0
$\begingroup$

The length of public-key matrix in McEliece cryptosystem is $n \times k$. Then the key size is $\frac{n \times k}{8}$ bytes. But I find that the key size is $\frac{(n-k) \times k}{8}$ bytes. why ?

$\endgroup$
3
$\begingroup$

Let $G$ be the public key matrix. With Gaussian elimination, you can find a Matrix $G'$ with $G = (E_k | G')$, where $E_k$ is the identity matrix with dimension $k$. Then, you only have to store $G'$. And $G'$ has the dimension $(n-k) \times k$.

Source: German Wikipedia (https://de.wikipedia.org/wiki/McEliece-Kryptosystem)

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.