多次一密（Many Time Pad Attack）

用同一个密钥去加密多条明文，当密文条数较多时就很容易被攻击，例如Many Time Pad。

这个攻击的原理是 c1⊕c2=m1⊕m2c1⊕c2=m1⊕m2，而通过 m1⊕m2m1⊕m2 可以分析出 m1⊕m2m1⊕m2，因此 m1⊕m2m1⊕m2 不再安全。

#!/usr/bin/python
## OTP - Recovering the private key from a set of messages that were encrypted w/ the same private key (Many time pad attack) - crypto100-many_time_secret @ alexctf 2017
# Original code by jwomers: https://github.com/Jwomers/many-time-pad-attack/blob/master/attack.py)

import string
import collections
import sets, sys

# 11 unknown ciphertexts (in hex format), all encrpyted with the same key
c1='daaa4b4e8c996dc786889cd63bc4df4d1e7dc6f3f0b7a0b61ad48811f6f7c9bfabd7083c53ba54'
c2='c5a342468c8c7a88999a9dd623c0cc4b0f7c829acaf8f3ac13c78300b3b1c7a3ef8e193840bb'
c3='dda342458c897a8285df879e3285ce511e7c8d9afff9b7ff15de8a16b394c7bdab920e7946a05e9941d8308e'
c4='d9b05b4cd5ce7c8f938bd39e24d0df191d7694dfeaf8bfbb56e28900e1b8dff1bb985c2d5aa154'
c5='d9aa4b00c88b7fc79d99d38223c08d54146b88d3f0f0f38c03df8d52f0bfc1bda3d7133712a55e9948c32c8a'
c6='c4b60e46c9827cc79e9698936bd1c55c5b6e87c8f0febdb856fe8052e4bfc9a5efbe5c3f57ad4b9944de34'
c7='d9aa5700da817f94d29e81936bc4c1555b7b94d5f5f2bdff37df8252ffbecfb9bbd7152a12bc4fc00ad7229090'
c8='c4e24645cd9c28939a86d3982ac8c819086989d1fbf9f39e18d5c601fbb6dab4ef9e12795bbc549959d9229090'
c9='d9aa4b598c80698a97df879e2ec08d5b1e7f89c8fbb7beba56f0c619fdb2c4bdef8313795fa149dc0ad4228f'
c10='cce25d48d98a6c8280df909926c0de19143983c8befab6ff21d99f52e4b2daa5ef83143647e854d60ad5269c87'
c11='d9aa4b598c85668885df9d993f85e419107783cdbee3bbba1391b11afcf7c3bfaa805c2d5aad42995ede2cdd82977244'
ciphers = [c1, c2, c3, c4, c5, c6, c7, c8, c9, c10, c11]
# The target ciphertext we want to crack

# XORs two string
def strxor(a, b):     # xor two strings (trims the longer input)
    return "".join([chr(ord(x) ^ ord(y)) for (x, y) in zip(a, b)])

def target_fix(target_cipher):
    print '-------begin-------'

    # To store the final key
    final_key = [None]*150
    # To store the positions we know are broken
    known_key_positions = set()

    # For each ciphertext
    for current_index, ciphertext in enumerate(ciphers):
        counter = collections.Counter()
        # for each other ciphertext
        for index, ciphertext2 in enumerate(ciphers):
            if current_index != index: # don't xor a ciphertext with itself
                for indexOfChar, char in enumerate(strxor(ciphertext.decode('hex'), ciphertext2.decode('hex'))): # Xor the two ciphertexts
                    # If a character in the xored result is a alphanumeric character, it means there was probably a space character in one of the plaintexts (we don't know which one)
                    if char in string.printable and char.isalpha(): counter[indexOfChar] += 1 # Increment the counter at this index
        knownSpaceIndexes = []

        # Loop through all positions where a space character was possible in the current_index cipher
        for ind, val in counter.items():
            # If a space was found at least 7 times at this index out of the 9 possible XORS, then the space character was likely from the current_index cipher!
            if val >= 7: knownSpaceIndexes.append(ind)
        #print knownSpaceIndexes # Shows all the positions where we now know the key!

        # Now Xor the current_index with spaces, and at the knownSpaceIndexes positions we get the key back!
        xor_with_spaces = strxor(ciphertext.decode('hex'),' '*150)
        for index in knownSpaceIndexes:
            # Store the key's value at the correct position
            final_key[index] = xor_with_spaces[index].encode('hex')
            # Record that we known the key at this position
            known_key_positions.add(index)

    # Construct a hex key from the currently known key, adding in '00' hex chars where we do not know (to make a complete hex string)
    final_key_hex = ''.join([val if val is not None else '00' for val in final_key])
    # Xor the currently known key with the target cipher
    output = strxor(target_cipher.decode('hex'),final_key_hex.decode('hex'))

    print "Fix this sentence:"
    print ''.join([char if index in known_key_positions else '*' for index, char in enumerate(output)])+"\n"

    # WAIT.. MANUAL STEP HERE 
    # This output are printing a * if that character is not known yet
    # fix the missing characters like this: "Let*M**k*ow if *o{*a" = "cure, Let Me know if you a"
    # if is too hard, change the target_cipher to another one and try again
    # and we have our key to fix the entire text!

    #sys.exit(0) #comment and continue if u got a good key

    print '------end------'
    
for i in ciphers:
    target_fix(i)

/**
* 复制并使用代码请注明引用出处哦~
* Lazzaro @ https://lazzzaro.github.io
*/