Summary

Today we define the notion of computational zero knowledge and show that the simulator we described in the last lecture establishes the computational zero knowledge property of the 3-coloring protocol.

1. The Protocol and the Simulator

Recall that we use a commitment scheme {(C,O)} for messages in {\{1,2,3\}}, and that the common input to the prover and the verifier is a graph {G=([n],E)}, where {[n]:= \{1,2,\ldots,n\}}. The prover, in addition, is given a valid 3-coloring {\alpha : [n] \rightarrow \{1,2,3\}} of {G}.

The protocol is defined as follows:

  • The prover picks a random permutation {\pi: \{1,2,3\} \rightarrow \{1,2,3\}} of the set of colors, and defines the 3-coloring {\beta(v) := \pi(\alpha(v))}. The prover picks {n} keys {K_1,\ldots,K_n} for {(C,O)}, constructs the commitments {c_v := C(K_v,\beta(v))} and sends {(c_1,\ldots,c_n)} to the verifier;
  • The verifier picks an edge {(u,v) \in E} uniformly at random, and sends {(u,v)} to the prover;
  • The prover sends back the keys {K_u,K_v};
  • If {O(K_u,c_u)} and {O(K_v,c_v)} are the same color, or if at least one of them is equal to {FAIL}, then the verifier rejects, otherwise it accepts

For every verifier algorithm {V^*}, we defined a simulator algorithm {S^*} which repeats the following procedure until the output is different from {FAIL}:

Algorithm {S_{1round}^*}

  • Input: graph {G=([n],E)}
  • Pick random coloring {\gamma : [n] \rightarrow \{1,2,3\}}.
  • Pick {n} random keys {K_1,\ldots,K_n}
  • Define the commitments {c_i := C(K_i, \gamma(i))}
  • Let {(u,v)} be the 2nd-round output of {V^*} given {G} as input and {c_1,\ldots,c_n} as first-round message
  • If {\gamma(u) = \gamma(v)}, then output FAIL
  • Else output {((c_1,\ldots,c_n),(u,v),(K_u,K_v))}

We want to show that this simulator construction establishes the computational zero knowledge property of the protocol, assuming that {(C,O)} is secure. We give the definition of computational zero knowledge below.

Definition 1 (Computational Zero Knowledge) We say that a protocol {(P,V)} for 3-coloring is {(t,\epsilon)} computational zero knowledge with simulator overhead {so(\cdot)} if for every verifier algorithm {V^*} of complexity {\leq t} there is a simulator {S^*} of complexity {\leq so(t)} on average such that for every algorithm {D} of complexity {\leq t}, every graph {G} and every valid 3-coloring {\alpha} we have

\displaystyle  | \mathop{\mathbb P} [ D(P(G,\alpha) \leftrightarrow V^*(G)) =1] - \mathop{\mathbb P} [ D(S^*(G)) =1] | \leq \epsilon

Theorem 2 Suppose that {(C,O)} is {(2t+O(nr),\epsilon/(4\cdot |E|\cdot n))}-secure and that {C} is computable in time {\leq r}.

Then the protocol defined above is {(t,\epsilon)} computational zero knowledge with simulator overhead at most {1.6 \cdot t+O(nr)}.

2. Proving that the Simulation is Indistinguishable

In this section we prove Theorem 2.

Suppose that the Theorem is false. Then there is a graph {G}, a 3-coloring {\alpha}, a verifier algorithm {V^*} of complexity {\leq t}, and a distinguishing algorithm {D} also of complexity {\leq t} such that

\displaystyle  | \mathop{\mathbb P} [ D(P(G,\alpha) \leftrightarrow V^*(G)) =1 - \mathop{\mathbb P} [ D(S^*(G))=1] | \geq \epsilon

Let {2R_{u,v}} be the event that the edge {(u,v)} is selected in the second round; then

\displaystyle  \begin{array}{rcl}  \epsilon &\leq & | \mathop{\mathbb P} [ D(P(G,\alpha) \leftrightarrow V^*(G)) =1 ] - \mathop{\mathbb P} [ D(S^*(G))=1] | \\ & = & \left| \sum_{(u,v) \in E} \mathop{\mathbb P} [ D(P(G,\alpha) \leftrightarrow V^*(G)) =1 \wedge 2R_{u,v} ]\right. \\ && - \left. \sum_{(u,v) \in E} \mathop{\mathbb P} [ D(S^*(G))=1 \wedge 2R_{u,v} ] \right| \\ & \leq & \sum_{(u,v) \in E} | \mathop{\mathbb P} [ D(P(G,\alpha) \leftrightarrow V^*(G)) =1 \wedge 2R_{u,v} ]\\ && - \mathop{\mathbb P} [ D(S^*(G))=1 \wedge 2R_{u,v} ] | \end{array}

So there must exist an edge {(u^*,v^*)\in E} such that

\displaystyle  | \mathop{\mathbb P} [ D(P \leftrightarrow V^*) =1 \wedge 2R_{u^*,v^*} ] - \mathop{\mathbb P} [ D(S^*)=1 \wedge 2R_{u^*,v^*} ] | \geq \frac \epsilon {|E|} \ \ \ \ \ (1)

(We have omitted references to {G,\alpha}, which are fixed for the rest of this section.)

Now we show that there is an algorithm {A} of complexity {2t + O(nr)} that is able to distinguish between the following two distributions over commitments to {3n} colors:

  • Distribution (1) commitments to the {3n} colors {1,2,3,1,2,3,\ldots,1,2,3};
  • Distribution (2) commitments to {3n} random colors

Algorithm {A}:

  • Input: 3n commitments {d_{a,i}} where {a\in \{1,2,3\}} and {i\in \{1,\ldots,n\}};
  • Pick a random permutation {\pi: \{1,2,3\} \rightarrow \{1,2,3\}}
  • Pick random keys {K_{u^*}}, {K_{v^*}}
  • Construct the sequence of commitments {c_1,\ldots,c_n} by setting:

    • {c_{u^*} := C(K_{u^*} , \pi(\alpha(u^*))}
    • {c_{v^*} := C(K_{v^*} , \pi(\alpha(v^*))}
    • for every {w\in [n] - \{u^*,v^*\}}, {c_w := d_{\pi(\alpha(w)),w}}
  • If the 2nd round output of {V^*} given {G} and {c_1,\ldots,c_n} is different from {(u^*,v^*)} output 0
  • Else output {D((c_1,\ldots,c_n),(u^*,v^*),(K_{u^*},K_{v^*}))}

First, we claim that

\displaystyle  \mathop{\mathbb P} [ A( \mbox{Distribution 1} ) = 1 ] = \mathop{\mathbb P} [ D(P \leftrightarrow V^*) =1 \wedge 2R_{u^*,v^*} ] \ \ \ \ \ (2)

This follows by observing that {A} on input Distribution (1) behaves exactly like the prover given the coloring {\alpha}, and that {A} accepts if and only if the event {2R_{u^*,v^*}} happens and {D} accepts the resulting transcript.

Next, we claim that

\displaystyle   | \mathop{\mathbb P} [ A( \mbox{Distribution 2} ) = 1 ] - \mathop{\mathbb P} [ D(S^*)=1 \wedge 2R_{u^*,v^*} ] | \leq \frac \epsilon {2|E|} \ \ \ \ \ (3)

To prove this second claim, we introduce, for a coloring {\gamma}, the quantity {DA(\gamma)}, defined as the probability that the following probabilistic process outputs 1:

  • Pick random keys {K_1,\ldots,K_n}
  • Define commitments {c_u:= C(K_u,\gamma(u))}
  • Let {(u,v)} be the 2nd round output of {V^*} given the input graph {G} and first round message {c_1,\ldots,c_n}
  • Output 1 iff {(u,v) = (u^*,v^*)}, {\gamma(u^*) \neq \gamma(v^*)}, and

    \displaystyle  D((c_1,\ldots,c_n),(u^*,v^*),(K_{u^*}, K_{v^*})) = 1

Then we have

\displaystyle  \mathop{\mathbb P} [ A( \mbox{Distribution 2} ) = 1 ] = \sum_{\gamma: \gamma(u^*) \neq \gamma(v^*)} \frac 32 \cdot \frac 1 {3^n} \cdot DA(\gamma) \ \ \ \ \ (4)

Because {A}, on input Distribution 2, first prepares commitments to a coloring chosen uniformly at random among all {1/(6 \cdot 3^{n-2})} colorings such that {\gamma(u^*) \neq \gamma(v^*)} and then outputs 1 if and only if, given such commitments as first message, {V^*} replies with {(u^*,v^*)} and the resulting transcript is accepted by {D}.

We also have

\displaystyle  \mathop{\mathbb P} [ D(S^*)=1 \wedge 2R_{u^*,v^*} ]

\displaystyle   = \frac {1}{\mathop{\mathbb P} [ S^*_{1Round} \neq FAIL]} \cdot \sum_{\gamma: \gamma(u^*) \neq \gamma(v^*)} \frac 1 {3^n} \cdot DA(\gamma) \ \ \ \ \ (5)

To see why Equation (5) is true, consider that the probability that {S^*} outputs a particular transcript is exactly {1/\mathop{\mathbb P} [ S^*_{1Round} \neq FAIL]} times the probability that {S^*_{1Round}} outputs that transcript. Also, the probability that {S^*_{1Round}} outputs a transcript which involves {(u^*,v^*)} at the second round and which is accepted by {D()} conditioned on {\gamma} being the coloring selected at the beginning is {DA(\gamma)} if {\gamma} is a coloring such that {\gamma(u^*) \neq \gamma(v^*)}, and it is zero otherwise. Finally, {S^*_{1Round}} selects the initial coloring uniformly at random among all possible {3^n} coloring.

From our security assumption on {(C,O)} and from Lemma 6 in Lecture 27 we have

\displaystyle   \left| \mathop{\mathbb P} [ S^*_{1Round} \neq FAIL] - \frac 23 \right| \leq \frac{\epsilon}{4|E|} \ \ \ \ \ (6)

and so the claim we made in Equation (3) follows from Equation (4), Equation (5), Equation (6) and the fact that if {p,q} are quantities such that {\frac 32 p \leq 1}, {\frac 1q \cdot p \leq 1}, and {\left|q - \frac 23 \right| \leq \delta \leq \frac 16} (so that {q \geq 1/2}), then

\displaystyle  \left| \frac 32 p - \frac 1q p \right| = \frac 32 \cdot p \cdot \frac 1q \cdot \left| q- \frac 23 \right| \leq 2 \delta

(We use the above inequality with {q = \mathop{\mathbb P} [ S^*_{1Round} \neq FAIL]}, {\delta = \epsilon/4|E|}, and {p = \sum_{\gamma: \gamma(u^*) \neq\gamma(v^*)}\frac 1 {3^n} DA(\gamma)}.)

Having proved that Equation (3) holds, we get

\displaystyle  | \mathop{\mathbb P} [ A( \mbox{Distribution 1} ) = 1 ] - \mathop{\mathbb P} [ A( \mbox{Distribution 2} ) = 1 ] | \geq \frac \epsilon {2|E|}

where {A} is an algorithm of complexity at most {2t + O(nr)}. Now by a proof similar to that of Theorem 3 in Lecture 27, we have that {(C,O)} is not {(2t +O(nr), \epsilon/(2|E|n))} secure.

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