CTFs and solutions

decrypt-the-flag/attack.py

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+#!/usr/bin/env python3
+
+from pwn import *
+import random
+from Cryptodome.Cipher import ChaCha20
+from Cryptodome.Util.number import long_to_bytes
+
+HOST = "130.192.5.212"
+PORT = "6561"
+#Using the same seed I generate the same random numbers
+#in the same order
+# Repeating a nonce with the same key
+# reveals the XOR of two different messages, which allows decryption.
+seed = 123
+nlen = 12*8
+random.seed(seed)
+nonce = random.getrandbits(nlen)
+print(f"Nonce:{nonce}")
+random.seed(seed)
+nonce1 = random.getrandbits(nlen)
+#Use this nonce
+print(f"Nonce1:{nonce1}")
+print(long_to_bytes(nonce1).hex())
+# Used nonce
+flag="81d36783bb44a32f060a30aa0551f71c12d81a888dfdd8c317dd3afd0905db796357dbb8642a2c9eae2ab1db2eb7"
+flag = bytes.fromhex(flag)
+amsg="83c07f92ae4ad05b3c7e10dd7472856c63b43df8f588b4b660aa4a917170ab5a0f73fb9b120e5ce78b08c0ad5c8b"
+amsg = bytes.fromhex(amsg)
+apayload = b'A'*46
+ks = bytes(m ^ a for m,a in zip(amsg,apayload))
+fflag = bytes(f ^ k for f,k in zip(flag,ks))
+print(fflag)
+#ks= bytes([f ^ a for f,a in zip(bytes.fromhex(b'A'*46),bytes.fromhex(amsg))])
+#print(bytes([f ^ a for f,a in zip(flag,ks)]))