D
defcon8
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>
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>
