Let m be a message consisting of ℓ aes blocks (say ℓ=100). alice encrypts m using randomized counter mode and transmits the resulting ciphertext to bob. due to a network error, ciphertext block number ℓ/2 is corrupted during transmission. all other ciphertext blocks are transmitted and received correctly. once bob decrypts the received ciphertext, how many plaintext blocks will be corrupted?

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Computers & Technology

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19 January, 08:34

Let m be a message consisting of ℓ aes blocks (say ℓ=100). alice encrypts m using randomized counter mode and transmits the resulting ciphertext to bob. due to a network error, ciphertext block number ℓ/2 is corrupted during transmission. all other ciphertext blocks are transmitted and received correctly. once bob decrypts the received ciphertext, how many plaintext blocks will be corrupted?

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Jadiel Greer · Answer 1

Only 1.

Explanation:

One of the merits of counter mode encryption is that the blocks can be decrypted independently of each other. There is no chaining involved, where one transmission error would propagate into all following blocks.

The "randomized" aspect doesn't have anything to do with it (I think). It applies only to the initialization vector and/or the counter start value, which has to be shared with the receiver in plaintext.

The initialization vector is a combination of a random value and a part where the counter is stored.