Langton’s Ant

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Langton’s Ant is a turmite governed by simple rules whose outcome is both unpredictable and intresting. The path taken by the ant generates some surprising shapes, never appearing when you would expect them to, but a seemingly random moments. This article describes the rules behind Langton’s Ant, shows some of the images formed and provides a Python programme to simulate the ant.

Langton’s Ant

Chris Langton is a biologist, and the founder of the field of artificial life. One of his simplest and most intresting creations is Langton’s Ant. It is quite important theoretically, but it also has some intresting practical applications.

The ant’s world consists of a grid, possibly infinite, but limited in our example:

The ant always moves through this grid one step at a time. It’s direction and it’s effect on the grid is defined by 3 simple rules:

1. If the ant is on a black square, it turns right 90 degrees.
2. If the ant is on a white square, it turns left 90 degrees.
3. When the ant leaves a square, it inverts colour.

It is fairly obvious that the ant’s movement will leave a coloured trail. When asked, most people would guess that either no patterns show or that some basic symmetric images appear. What actually happens is this.

For the first 50 or so steps, it just seems to move around randomly, and then at step 52, you get a heart:

For the next 300 or so steps, the ant seems to draw random shapes, ofter erasing what it had drawn before. At about step 383 you get something that looks like this:

Next you get what’s technically known as a mess. The ant’s movement degenerates into chaos. No more patterns are observed:

By now, you are probably thinking: “Well yes. Simple rules, so at first you get simple patterns, but as the board gets more and more complex, the simple rules can’t handle it, so the ant moves randomly”. Well that’s right, up until about step 10000; that is when you start getting highways.

The highways the ant draws are intresting shapes. You can see one at the bottom of the next image. It is the diagonal bit going down and right. To build then, the ant seems to move in circles, the last row forcing it to draw the next. Thus, the ant continues to draw until something gets in its way. In this simulation, because the screen wraps around, the ant eventually gets back to the mess, where it starts moving randomly again.

What’s intresting is that, after some time, the ant starts building highways again. Actually, there is no known way to stop it. The only hindrance here is that the board is finite so it has a tendency to fill up. Anyway, here’s what the board looks like at step 97049:

The Programme

The following is a simple programme written in Python and using PyGame.

The variables that control the display and the board are at the top of the programme. They are heavily commented so there shouldn’t be any problems.

Here’s the code in Python (langton.py):

#************************************************
# Rules of the game
#	1. If the ant is on a black square, it turns
#		right 90 and moves forward one unit
#	2. If the ant is on a white square, it turns
# 		left 90 and moves forward one unit
#	3. When the ant leaves a square, it inverts
#		colour
#
# SEE: http://mathworld.wolfram.com/LangtonsAnt.html
#************************************************

import sys, pygame
from pygame.locals import *
import time

dirs = (
		(-1, 0),
		(0, 1),
		(1, 0),
		(0, -1)
		)

cellSize = 12 # size in pixels of the board (4 pixels are used to draw the grid)
numCells = 64 # length of the side of the board
background = 0, 0, 0 # background colour; black here
foreground = 23, 23, 23 # foreground colour; the grid's colour; dark gray here
textcol = 177, 177, 177 # the colour of the step display in the upper left of the screen
antwalk = 44, 88, 44 # the ant's trail; greenish here
antant = 222, 44, 44 # the ant's colour; red here
fps = 1.0 / 40 # time between steps; 1.0 / 40 means 40 steps per second

def main():
	pygame.init()

	size = width, height = numCells * cellSize, numCells * cellSize

	pygame.display.set_caption("Langton's Ant")

	screen = pygame.display.set_mode(size) # Screen is now an object representing the window in which we paint
	screen.fill(background)
	pygame.display.flip() # IMPORTANT: No changes are displayed until this function gets called

	for i in xrange(1, numCells):
		pygame.draw.line(screen, foreground, (i * cellSize, 1), (i * cellSize, numCells * cellSize), 2)
		pygame.draw.line(screen, foreground, (1, i * cellSize), (numCells * cellSize, i * cellSize), 2)
	pygame.display.flip() # IMPORTANT: No changes are displayed until this function gets called

	font = pygame.font.Font(None, 36)

	antx, anty = numCells / 2, numCells / 2
	antdir = 0
	board = [[False] * numCells for e in xrange(numCells)]

	step = 1
	pause = False
	while True:
		for event in pygame.event.get():
				if event.type == QUIT:
					return
				elif event.type == KEYUP:
					if event.key == 32: # If space pressed, pause or unpause
						pause = not pause
					elif event.key == 115:
						pygame.image.save(screen, "Step%d.tga" % (step))

		if pause:
			time.sleep(fps)
			continue

		text = font.render("%d " % (step), True, textcol, background)
		screen.blit(text, (10, 10))

		if board[antx][anty]:
			board[antx][anty] = False # See rule 3
			screen.fill(background, pygame.Rect(antx * cellSize + 1, anty * cellSize + 1, cellSize - 2, cellSize - 2))
			antdir = (antdir + 1) % 4 # See rule 1
		else:
			board[antx][anty] = True # See rule 3
			screen.fill(antwalk, pygame.Rect(antx * cellSize + 1, anty * cellSize + 1, cellSize - 2, cellSize - 2))
			antdir = (antdir + 3) % 4 # See rule 2

		antx = (antx + dirs[antdir][0]) % numCells
		anty = (anty + dirs[antdir][1]) % numCells

		# The current square (i.e. the ant) is painted a different colour
		screen.fill(antant, pygame.Rect(antx * cellSize + 1, anty * cellSize + 1, cellSize -2, cellSize -2))

		pygame.display.flip() # IMPORTANT: No changes are displayed until this function gets called

		step += 1
		time.sleep(fps)

if __name__ == "__main__":
	main()

It is intresting to note that although you know all the rules that govern the ant's universe, you cannot predict its movement without resorting to simulation. This just goes to show that knowing the rules of the components at the lowest level might not help you understand the system as a whole.

That’s it. Have fun with the code. Always open to comments.

 

Finding All Paths of Minimum Length to a Node Using Dijkstra’s Algorithm

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In this article I describe a way of modifying Dijkstra’s Alogrithm in order to find all the shortest path from a source to a node.

This article assumes you know how Dijkstra’s Algorithm works. If you don’t, see my previous post or the Wikipedia article.

The Problem

You know how to use Dijkstra’s algorithm to find the length of the shortest path to a node. You’ve even figured out how to record the path to each node. But you what you really need are all the shortest paths leading to a node.

The Idea

I can help, but to be honest, this is obvious.

In order to record the path to each node, I used an array to record which node comes before each other node in the shortest path. That is to say: prev[i] was the node that comes just before node i in the shortest path from the source to node i.

To record all the shortest paths that lead to a node, I just turned prev into a matrix with the following meaning: prev[i][0] is the number of nodes that could come before node i on a path of minimum length; prev[i][1..] are the nodes that could come before node i on path of minimum length.

The Programme

Here’s the code in C (dijkstraAll.c):

#include <stdio.h>

#define GRAPHSIZE 2048
#define INFINITY GRAPHSIZE*GRAPHSIZE
#define MAX(a, b) ((a > b) ? (a) : (b))

int e; /* The number of nonzero edges in the graph */
int n; /* The number of nodes in the graph */
long dist[GRAPHSIZE][GRAPHSIZE]; /* dist[i][j] is the distance between node i and j; or 0 if there is no direct connection */
long d[GRAPHSIZE]; /* d[i] is the length of the shortest path between the source (s) and node i */
int prev[GRAPHSIZE][GRAPHSIZE + 1]; /* prev[i] holds the nodes that could comes right before i in the shortest path from the source to i;
		prev[i][0] is the number of nodes and prev[i][1..] are the nodes */

void printD() {
	int i;

	printf("Distances:\n");
	for (i = 1; i <= n; ++i)
		printf("%10d", i);
	printf("\n");
	for (i = 1; i <= n; ++i) {
		printf("%10ld", d[i]);
	}
	printf("\n");
}

/*
 * Prints the shortest path from the source to dest.
 *
 * dijkstra(int) MUST be run at least once BEFORE
 * this is called
 */
void printPath(int dest, int depth) {
	int i, j;

	printf("-%d\n", dest);
	for (i = 1; i <= prev[dest][0]; ++i) {
		for (j = 0; j <= depth; ++j)
			printf(" |");
		printPath(prev[dest][i], depth + 1);
	}
}

void dijkstra(int s) {
	int i, k, mini;
	int visited[GRAPHSIZE];

	for (i = 1; i <= n; ++i) {
		d[i] = INFINITY;
		prev[i][0] = 0; /* no path has yet been found to i */
		visited[i] = 0; /* the i-th element has not yet been visited */
	}

	d[s] = 0;

	for (k = 1; k <= n; ++k) {
		mini = -1;
		for (i = 1; i <= n; ++i)
			if (!visited[i] && ((mini == -1) || (d[i] < d[mini])))
				mini = i;

		visited[mini] = 1;

		for (i = 1; i <= n; ++i)
			if (dist[mini][i]) {
				if (d[mini] + dist[mini][i] < d[i]) { /* a shorter path has been found */
					d[i] = d[mini] + dist[mini][i];
					prev[i][0] = 1;
					prev[i][1] = mini;
				} else if (d[mini] + dist[mini][i] == d[i]) { /* a path of the same length has been found */
					++prev[i][0];
					prev[i][prev[i][0]] = mini;
				}
			}
	}
}

int main(int argc, char *argv[]) {
	int i, j;
	int u, v, w;

	FILE *fin = fopen("dist.txt", "r");
	fscanf(fin, "%d", &e);
	for (i = 0; i < e; ++i)
		for (j = 0; j < e; ++j)
			dist[i][j] = 0;
	n = -1;
	for (i = 0; i < e; ++i) {
		fscanf(fin, "%d%d%d", &u, &v, &w);
		dist[u][v] = w;
		n = MAX(u, MAX(v, n));
	}
	fclose(fin);

	dijkstra(1);

	printD();

	printf("\n");
	for (i = 1; i <= n; ++i) {
		printf("Path to %d:\n", i);
		printPath(i, 0);
		printf("\n");
	}

	return 0;
}

And here’s an input file: dist.txt.

10
1 2 5
1 4 3
2 3 1
2 4 3
3 5 6
4 2 2
4 3 9
4 5 2
5 1 7
5 3 4

The input file describes this graph:

As you can see, there are two paths from node 1 to node 3: 1 -> 2 -> 3 and 1 -> 4 -> 2 -> 3 both of length 6.

Now, what does the programme output?

Distances:
1 2 3 4 5
0 5 6 3 5

Path to 1:
-1

Path to 2:
-2
|-1
|-4
| |-1

Path to 3:
-3
|-2
| |-1
| |-4
| | |-1

Path to 4:
-4
|-1

Path to 5:
-5
|-4
| |-1

It first outputs the distances, and… yes! They’re correct.

Next, it prints those ASCII art drawings. They not drawings. They’re trees with the destination as root and the leafs as the source. To read a path from such a tree, start at a leaf (always 1) and go left, reading the first numbers you can see above.

Let’s find the paths to node 3. There are two leafs, so there are two paths of minimal length. The first one is 1 -> 4 -> 2 -> 3. The second one is 1 -> 2 -> 3. Check on the graph.

That’s it. If you’re up to a challenge, implement prev as an array of linked lists.

Good luck. Always open to comments.