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  1. #!/usr/bin/env python 
  2. import string
  3. import collections
  4. import sets
  5.  
  6. # XORs two string
  7. def strxor(a, b):     # xor two strings (trims the longer input)
  8.     return "".join([chr(ord(x) ^ ord(y)) for (x, y) in zip(a, b)])
  9.  
  10. # 10 unknown ciphertexts (in hex format), all encrpyted with the same key
  11. c1 = "315c4eeaa8b5f8aaf9174145bf43e1784b8fa00dc71d885a804e5ee9fa40b16349c146fb778cdf2d3aff021dfff5b403b510d0d0455468aeb98622b137dae857553ccd8883a7bc37520e06e515d22c954eba5025b8cc57ee59418ce7dc6bc41556bdb36bbca3e8774301fbcaa3b83b220809560987815f65286764703de0f3d524400a19b159610b11ef3e"
  12. c2 = "234c02ecbbfbafa3ed18510abd11fa724fcda2018a1a8342cf064bbde548b12b07df44ba7191d9606ef4081ffde5ad46a5069d9f7f543bedb9c861bf29c7e205132eda9382b0bc2c5c4b45f919cf3a9f1cb74151f6d551f4480c82b2cb24cc5b028aa76eb7b4ab24171ab3cdadb8356f"
  13. c3 = "32510ba9a7b2bba9b8005d43a304b5714cc0bb0c8a34884dd91304b8ad40b62b07df44ba6e9d8a2368e51d04e0e7b207b70b9b8261112bacb6c866a232dfe257527dc29398f5f3251a0d47e503c66e935de81230b59b7afb5f41afa8d661cb"
  14. c4 = "32510ba9aab2a8a4fd06414fb517b5605cc0aa0dc91a8908c2064ba8ad5ea06a029056f47a8ad3306ef5021eafe1ac01a81197847a5c68a1b78769a37bc8f4575432c198ccb4ef63590256e305cd3a9544ee4160ead45aef520489e7da7d835402bca670bda8eb775200b8dabbba246b130f040d8ec6447e2c767f3d30ed81ea2e4c1404e1315a1010e7229be6636aaa"
  15. c5 = "3f561ba9adb4b6ebec54424ba317b564418fac0dd35f8c08d31a1fe9e24fe56808c213f17c81d9607cee021dafe1e001b21ade877a5e68bea88d61b93ac5ee0d562e8e9582f5ef375f0a4ae20ed86e935de81230b59b73fb4302cd95d770c65b40aaa065f2a5e33a5a0bb5dcaba43722130f042f8ec85b7c2070"
  16. c6 = "32510bfbacfbb9befd54415da243e1695ecabd58c519cd4bd2061bbde24eb76a19d84aba34d8de287be84d07e7e9a30ee714979c7e1123a8bd9822a33ecaf512472e8e8f8db3f9635c1949e640c621854eba0d79eccf52ff111284b4cc61d11902aebc66f2b2e436434eacc0aba938220b084800c2ca4e693522643573b2c4ce35050b0cf774201f0fe52ac9f26d71b6cf61a711cc229f77ace7aa88a2f19983122b11be87a59c355d25f8e4"
  17. c7 = "32510bfbacfbb9befd54415da243e1695ecabd58c519cd4bd90f1fa6ea5ba47b01c909ba7696cf606ef40c04afe1ac0aa8148dd066592ded9f8774b529c7ea125d298e8883f5e9305f4b44f915cb2bd05af51373fd9b4af511039fa2d96f83414aaaf261bda2e97b170fb5cce2a53e675c154c0d9681596934777e2275b381ce2e40582afe67650b13e72287ff2270abcf73bb028932836fbdecfecee0a3b894473c1bbeb6b4913a536ce4f9b13f1efff71ea313c8661dd9a4ce"
  18. c8 = "315c4eeaa8b5f8bffd11155ea506b56041c6a00c8a08854dd21a4bbde54ce56801d943ba708b8a3574f40c00fff9e00fa1439fd0654327a3bfc860b92f89ee04132ecb9298f5fd2d5e4b45e40ecc3b9d59e9417df7c95bba410e9aa2ca24c5474da2f276baa3ac325918b2daada43d6712150441c2e04f6565517f317da9d3"
  19. c9 = "271946f9bbb2aeadec111841a81abc300ecaa01bd8069d5cc91005e9fe4aad6e04d513e96d99de2569bc5e50eeeca709b50a8a987f4264edb6896fb537d0a716132ddc938fb0f836480e06ed0fcd6e9759f40462f9cf57f4564186a2c1778f1543efa270bda5e933421cbe88a4a52222190f471e9bd15f652b653b7071aec59a2705081ffe72651d08f822c9ed6d76e48b63ab15d0208573a7eef027"
  20. c10 = "466d06ece998b7a2fb1d464fed2ced7641ddaa3cc31c9941cf110abbf409ed39598005b3399ccfafb61d0315fca0a314be138a9f32503bedac8067f03adbf3575c3b8edc9ba7f537530541ab0f9f3cd04ff50d66f1d559ba520e89a2cb2a83"
  21. ciphers = [c1, c2, c3, c4, c5, c6, c7, c8, c9, c10]
  22. # The target ciphertext we want to crack
  23. target_cipher = "32510ba9babebbbefd001547a810e67149caee11d945cd7fc81a05e9f85aac650e9052ba6a8cd8257bf14d13e6f0a803b54fde9e77472dbff89d71b57bddef121336cb85ccb8f3315f4b52e301d16e9f52f904"
  24.  
  25. # To store the final key
  26. final_key = [None]*150
  27. # To store the positions we know are broken
  28. known_key_positions = set()
  29.  
  30. # For each ciphertext
  31. for current_index, ciphertext in enumerate(ciphers):
  32.  
  33. 	counter = collections.Counter()
  34. 	# for each other ciphertext
  35. 	for index, ciphertext2 in enumerate(ciphers):
  36. 		if current_index != index: # don't xor a ciphertext with itself
  37. 			for indexOfChar, char in enumerate(strxor(ciphertext.decode('hex'), ciphertext2.decode('hex'))): # Xor the two ciphertexts
  38. 				# 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)
  39. 				if char in string.printable and char.isalpha(): counter[indexOfChar] += 1 # Increment the counter at this index
  40. 	knownSpaceIndexes = []
  41.  
  42. 	# Loop through all positions where a space character was possible in the current_index cipher
  43. 	for ind, val in counter.items():
  44. 		# 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!
  45. 		if val >= 7: knownSpaceIndexes.append(ind)
  46. 	#print knownSpaceIndexes # Shows all the positions where we now know the key!
  47.  
  48. 	# Now Xor the current_index with spaces, and at the knownSpaceIndexes positions we get the key back!
  49. 	xor_with_spaces = strxor(ciphertext.decode('hex'),' '*150)
  50. 	for index in knownSpaceIndexes:
  51. 		# Store the key's value at the correct position
  52. 		final_key[index] = xor_with_spaces[index].encode('hex')
  53. 		# Record that we known the key at this position
  54. 		known_key_positions.add(index)
  55.  
  56. # 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)
  57. final_key_hex = ''.join([val if val is not None else '00' for val in final_key])
  58. # Xor the currently known key with the target cipher
  59. output = strxor(target_cipher.decode('hex'),final_key_hex.decode('hex'))
  60. # Print the output, printing a * if that character is not known yet
  61. print ''.join([char if index in known_key_positions else '*' for index, char in enumerate(output)])
  62.  
  63. '''
  64. Manual step
  65. '''
  66. # From the output this prints, we can manually complete the target plaintext from:
  67. # The secuet-mes*age*is: Wh** usi|g **str*am cipher, nev***use th* k*y *ore than onc*
  68. # to:
  69. # The secret message is: When using a stream cipher, never use the key more than once
  70.  
  71. # We then confirm this is correct by producing the key from this, and decrpyting all the other messages to ensure they make grammatical sense
  72. target_plaintext = "The secret message is: When using a stream cipher, never use the key more than once"
  73. print target_plaintext
  74. key = strxor(target_cipher.decode('hex'),target_plaintext)
  75. for cipher in ciphers:
  76. 	print strxor(cipher.decode('hex'),key)
Success #stdin #stdout 0.01s 23696KB
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The secuet-mes*age*is: Wh** usi|g **str*am cipher, nev***use th* k*y *ore than onc*
The secret message is: When using a stream cipher, never use the key more than once
We can factor the number 15 with quantum computers. We can also factor the number 1
Euler would probably enjoy that now his theorem becomes a corner stone of crypto - 
The nice thing about Keeyloq is now we cryptographers can drive a lot of fancy cars
The ciphertext produced by a weak encryption algorithm looks as good as ciphertext 
You don't want to buy a set of car keys from a guy who specializes in stealing cars
There are two types of cryptography - that which will keep secrets safe from your l
There are two types of cyptography: one that allows the Government to use brute for
We can see the point where the chip is unhappy if a wrong bit is sent and consumes 
A (private-key)  encryption scheme states 3 algorithms, namely a procedure for gene
 The Concise OxfordDictionary (2006) defines crypto as the art of  writing o r sol
https://ideone.com/eVV2v2
language:
Python (cpython 2.7.16)
created:
4 years ago
visibility:
public

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