# Tag Info

## Hot answers tagged hill-cipher

3

No, because matrix inversion can be done efficiently. Namely, if encryption is multiplication by a matrix $A$, then you can define decryption with $A$ as first computing the inverse of $A$ and then multiplying. An essential property of public-key cryptosystems is that it should not be possible to efficiently derive the (private) decryption key from the ...

1

Indeed. If the matrix is $n$ by $n$, and you have $n$ many known plaintext blocks with corresponding ciphertext, you get $n^2$ linear equations in $n^2$ unknowns (the matrix elements) (modulo the alphabet size), which very often can be solved uniquely.

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