The interactive protocol for GIso of Example 10.1 is not perfect zero-knowledge, because a ver

Step 1: **Understand the Problem** u000AThe problem involves the interactive protocol for the compl

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The interactive protocol for $\overline{\text { GIso }}$ of Example 10.1 is not perfect zero-knowledge, because a verifier $M_{v^{\prime}}$ may send a graph $G^{\prime}$ to the prover that is not randomly generated from $G_1$ or $G_2$ and gain information of whether $G^{\prime}$ is isomorphic to $G_1$ or $G_2$. Modify this interactive protocol to a perfect zero-knowledge proof system for $\overline{\text { GIso. }}$

   The interactive protocol for $\overline{\text { GIso }}$ of Example 10.1 is not perfect zero-knowledge, because a verifier $M_{v^{\prime}}$ may send a graph $G^{\prime}$ to the prover that is not randomly generated from $G_1$ or $G_2$ and gain information of whether $G^{\prime}$ is isomorphic to $G_1$ or $G_2$. Modify this interactive protocol to a perfect zero-knowledge proof system for $\overline{\text { GIso. }}$

Theory of computational complexity

Ding-Zhu Du, Ker-I… 1st Edition

Chapter 10, Problem 16

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The issue is that the protocol is not perfect zero-knowledge because a verifier could potentially send a graph $G'$ that is not a random permutation of $G_1$ or $G_2$, thereby gaining information about the isomorphism status of $G'$ with respect to $G_1$ or $G_2$. Show more…

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The interactive protocol for $\overline{\text { GIso }}$ of Example 10.1 is not perfect zero-knowledge, because a verifier $M_{v^{\prime}}$ may send a graph $G^{\prime}$ to the prover that is not randomly generated from $G_1$ or $G_2$ and gain information of whether $G^{\prime}$ is isomorphic to $G_1$ or $G_2$. Modify this interactive protocol to a perfect zero-knowledge proof system for $\overline{\text { GIso. }}$

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