Rating: 5.0

## Challenge name: Random ECB

### Description:
> nc ecb.utctf.live 9003
> P.S: given file: *server.py*

### Solution:

We have a server with it's application. Here I explain how it works:

When you connect to server, first of all, it choose 16bytes of random, for **AES** key, then it asks you to "*Input a string to encrypt (input 'q' to quit):*". After that, it will concat your string with **flag** and then encrypt it using **AES ECB mode**. And in 50% of cases, it will put a character ('*A*') at the first. If you read about **ECB mode** you realize that encrypting each block in this mode is independent to each other. So what now ?!

Assume we give the server these 15 characters (e.g. *'aaaaaaaaaaaaaaa'*), then the first block will become on of these:

aaaaaaaaaaaaaaa? ('?' is the first character of flag)

and because the key is same for this connection, we will only 2 encryption text for this input. Now if we use bruteforce, and try all characters in the '*?*' place, and compare it to encrypted text received by the '*aaaaaaaaaaaaaaa*' input, you can find the first character of the flag! (Sorry, I'm not so good in explaining :P). Then Use this technique for finding all characters of the flag.

**Bruteforce function**

def bf():
global flag
pf = 'a' * (31 - len(flag))
print('\t|--Poison data: "%s"' % (pf))
resp1 = f_send(pf)[:64]
resp2 = ''
while True:
resp2 = f_send(pf)[:64]
if resp1 != resp2:
print('\t\t|--resp1: %s' % (resp1))
print('\t\t|--resp2: %s' % (resp2))

for c in pchar:
pd = pf + flag + c
print('\t\t|--Test on "%s"' % (pd))
res1 = f_send(pd)[:64]
res2 = ''
while True:
res2 = f_send(pd)[:64]
if res2 != res1:

print('\t\t\t|--res1: %s' % (res1))
print('\t\t\t|--res2: %s' % (res2))
if (res1 == resp1 and res2 == resp2) or (res1 == resp2 and res2 == resp1):
flag += c
print('\t\t\t|--Character found: %c' % (c))
return 0

print ('\t\t\t|--not found!')
return -1

We know The flag size is between 16bytes and 32bytes. So i start with sending 31 characters and comparing 2bytes!

[final script.py](./script.py)

**The Flag**


Original writeup (https://github.com/RaaCT0R/CTF-Challenges/blob/master/cryptography/utctf2020/Random-ECB/WRITEUP.md).