Rating: 2.5
## flagcrypt 100 (crypto)
### Solution
Send a message with length >= 20 to the server, and the server will reponse with the encrypted message and its length.
Through observing the provided code, the length of encrypted message will be shorter if there is repeated pattern (length >= 3) occur in the plaintext, thus the encryption is vulnerable to [compression side channel attacks](https://www.sjoerdlangkemper.nl/2016/08/23/compression-side-channel-attacks/).
The charset was told in the description, this exploit simply bruteforce the products of these characters with length = 3 first. Once a pattern found, it will try to extended it with single character at once.
A problem is there are 3 bytes patterns with common prefix or subfix ("me_", "ve_") and ("e\_d", "e\_a"), thus I manually set the flag variable and run the script again.
```python
#!/usr/bin/env python3
from pwn import *
import itertools
charset = set("abcdefghijklmnopqrstuvwxyz_")
r = remote("crypto.chal.csaw.io", 8041)
payload = "^" * 20
r.writelineafter("service\n", payload + "///")
data = r.recvline().strip()
data, sz = data[:-1], int(data[-1])
minsz = sz
flag = ""
def front_extend():
global flag
p = flag[:2]
lst = []
for c in charset:
guess = payload + c + p
r.writelineafter("service\n", guess)
data = r.recvline().strip()
data, sz = data[:-1], int(data[-1])
lst.append((sz, c))
minsz, c = min(lst, key=lambda t: t[0])
maxsz, _ = max(lst, key=lambda t: t[0])
lst = list(filter(lambda t: t[0] == minsz, lst))
if minsz == maxsz:
return
if len(lst) > 1:
print("Multiple possibilities :", lst)
else:
flag = c + flag
print("flag : %s ... " % flag)
front_extend()
def back_extend():
global flag
p = flag[-2:]
lst = []
for c in charset:
guess = payload + p + c
r.writelineafter("service\n", guess)
data = r.recvline().strip()
data, sz = data[:-1], int(data[-1])
lst.append((sz, c))
minsz, c = min(lst, key=lambda t: t[0])
maxsz, _ = max(lst, key=lambda t: t[0])
lst = list(filter(lambda t: t[0] == minsz, lst))
if minsz == maxsz:
return
if len(lst) > 1:
print("Multiple possibilities :", lst)
else:
flag += c
print("flag : %s ... " % flag)
back_extend()
for p in itertools.product(charset, repeat=3):
p = "".join(p)
guess = payload + p
r.writelineafter("service\n", guess)
data = r.recvline().strip()
data, sz = data[:-1], int(data[-1])
print(guess, sz)
if sz < minsz:
flag = p
break
print("found %s, extending..." % flag)
front_extend()
back_extend()
```
### Flag
```
crime_doesnt_have_a_logo
```
