munintrollet

Author: Zukane

We intercepted some encrypted communication we think is very important. But it appears they are using uncrackable OTP? We have two ciphertexts from the communication, and also the plaintext for the first. Can you help decrypt the second?

Challenge Overview

In this crypto challenge, we are given a intercept.txt textfile containing the following:

# Known plaintext 1:
"Always use the one master OTP for all secure communications, it is uncrackable. Now listen carefully"

# Intercepted ciphertext 1 (hex):
c607eea76fc53ada95c9e7321434c264822158f8cc8ae2517deb4cf1ef6687bb4eecff4ac57c5bfcb1f39d5265d89a5e451171f2e3e1d3ffd8916f328e1ea7b3b073466a7e8eb7b56255051de269996654b85254cff2e49b71faec792c908d318ad42c1d

# Plaintext 2: 
🤷♂

# Intercepted ciphertext 2 (hex):
cb02f4af629663c093dee72b1922916a8b210bb5d996b6053ffb60cdde3492f448a3be40c02808edbae3cf5a24c881565a4450cfd0ae87d8d8883c6bdd12f3e7b169153f6388a6a6644a5e5fcb5ce33d6de4491793f6a4b060a4937734bd8325ceda6119    

Essentially, we are given C1, C2 and P1 where P2 is presumably the flag. A One-Time-Pad can easily be broken if the same key is used multiple times. The messages are encrypted by performing:

$$C1 = P1 \oplus K$$

$$C2 = P2 \oplus K$$

Since the XOR has the inverse property, we can retrieve K by performing

$$K = C1 \oplus P1$$

and with the key, we can simply XOR with C2 to decrypt:

$$C1 \oplus K = P2$$

Solution

def xor_bytes(bytes1, bytes2):
    return bytes([b1 ^ b2 for b1, b2 in zip(bytes1, bytes2)])

plaintext1 = b"Always use the one master OTP for all secure communications, it is uncrackable. Now listen carefully"
ciphertext1 = bytes.fromhex("c607eea76fc53ada95c9e7321434c264822158f8cc8ae2517deb4cf1ef6687bb4eecff4ac57c5bfcb1f39d5265d89a5e451171f2e3e1d3ffd8916f328e1ea7b3b073466a7e8eb7b56255051de269996654b85254cff2e49b71faec792c908d318ad42c1d")
ciphertext2 = bytes.fromhex("cb02f4af629663c093dee72b1922916a8b210bb5d996b6053ffb60cdde3492f448a3be40c02808edbae3cf5a24c881565a4450cfd0ae87d8d8883c6bdd12f3e7b169153f6388a6a6644a5e5fcb5ce33d6de4491793f6a4b060a4937734bd8325ceda6119")
key = key[:len(ciphertext2)]

plaintext2 = xor_bytes(ciphertext2, key).decode()
print(plaintext2)

This gives us the output:

Limit your messages to 100chars to fit the master OTP. Now use this secret: EPT{w3lc0m3_t0_my_kr1b!}

flag: EPT{w3lc0m3_t0_my_kr1b!}