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In some papers, server generates keys for users as follows: $$sk_i=msk+H(ID_i)v_i\ (mod \ p)$$ where $msk=$ the secret key of server

$v_i$ is randomly selected from $Z_p$ for i-th user , $p$ is a prime number

$ID_i=$ the identity of i-th user

$sk_i=$ the secret key for i-thuser

$H$ is the hash function

If there is a user who knows more than one secret key, does that key generation scheme prevent to reveal the value of $msk$ from users.

How to prove the security?

I have forgotten paper references. Sorry for that.

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    $\begingroup$ Can we assume that $v$ and $p$ are generated once for the server? Please explain how they are used in the protocol. You've specified the algorithm, which is great, but I'm missing the context and time factor. $\endgroup$
    – Maarten Bodewes
    Commented Jan 24, 2017 at 8:27
  • $\begingroup$ @MaartenBodewes. $v$ is uniformly randomly selected for each users. Therefore, we can assume $v_i$ for i-th user. $p$ is generated once for the server. $\endgroup$ Commented Jan 24, 2017 at 10:12
  • $\begingroup$ Do the users learn their $v_i$ directly at any point or are they just being given their $sk_i$? $\endgroup$
    – SEJPM
    Commented Jan 24, 2017 at 12:10
  • $\begingroup$ @SEJPM. No, server gives $v_iP$ and $sk_i$ to user. Users do not learn $v_i$ directly. $\endgroup$ Commented Jan 25, 2017 at 1:48

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Not really worked with these kinds of schemes, but here's my viewpoint.

Write $w_i=H(ID_i)v_i$. The value $H(ID_i)$ is some fixed public value, so we can kind of forget about it. That is, if $v_i$ is uniformly random, then so is $w_i$.

Now the key is generated as $sk_i=msk+w_i\pmod{p}$. Since $w_i$ is uniformly random, this is simply a generalisation of the one-time pad. In other words, $sk_i$ reveals no information whatsoever about $msk$. Therefore however many times you repeat this, the user will still learn nothing about $msk$.

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  • $\begingroup$ This is a good answer considering the information about the problem that is given and assuming that when the user gets $v_iP$ the $v_i$ is protected by DLOG. $\endgroup$
    – Elias
    Commented Jan 25, 2017 at 17:43

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