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puzzles/unfinished/path.c
Simon Tatham c9b3c3896c unfinished/path: some jottings towards a solver. 
I still don't actually have a solver design, but a recent conversation
sent my brain in some more useful directions than I've come up with
before, so I might as well at least note it all down.
2020-05-12 22:05:52 +01:00

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/*
* Experimental grid generator for Nikoli's `Number Link' puzzle.
*/
#include <stdio.h>
#include <stdlib.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_init("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
#ifdef TEST_GENERAL
#include <stdarg.h>
void fatal(const char *fmt, ...)
{
va_list ap;
fprintf(stderr, "fatal error: ");
va_start(ap, fmt);
vfprintf(stderr, fmt, ap);
va_end(ap);
fprintf(stderr, "\n");
exit(1);
}
#endif
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