> Cellular Automata>> 148>> It's hard to reverse a step in a cellular automata, but solvable if done right.> > https://cellularautomata.web.ctfcompetition.com/
As the [rules](https://cellularautomata.web.ctfcompetition.com/) state we are dealing with a [Rule 126 automata](http://mathworld.wolfram.com/Rule126.html). Patterns ```000``` and ```111``` produce a ```0``` bit while all others produce ```1```.
The problem with reversing cellular automata is that a lot of different steps correspond to the step you are trying to reverse. Straight bruteforcing will take forever - 64-bit steps are just too bit. However, the bits in the reverse step are not random, and are limited by 2 rules:
1. ```0``` has to reverse to patterns ```000``` or ```111```, and ```1``` to all others2. If a pattern is selected the patterns that follow it are limited to 2 possibilities (because they have to overlap by 2 bits). For example, pattern ```000``` can only be followed by patterns ```000``` and ```001```.
If we apply these 2 rules we can generate all possibilities very quickly with the help of the following script:
```pythonimport sys
if len(sys.argv) != 2:	print "Please supply a hex number on the command line"	quit() hexval = sys.argv[1]bit_size = len(hexval)*4
# convert hex value to bit stringbitstr = (bin(int(hexval,16))[2:]).zfill(bit_size)
# map from bits to patterns that generate itpatterns_generating_bit = {"0":[0,7], "1":[1,2,3,4,5,6]}
# valid patterns that can follow each pattern; for example, pattern 010 can be followed #  only by 100 and 101 because they must overlap with its last 2 digits (10)valid_next_patterns = {0:[0,1], 1:[2,3], 2:[4,5], 3:[6,7], 4:[0,1], 5:[2,3], 6:[4,5], 7:[6,7]}
# mid bits in each patternpattern_mid_bits = {0:"0", 1:"0", 2:"1", 3:"1", 4:"0", 5:"0", 6:"1", 7:"1"}
def reverse_rule126(bitstr, depth, valid_patterns, patterns_in_step): 	# walk through all patterns that generate the current bit	for pattern in patterns_generating_bit[bitstr[depth]]:		# make sure the pattern is valid based on previously seen patterns		if pattern in valid_patterns:  			# if we are not at the last bit - keep going recursively			if depth < (bit_size-1):				reverse_rule126(bitstr, depth+1, valid_next_patterns[pattern], patterns_in_step+[pattern])
			# if we are at the last bit...			if depth == (bit_size-1):				# ...and the last pattern wraps around properly to the beginning of the step string 				if patterns_in_step[0] in valid_next_patterns[pattern]: 					# generate the full bitstring for the step and print it out					found_step = ""					for x in patterns_in_step:						found_step += pattern_mid_bits[x]					found_step += pattern_mid_bits[pattern] 					print hex(int(found_step,2))[2:]
reverse_rule126(bitstr, 0, [0,1,2,3,4,5,6,7], [])```
The script generates about 10K possibilities, which we can try to determine if there is a flag in the output:
```bash#!/bin/sh
for i in $(python solve.py 66de3c1bf87fdfcf); do    echo "$i" > /tmp/plain.key; xxd -r -p /tmp/plain.key > /tmp/enc.key    echo "U2FsdGVkX1/andRK+WVfKqJILMVdx/69xjAzW4KUqsjr98GqzFR793lfNHrw1Blc8UZHWOBrRhtLx3SM38R1MpRegLTHgHzf0EAa3oUeWcQ=" | openssl enc -d -aes-256-cbc -pbkdf2 -md sha1 -base64 --pass file:/tmp/enc.key 2>/dev/null | grep CTFdone
```
When we run the script we quickly get the flag decoded:
```sh$ ./solve.shCTF{reversing_cellular_automatas_can_be_done_bit_by_bit}```
The flag is ```CTF{reversing_cellular_automatas_can_be_done_bit_by_bit}```.
