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
if padding != 0:
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
if padding != 0:
data = data + "\x00" * padding
print data
i = 0
padding = 4
output = ''
result = ''
while i < len(data):
junk = data[i:i + padding]
print i, padding
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}
```
