Obfuscated Python writeup

Obfuscated Python writeup:
solved by AuPhishYellow member cookthebook for sCTF 2015

Obfuscated python is actually a simple crypto challenge, not a python exploit. If you connect to the server and send 'a', you will be given a number. Then if you send 'b', you will get the number that is one higher than 'a' was. You should also notice that if you send 'aa', the first number you get does not change. Thus, we can assume the following:

1. Each byte is encrypted separately
2. The digits you get are hex encodings of ascii

Thus, all we have to do is calculate what "flag.txt" would be, since we want that file to be printed. The following script checks each letter one at a time. If I were to rewrite it, I would simply add a given value to the byte I was sending based upon what I got, but I used this script first to solve the challenge, and it works even if it is a bit slower:

---------------------------------------------------------------------
import socket, time

total = ''

for j in range(len("flag.txt")):
	print "flag.txt"[0:j+1]
	for i in range(256):
		send = hex(i).replace('0x','')
		if len(send) == 1:
			send = '0' + send
		sendme = total + send

		s = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
		s.connect(("python.sctf.io", 25566))
		s.recv(1024)
		s.send(sendme)
		data = s.recv(1024)
		print data
		if data[-j-2:-1] == "flag.txt"[0:j+1]:
			total += send
			print total
			break
---------------------------------------------------------------------

If you run this, it will slowly go through each value for "i", and eventually print the flag.