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I ask about represent message in Menezes–Vanstone elliptic curve cryptography I now encrypt function as follow

$C_1 = (M_1 * K_1) mod\ P$

$C_2=(M2 * k_2) mod\ P$

My question is about how much the size of $M_1$ and $M 2$, if we assume we are working on the 256 curve. Does the size of $M_1$ is one character and $M_2$ is also one character or $M_1$ is 32 byte and $M_2$ is 32 bytes. I hope anyone clarification about my problem.

I need any proposal.

Mhsz
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  • You should not use it... http://crypto.stackexchange.com/questions/16596/what-differences-between-menezes-vanstone-ecc-and-elgamal-ecc – DrLecter Dec 05 '14 at 23:23
  • Ok thanks for your response. But I have something I did not understand. When use Elgamal EC 160 bit. I can encrypt text file character by character then each block is character. While if we used MVECC we can encrypt text file 159 bit each block. If that right then MVECC is much faster EC ELGAMAL. Is that right. – Mhsz Dec 05 '14 at 23:43
  • No one ever would use a public key encryption scheme to directly encrypt a message. Please look at the hybrid encryption paradigm and at ECIES in particular when working with EC groups. – DrLecter Dec 05 '14 at 23:46
  • Well I was just wondering about the scientific theory. In other words, Is size of the block block has to do with the security of the algorithm or not. – Mhsz Dec 05 '14 at 23:52
  • Ok, M1 and M2 are elements of $Z_p^*$, i.e., in the set ${1,\ldots,p-1}$. The size of $p$ has to do with the security level...but thats not in the question. – DrLecter Dec 05 '14 at 23:57

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