index];
ctx->grid[j] = c;
}
return true;
}
return false;
}
/*
* The main grid generation loop.
*/
static void gridgen_mainloop(struct genctx *ctx, random_state *rs)
{
int w = ctx->w, h = ctx->h;
int i, n;
/*
* The generation algorithm doesn't always converge. Loop round
* until it does.
*/
while (1) {
for (i = 0; i < w*h; i++)
ctx->grid[i] = -1;
ctx->npaths = 0;
while (1) {
/*
* See if the grid is full.
*/
for (i = 0; i < w*h; i++)
if (ctx->grid[i] < 0)
break;
if (i == w*h)
return;
#ifdef GENERATION_DIAGNOSTICS
{
int x, y;
for (y = 0; y < h; y++) {
printf("|");
for (x = 0; x < w; x++) {
if (ctx->grid[y*w+x] >= 0)
printf("%2d", ctx->grid[y*w+x]);
else
printf(" .");
}
printf(" |\n");
}
}
#endif
/*
* Try adding a path.
*/
if (add_path(ctx, rs)) {
#ifdef GENERATION_DIAGNOSTICS
printf("added path\n");
#endif
continue;
}
/*
* Try extending a path. First list all the possible
* extensions.
*/
for (i = 0; i < ctx->npaths * 8; i++)
ctx->extends[i] = i;
n = i;
/*
* Then shuffle the list.
*/
for (i = n; i-- > 1 ;) {
int k = random_upto(rs, i+1);
if (k != i) {
int t = ctx->extends[i];
ctx->extends[i] = ctx->extends[k];
ctx->extends[k] = t;
}
}
/*
* Now try each one in turn until one works.
*/
for (i = 0; i < n; i++) {
int p, d, e;
p = ctx->extends[i];
d = p % 4;
p /= 4;
e = p % 2;
p /= 2;
#ifdef GENERATION_DIAGNOSTICS
printf("trying to extend path %d end %d (%d,%d) in dir %d\n", p, e,
ctx->pathends[p*2+e] % w,
ctx->pathends[p*2+e] / w, d);
#endif
if (extend_path(ctx, p, e, d)) {
#ifdef GENERATION_DIAGNOSTICS
printf("extended path %d end %d (%d,%d) in dir %d\n", p, e,
ctx->pathends[p*2+e] % w,
ctx->pathends[p*2+e] / w, d);
#endif
break;
}
}
if (i < n)
continue;
break;
}
}
}
/*
* Wrapper function which deals with the boring bits such as
* removing the solution from the generated grid, shuffling the
* numeric labels and creating/disposing of the context structure.
*/
static int *gridgen(int w, int h, random_state *rs)
{
struct genctx *ctx;
int *ret;
int i;
ctx = new_genctx(w, h);
gridgen_mainloop(ctx, rs);
/*
* There is likely to be an ordering bias in the numbers
* (longer paths on lower numbers due to there having been more
* grid space when laying them down). So we must shuffle the
* numbers. We use ctx->pathspare for this.
*
* This is also as good a time as any to shift to numbering
* from 1, for display to the user.
*/
for (i = 0; i < ctx->npaths; i++)
ctx->pathspare[i] = i+1;
for (i = ctx->npaths; i-- > 1 ;) {
int k = random_upto(rs, i+1);
if (k != i) {
int t = ctx->pathspare[i];
ctx->pathspare[i] = ctx->pathspare[k];
ctx->pathspare[k] = t;
}
}
/* FIXME: remove this at some point! */
{
int y, x;
for (y = 0; y < h; y++) {
printf("|");
for (x = 0; x < w; x++) {
assert(ctx->grid[y*w+x] >= 0);
printf("%2d", ctx->pathspare[ctx->grid[y*w+x]]);
}
printf(" |\n");
}
printf("\n");
}
/*
* Clear the grid, and write in just the endpoints.
*/
for (i = 0; i < w*h; i++)
ctx->grid[i] = 0;
for (i = 0; i < ctx->npaths; i++) {
ctx->grid[ctx->pathends[i*2]] =
ctx->grid[ctx->pathends[i*2+1]] = ctx->pathspare[i];
}
ret = ctx->grid;
ctx->grid = NULL;
free_genctx(ctx);
return ret;
}
#ifdef TEST_GEN
#define TEST_GENERAL
int main(void)
{
int w = 10, h = 8;
random_state *rs = random_new("12345", 5);
int x, y, i, *grid;
for (i = 0; i < 10; i++) {
grid = gridgen(w, h, rs);
for (y = 0; y < h; y++) {
printf("|");
for (x = 0; x < w; x++) {
if (grid[y*w+x] > 0)
printf("%2d", grid[y*w+x]);
else
printf(" .");
}
printf(" |\n");
}
printf("\n");
sfree(grid);
}
return 0;
}
#endif
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