Automne's Shadow.

ASIS CTF 2019 A delicious soup WriteUp

python reverse
2019/04/23 Share

Crypto

时间有限，只做了这个比赛的密码学签到题，签到题都这个难度，我枯了。

给了flag.enc文件和加密文件。

flag.enc内容：

11d.3ilVk_d3CpIO_4nlS.ncnz3e_0S}M_kn5scpm345n3nSe_u_S{iy__4EYLP_aAAall

加密的脚本如下：

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#!/usr/bin/env python
#-*- coding:utf-8 -*-

import random
from flag import flag

def encrypt(msg, perm):
	W = len(perm)
	while len(msg) % (2*W):
		msg += "."
	msg = msg[1:] + msg[:1]
	msg = msg[0::2] + msg[1::2]
	msg = msg[1:] + msg[:1]
	res = ""
	for j in xrange(0, len(msg), W):
		for k in xrange(W):
			res += msg[j:j+W][perm[k]]
	msg = res
	return msg

def encord(msg, perm, l):
	for _ in xrange(l):
		msg = encrypt(msg, perm)
	return msg

W, l = 7, random.randint(0, 1337)
perm = range(W)
random.shuffle(perm)

enc = encord(flag, perm, l)
f = open('flag.enc', 'w')
f.write(enc)
f.close()

然后就是阅读代码，密文cipher其实是encord(flag, perm, l)的输出

flag未知
perm是range(7)生成的列表，再经过random.shuffle(perm)打乱顺序，这种组合其实不多，可以爆破
l是random.randint(0, 1337)生成的随机数，也是可以爆破的

接着再看encord()函数，这个函数其实就是执行了l次的encrypt(msg, perm)

分析这个函数可以看到首先判断如果不能被14整除就添加.

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while len(msg) % (2*7):
	msg += "."

然后做了一些替换操作

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msg = msg[1:] + msg[:1]
msg = msg[0::2] + msg[1::2]
msg = msg[1:] + msg[:1]

最后就是将上面生成的msg信息每7位一组，按照perm里的排序重新组合

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res = ""
for j in xrange(0, len(msg), 7):
	for k in xrange(7):
		res += msg[j:j+7][perm[k]]
msg = res
return msg

代码逻辑分析清楚后，整个过程其实除了加了.号之外，msg的长度其实没变，因此有flag.enc的长度为70，且包含了两个.可知，flag的长度为68位。

开始写逆向代码。

有几点需要注意，首先：

  1. msg = msg[1:] + msg[:1]的逆向过程是msg = msg[-1:] + msg[:-1]
  2. 调用iter = itertools.permutations(perm0,7)去生成7位的组合
  3. 使用perm.index(k)去获取k的索引

ok，最后给出爆破脚本：

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#encoding=utf-8

import itertools
import random
import time
import sys


def decrypt(cipher,perm):
	msg = ""
	msg2 = ""
	for j in xrange(0,70,7):
		for k in xrange(7):
			msg += cipher[j:j+7][perm.index(k)]
	#print msg + "$$$$$$$$$$$$$"
	#print len(msg)
	msg = msg[-1:] + msg[:-1]
	for m in xrange(35):
		msg2 += msg[m::35]
	msg3 = msg2[-1:] + msg2[:-1]
	#plaintext = msg[:-2]
	#print len(msg)
	return msg3


def brute():
	perm0 = range(7)
	list1 = []
	iter = itertools.permutations(perm0,7)
	list1.append(list(iter))
	for i in xrange(len(list1[0])):
		cipher = "11d.3ilVk_d3CpIO_4nlS.ncnz3e_0S}M_kn5scpm345n3nSe_u_S{iy__4EYLP_aAAall"
		#print n
		perm = list(list1[0][i])
		print perm
		for m in xrange(1338):
			#print cipher
			#print m
			cipher = decrypt(cipher,perm)
			print cipher
			if cipher.startswith('ASIS{') and cipher.endswith('}..'):
				print cipher
				print m
				print perm
				sys.exit(0)
		print "Nothing Found"
	print "All down."

if __name__ == "__main__":
	brute()

几分钟后即可得到flag：

automne

原文作者: Automne

原文链接: https://ce-automne.github.io/2019/04/23/ASIS-CTF-2019-ADeliciousSoup-WriteUp/

发表日期: April 23rd 2019, 10:46:00 pm

版权声明: 本文采用知识共享署名-非商业性使用 4.0 国际许可协议进行许可

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