Proving if a permutation cipher is perfectly secret?

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From what I've read, perfect secrecy in its most basic form, that the encrypted text reveals no information about the plaintext, be it structure or content.

A permutation cipher is easy for me to construe on paper as a reordering of the letters of a message, however I don't really know how to represent this reordering in a mathematically sensible way. e.g. The word "apart" becomes encrypted as "trapa".

As such, is there a mathematical way I can prove that a transposition cipher is/is not perfectly secret?

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I do have information after a permutation: I know exactly what the distribution of characters (bytes / alphabet members..) of the plain text was. If I see "trapa" and "olleh", I can certainly tell which one came from "apart"....

So it's pretty trivial to win the distinguisher game here (so no perfect secrecy).

Added:

Another, more formal, way to see this: one modern way to define perfect secrecy, is given an encryption scheme $\mathcal{E}$, with message space $\mathcal{M}$, ciphertext space $\mathcal{C}$ and keyspace $\mathcal{K}$, we have for all pairs of messages $m_1, m_2 \in \mathcal{M}$ and all $c \in \mathcal{C}$ that

$$P(\mathcal{E}(m_1,K) = c) = P(\mathcal{E}(m_2,K) = c)$$

where the distribution is taken over the uniform distribution of $K \in \mathcal{K}$ (which is the set of all permutations over some fixed length $n$ here). The message and cipher text space is just the set of all length $n$ strings over some non-trivial alphabet $A$.

Now take a message $m_1$ with $n$ many $a$'s, $m_2$ with $n$ many $b$'s (where $a \neq b \in A$) and $c = m_1$. Then the left probability is $1$ (for any permutation we get the same cipher text) and the right one is $0$. So no perfect secrecy in this definition.