README.md

Hyper Loop

Category

Crypto

Description

The author is aware that XOR alone is not sufficiently secure, but they have implemented a solution to address this issue. Use the provided Python script to recover the original flag.

Format : Hero{flag}
Author : xanhacks

Files

• hyper_loop.py

Write Up (Xanhacks)

The key is a random value repeated three times, example:

>> from os import urandom
>>> urandom(6) * 3
b'\xec\xab\xf6\x02s\x87\xec\xab\xf6\x02s\x87\xec\xab\xf6\x02s\x87'
# b'\xec\xab\xf6\x02s\x87  \xec\xab\xf6\x02s\x87  \xec\xab\xf6\x02s\x87'
>>> urandom(6) * 3
b'cW\xc4\xbe\x13\xf5cW\xc4\xbe\x13\xf5cW\xc4\xbe\x13\xf5'
# b'cW\xc4\xbe\x13\xf5  cW\xc4\xbe\x13\xf5  cW\xc4\xbe\x13\xf5'

The flag is XORed 32 times:

output[0] = flag[0] ^ key[0] ^ key[1] ^ ... ^ key[31]
output[1] = flag[1] ^ key[0] ^ key[1] ^ ... ^ key[31]
...
output[17] = flag[17] ^ key[0] ^ key[1] ^ ... ^ key[31]

However, its like the flag has been XORed only one time:

output[0] = flag[0] ^ X
with X = key[0] ^ key[1] ^ ... ^ key[31] 
output[1] = flag[1] ^ Y
with Y = key[0] ^ key[1] ^ ... ^ key[31] 

You can find the 5 first chars of the key using Hero{ and the last char using the last char of the flag }.

#!/usr/bin/env python3

flag = bytearray(b"Hero{????????????}")
output = bytearray(b'\x05p\x07MS\xfd4eFPw\xf9}%\x05\x03\x19\xe8')


flag[5] = output[5] ^ flag[17] ^ output[17]

for i in range(6):
	flag[6 + i] = output[6 + i] ^ flag[i] ^ output[i]

for i in range(5):
	flag[12 + i] = output[12 + i] ^ flag[i] ^ output[i]

print(flag)

Execution:

$ python3 solve.py     
bytearray(b'Hero{hyp3r_l00p!1}')

Write Up (Alol)

This challenge is a classical flag XOR key challenge with a little twist.

The flag is xored with not 1 but 32 keys. The keys are composed of 6 random bytes repeated 3 times. Thus we have something that looks like this, where we know enc and flag but not the values of k :

$$\begin{align*} enc_0 &= flag_0 \oplus k_{0,0} \oplus k_{0,1} \oplus ... \oplus k_{0,32} \\\ enc_1 &= flag_1 \oplus k_{1,0} \oplus k_{1,1} \oplus ... \oplus k_{1,32} \\\ ... \\\ enc_{17} &= flag_{17} \oplus k_{17,0} \oplus k_{17,1} \oplus ... \oplus k_{17,32} \\\ \end{align*}$$

Notice that we can simplify the equations as :

$$\begin{align*} enc_0 &= flag_0 \oplus K_{0} \\\ enc_1 &= flag_1 \oplus K_{1} \\\ ... \\\ enc_{17} &= flag_{17} \oplus K_{17} \\\ \end{align*}$$

Where :

$$K_{i} = k_{i,0} \oplus k_{i,1} \oplus ... \oplus k_{i,32}$$

We don't need to know the values of all the keys, we only need to find the values of K_i. Since we know 6 bytes of plaintext (Hero{ and }) we can find K and decrypt the flag.

xor = lambda a,b: bytes(x^y for x,y in zip(a,b))
enc = b'\x05p\x07MS\xfd4eFPw\xf9}%\x05\x03\x19\xe8'
K   = xor(b'Hero{', enc) + xor(b'}', enc[-1::])
print(xor(enc, K*3))

Flag

Hero{hyp3r_l00p!1}