Zero-Knowledge Proof System Slides by Ouzy Hadad, Yair Gazelle & Gil Ben-Artzi Adapted from Ely...

Zero-Knowledge Proof System

Zero-Knowledge Proof System

Slides by Slides by Ouzy HadadOuzy Hadad , Yair Gazelle & Gil Ben-, Yair Gazelle & Gil Ben-ArtziArtzi

Adapted from Adapted from Ely PoratEly Porat course lecture notes. course lecture notes.

Background and Motivation

Background and Motivation

The purpose of a traditional proof is to convince somebody, but typically the details of a proof give the verifier more info about the assertion.

A proof is a zero-knowledge if the verifier does not get from it anything that he can not compute by himself.

Background and Motivation (cont.)

Background and Motivation (cont.)

Whatever can be efficiently obtained by interacting with a prover, could also be computed without interaction, just by assuming that the assertion is true and conducting some efficient computation.

Let (P,V) be an interactive proof system for some language L. We say that (P,V), actually P, is zero-knowledge if for every probabilistic polynomial-time verifier V*, there exists a probabilistic polynomial-time machine M* s.t. for every xL holds

Machine M* is called the simulator for the interaction of V* with P.

Zero Knowledge (Definition)

Zero Knowledge (Definition)

LxLx xMxVP

)}({)}(,{

Perfect Zero Knowledge (Definition)

Perfect Zero Knowledge (Definition)

Let (P,V) be an interactive proof system for some language L. We say that (P,V), actually P, is perfect zero-knowledge (PZK) if for every probabilistic polynomial time verifier V*, there exists a probabilistic polynomial-time machine M* s.t. for every xL the distributions {<P,V*>(x)}xL and {M*(x)}xL are identical, i.e.,

LxLx xMxVP

)}({)}(,{

The distribution ensembles {Ax}xL and {Bx}xL are

statistically close or have negligible variationdistance if for every polynomial p(•) there exitsinteger N such that for every xL withholds:

Statistically close distributions (Definition)

Statistically close distributions (Definition)

Nx

xp

BobAob xx

1PrPr

Statistical zero-knowledge (Definition)

Statistical zero-knowledge (Definition)

Let (P,V) be an interactive proof system for some language L. We say that (P,V), actually P, is statistical zero knowledge (SZK) if for every probabilistic polynomial time verifier V* there exists a probabilistic polynomial-time machine M* s.t. the ensembles {<P,V*>(x)}xL and {M*(x)}xL are statistically close.

Computationally indistinguishable

(Definition)

Computationally indistinguishable

(Definition)Two ensembles {Ax}xL and {Bx}xL are

computationally indistinguishable if for everyprobabilistic polynomial time distinguisher D andfor every polynomial p(•) there exists an integerN such that for every xL with |x| N holds

|)(|

1|1),(Pr1,Pr|

xpBxDobAxDob xx

Computational zero-knowledge (Definition)Computational zero-

knowledge (Definition)Let (P,V) be an interactive proof system for some language L. (P,V), actually P, is computational zero knowledge (CZK) if for every probabilistic polynomial-time verifier V* there exists a probabilistic polynomial-time machine M* s.t. the ensembles {<P,V*>(x)}xL and {M*(x)}xL are computationally indistinguishable.

PZK by viewPZK by view

The pair <P,V> is PZK by view if for every p.p.t V*... (probability polynomial time machine) there exist p.p.t M* such that for every xL we have: {view(P,V*)(x)={M*(x)} where view(P,V*)(x) is the view of V* after running <P,V*> on the input x, and M*(x) is the output of M* on the input x.

IP is PZK iff PZK by viewIP is PZK iff PZK by viewLemma: An interactive proof system is perfectzero-knowledge iff it is perfect zero knowledgeby view.

Proof:

Let M* satisfy: {view<P,V*>(x)}xL {M*(x)}xL

for every xL. M* has on its work-tape thefinal view of V*. Hence, it is able to performthe last step of V* and output the result. Andso the modified M*(x) is identical to <P,V*>(x).

Proof of lemma (cont.)


Let M* satisfy: {<P,V*>(x)}xL {M*(x)}xL .

For a particular V*, let us consider a verifierV** that behaves exactly like V*, but outputsits whole view (at the end). There is a machineM** s.t.

LxLx

xMxVP

)()(,

Graph-IsomorphismGraph-Isomorphism

A pair of two graphs, WhereLets be an isomorphism between the input

graphs, namely is 1-1 and onto mapping of the vertex set V1 to the vertex set V2 so that

21 ))(),((),( EuviffEuv

.|V||V| 21

).E,(VG),E,(VG 222111

ZK proof for Graph Isomorphism

ZK proof for Graph Isomorphism

Prover’s first step(P1): Select random permutation over V1, construct the set , and send to the verifier.

Verifier’s first step gets H from P. select and send it to P.

P is supposed to answer with an isomorphism between and .

2,1

1v)(u,:(v)(u),:F E F),(VH 1

V:)(V1

V

0G H

ZK proof for Graph Isomorphism(cont.)ZK proof for Graph Isomorphism(cont.)

(P2): If =1, then send = to V. Otherwise send = -1 to V.

(V2): If is an isomorphism between G

and H then V output 1, otherwise itoutputs 0.

Construction (diagram)

Construction (diagram)

Prover Verifier

=Random Permotation

H G1 R{1,2}

If =1, send =

otherwise = -1

Accept iff

H = (G)

H

An example:An example:

22

55

11

44

33

GG11

33

11

22

GG2255

44

Common input: two graphs G1 and G2.

Only P knows

.

An example (cont.)An example (cont.)22

55

11

44

33

GG11

55

33

44

11

22

HH

33

11

22

55

44GG22

= -1

Only P knows .

P sends H to V. V gets

and accepts.

V sends

=2 to P.

Theorem: Graph isomorphism is in Zero-Knowledge

Theorem: Graph isomorphism is in Zero-Knowledge

Theorem 1:The construction above is aperfect zero-knowledgeinteractive proof system(with respect to statistical closeness).

Proof of Theorem 1Proof of Theorem 1Completeness:

If G1 G2 , V always accepts.

First, G’=(G1).

If =1 then = , Hence:

(G) = (G1) = (G1) = G’ .

If =2 then = -1, Hence:

(G) = -1(G2) = (G1) = G’ .

And hence V always accepts when G1 G2 .

Proof of Theorem 1 (cont.)


Soundness:Let P* be any prover.If it sends to V a graph not isomorphic neither to G1 nor to G2, then there is no isomorphism between G and G’. If G’ G1 then P* can convince V with probability at most 1/2 (V selects {1,2} uniformly).Hence: when G1 and G2 are non-isomorphic:

If we will run this several times we will get the desire probability.

21)G,G(V,PPr 21 accept

Zero Knowledge(Construction of a

simulator)


simulator)Let V* be any polynomial-time verifier, and let q(•)

be a polynomial bounding the running time of V*.

M* selects a string

01100…………011 =r

xq}1,0{r R

Construction of a Simulator (cont.)Construction of a Simulator (cont.)

M* selects R{1,2}.

M* selects a random permutation over V.

M* constructs G’’= (G).

25413

54321

55

33

44

11

22

G’G’’’

33

11

22

55

44

GG22

2

Construction of a Simulator (cont.)Construction of a Simulator (cont.)

M* runs V* with the latter’s strings set as follows:

Denote as V*‘s output.

M* halts with output (x,r,G’’,).

r

x

G’’

InputTape

RandomTape

MessageTape

2



Definition: Let (P,V) be an interactive proof system for L. (P,V) is perfect zero-knowledge by view iffor every probabilistic polynomial-time verifier V* there exists a probabilistic polynomial time machine M* s.t. for every xL holds:

{view<P,V*>(x)}xL {M*(x)}xL

where view<P,V*>(x) is the final view of V* after

running <P,V*> on input x.

view = all the data a machine possesses



Lemma: Then for every string r, graph H and permutation , it holds that:

Pr [view<P,V*>(x) = (x,r,H,)] = Pr [M*(x) = (x,r,H,) | M*(x) ]

Proof:Let m* describe M* conditioned on its not being .Define the 2 random variables: 1.v(x,r) - the last 2 elements of view(P,V*)(x) conditioned on the second element equals r. 2. (x,r) - the same with m*(x).

ISOGGx ),(Let 21



Let V* (x,r,H) denote the message sent by V*

for a fixed r and an incoming message H.We will show that v(x,r) and (x,r) areuniformly distributed over the set:

While running the simulator we have H=(G),and only the pairs satisfying =v*(x,r,H) lead toan output. Hence:

otherwise

GHifVHrxHrxV

0

)(|!|1

)),(),(Pr(),,(

1

.)(:, :),,(, HrxVrx GHHC



Consider v(x,r):

For each H (which is isomorphic to G1):

Observing thatand hence the lemma follows.

)),(G(

1))(Gr,(x,V)),(G(r)V(x,

12

11

otherwise

if

0

|!V|1

)),(r)(x,Pr(

H)r,(x,V1

1

otherwise

ifH

),,(V1

),,(V)(GH Hrx

Hrxiff



Corollary: view<P,V*>(x) and M*(x) are statistically close.Proof: A failure is output with probability If the simulator returns steps P1-P2 of theconstruction |x| times and at least once at stepP2 =, then output (x,r,G’’,). If in all |x|trials , then output rubbish.Hence, we got a statistical difference ofand so the corollary follows.

21

||2 x

Zero-Knowledge for NPZero-Knowledge for NP

NP Problem: A language L belongs to NP if and only if there exist a two-inputpolynomial-time algorithm A and constant Csuch that:

there exist a certificate y with

We say that algorithm A verifies language Lin polynomial time.

:1,0 xL

1),()|(| yxAthatsuchxOy c

IP for NPIP for NP

Lets L language belong to NP, and x L , P should prove V that he know the solution for x.

(P1): P guess the solution y for the problem x.(V1) V verify in polynomial time that A(x,y)=1.

We will give ZK interactive proof system for NP complete problem (G3C), which implies that for every NP problem, we have ZK proof.

G3CG3C

Common Input: A graph 12

3 4

5

12

3 4

5

P can paint the graph in 3 colors.

P must keep the coloring a secret.

12

3 4

5

12

3 4

5

12

3 4

5

G3C is in Zero-Knowledge

G3C is in Zero-Knowledge

P chooses a random color permutation.

He puts all the nodes inside envelopes.

And sends them to the verifier.

Construction (ZK IP for G3C):

G3C is in ZK (cont.)G3C is in ZK (cont.)Verifier receives a 3-colored

graph, but colors are hidden.1

2

3 4

5

12

3 4

5

He chooses an edge at random.

And asks the prover to open the 2 envelopes.

G3C is in ZK (cont.)

G3C is in ZK (cont.)

Prover opens the envelopes, revealing the colors.

12

3 4

5

12

3Verifier accepts if the colors are different.

Formally,Formally,G = (V,E) is 3-colorable if there exists a mapping

for every .Let be a 3-coloring of G, and let be a

permutation over {1,2,3} chosen randomly.Define a random 3-coloring.Put each (v) in a box with v marked on it.Send all the boxes to the verifier.

)()(}3,2,1{: vuthatsoV Evu ),(

))(()( vv

Formally, (cont.)Formally, (cont.)Verifier selects an edge at random

asking to inspect the colors.Prover sends the keys to boxes u and v.Verifier uses the keys to open the boxes.If he finds 2 different colors from {1,2,3} - Accept.Otherwise - Reject.

Evue R ),(

G3C (diagram)G3C (diagram)

(1) (n)(2)1 2 n

P V

P V

Keyu , keyv

P V

Evue R ),(

The construction is in ZK:

The construction is in ZK:Completeness:

If G is 3-colorable and both P and V follow the rules, V will accept.

Soundness:Suppose G is not 3-colorable and P* tries to cheat. Then at least one edge (u,v) will be colored badly: (u) = (v).V will pick a bad edge with probability which can be increased to by repeating the protocol sufficiently many times. 3

2||

1E


simulator)


simulator)Let V* be any polynomial-time verifier, and let q(•)

be a polynomial bounding the running time of V*.

M* selects a string

110.......11010r

|)(|1,0 xqRr

Construction of a Simulator (cont.)

Construction of a Simulator (cont.)

M* selects e’=(u’,v’) R E.M* sends to V* boxes filled with garbage, except

for the boxes of u’ and v’, colored as follows:

c d

u’ v’

Otherwise, the simulation fails.

C R {1,2,3} d R {1,2,3}\{c} If V* picks (u’,v’), M* sends V* their

keys and the simulation is completed.

Analysis of the Simulation

Analysis of the Simulation

For every GG3C, the distribution ofm*(<G>) = M*(<G>) | (M*(<G>) ) is identical to <P,V*>(<G>).Since V* can’t tell e’ from other edges bylooking at the boxes, he picks e’ withprobability 1/|E|, which can be increasedto a constant by repeating M* sufficientlymany times.So if the boxes are perfectly sealed,G3CPZK.

ZK for Finding square modulo n

ZK for Finding square modulo n

Input: x2 modulo n .output: x modulo n.The prover need to prove that he know the

output.

ZK for Finding square modulo n (cont.)

ZK for Finding square modulo n (cont.)

(P1): P find two large prime number p,q,where n=p·q. He also choose randomlyr [n, n4].

P send n, x2 mod n and r2 mod n to V.(V1): V has two possibilities (a) Ask r. check the value of r2 mod n.(b) Ask for x ·r. check the value of x2 r2 mod n

Analysis of the Protocol - square modulo n

Analysis of the Protocol - square modulo n

Soundness: If P does not know x, then in probability of 50% V will catch him, if we will run this several times we will get the Vwill reject in probability larger then 2/3.

Completeness: If P know x, V always accept.

Analysis of the Protocol - square modulo n (cont.)Analysis of the Protocol - square modulo n (cont.)

This protocol is computational ZK.The Protocol give the value x2 mod n but the

verifier can't calculate x from it .If the verifier ask option 1 from the prover, he get

no additional info.If the verifier ask option 2 from the prover, he get

xr which is random.

CO-NP ZKCO-NP ZK

In order to prove the above it’s enough to show that CO-NP complete problem is in IP

We will show that CO-SAT belongs to IP. Than we can show that CO-SAT belongs to ZK. Reminder: CO-SAT means that there are no truth

assignment for an equation. We can treat it as a specific case of proving that for an

equation there are exactly K truth assignments (In this case , K=0)

CO-SAT IPCO-SAT IP Lemma

1. (x1,x2,x3,…,Xn) has exactly Kn truth assignments k0,k1 : Kn=k0+k1

2. (0,x2,x3,…Xn) = 0(x2,x3,…Xn) has exactly k0 truth assignments

3. (1,x2,x3,…Xn) = 1(x2,x3,…Xn) has exactly k1 truth assignments

Informal explanation By setting a variable in the original equations we create a new

equation with a special relation to the original one. Each new equation must have a specific number of

assignments which can be pre-calculate.

CO-SAT IPCO-SAT IP We can now construct a solution based upon the

previous lemma Prover will send verifier k0,k1 for (n) Verifier will check that for (n-1) , condition 1 of lemma

is true ( Kn=k0+k1) Verifier will create randomly a new equation (n+1), by

assigning 1 or 0 to the first variable of n If we assign 1 , the number of solutions should be K0 ,

otherwise k0 Verifier will send to prover the new equation

CO-SAT IPCO-SAT IP

Now prover will send the new k0n,k1n for the new (n+1)

Verifier remember previous k1 and can check if k1=k0n+k1n , so the prover cannot cheat him

Each stage we reduced one variable from equation by assign a value to it

Now let’s prove completeness & soundness

CO-SAT IPCO-SAT IP Completeness

If prover does not cheat , each new equation will have the appropriate relation to the previous one and verifier will be convinced

Soundness If prover cheat i.e. send k0 as a false one, the new equation

should be based upon assignment of 0 to first element in order to see it (remember that we check only one of k0/k1 – it’s deepened on the assignment). We have a probability of ½ to do this , and we should always peek the right assignment down the road. Total probability (in the worst case) is (½)^n

Huston, we have a problem ! ( no soundness )

CO-SAT IP, Solution 2CO-SAT IP, Solution 2

We will expand the range variables of to a field (F) such that |F| > (2)^n

Each variable can get now not just 0 or 1 but a value from the field

We will construct a new equation `: F0 , T positive integer ^ * , +(p)`p , (~p)` 1-p`(p^q)`p`q` , (pq)` ~(~(~p ^ ~q))’

CO-SAT IP, Solution 2CO-SAT IP, Solution 2 We got now ` that is a polynomial of (x1,

…,Xn) over field F. Prover should now prove that

Note that1. Number of root for [p1(0)+p1(1)]= p0().2. Polynomials have the same number of roots for

[p1()-p2()] = 0

1,01,01 1,02

),...1`(.....xnx x

KXnx


Prover will send the polynom [P1], and the number of roots (K) for this polynom

Verifier will check that K=p1(0)+p1(1), choose a random value F and send it to prover

Prover will now construct a new polynom P2 = P1(), calculate the number of roots for the new one and send it to verfier

This process continue until all variable has been assign (2n iterations)


Completeness is clear.Soundness

In order to lie , the prover should send the verifier a false polynom. This polynom should have the same roots as the correct one. Since we have a field of elements ,The probability for this is n/|F|. The probability not found this is (1-n/|F|) > 2/3

We proved that CO-NP is in IP

CO-NP ZKCO-NP ZK

It’s enough to show that CO-SAT is in ZKThe problem in the previous solution is that the

verifier can see each stage the solution of the previous.

He can use it to get some other information from prover

CO-SAT ZKCO-SAT ZK

The prover can now send the polynom in an envelops , just like in G3C

The verifier should now check that the prover has not mislead him

We have got now a new problem : How can we open the envelops without gaining any information from the prover

CO-SAT ZKCO-SAT ZK

The problem of opening an envelops is in NP , since the oracle can guess the keys and we can verify in a polynomial time that indeed we have the appropriate keys

Since NP ZK , we can now make a reduction and solve the above problem

CO-SAT ZK CO-NP ZK !

Zero-Knowledge Proof System Slides by Ouzy Hadad, Yair Gazelle & Gil Ben-Artzi Adapted from Ely...

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