Cellular Automata Method for Generating Random Cave-Like Levels
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By Jim Babcock
It is an old and fairly well documented trick to use cellular automata to generate cave-like structures. The basic idea is to fill the map randomly, then repeatedly apply the 4-5 rule: a tile is a wall if it is a wall and has 4 neighbors that are walls, or if it is not a wall and has 5 neighbors that are. This rule can be stated more simply: a tile becomes or remains a wall if the 3x3 region centered on it contains at least 5 walls. (Note: It is important to do this for each tile simultaneously. If you update one, then use its value when you update the next, your results won't look as good, and the tricks described later won't work.)
If the map initially contains 45% walls, and the process above is repeated 5 times, the output looks like (for example)
############################################################ ###....####################################.....############ ##......######################..#########.........########## ##......#####################....#######...........####.#### ##......###################.........................##...### ##......##################..........................###...## #........##############.............................###...## #........#############...............................#....## ##.......##############..................................### ##.......###..############..............................#### ##.......##....############.............................#### #..............############...###........................### #...............###########..#####...............##.......## #................#################...............##.......## ##.....#####..........###########....#..........###.......## ##....#######...........########....###.........####......## ##....#######............######....####........#####......## ##....#######.............####....#####.......#####......### #......######..............###....####........####......#### #.......######.............###...####.........###.......#### #........#####.............###..####.....................### ##........####..............#...####.....................### #####......##...................####.....................### ######...........................##.....................#### ######..................................................#### ######.........###.....................####.............#### ######......#########.................######............#### #######....#############.......##############.....###..##### ##############################################..############ ############################################################
The problem is, the results of the algorithm are very inconsistent. Not only is it prone to generating disjoint (not connected) maps, like in this example, with the same parameters:
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it also sometimes generates maps which consist of basically one huge open space, like this one:
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We can fix the disjoint segments problem in one of three ways. Either throw away maps that have disjoint segments in them, connect up the segments after the fact, or fill in all but the biggest segment. We can't just retry when we get a disjoint map, because if the map is big then, statistically, that will be almost 100% of the time. Filling in all but the biggest segment will tend to produce a small area in a map that was supposed to be big. Connecting up the regions works, but it tends to look unnatural, as in the example from above, now connected:
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The solution to both problems, as it turns out, is to revisit the original cellular automata rules. Recall that the original rule was
- There is a wall initially at P with 45% probability
- In the next generation, there is a wall at spot P if the number of tiles around P which are walls is at least 5
Or, in more compact notation:
- Winit(p) = rand[0,100) < 45
- R(p) = the number of tiles within 1 step of p which are walls
- W?(p) = R(p) ? 5
Now, one of the problems was that we tend to get big, open areas. So why not try filling those areas in? Change the rule to
- W?(p) = R(p) ? 5 or p is in the middle of an open space
Or more formally,
- Rn(p) = the number of tiles within n steps of p which are walls
- W?(p) = R1(p)?5 || R2(p)=0
So how does it look?
Winit(p) = rand[0,100) < 45 Repeat 5: W?(p) = R1(p) ? 5 || R2(p) ? 1
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This is more interesting - it doesn't have any big open areas, it has a decent layout. It's almost fully connected. Still, it has some new problems: there are isolated single-tile walls in places, and in general it's not very smooth. But with a little tweaking:
Winit(p) = rand[0,100) < 40 Repeat 4: W?(p) = R1(p) ? 5 || R2(p) ? 2 Repeat 3: W?(p) = R1(p) ? 5
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Notice that the initial fill percentage is a little lower, the cutoffs are different, and we switch rules after a few generations. This is more like the desired result. So, with these parameters, I give you some more samples, at various sizes.
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There's still no guarantee of connectedness, though. However, it's now consistently almost-connected, so that you can reasonably just drop the isolated chunks.
Finally, here is the C program I used to try out different parameters. Before putting this into an actual game, handling of disconnected regions is needed.
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#define TILE_FLOOR 0
#define TILE_WALL 1
typedef struct {
int r1_cutoff, r2_cutoff;
int reps;
} generation_params;
int **grid;
int **grid2;
int fillprob = 40;
int r1_cutoff = 5, r2_cutoff = 2;
int size_x = 64, size_y = 20;
generation_params *params;
generation_params *params_set;
int generations;
int randpick(void)
{
if(rand()%100 < fillprob)
return TILE_WALL;
else
return TILE_FLOOR;
}
void initmap(void)
{
int xi, yi;
grid = (int**)malloc(sizeof(int*) * size_y);
grid2 = (int**)malloc(sizeof(int*) * size_y);
for(yi=0; yi<size_y; yi++)
{
grid [yi] = (int*)malloc(sizeof(int) * size_x);
grid2[yi] = (int*)malloc(sizeof(int) * size_x);
}
for(yi=1; yi<size_y-1; yi++)
for(xi=1; xi<size_x-1; xi++)
grid[yi][xi] = randpick();
for(yi=0; yi<size_y; yi++)
for(xi=0; xi<size_x; xi++)
grid2[yi][xi] = TILE_WALL;
for(yi=0; yi<size_y; yi++)
grid[yi][0] = grid[yi][size_x-1] = TILE_WALL;
for(xi=0; xi<size_x; xi++)
grid[0][xi] = grid[size_y-1][xi] = TILE_WALL;
}
void generation(void)
{
int xi, yi, ii, jj;
for(yi=1; yi<size_y-1; yi++)
for(xi=1; xi<size_x-1; xi++)
{
int adjcount_r1 = 0,
adjcount_r2 = 0;
for(ii=-1; ii<=1; ii++)
for(jj=-1; jj<=1; jj++)
{
if(grid[yi+ii][xi+jj] != TILE_FLOOR)
adjcount_r1++;
}
for(ii=yi-2; ii<=yi+2; ii++)
for(jj=xi-2; jj<=xi+2; jj++)
{
if(abs(ii-yi)==2 && abs(jj-xi)==2)
continue;
if(ii<0 || jj<0 || ii>=size_y || jj>=size_x)
continue;
if(grid[ii][jj] != TILE_FLOOR)
adjcount_r2++;
}
if(adjcount_r1 >= params->r1_cutoff || adjcount_r2 <= params->r2_cutoff)
grid2[yi][xi] = TILE_WALL;
else
grid2[yi][xi] = TILE_FLOOR;
}
for(yi=1; yi<size_y-1; yi++)
for(xi=1; xi<size_x-1; xi++)
grid[yi][xi] = grid2[yi][xi];
}
void printfunc(void)
{
int ii;
printf("W[0](p) = rand[0,100) < %i\n", fillprob);
for(ii=0; ii<generations; ii++)
{
printf("Repeat %i: W'(p) = R[1](p) >= %i",
params_set[ii].reps, params_set[ii].r1_cutoff);
if(params_set[ii].r2_cutoff >= 0)
printf(" || R[2](p) <= %i\n", params_set[ii].r2_cutoff);
else
putchar('\n');
}
}
void printmap(void)
{
int xi, yi;
for(yi=0; yi<size_y; yi++)
{
for(xi=0; xi<size_x; xi++)
{
switch(grid[yi][xi]) {
case TILE_WALL: putchar('#'); break;
case TILE_FLOOR: putchar('.'); break;
}
}
putchar('\n');
}
}
int main(int argc, char **argv)
{
int ii, jj;
if(argc < 7) {
printf("Usage: %s xsize ysize fill (r1 r2 count)+\n", argv[0]);
return 1;
}
size_x = atoi(argv[1]);
size_y = atoi(argv[2]);
fillprob = atoi(argv[3]);
generations = (argc-4)/3;
params = params_set = (generation_params*)malloc( sizeof(generation_params) * generations );
for(ii=4; ii+2<argc; ii+=3)
{
params->r1_cutoff = atoi(argv[ii]);
params->r2_cutoff = atoi(argv[ii+1]);
params->reps = atoi(argv[ii+2]);
params++;
}
srand(time(NULL));
initmap();
for(ii=0; ii<generations; ii++)
{
params = ¶ms_set[ii];
for(jj=0; jj<params->reps; jj++)
generation();
}
printfunc();
printmap();
return 0;
}
By Provolik
I used this methon on my game (Arcan Myth RL but only in developing version) but I limited the repetition cycle to 3: I like the result very much and I don't have the problem about big isolated cave. However very good method. Thank you :-)