pycrypto AES CBC











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I have written some code using the pycrypt library and I must be doing something wrong, but I can't figure out what it is that I am doing wrong. I can (nearly) decrypt messages with the wrong initialization vector even though I believe I am following their examples.



from Crypto.Cipher import AES
import os
from string import ascii_letters

key, iv = os.urandom(32), os.urandom(16)
cipher = AES.new(key, AES.MODE_CBC, iv)
plaintext = ascii_letters
plaintext += ' ' * (-len(plaintext) % 16) # Padding
ciphertext = cipher.encrypt(plaintext)

cipher = AES.new(key, AES.MODE_CBC, os.urandom(16))
text = cipher.decrypt(ciphertext)
# text[16:] = b'qrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ '


Now aside from the first 16 bytes of the decrypted text, you have everything decrypted correctly with a totally random choice of IV. Can someone help me figure out where I am going wrong?










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    up vote
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    down vote

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    I have written some code using the pycrypt library and I must be doing something wrong, but I can't figure out what it is that I am doing wrong. I can (nearly) decrypt messages with the wrong initialization vector even though I believe I am following their examples.



    from Crypto.Cipher import AES
    import os
    from string import ascii_letters

    key, iv = os.urandom(32), os.urandom(16)
    cipher = AES.new(key, AES.MODE_CBC, iv)
    plaintext = ascii_letters
    plaintext += ' ' * (-len(plaintext) % 16) # Padding
    ciphertext = cipher.encrypt(plaintext)

    cipher = AES.new(key, AES.MODE_CBC, os.urandom(16))
    text = cipher.decrypt(ciphertext)
    # text[16:] = b'qrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ '


    Now aside from the first 16 bytes of the decrypted text, you have everything decrypted correctly with a totally random choice of IV. Can someone help me figure out where I am going wrong?










    share|improve this question
























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      I have written some code using the pycrypt library and I must be doing something wrong, but I can't figure out what it is that I am doing wrong. I can (nearly) decrypt messages with the wrong initialization vector even though I believe I am following their examples.



      from Crypto.Cipher import AES
      import os
      from string import ascii_letters

      key, iv = os.urandom(32), os.urandom(16)
      cipher = AES.new(key, AES.MODE_CBC, iv)
      plaintext = ascii_letters
      plaintext += ' ' * (-len(plaintext) % 16) # Padding
      ciphertext = cipher.encrypt(plaintext)

      cipher = AES.new(key, AES.MODE_CBC, os.urandom(16))
      text = cipher.decrypt(ciphertext)
      # text[16:] = b'qrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ '


      Now aside from the first 16 bytes of the decrypted text, you have everything decrypted correctly with a totally random choice of IV. Can someone help me figure out where I am going wrong?










      share|improve this question













      I have written some code using the pycrypt library and I must be doing something wrong, but I can't figure out what it is that I am doing wrong. I can (nearly) decrypt messages with the wrong initialization vector even though I believe I am following their examples.



      from Crypto.Cipher import AES
      import os
      from string import ascii_letters

      key, iv = os.urandom(32), os.urandom(16)
      cipher = AES.new(key, AES.MODE_CBC, iv)
      plaintext = ascii_letters
      plaintext += ' ' * (-len(plaintext) % 16) # Padding
      ciphertext = cipher.encrypt(plaintext)

      cipher = AES.new(key, AES.MODE_CBC, os.urandom(16))
      text = cipher.decrypt(ciphertext)
      # text[16:] = b'qrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ '


      Now aside from the first 16 bytes of the decrypted text, you have everything decrypted correctly with a totally random choice of IV. Can someone help me figure out where I am going wrong?







      python encryption aes pycrypto






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      asked Nov 10 at 0:21









      Lars

      61




      61
























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          That's normal for cipher block chaining. In CBC decryption, the IV is only necessary to reconstruct the first block of plaintext. The computation of other blocks of plaintext doesn't actually involve the IV. Here's a diagram (source: Wikimedia user WhiteTimberwolf, public domain):



          CBC decryption diagram






          share|improve this answer





















          • So I guess I was misunderstanding the role of the IV. It is only there to make it more difficult to unravel the encryption by adding additional mixing, but not to add any additional element of security beyond the secrecy of the key. Thanks!
            – Lars
            Nov 10 at 11:49










          • @Lars: The IV does improve security; attacks that would be possible if CBC did not have an IV are not possible due to the IV. However, it is not supposed to be a secret, and the fact that most of the plaintext can be recovered without the IV does not reduce the system's security.
            – user2357112
            Nov 10 at 15:01











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          up vote
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          That's normal for cipher block chaining. In CBC decryption, the IV is only necessary to reconstruct the first block of plaintext. The computation of other blocks of plaintext doesn't actually involve the IV. Here's a diagram (source: Wikimedia user WhiteTimberwolf, public domain):



          CBC decryption diagram






          share|improve this answer





















          • So I guess I was misunderstanding the role of the IV. It is only there to make it more difficult to unravel the encryption by adding additional mixing, but not to add any additional element of security beyond the secrecy of the key. Thanks!
            – Lars
            Nov 10 at 11:49










          • @Lars: The IV does improve security; attacks that would be possible if CBC did not have an IV are not possible due to the IV. However, it is not supposed to be a secret, and the fact that most of the plaintext can be recovered without the IV does not reduce the system's security.
            – user2357112
            Nov 10 at 15:01















          up vote
          1
          down vote













          That's normal for cipher block chaining. In CBC decryption, the IV is only necessary to reconstruct the first block of plaintext. The computation of other blocks of plaintext doesn't actually involve the IV. Here's a diagram (source: Wikimedia user WhiteTimberwolf, public domain):



          CBC decryption diagram






          share|improve this answer





















          • So I guess I was misunderstanding the role of the IV. It is only there to make it more difficult to unravel the encryption by adding additional mixing, but not to add any additional element of security beyond the secrecy of the key. Thanks!
            – Lars
            Nov 10 at 11:49










          • @Lars: The IV does improve security; attacks that would be possible if CBC did not have an IV are not possible due to the IV. However, it is not supposed to be a secret, and the fact that most of the plaintext can be recovered without the IV does not reduce the system's security.
            – user2357112
            Nov 10 at 15:01













          up vote
          1
          down vote










          up vote
          1
          down vote









          That's normal for cipher block chaining. In CBC decryption, the IV is only necessary to reconstruct the first block of plaintext. The computation of other blocks of plaintext doesn't actually involve the IV. Here's a diagram (source: Wikimedia user WhiteTimberwolf, public domain):



          CBC decryption diagram






          share|improve this answer












          That's normal for cipher block chaining. In CBC decryption, the IV is only necessary to reconstruct the first block of plaintext. The computation of other blocks of plaintext doesn't actually involve the IV. Here's a diagram (source: Wikimedia user WhiteTimberwolf, public domain):



          CBC decryption diagram







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered Nov 10 at 0:34









          user2357112

          148k12151240




          148k12151240












          • So I guess I was misunderstanding the role of the IV. It is only there to make it more difficult to unravel the encryption by adding additional mixing, but not to add any additional element of security beyond the secrecy of the key. Thanks!
            – Lars
            Nov 10 at 11:49










          • @Lars: The IV does improve security; attacks that would be possible if CBC did not have an IV are not possible due to the IV. However, it is not supposed to be a secret, and the fact that most of the plaintext can be recovered without the IV does not reduce the system's security.
            – user2357112
            Nov 10 at 15:01


















          • So I guess I was misunderstanding the role of the IV. It is only there to make it more difficult to unravel the encryption by adding additional mixing, but not to add any additional element of security beyond the secrecy of the key. Thanks!
            – Lars
            Nov 10 at 11:49










          • @Lars: The IV does improve security; attacks that would be possible if CBC did not have an IV are not possible due to the IV. However, it is not supposed to be a secret, and the fact that most of the plaintext can be recovered without the IV does not reduce the system's security.
            – user2357112
            Nov 10 at 15:01
















          So I guess I was misunderstanding the role of the IV. It is only there to make it more difficult to unravel the encryption by adding additional mixing, but not to add any additional element of security beyond the secrecy of the key. Thanks!
          – Lars
          Nov 10 at 11:49




          So I guess I was misunderstanding the role of the IV. It is only there to make it more difficult to unravel the encryption by adding additional mixing, but not to add any additional element of security beyond the secrecy of the key. Thanks!
          – Lars
          Nov 10 at 11:49












          @Lars: The IV does improve security; attacks that would be possible if CBC did not have an IV are not possible due to the IV. However, it is not supposed to be a secret, and the fact that most of the plaintext can be recovered without the IV does not reduce the system's security.
          – user2357112
          Nov 10 at 15:01




          @Lars: The IV does improve security; attacks that would be possible if CBC did not have an IV are not possible due to the IV. However, it is not supposed to be a secret, and the fact that most of the plaintext can be recovered without the IV does not reduce the system's security.
          – user2357112
          Nov 10 at 15:01


















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