Complexity 22 1 Probabilistic Algorithms Complexity Andrei Bulatov

Complexity 22 -1 Probabilistic Algorithms Complexity Andrei Bulatov

Complexity Non-Deterministic vs. Probabilistic All algorithms we nave seen so far are either deterministic or impractical (non-deterministic) To make non-deterministic algorithms more practical we introduce probabilistic algorithms A probabilistic algorithm (Turing Machine) is a nondeterministic algorithm that makes non-deterministic choices randomly, e. g. by flipping a coin This is still not practical, because sometimes the algorithm should be extremely lucky to solve problems 22 -2

Complexity 22 -3 Interactive Proofs Prover Verifier Has unlimited computational power Can perform polynomial time computations Wants to convince Verifier in something Accepts or rejects after performing some computation They can exchange massages

Complexity Proofs for Problems in NP SAT Prover and Verifier get an instance of SAT Prover solves the instance using his unlimited computational power and send a satisfying assignment to Verifier checks (in polynomial time) if what obtained is a satisfying assignment, and accepts if it is or rejects otherwise 22 -4

Complexity 22 -5 Problems from co. NP Graph Non-Isomorphism Instance: Graphs G and H. Question: Are G and H isomorphic? This problem belongs to co. NP, but is not believed to be co. NP-co Apparently, there is no way to prove interactively that two graphs are not isomorphic

Complexity 22 -6 Randomized Verifier Now suppose that Verifier has a fair coin Given graphs G and H • Verifier choose one of G and H by flipping a coin • Verifier then rename somehow the vertices of the chosen graph and send it to Prover • Prover decides which graph it received • Prover send the answer to Verifier • Repeat the procedure

Complexity 22 -7 Analysis If the graphs are not isomorphic then Prover always gives the right answer If the graphs are isomorphic then Prover gives a correct answer with probability 1/2 Therefore if Prover is wrong we conclude that the graphs are isomor If after n repetitions of the protocol, Prover gives only right answers, then Verifier can conclude with probability that graphs are not isomorphic

Complexity Probabilistic Turing Machines Definition A Probabilistic Turing Machine is a nondeterministic polynomial time Turing Machine PT such that • from each configuration of PT, there at most two possible next configurations • PT chooses which of the two possible next Thus, with each computational path, we can associate the configurations probability of taking this path. This probability is equal to to take by flipping a fair coin where k is the number of coin flips made along this path Denote this probability by Pr[p] 22 -8

Complexity Define the probability that PT accepts w to be Clearly 22 -9

Complexity 22 -10 Class BPP Definition A Probabilistic Turing Machine PT recognizes language L with error probability if • w L implies Pr[PT accepts w] 1 – • w L implies Pr[PT rejects w] 1 – We say that PT operates with error probability if the above inequalities hold for where n is th length of w Definition BPP is the class of languages that are recognizable by probabilistic Turing Machines with error probability of 1/3

Complexity Amplification The error probability 1/3 may seem random Actually, we can choose any value 0 1 Amplification Lemma Let 0 1. Then for any polynomial p(n) and a probabilistic TM that operates with error probability , there is a probabilistic TM that operates with an error probability The main idea is to run many times and then output the majority of votes 22 -11

Complexity Math Prerequisites Let be a series of independent experiments (for example, coin flips) such that the probability of success in each of them is p Theorem (Chernoff Bound) for some , then the probability If that the number of successes in a series of n experiments is less than is at most 22 -12

Complexity 22 -13 Proof of Amplification Lemma Machine works as follows On input w • for i = 1 to t(|w|) do - • simulate if most runs of on w accept, then accept; otherwise reject

Complexity Analysis The number t(|w|) must be such that 22 -14

Complexity 22 -15 Primes Complexity Andrei Bulatov

Complexity 22 -16 The Problem Primes Instance: A positive integer k. Question: Is k prime? The complement of Primes, the Composite problem, belongs to NP. Therefore Primes is in co. NP Recently M. Agarwal et al. Proved that Primes can be solved in polynomial time (see http: //www. cse. iitk. ac. in/news/primality. html) However, the probabilistic algorithm we are going describe is far mo

Complexity 22 -17 Residues For a positive integer n, we denote • the set {0, 1, 2, …, n – 1} • the set {1, 2, …, n – 1} • addition, multiplication and exponentiation modulo together with these operations is called the set of residues m Every integer m, positive or negative, has a corresponding residue — m mod n For example, 17 mod 5 = 2 20 mod 5 = 0 -1 mod 5 = 4

Complexity 22 -18 Complexity of Arithmetic Given two integers, a and b, we can compute • • a + b in O(max(log a, log b)) a b in O(log a log b) cannot be computed in polynomial time, because the size of this number is blog a It is possible modulo n Let be the binary representation of b (k = log b) Then that implies First, we consecutively compute Then we compute the product again in in

Complexity 22 -19 Prime and Coprime Integers a and b are called coprime if their greatest common div For example, 16 and 27 are coprime, and 15 and 18 are not Theorem (Chinese Remainder Theorem) If p and q are coprime then, for any a and b, there is x such that For example, if p = 5, q = 3, and a = 2, b = 1, then x can be chosen to be 7

Complexity 22 -20 Fermat’s Theorem (Fermat’s Little Theorem) If p is prime then, for any we have If the converse were true, we could use it for a probabilistic primality • Choose k residues modulo n; • Compute their n – 1 powers; • Accept if all results are 1 (mod n), reject otherwise

Complexity 22 -21 Carmichael Numbers Unfortunately, the converse is true just “almost” Definition A number n passes Fermat’s test if for all a coprime with n A number that passes Fermat’s test is called pseudo-prime One can straightforwardly check that, for any 561 is a Carmichael number n is said to be a Carmichael number if, for any prime divisor p of n, p – 1 | n – 1 Pseudo-prime = Prime + Carmichael , coprime w

Complexity 22 -22 Roots of 1 A square root of 1 modulo n is a number a such that Clearly, 1 and -1 (that is n – 1) are always roots of 1, but if n is composite, then it may have more than two roots of 1 For example, 8 has four roots of 1: 1, -1, 3, and 5 561 has eight: 1, -1, 188, 373 (find the remaining four) Lemma Any Carmichael number has at least 8 roots of 1

Complexity 22 -23 Algorithm On input n • if n is even, then if n = 2 accept, otherwise reject • select randomly • for i = 1 to k do - if then reject - let n – 1 = st where s is odd and - compute the sequence - if then let j be the maximal with this property if • accept then reject is a power o modulo n

Complexity 22 -24 Analysis First we show that the algorithm does not give false negatives, that is it accepts all prime numbers If n = 2 then n is accepted. Let n be an odd prime number Then n passes Fermat test n cannot be rejected in the last line, because n has only two roots o

Complexity Next we show that if n is composite, then Pr[n accepted] A number such that a does not pass either Fermat test or the square root test, is called a witness It is enough to prove that Pr[a is a witness] 1/2, or, in other words, that at least half of the elements of are witnesses For every nonwitness d we find a witness d´ such that if then For a nonwitness a the sequence either contains 1 s only, or it contains -1 followed by 1 s Nonwitnesses of both types are present: 1 is a nonwitness of the first type, and -1 is a nonwitness of the second type 22 -25

Complexity 22 -26 Let d be a nonwitness of the second type such that the – 1 appears in the largest position in the sequence Let and Since n is composite, n = qr for some coprime q and r Note that and By the Chinese Reminder Theorem, there is t such that therefore Hence t is a witness, because but

Complexity 22 -26 Now, for every nonwitness a we set a´ = a · t • a´ is a witness, because and but • if then Assume the contrary Then, since Finally, we have