Wave function collapse
The Wave Function Collapse algorithm is a heuristic for generating tiled images.
The algorithm begins with a collection of equal sized image blocks and randomly places them, one at a time, within a grid subject to the tiling constraint and an entropy constraint, and it wraps (the top row of blocks in the grid is treated as adjacent to the bottom row of blocks, and similarly the left and right columns of blocks are treated as adjacent to each other).
The blocks are tiled within the grid. Tiled means they are placed with a one pixel overlap and the tiling constraint requires that the pixels overlapping border between two adjacent blocks match.
Entropy, here, means the number of blocks eligible to be placed in an unassigned grid location. The entropy constraint here is that each block is placed in a grid location with minimum entropy. (Placing a block may constrain the entropy of its four nearest neighbors -- up, down, left, right.)
For this task, we start with five blocks of 3x3 pixels and place them in an 8x8 grid to form a 17x17 tile. A tile is a block which may be tiled with itself. Here, we show these five blocks adjacent but not tiled:
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Note that this algorithm sometimes does not succeed. If an unassigned grid location has an entropy of 0, the algorithm fails and returns an empty or null result. We'll ignore those failure cases for this task.
- Reference WFC explained and another WFC explained
C
<lang C>#include <stdio.h>
- include <stdlib.h>
- include <string.h>
- include <time.h>
- define XY(row, col, width) ((col)+(row)*(width))
- define XYZ(page, row, col, height, width) XY(XY(page, row, height), col, width)
char blocks[5][3][3]= {
{
{0, 0, 0},
{0, 0, 0},
{0, 0, 0}
},{
{0, 0, 0},
{1, 1, 1},
{0, 1, 0}
},{
{0, 1, 0},
{0, 1, 1},
{0, 1, 0}
},{
{0, 1, 0},
{1, 1, 1},
{0, 0, 0}
},{
{0, 1, 0},
{1, 1, 0},
{0, 1, 0}
}
};
/* avoid problems with slightly negative numbers and C's X%Y */
- define MOD(X,Y) ((Y)+(X))%(Y)
char *wfc(char *blocks, int *bdim /* 5,3,3 */, int *tdim /* 8,8 */) {
int N= tdim[0]*tdim[1], td0= tdim[0], td1= tdim[1];
int *adj= calloc(N*4, sizeof (int)); /* indices in R of the four adjacent blocks */
for (int i= 0; i<td0; i++) {
for (int j=0; j<td1; j++) {
adj[XYZ(i,j,0,td1,4)]= XY(MOD(i-1, td0), MOD(j, td1), td1); /* above (index 1 in a 3x3 grid) */
adj[XYZ(i,j,1,td1,4)]= XY(MOD(i, td0), MOD(j-1, td1), td1); /* left (index 3 in a 3x3 grid) */
adj[XYZ(i,j,2,td1,4)]= XY(MOD(i, td0), MOD(j+1, td1), td1); /* right (index 5 in a 3x3 grid) */
adj[XYZ(i,j,3,td1,4)]= XY(MOD(i+1, td0), MOD(j, td1), td1); /* below (index 7 in a 3x3 grid) */
}
}
int bd0= bdim[0], bd1= bdim[1], bd2= bdim[2];
char *horz= malloc(bd0*bd0); /* blocks which can sit next to each other horizontally */
for (int i= 0; i<bd0; i++) {
for (int j= 0; j<bd0; j++) {
horz[XY(i,j,bd0)]= 1;
for (int k= 0; k<bd1; k++) {
if (blocks[XYZ(i, k, 0, bd1, bd2)] != blocks[XYZ(j, k, bd2-1, bd1, bd2)]) {
horz[XY(i, j, bd0)]= 0;
}
}
}
}
char *vert= malloc(bd0*bd0); /* blocks which can sit next to each other vertically */
for (int i= 0; i<bd0; i++) {
for (int j= 0; j<bd0; j++) {
vert[XY(i,j,bd0)]= 1;
for (int k= 0; k<bd2; k++) {
if (blocks[XYZ(i, 0, k, bd1, bd2)] != blocks[XYZ(j, bd1-1, k, bd1, bd2)]) {
vert[XY(i, j, bd0)]= 0;
break;
}
}
}
}
char *allow= malloc(4*(bd0+1)*bd0); /* all block constraints, based on neighbors */
memset(allow, 1, 4*(bd0+1)*bd0);
for (int i= 0; i<bd0; i++) {
for (int j= 0; j<bd0; j++) {
allow[XYZ(0, i, j, bd0+1, bd0)]= vert[XY(j, i, bd0)]; /* above (north) */
allow[XYZ(1, i, j, bd0+1, bd0)]= horz[XY(j, i, bd0)]; /* left (west) */
allow[XYZ(2, i, j, bd0+1, bd0)]= horz[XY(i, j, bd0)]; /* right (east) */
allow[XYZ(3, i, j, bd0+1, bd0)]= vert[XY(i, j, bd0)]; /* below (south) */
}
}
free(horz);
free(vert);
int *todo= calloc(N, sizeof (int));
char *wave= malloc(N*bd0);
int *entropy= calloc(N, sizeof (int));
int *indices= calloc(N, sizeof (int));
int min;
int *possible= calloc(bd0, sizeof (int));
int *R= calloc(N, sizeof (int)); /* tile expressed as list of block indices */
for (int i= 0; i<N; i++) R[i]= bd0;
while (1) {
int c= 0;
for (int i= 0; i<N; i++)
if (bd0==R[i])
todo[c++]= i;
if (!c) break;
min= bd0;
for (int i= 0; i<c; i++) {
entropy[i]= 0;
for (int j= 0; j<bd0; j++) {
int K= 4*todo[i];
entropy[i]+=
wave[XY(i, j, bd0)]=
allow[XYZ(0, R[adj[XY(todo[i],0,4)]], j, bd0+1, bd0)] &
allow[XYZ(1, R[adj[XY(todo[i],1,4)]], j, bd0+1, bd0)] &
allow[XYZ(2, R[adj[XY(todo[i],2,4)]], j, bd0+1, bd0)] &
allow[XYZ(3, R[adj[XY(todo[i],3,4)]], j, bd0+1, bd0)];
}
if (entropy[i] < min) min= entropy[i];
}
if (!min) {
free(R);
R= NULL;
break;
}
int d= 0;
for (int i= 0; i<c; i++) {
if (min==entropy[i]) indices[d++]= i;
}
int ndx= indices[random()%d];
int ind= ndx*bd0;
d= 0;
for (int i= 0; i<bd0; i++) {
if (wave[ind+i]) possible[d++]= i;
}
R[todo[ndx]]= possible[random()%d];
}
free(adj);
free(allow);
free(todo);
free(wave);
free(entropy);
free(indices);
free(possible);
if (!R) return NULL;
char *tile= malloc((1+td0*(bd1-1))*(1+td1*(bd2-1)));
for (int i0= 0; i0<td0; i0++)
for (int i1= 0; i1<bd1; i1++)
for (int j0= 0; j0<td1; j0++)
for (int j1= 0; j1<bd2; j1++)
tile[XY(XY(j0, j1, bd2-1), XY(i0, i1, bd1-1), 1+td1*(bd2-1))]=
blocks[XYZ(R[XY(i0, j0, td1)], i1, j1, bd1, bd2)];
free(R);
return tile;
}
int main() {
int bdims[3]= {5,3,3};
int size[2]= {8,8};
time_t t;
srandom((unsigned) time(&t));
char *tile= wfc((char*)blocks, bdims, size);
if (!tile) exit(0);
for (int i= 0; i<17; i++) {
for (int j= 0; j<17; j++) {
printf("%c ", " #"[tile[XY(i, j, 17)]]);
}
printf("\n");
}
free(tile);
exit(0);
}</lang>
Note: here we use R where J used i, because we use i as an index/loop counter (other than m, y and i, the comments on the j implementation should be directly relevant here). Also, when assembling the result at the end, it was convenient to treat the block overlap issue during indexing.
For simplicity, we use char as our pixel datatype (and for truth values), and int for indices (C offers a variety of similar datatypes but nothing we are doing here is big enough for that to be a concern).
- Output:
# # # #
# # # # # # # # # #
# # #
# # # # # # # # # #
# # # #
# # # # # # # # # # # # # # # #
# # # # # #
# # # # # # # # # # # # # # # #
# # # #
# # # # # # # # # # # #
# # # # # # #
# # # # # # # # # # #
# # # # # #
# # # # # # # # # # # # # # #
# # # # # #
# # # # # # # # # # # # # # # #
# # # #
J
Implementation:<lang J>blocks=: 0,(|.@|:)^:(i.4)0,1 1 1,:0 1 0 wfc=: Template:Adj=: y</lang>
We work with the 3x3 partial tiles, and the larger 17x17 tile which we are randomly generating. (17x17 because every 3x3 block contributes 2x2 pixels to the result and along a horizontal and vertical edge row and column of the tile, the 3x3 blocks contribute an additional row and column of pixels.)
Here, m is the list of tiles, and i represents an 8x8 list of indexes into that list (or, conceptually whatever dimensions were specified by y, the right argument to wfc -- but for this task y will always be 8 8), with _1 being a placeholder for the case where the index hasn't been choosen -- initially, we pick a random location in i and assign an arbitrarily picked tile to that location.
adj indexes into i -- for each item in i, adj selects that item, the item "above" it, the item to the "left" of it, the item to the "right" of it and the item "below" it (with scare quotes because the tile represented by i "wraps around" on all sides). And, allow lists the allowable tiles corresponding to each of those adj constraints.
To build allow we first matched the left side of each tile with the right side of each tile (cartesian product) forming horz and similarly matched the tops and bottoms of the tiles forming vert. Then we build north which limits tiles based on the tile above it, and similarly for west, east, and south (when the adjacent tile is a _1 tile, no limit is imposed).
Once we're set up, we drop into a loop: todo selects the unchosen tile locations, wavelists for each of the unchoosen tiles (for each todo value in i we select the tiles allowed by each of its adjacent locations and find the set intersection of all of those), entropy counts how many tiles are eligible for each of those location, and min is the smallest value in entropy. ndx is a randomly picked index into todo with minimal entropy and for that location we randomly pick one of the options and update i with it. (When there's only one option remaining, "randomly pick" here means we pick that option.)
Once we've assigned a tile to every location in i, we use those indices to assemble the result (note that the 3x3 tiles overlap at their borders so we introduce a mechanism to discard the redundant pixels).
For task purposes here, we will use space to represent a white pixel and "#" to represent a black pixel. Also, because characters are narrow, we will insert a space between each of these "pixels" to better approximate a square aspect ratio.
Task example (the initial tiles and three runs of wave function collapse (three, to illustrate randomness):<lang J> (<"2) 1j1#"1 ' #'{~ tiles ┌──────┬──────┬──────┬──────┬──────┐ │ │ │ # │ # │ # │ │ │# # # │ # # │# # # │# # │ │ │ # │ # │ │ # │ └──────┴──────┴──────┴──────┴──────┘
task=: Template:1j1 task&.>0 0 0
┌──────────────────────────────────┬──────────────────────────────────┬──────────────────────────────────┐ │ # # # # │ # # # # # # │ # # # # # │ │# # # # # # # # # # # # # # │ # # # # # # # # # # # # # # │ # # # # # # # # # # # # │ │ # # # # # │ # # # # # # │ # # # # # # │ │ # # # # # # # # # # # # │ # # # # # # # # # # # # # # │# # # # # # # # # # # # # # # │ │ # # # # # # │ # # # # # # │ # # # # # # │ │# # # # # # # # # # # # # # # │# # # # # # # # # # # │# # # # # # # # # # # # # # # │ │ # # # # # # │ # # # # │ # # # # # # │ │# # # # # # # # # # # # # # # │ # # # # # # # # # # # │ # # # # # # # # # # # │ │ # # # # # # │ # # # # │ # # # # # # # │ │# # # # # # # # # # # # # # # │# # # # # # # # # # # │# # # # # # # # # # # # # │ │ # # # # # # │ # # # # # # │ # # # # │ │ # # # # # # # # # # # # │# # # # # # # # # # │ # # # # # # # # │ │ # # # # # │ # # # # # # │ # # # # # │ │# # # # # # # # # # # # # # │# # # # # # # # # # # # # # # │# # # # # # # # # # # │ │ # # # # │ # # # # # # │ # # # # # │ │ # # # # # # │# # # # # # # # # # # # # # # │# # # # # # # # # # # │ │ # # # # │ # # # # # # │ # # # # # │ └──────────────────────────────────┴──────────────────────────────────┴──────────────────────────────────┘</lang>
Phix
You can [NOT YET!] run this online here.
NOTE: For discussion purposes only. I will not lose a second's sleep should this be deleted.
Marked as incorrect to dissuade anyone slavishly copying this.
Primarily intended so you can see the results, rather than this being particularly interesting code.
For the purposes of discussing this task, ignore everything below "--<boilerplate code:>".
Wren
Well, I don't know whether this task is going to be deleted or not though, given the effort Rdm has put into understanding it, I'd be in favor now of letting it stand. It is after all an interesting application of an algorithm inspired by quantum mechanics to generating images.
The following is a translation of his C version. <lang ecmascript>import "random" for Random
var rand = Random.new()
var blocks = [
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 1, 0, 0, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 1, 0, 0, 0, 0, 1, 0, 1, 1, 0, 0, 1, 0
]
var wfc = Fn.new { |blocks, tdim, target|
var N = target[0] * target[1]
var t0 = target[0]
var t1 = target[1]
var adj = List.filled(4*N, 0)
for (i in 0...t0) {
for (j in 0...t1) {
var k = j + t1*i
var m = 4 * k
adj[m ] = j + t1*((t0+i-1)%t0) /* above (1) */
adj[m+1] = (t1+j-1)%t1 + t1* i /* left (3) */
adj[m+2] = ( j+1)%t1 + t1* i /* right (5) */
adj[m+3] = j + t1*(( i+1)%t0) /* below (7) */
}
}
var td0 = tdim[0]
var td1 = tdim[1]
var td2 = tdim[2]
var horz = List.filled(td0*td0, 0)
for (i in 0...td0) {
for (j in 0...td0) {
horz[j+i*td0] = 1
for (k in 0...td1) {
if (blocks[0+td2*(k+td1*i)] != blocks[(td2-1)+td2*(k+td1*j)]) {
horz[j+i*td0] = 0
break
}
}
}
}
var vert = List.filled(td0*td0, 0)
for (i in 0...td0) {
for (j in 0...td0) {
vert[j+i*td0]= 1
for (k in 0...td2) {
if (blocks[k+td2*(0+td1*i)] != blocks[k+td2*((td2-1)+td1*j)]) {
vert[j+i*td0]= 0
break
}
}
}
}
var stride = (td0+1) * td0
var allow = List.filled(4 * stride, 1)
for (i in 0...td0) {
for (j in 0...td0) {
allow[ (i*td0)+j] = vert[(j*td0)+i] /* above (north) */
allow[ stride +(i*td0)+j] = horz[(j*td0)+i] /* left (west) */
allow[(2*stride)+(i*td0)+j] = horz[(i*td0)+j] /* right (east) */
allow[(3*stride)+(i*td0)+j] = vert[(i*td0)+j] /* below (south) */
}
}
var R = List.filled(N, td0)
var todo = List.filled(N, 0)
var wave = List.filled(N*td0, 0)
var entropy = List.filled(N, 0)
var indices = List.filled(N, 0)
var min = 0
var possible = List.filled(td0, 0)
while (true) {
var c = 0
for (i in 0...N) {
if (td0 == R[i]) {
todo[c] = i
c = c + 1
}
}
if (c == 0) break
min = td0
for (i in 0...c) {
entropy[i] = 0
for (j in 0...td0) {
var K = 4*todo[i]
wave[i*td0 + j] = allow[ td0*R[adj[K ]]+j] & /* above */
allow[ stride +td0*R[adj[K+1]]+j] & /* left */
allow[(2*stride)+td0*R[adj[K+2]]+j] & /* right */
allow[(3*stride)+td0*R[adj[K+3]]+j] /* below */
entropy[i] = entropy[i] + wave[i*td0 + j]
}
if (entropy[i] < min) min = entropy[i]
}
if (min == 0) {
R = null
break
}
var d = 0
for (i in 0...c) {
if (min == entropy[i]) {
indices[d] = i
d = d + 1
}
}
var ndx = indices[rand.int(0, d)]
var ind = ndx * td0
d = 0
for (i in 0...td0) {
if (wave[ind+i] != 0) {
possible[d] = i
d = d + 1
}
}
R[todo[ndx]] = possible[rand.int(0, d)]
}
if (!R) return null
var tile = List.filled((1+t0*(td1-1))*(1+t1*(td2-1)), 0)
for (i0 in 0...t0) {
for (i1 in 0...td1) {
for (j0 in 0...t1) {
for (j1 in 0...td2) {
var t = j1 + (td2-1)*j0 + (1+t1*(td2-1))*(i1 + (td1-1)*i0)
tile[t] = blocks[j1 + td2*(i1 + td1*R[j0+t1*i0])]
}
}
}
}
return tile
}
var tdims = [5, 3, 3] var size = [8, 8] var tile = wfc.call(blocks, tdims, size) if (!tile) return for (i in 0..16) {
for (j in 0..16) {
System.write("%(" #"[tile[j+i*17]]) ")
}
System.print()
}</lang>
- Output:
Sample output:
# # # # #
# # # # # # # # # # #
# # # # #
# # # # # # # # # # # # # #
# # # #
# # # # # # # # # #
# # # # # #
# # # # # # # # # # # # #
# # # # #
# # # # # # # # # # # # # # # #
# # # # #
# # # # # # # # # # # # #
# # # #
# # # # # # # # # # # # # #
# # # # #
# # # # # # # # # # #
# # # # #