Tags: crypto

Rating:

### XOR Crypter - Crypto 200pts

## Problem

Description: The state of art on encryption, can you defeat it?
CjBPewYGc2gdD3RpMRNfdDcQX3UGGmhpBxZhYhFlfQA=

## Solution

We get Python script used for encrypt the flag. Output of this script is Base64 string contains encrypted flag (CjBPewYGc2gdD3RpMRNfdDcQX3UGGmhpBxZhYhFlfQA=)

python
import struct
import sys
import base64

if len(sys.argv) != 2:
print "Usage: %s data" % sys.argv[0]
exit(0)

data = sys.argv[1]
padding = 4 - len(data) % 4
data = data + "\x00" * padding

result = []
blocks = struct.unpack("I" * (len(data) / 4), data)
for block in blocks:
result += [block ^ block >> 16]

output = ''
for block in result:
output += struct.pack("I", block)

print base64.b64encode(output)


To resolve this, we have to create "decrypter".

Here's my sample solution for this, maybe not state-of-the-art, but I was able to get the flag :)

python
#!/usr/bin/env python

import struct
import sys
import base64

data = base64.b64decode(sys.argv[1])
padding = 4 - len(data) % 4

data = data + "\x00" * padding

print data

i = 0
output = ''
result = ''
while i < len(data):
output = struct.unpack("I", junk)
for s in output:
r = s ^ s >> 16
result += struct.pack("I", r)
i += 4

print result



And here's output with flag:


\$ ./test.py CjBPewYGc2gdD3RpMRNfdDcQX3UGGmhpBxZhYhFlfQA=

0O{♠♠sh↔☼ti1‼_t7►_u♠→hi▬ab◄e}
0 4
4 4
8 4
12 4
16 4
20 4
24 4
28 4
32 4
EKO{unshifting_the_unshiftable}


Original writeup (https://github.com/bl4de/ctf/blob/master/2015/Ekoparty_CTF_2015/XOR_Crypter_Crypto200.md).