Can attacker create encryption of message m XOR 1^n given the ciphertext c, in nonce-based counter mode?

Knightoforous

11/20/23, 7:44 AM

The following question is from Stanford cryptography course final exam paper.

Suppose an attacker intercepts a ciphertext c which is the encryption of a message m ε {0, 1}^n under nonce-based counter mode. Can the attacker create the encryption of m XOR 1^n just given c? If so, explain how. If not, explain why not.

I cannot prove why that wouldn't be possible, although intuitively given it is a counter mode, I feel the attacker cannot create the encryption.

1 + 0

aes

block-cipher

modes-of-operation

nonce

Score:0

Crypto

Sezzart

11/20/23, 8:26 AM

Yes, the ciphertext is plaintext xor pad. Since in case we want a ciphertext for a changed plaintext, the pad will not be altered since its not text-dependent. Therefore, consider how would the ciphertext change if You add 1^n to the plaintext

+ 7

Knightoforous

11/20/23, 9:10 AM

It is still unclear to me. Could you elaborate please?

Sezzart

11/20/23, 9:11 AM

Which part is unclear?

Knightoforous

11/20/23, 9:34 AM

Everything that you said. I understand that in CTR mode, we have ciphertext C = M XOR E(Nonce||Counter), where E is the encryption function. Now we have been given C and then how do we find out E(M XOR 1^n) or is it asking us to find E(M) XOR 1^n?

Sezzart

11/20/23, 10:10 AM

You get C=Enc(M)=AES(Nonce||Counter) + M. You want to find C' = Enc(M+1^n) which will be....?

Knightoforous

11/20/23, 12:52 PM

C'= Enc(M + 1^n) = AES(Nonce||Counter) + M + 1^n?

Sezzart

11/20/23, 7:57 PM

yes, which means what for Your task of trying to predict what the cipher of M + 1^n will be?

Knightoforous

11/21/23, 5:37 AM

Yes, so the cipher of M + 1^n will be C' = C + 1^n

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EN: Can attacker create encryption of message m XOR 1^n given the ciphertext c, in nonce-based counter mode?

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