# scs7

Author: trupplesContest: TokyoWesterns CTF 4th 2018

## Proof of Flag

```
TWCTF{67ced5346146c105075443add26fd7efd72763dd}
```

## Summary

Base59 with a shuffled alphabet. We can figure out the alphabet by sending it bytes from 59 to 117, observing the last encrypted character, and then using the observed alphabet to decrypt the flag.

## Proof of solving

I initially tried sending `TWCTF{`

+ each character and seeing which one would
match the largest prefix. The problem with that is that as they use base59, the
symbol boundaries don’t really line up with encodings with power-of-two symbols
so I got many false positives and false negatives.

Here’s the successful solution:

I tried sending all characters from 0 to 99 and seeing the result:

```
from pwn import *
def request(plaintext):
r.recvuntil("message: ")
r.sendline(plaintext)
r.recvuntil("ciphertext: ")
return r.recvline()[:-1]
r = remote("crypto.chal.ctf.westerns.tokyo", 14791)
requestcount = 0
r.recvuntil("encrypted flag: ")
encflag = r.recvline().strip()
print("Target flag: \n" + encflag)
for i in range(0, 256):
print(i, request(chr(i)))
```

And that made the encryption algorithm obvious. Here’s an example output of that program:

```
Target flag:
0KhpGLHUSc3D7B449JsxM4L7etCnYW9HH5jBpo72kPgrgXeWzbKxbhyrkN29gb64
(0, '')
(1, 'S')
(2, 'm')
(3, 'h')
...
(38, 'x')
(39, 'f')
(40, 'E')
...
(57, 's')
(58, '6')
(59, 't')
(60, 'ST')
(61, 'SS')
(62, 'Sm')
(63, 'Sh')
...
(97, 'Sx')
(98, 'Sf')
(99, 'SE')
```

It is clear that the last letter repeats every 59 values. We could call this a base 59 encoding, but the symbols are not in alphabetical order so the output changes every time. It’s safe to assume that the alphabet is shuffled for every connection.

In this case we only need to observe 59 encryptions to reconstruct the alphabet and decrypt the flag:

```
from pwn import *
r = remote("crypto.chal.ctf.westerns.tokyo", 14791)
r.recvuntil("encrypted flag: ")
encflag = r.recvline().strip()
print("Target flag: \n" + encflag)
alphabet = ["?"] * 59
for i in range(60, 60+59):
r.recvuntil("message: ")
r.sendline(chr(i))
r.recvuntil("ciphertext: ")
ct = r.recvline().strip()
alphabet[i % 59] = ct[-1]
print(i-60) # to know the progress
plaintext_num = 0
for c in encflag:
plaintext_num = plaintext_num * 59 + alphabet.index(c)
print(hex(plaintext_num))
# => 0x54574354467b363763656435333436313436633130353037353434336164643236666437656664373237363364647d
# => TWCTF{67ced5346146c105075443add26fd7efd72763dd}
```