# CS276 Lecture 25 (draft)

Summary

Today we show that the graph isomorphism protocol we defined last time is indeed a zero-knowledge protocol. Then we discuss the quadratic residuosity problem modulo a composite, and define a protocol for proving quadratic residuosity. (We shall prove that the protocol is zero knowledge next time.)

1. The Graph Isomorphism Protocol

Last time we considered the following protocol for the graph isomorphism protocol.

• Verifier’s input: two graphs ${G_1=(V,E_1)}$, ${G_2= (V,E_2)}$;
• Prover’s input: ${G_1,G_2}$ and permutation ${\pi^*}$ such that ${\pi^*(G_1) = G_2}$; the prover wants to convince the verifier that the graphs are isomorphic
• The prover picks a random permutation ${\pi_R:V\rightarrow V}$ and sends the graph ${G:= \pi_R(G_2)}$
• The verifier picks at random ${b\in \{1,2\}}$ and sends ${b}$ to the prover
• The prover sends back ${\pi_R}$ if ${b=2}$, and ${\pi_R(\pi^*(\cdot))}$ otherwise
• The verifier cheks that the permutation ${\pi}$ received at the previous round is such that ${\pi(G_b) = G}$, and accepts if so.

In order to prove that this protocol is zero knowledge, we to show the existence of an efficient simulator.

Theorem 1 For every verifier algorithm ${V^*}$ of complexity ${t}$ there is a simulator algorithm ${S^*}$ of expected complexity ${\leq 2t + O(n^2)}$ such that, for every two isomorphic graphs ${G_1,G_2}$, and for every isomorphism ${\pi}$ between them, the distributions of transcripts $\displaystyle P(\pi,G_1,G_2) \leftrightarrow V^*(G_1,G_2) \ \ \ \ \ (1)$

and $\displaystyle S^* (G_1,G_2) \ \ \ \ \ (2)$

are identical.

If ${N=p\cdot q}$ is the product of two distinct odd primes, and ${{\mathbb Z}^*_N}$ is the set of all numbers in ${\{1,\ldots,N-1\}}$ having no common factor with ${N}$, then we have the following easy consequences of the Chinese remainder theorem:

• ${{\mathbb Z}^*_N}$ has ${(p-1) \cdot (q-1)}$ elements, and is a group with respect to multiplications;
• If ${r = x^2 \bmod N}$ is a quadratic residue, and is an element of ${{\mathbb Z}^*_N}$, then it has exactly 4 square roots in ${{\mathbb Z}^*_N}$
• Precisely ${(p-1)\cdot (q-1) / 4}$ elements of ${{\mathbb Z}^*_N}$ are quadratic residues
• Knowing the factorization of ${N}$, there is an efficient algorithm to check if a given ${y\in {\mathbb Z}^*_N}$ is a quadratic residue and, if so, to find a square root.

It is, however, believed to be hard to find square roots and to check residuosity modulo ${N}$ if the factorization of ${N}$ is not known.

Indeed, we can show that from any algorithm that is able to find square roots efficiently mod ${N}$ we can derive an algorithm that factors ${N}$ efficiently

We consider the following protocol for proving 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.

We show that:

• If ${r}$ is a quadratic residue, the prover is given a square root ${x}$, and the parties follow the protocol, then the verifier accepts with probability 1;
• If ${r}$ is not a quadratic residue, then for every cheating prover strategy ${P^*}$, the verifier rejects with probability ${\geq 1/2}$.

Next time we shall prove that the protocol is zero knowledge.