Following the same definition in this question for perfect secrecy for two messages $m,m' \in \mathcal{M}$. I don't understand how the accepted answer produces a secure system? I mean The adversary would always be able to xor the two ciphertexts to obtain the xor of the original message s thus changing the distribution of $\mathcal{M}$ and violating the definition.

I tried to come up with another scheme where $\mathcal{M}=\{0,1\}^l,\mathcal{K}=\{0,1\}^{l2^l}$, where each key $k$ would the concatenation of smaller keys $k_0,...,k_{2^l-1}$ of size $l$ and to encrypt a message $m \in \mathcal{M}$ we xor it with $k_{m}$. This way different message will be encrypted using different small keys thus maintaining perfect secrecy.

However, this will not allow decryption by xoring back with $k_{c}$ because it might be that $k_{c}\neq k_{m}$. After thinking for a while I start to suspect if one-time pad can be extended to satisfy the above definition or am I doing something wrong?


I know a comment on the original question would be more relevent, but couldn't do that due to reputation restrictions.

  • $\begingroup$ xor'ing ciphertexts creates a new random variable. it does not change the distribution on $\cal M$ which is given and fixed. $\endgroup$ – kodlu Feb 15 '18 at 20:57
  • $\begingroup$ By that, I meant some messages will be more probable than others after seeing the two ciphertexts $\endgroup$ – Shadowfirex Feb 16 '18 at 3:10

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