2
$\begingroup$

enter image description here

Why randomness space must be significantly larger than the commitment space |R|>|C|? the picture is from https://youtu.be/IkNZWJFcfcU?t=236

$\endgroup$

1 Answer 1

1
$\begingroup$

For the hiding property we require that the commitment value provides no information about the message. In particular we hope that for any given message $m$ it is possible for $H(m,r)$ to take all possible values in $C$ (otherwise one could exclude some messages as corresponding to some commitments). If $H$ behaves like a random function, then it unlikely to have this surjective property unless $|R|>|C|\log|C|$.

More powerfully we probably want the values of $H(m,r)$ for any fixed $m$ to be distributed uniformly across the values of $C$.

Note that these are information theoretic "zero knowledge" requirements rather than complexity/computation bounded requirements.

$\endgroup$
3
  • 1
    $\begingroup$ For any secret value the random blinding of a Pedersen commitment covers the whole image space uniformly. It can do this because the random blinding is a random permutation rather than a random function. $\endgroup$
    – Daniel S
    Feb 26, 2022 at 14:05
  • $\begingroup$ "In particular we hope that for any given message $m$ it is possible for $H(m,r)$ to take all possible values in $C$"; while this would be sufficient, this is not actually necessary. What is necessary is that any plausible adversary be unable to obtain any information about $m$ from the commitment; however, because computationally unbounded adversaries are not plausible, we can consider lesser goals (such as deducing information about $m$ computationally infeasible..) $\endgroup$
    – poncho
    Feb 26, 2022 at 17:26
  • $\begingroup$ @poncho: There are three levels of hiding in commitment schemes perfect, statistical, and computational. The lecturer is tacitly requiring statistical hiding (hence my reference to information theory) and the requirement $|R|\gg |C|$ is necessary to meet this. $\endgroup$
    – Daniel S
    Feb 26, 2022 at 17:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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