I thought people would be interested in this little script I wrote to

reproduce the 256 simple automata that is shown in the first chapters

of "A New Kind of Science". You can see a few results without running

the script, at

http://cooper-j.blogspot.com . And here is the code (You

will need PIL (Python Imaging Library)):

<code>

import Image

# Contract:

# To simulate simple cellular automata.

class calculations:

def __init__(self,which):

self.against = self.array_maker_2()[which]

self.check = self.array_maker_1()

self.array = self.array_maker_3(311)

self.calculator()

def binary(self, n, size): ## This is the Int -> str(BINARY)

converter

assert n >= 0

bits = []

while n:

bits.append('01'[n&1])

n >>= 1

bits.reverse()

result = ''.join(bits) or '0'

for iteration in range(len(result),size):

result = "0" + result

return result

def array_maker_1(self): # This makes the array that represents the

8 different permutations of 3 cells. Itself, its left and its right.

return [self.binary(n, 3) for n in range(8)]

def array_maker_2(self): # This makes the array that represents

every single different rule. If for instance the second element in one

# of these rules is 1, then the corresponding permutation that may

be found in the result array (array_maker_3), will be 1 (black).

return [self.binary(n, 8) for n in range(256)]

def array_maker_3(self, y): # This is the array for all the

results. The automaton starts from the middle of the first row

x = [["0" for x in range((2*y)+1)] for n in range(y)]

x[0][(2*y+1)/2] = "1"

return x

def calculator(self): # This cycles over all of the cells, and

scans one row at a time, and changes the next row according to the

current cell.

self.buff_result = ["0","0","0"] # This is the current permutation

buffer to be checked against the corresponding arrays.

for i in range(len(self.array)-1):

for j in range(1, len(self.array[0])-1):

self.step1(j,i)

self.step2(j,i)

self.step3(j,i)

y = self.check.index(''.join(self.buff_result))

self.array[i+1][j] = self.against[y]

# The steps update the result buffer.

def step1(self, step, y):

self.buff_result[0] = self.array[y][step-1]

def step2(self, step, y):

self.buff_result[1] = self.array[y][step]

def step3(self, step, y):

self.buff_result[2] = self.array[y][step+1]

for number in range(256):

objo = calculations(number)

x = objo.array

y = []

for num,zo in enumerate(x):

for com,wo in enumerate(zo):

x[num][com] = int(wo)

nim = Image.new("1", (623,311))

for n in x: #converting the array of arrays into a single array so

putdata can take it.

for p in n:

y.append(p)

nim.putdata(y)

nim.resize((6230/2,3110/2)).save("output" + str(number) + ".png")

print number

</code>