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Let's assume all operations are done on $\mathbb{Z}_p$ where $p$ is a large non-prime number.

To mask a value $a$, we do the following:

  1. Pick a uniformly random value: $r$, from the ring.

  2. Do as follows: $c= r+a \bmod p$.

Question: Is the above one-time pad secure?

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    $\begingroup$ This looks like homework. What have you done to answer this question yourself? Where are you stuck? $\endgroup$ – yyyyyyy Apr 4 at 14:32
  • $\begingroup$ Asides from the question seemingly being homework, what do you mean by 'secure'? Perfectly secure? Computationally secure? $\endgroup$ – ElectronicToothpick Apr 4 at 14:35
  • $\begingroup$ my intuition is that it is secure at least computationally. $\endgroup$ – user153465 Apr 4 at 14:36
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    $\begingroup$ Hint: Look at your favourite proof of security for the OTP and see if you can adapt it to work with the new set / structure. $\endgroup$ – SEJPM Apr 4 at 14:38
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    $\begingroup$ Calling an explicitly non-prime integer variable $p$ is slightly confusing. I would suggest using a different character for it. Regarding the question, keep in mind that the addition of a ring is always a group, but not necessarily abelean. And thus they are always closed and inverse elements exist. Add to that, that uniform distributions only make sense for finite structures. $\endgroup$ – tylo Apr 4 at 15:13
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The classical xor-based one-time pad can be generalized to finite groups.

Let $(G,*)$ be such group with order $p$ and $*$ is the group operation(like the xor). The message, the pad and the ciphertext are elements of $G$.

Now to encrypt a message $m \in G$, choose $k \in G$ uniformly at random and set $c = m * k$. One of the security proofs of the one-time pad consists of showing that $c$ does not give any information on $m$(i.e to find $m$ we might as well pick a $c'$ at random and ignore $c$ completely).

More precisely, if $M$ is a random variable for messages distributed somehow, $K$ is a uniform random variable for the keys, and $C = M*K$ the random variables for the ciphertexts.

What we need to show is that $C = M*K$ is independent of $M$. i.e $C$ does not give any information on $M$.

Proof: We want to show that $P_{C|M}[c|m] = P_C[c]$. First, It's easy to see that $C$ defined as above is uniform(i.e $P_C[c] = \frac{1}{|G|}$). Next observe that $P_{C|M}[c|m] = P_{K|M}[c*(m)^{-1}| m] = P_K[c*(m)^{-1}] = \frac{1}{|G|}$. This follows form the fact that $K$ and $M$ are independent.

Therefore we showed the 'perfect secrecy' property of this constriction.

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