CS276 Lecture 26 (draft)

Summary

After showing that last week’s protocol for quadratic residuosity is indeed zero-knowledge, we move on to the formal definition of proof of knowledge, and we show that the quadratic residuosity protocol is also a proof of knowledge.

1. The Quadratic Residuosity Protocol

Last time we considered the following protocol for quadratic residuosity:

Verifier’s input: an integer ${N}$ (product of two unknown odd primes) and a integer ${r \in {\mathbb Z}^*_N}$ ;
Prover’s input: ${N,r}$ and a square root ${x\in Z^*_N}$ such that ${x^2 \bmod N = r}$ .
The prover picks a random ${y \in Z_N^*}$ and sends ${a := y^2 \bmod N}$ to the verifier
The verifier picks at random ${b\in \{0,1\}}$ and sends ${b}$ to the prover
The prover sends back ${c:= y}$ if ${b=0}$ or ${c:= y\cdot x \bmod N}$ if ${b=1}$
The verifier cheks that ${c^2 \bmod N = a}$ if ${b=0}$ or that ${c^2 \equiv a \cdot r \pmod N}$ if ${b=1}$ , and accepts if so.

Today we show that it is zero knowledge, that is,

Theorem 1 For every verifier algorithm ${V^*}$ of complexity ${\leq t}$ there is a simulator algorithm of average complexity ${\leq 2t + (\log N)^{O(1)}}$ such that for every odd composite ${N}$ , every ${r}$ which is a quadratic residue ${\pmod N}$ and every square root of ${x}$ of ${r}$ , the distributions

$\displaystyle S^*(N,r) \ \ \ \ \ (1)$

and

$\displaystyle P(N,r,x) \leftrightarrow V^*(N,r) \ \ \ \ \ (2)$

are identical.

2. Proofs of Knowledge

Suppose that ${L}$ is a language in NP; then there is an NP relation ${R_L(\cdot,\cdot)}$ computable in polynomial time and polynomial ${p(\cdot)}$ such that ${x\in L}$ if and only if there exists a witness ${w}$ such that ${|w| \leq p(|x|)}$ (where we use ${|z|}$ to denote the length of a bit-string ${z}$ ) and ${R(x,w) = 1}$ .

Recall the definition of soundness of a proof system ${(P,V)}$ for ${L}$ : we say that the proof system has soundness error at most ${\epsilon}$ if for every ${x\not\in L}$ and for every cheating prover strategy ${P^*}$ the probability that ${P^*(x) \leftrightarrow V(x)}$ accepts is at most ${\epsilon}$ . Equivalently, if there is a prover strategy ${P^*}$ such that the probability that ${P^*(x) \leftrightarrow V(x)}$ accepts is bigger than ${\epsilon}$ , then it must be the case that ${x\in L}$ . This captures the fact that if the verifier accepts then it has high confidence that indeed ${x\in L}$ .

In a proof-of-knowledge, the prover is trying to do more than convince the verifier that a witness exists proving ${x\in L}$ ; he wants to convince the verifier that he (the prover) knows a witness ${w}$ such that ${R(x,w)=1}$ . How can we capture the notion that an algorithm “knows” something?

Definition 2 (Proof of Knowledge) A proof system ${(P,V)}$ for an NP relation ${R_L}$ is a proof of knowledge with knowledge error at most ${\epsilon}$ and extractor slowdown ${es}$ if there is an algorithm ${K}$ (called a knowledge extractor) such that, for every verifier strategy ${P^*}$ of complexity ${\leq t}$ and every input ${x}$ , if

$\displaystyle \mathop{\mathbb P} [ P^*(x) \leftrightarrow V(x) \mbox { accepts} ] \geq \epsilon + \delta$

then ${K(P^*,x)}$ outputs a ${w}$ such that ${R(x,w)=1}$ in average time at most

$\displaystyle es \cdot (n^{O(1)} + t ) \cdot \delta^{-1}$

In the definition, giving ${P^*}$ as an input to ${K}$ means to give the code of ${P^*}$ to ${K}$ . A stronger definition, which is satisfied by all the proof systems we shall see, is to let ${K}$ be an oracle algorithm of complexity ${\delta^{-1}\cdot es \cdot poly(n)}$ , and allow ${K}$ to have oracle access to ${P^*}$ . In such a case, “oracle access to a verifier strategy” means that ${K}$ is allowed to select the randomness used by ${P^*}$ , to fix an initial part of the interaction, and then obtain as an answer what the next response from ${P^*}$ would be given the randomness and the initial interaction.

Theorem 3 The protocol for quadratic residuosity of the previous section is a proof of knowledge with knowledge error ${1/2}$ and extractor slowdown 2.

Enter your comment here...

Fill in your details below or click an icon to log in:

Email (required) (Address never made public)

Name (required)

Website

You are commenting using your WordPress.com account. ( Log Out / Change )

You are commenting using your Google account. ( Log Out / Change )

You are commenting using your Twitter account. ( Log Out / Change )

You are commenting using your Facebook account. ( Log Out / Change )

Cancel

Connecting to %s

in theory

"Marge, I agree with you – in theory. In theory, communism works. In theory." — Homer Simpson

CS276 Lecture 26 (draft)

Leave a Reply

Share this:

Like this:

Related

Leave a Reply