I can't understand how Niederreiter cryptosystem works. If $c=mH^{'T}$ than why we cannot compute $m$ directly by multiplying $c$ with the $(H^{'T})^{-1}$? Can you give me an example of a "fast decoding algorithm"?
Thank you!
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$\begingroup$ did you consider the noise removal by $D$? $\endgroup$– kelalakaFeb 8, 2019 at 11:43
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$\begingroup$ What noise? That's confuses me. In the McEliece cryptosystem we add some error $e$ but in all Niederreiter documentations I didn't see any error adding to the plaintext message $\endgroup$– mipFeb 8, 2019 at 11:46
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2 Answers
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In the Niederreiter system, the plaintext is mapped to some error vector of weight $t$, where the code correction capability is $d=2t+1.$
With the trapdoor information (permutation) this can be decoded by the legitimate receiver by syndrome decoding.
Without the trapdoor information, this is equivalent to decoding a random vector, which is hard, as in the McEliece cryptosystem.
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$\begingroup$ It is always guaranteed that the matrix H' doesn't has an inverse? $\endgroup$– mipFeb 10, 2019 at 6:44
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1$\begingroup$ @kelalaka If you took $d=t+k+1$ you might be unlucky and the actual codeword might only be Hamming distance $k+1$ away which can be found with $O(2^{k+1})$ effort. $\endgroup$– kodluFeb 10, 2019 at 20:46
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Since the parity check matrix H' is not squared (it has dimensions $n-k \times n$), one can not ouput the message $m$ from the ciphertext $c$.
Nevertheless, there is an attack named Lee-Brickell, which defines the security McEliece, but was later found that can also be adapted to Niederreiter, showcasing the equivalence in security of both cryptosystems. The Lee-Brickell attack is based on extracting a full rank submatrix $H_{n-k} \in \mathbb{F}_2^{n-k \times n-k} $ which can be inversed so that the message is find randomly in a much smaller configuration space in $\mathbb{F}_2^{k}$.
