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Complexity 26-1 Complexity Andrei Bulatov Interactive Proofs

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Complexity 26-2 Interactive Proofs ProverVerifier Has unlimited computational power Can perform polynomial time computations They can exchange messages Wants to convince Verifier in something Accepts or rejects after performing some computation

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Complexity 26-3 Formal Model Prover and Verifier are represented by functions: Verifier (V) Input: an input string, w a random string, r message history Output: A message, that is a string, or accept or reject Prover (P) Input: an input string, w message history Output:A message, that is a string

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Complexity 26-4 A Prover’s move is the message A Verifier’s move is the message We write if for some k and if for some k We assume that the length of messages, the length of r, and k are bounded with a polynomial p(|w|) Finally, we denote where r is a random string of length p(|w|)

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Complexity 26-5 The Class IP Definition A language L belongs to IP if some polynomial time function V and arbitrary function P exist, such that for every function R and string w w L implies Pr[V P accepts w] w L implies Pr[V R accepts w]

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Complexity 26-6 Examples Instance: Graphs G and H. Question: Are G and H non-isomorphic? Graph Non-Isomorphism Instance: A formula in CNF and a number k Question: Is it true that has exactly k satisfying assignments? #SAT Theorem #SAT IP

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Complexity 26-7 Proof Given a formula and a number k. Let be the variables of The first stage is “arithmetization” of : We construct a polynomial such that

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Complexity 26-8 By induction for X define the corresponding polynomial to be x for X define the corresponding polynomial to be 1 – x for with corresponding polynomials g and h, define the corresponding polynomial to be gh (the product) Arithmetization for with corresponding polynomials g and h, define the corresponding polynomial to be 1 – (1 – g)(1 – h) for with corresponding polynomial g, define the corresponding polynomial to be 1 – g

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Complexity 26-9 Protocol Let be the result of arithmetization of Then The protocol will go in n rounds On every round, we will construct from an equality an equality of the form

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Complexity 26-10 The property we are trying to achieve is if and only if If this is possible, starting off with after n rounds we get an equality which is easy to verify The last equality is true if and only if the first one is true

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If was true before the this round, then Complexity 26-11 Randomness Unfortunately, the condition above is impossible to achieve Instead, we use a probabilistic one is true after this round with probability 1 If was false before the this round, then is false after this round with probability 1–

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Complexity 26-12 Analysis Thus, after n rounds If we started off with a correct equality, then after n rounds we have a correct equality with probability 1 If we started off with an incorrect equality, then after n rounds the probability that we end up with a correct equality is the probability that the error was made in one of the n rounds, which is at most n . Thus, with probability at least 1 – n , we will have an incorrect equality We need to organize every round such that

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Complexity 26-13 Round Having the equality Verifier sends to Prover the polynomial g Prover returns the polynomial or what he pretends this polynomial is Verifier checks if the degree of h is not higher than that of g Verifier checks if Verifier randomly chooses a number r, such that where m is the number of clauses in , and replace the original equality with

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Complexity 26-14 Analysis If the original equality is true then Prover should behave honestly, that is return actual coefficients of h(x). Then - The degree of h is not higher than that of g - h(0) + h(1) = k - For any r,

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Complexity 26-15 If the original equation is not true then consider two cases - if Prover is honest that is he returns the actual coefficients of h, then h(0) + h(1) k - if Prover is dishonest that is he returns some polynomial p(x) such that the degree of p is not higher than that of g and p(0) + p(1) = k In this case p and h are different. Therefore their values can be equal for at most d = max{deg(p), deg(h)} values of x. The probability that for a randomly chosen is

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Complexity 26-16 Note that the degree of f is at most mn. Therefore

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