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As we know, breaking a Transposition Cipher is quite trivial and can be easily done by either exploiting single letter frequencies (not $d$-gram frequencies though, given that $2 < d < L$), or by employing a simple CPA, thus deriving the snippets length $L$ and the transposition matrix (key).

If we imagine to encrypt the plaintext twice, employing two different keys, would this help improving security?

If we employ the same snippet length $L_1 = L_2$, this clearly won't improve security as an attacker could easily find the key $k_3 = k_1 \times k_2$ (where $\times$ is the matrix product), without even bothering about $k_1$ and $k_2$.

However, that's not the case if $L_1 \neq L_2$. So I guess there's some kind of improvement; can you show me if and how this is true?

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First determine the permutation domain. If $L_2 > L_1$, then the cipher is broken by $\lceil log_2 (L_2) \rceil$ pairs of plaintext/ciphertect. The security improvement is small, as the attacker only needs to use $\lceil log_2 (L_2) \rceil - \lceil log_2 (L_1) \rceil$ additional pairs to find the key.

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