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Complexity 1 coNP Having Proofs for Incorrectness

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Complexity 2 Introduction Objectives: –To introduce the complexity class coNP –To explore the primality problem. Overview: –coNP: Definition and examples –coNP=NP? and NP=P? –PRIMES and Pratt’s theorem

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Complexity 3 CoNP Def: CoNP is the class of problems that have succinct non-membership witnesses.

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Complexity 4 VALIDITY Instance: A Boolean formula Problem: To decide if the formula is valid (i.e satisfiable by all possible assignments) A valid Boolean formula: An invalid Boolean formula:

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Complexity 5 VALIDITY is in coNP Guess an assignment Verify it doesn’t satisfy the formula (x)=F x Indeed it doesn’t satisfy x!

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Complexity 6 Using what we Know about NP By definition, the complement of every NP language is in coNP. The complement of a coNP language is NP. VALIDITY is in coNP! Since SAT is in NP...

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Complexity 7 NP and coNP P NP coNP P coNP: As coP = P, and P NP

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Complexity 8 NP-Complete & coNP-Complete L NP-Complete L c coNP-Complete. A c NP L NP- Complet e A coNP R R L c coNP -Complete

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Complexity 9 NP=P? & coNP=NP? Claim: P=NP implies coNP=NP. Proof: P=coP, hence if P=NP, NP=coNP. Does the opposite direction also hold?

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Complexity 10 coNP=NP? & Completeness in coNP Claim: If a coNP-Complete problem L is in NP, under Karp reduction, then coNP=NP. A coNP L coNP- Complete R L NP A NP Proof: in that case, any A coNP, must be in NP

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Complexity 11 What’s coNP’s Proper Position? P NP

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Complexity 12 Here It Is! P Open question: Are NP\coNP, coNP\NP actually empty? NPcoNP

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Complexity 13 PRIMES Instance: A number in binary representation. Problem: To decide if this number is prime Yes instance: No instance:10110

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Complexity 14 Is Primes in P ?! What’s the problem with the following algorithm? Input: a number N Output: is N prime? for i in 2.. N do for j in 2.. N do if i*j=N, return FALSE return TRUE

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Complexity ## PRIMES is in coNP Given a number N ### Guess two numbers i and j Verify i*j=N Don’t forget to make sure this takes polynomial time

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Complexity 16 Is PRIMES in NP? Claim: A number p > 2 is prime iff a number 1

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Complexity 17 Pratt’s Theorem Pratt’s Theorem: PRIMES is in NP coNP. Proof: Assuming the above claim we need to find some type of a guess that can be easily verify...

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Complexity 18 What Can We Get By Guessing r? We first need to verify r p-1 =1 (mod p) r p-1 can be super- exponential! BUT r p-1 mod p requires only poly-space

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Complexity 19 Performing p-1 multiplications is not polynomial! What Can We Get By Guessing r? We first need to verify r p-1 =1 (mod p) But you can start with r and square log(p-1) times!

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Complexity 20 Verifying the Second Requirement Next we need to verify, that prime divisor q of p-1: r (p-1)/q 1 (mod p) Lemma: Any n>1 has k logn prime divisors. Proof: Denote the prime divisors of n by q 1,...,q k. Note that n q 1 ·... ·q k and all q i 2. Thus n 2 k, i.e - k logn.

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Complexity 21 Verifying the Second Requirement Next we need to verify, that prime divisor q of p-1: r (p-1)/q 1 (mod p) How would you find the prime divisors of p-1? Obviously I wouldn’t! I’d just guess them!

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Complexity 22 Verifying the Second Requirement Next we need to verify, that prime divisor q of p-1: r (p-1)/q 1 (mod p) How would you verify they are prime? Exactly the same way!

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Complexity 23 Claim Theorem The certificate that a natural p is a prime is the following: p=2C(p)=() p>2C(p)=(r,q 1,C(q 1 ),...,q k,C(q k )) Make sure it’s succinct

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Complexity 24 The Verification 1. If p=2, accept 2. Otherwise, verify r p-1 =1 (mod p). 3. Check that p can be reduced to 1 by repeated divisions by the q i ’s. 4. Check r (p-1)/q i 1 (mod p) for all the q i ’s. 5. Recursively apply this algorithm upon every q i,C(q i ) Make sure it takes poly-time

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Complexity 25 Proof of Claim Need to show that every prime satisfies both conditions and that any number satisfying both conditions is a prime

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Complexity 26 Euler’s Function (n) = { m | 1 m < n AND gcd(m,n)=1 } Euler’s function: (n)=| (n)| (12)={1,2,3,4,5,6,7,8,9,10,11} (12)=4 Example: Observe: For any prime p, (p)={1,...,p-1}

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Complexity 27 Fermat’s Little Theorem Fermat’s Little Theorem: Let p be a prime number 0 < a < p, a p-1 =1 (mod p) p=5; a= mod 5 = 16 mod 5 = 1 p=5; a= mod 5 = 16 mod 5 = 1 Example:

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Complexity 28 Observation 0

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Complexity 29 Fermat’s Theorem: Proof Therefore, for any 0

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Complexity 30 Generalization Claim: For all a (n), a (n) =1 (mod n). n=8, (8) = {1,3,5,7} 3 4 =1 (mod 8) n=8, (8) = {1,3,5,7} 3 4 =1 (mod 8) Example:

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Complexity 31 Generalization: Proof (8) Example: * (mod 8) Again: For any a (n), a· (n)= (n) Again: m (n) m 0 (mod n) And the claim follows.

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Complexity 32 What have we got So Far We know if p is prime condition (1) holds for all a For non prime n, condition (1) may hold for some a but then a (n) =1 (mod n) as well, hence a n-1- (n) =1 (mod n)

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Complexity 33 Exponents Def: If m (p), the exponent of m is the smallest integer k > 0 such that m k =1 (mod p). p=7, m=4 (7), the exponent of 4 is 3. p=7, m=4 (7), the exponent of 4 is 3. Example:

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Complexity 34 All Residues Have Exponents Let s (p). j > i N that satisfy s i =s j (mod p). s i is indivisible by p. s j-i =1 (mod p).

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Complexity 35 Regarding Exponents Observation: The only powers of m that are 1 (mod p) are multiplies of its exponent! Assuming r p-1 = 1 (mode p), by Fermat’s theorem, r’s exponent divides p-1

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Complexity 36 Non Primes Must Fail For a non prime n: It must be that (p) < p-1. Assume there is r s.t r p-1 =1 (mod p) We’ve shown r (p) =1 (mod p) So there is also a prime divisor q of p-1, s.t r (p-1)/q =1 mod p. We may conclude: if both conditions hold p is prime!

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Complexity 37 An Equivalent Definition of Euler’s Function Using Prime Divisors Let p be a prime divisor of n. The probability p divides a candidate is 1/p. Thus:... all the residues modulo n are candidates for (n)

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Complexity 38 Corollaries Corollary: If gcd(m,n)=1, (mn)= (m) (n). Proof: (6)=|{1,5}|=2 (2)=|{1}|=1 (3)=|{1,2}|=2 (6)=|{1,5}|=2 (2)=|{1}|=1 (3)=|{1,2}|=2

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Complexity 39 21=7·3 (21)={1,2,4,5,8,10,11,13,16,17,19,20} (3) ={1,2} (7) ={1,2,3,4,5,6} 21=7·3 (21)={1,2,4,5,8,10,11,13,16,17,19,20} (3) ={1,2} (7) ={1,2,3,4,5,6} The Chinese Remainder Theorem The Chinese Remainder Theorem: If n is the product of distinct primes p 1,...,p k, for each k-tuple of residues (r 1,...,r k ), where r i (p i ), there is a unique r (n), where r i =r mod p i for every 1 i k.

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Complexity 40 The Chinese Remainder Theorem Proof: If n is the product of distinct primes p 1,...,p k, then (n)= 1 i k (p i -1). This means | (n)|=| (p 1 ) ... (p k )|. The following is a 1-1 correspondence between the two sets: r (r mod p 1,...,r mod p k )

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Complexity 41 Another Property of the Euler Function Claim: m|n (m)=n. m|12 (m)= (1) + (2) + (3) + (4) + (6) + (12)= |{1}| + |{1}| + |{1,2}| + |{1,3}| + |{1,5}| + |{1,5,7,11}|= = 12 m|12 (m)= (1) + (2) + (3) + (4) + (6) + (12)= |{1}| + + |{1,2}| + |{1,3}| + |{1,5}| + |{1,5,7,11}|= = 12 Example:

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Complexity 42 Another Property of the Euler Function Claim: m|n (m)=n. Proof: Let 1 i l p i k i be the prime factorization of n. (n)=n p|n (1-1/p) telescopic sum m|n (m)= Since ( ab )= ( a ) ( b ) ( ab )= ( a ) ( b )

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Complexity 43 Group together Residues with Same Exponent Fix a p and let R(k) denote the number of residues with exponent k. If k does not divide p-1, R(k)=0. Can you upper bound R(k)?

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Complexity 44 Polynomials Have Few Roots Claim: Any polynomial of degree k that is not identically zero has at most k distinct roots modulo p. Proof: By induction on k. Trivially holds for k=0. Suppose it also holds for some k-1. By way of contradiction, assume x 1,...,x k+1 are roots of (x)=a k x k +...+a 0. ’(x)= (x)-a k 1 i k (x-x i ) is of degree k-1 and not identically zero. x 1,...,x k are its roots - Contradiction!

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Complexity 45 How Many Residues Can Share an Exponent? Conclusion: There are at most k residues of exponent k. Claim: R(k) ≤ (k) Proof: –Let s be a residue of exponent k. –(1,s,s 2,…,s k-1 ) are k distinct solutions of x k =1 (mod p) (why?) –If s l has exponent k, l (k) (otherwise its exponent is lower).

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Complexity 46 Summing Up p-1 = = p-1 All p-1 residues have exponents m|n (m)=n

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Complexity 47 Summing Up R(k)= (k) for all divisors of p-1 R(p-1) = (p-1) > 0 p has at least one primitive root

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Complexity 48 Where Do We Stand? We’ve shown every prime has a primitive root. Hence any prime satisfied both conditions We’ve previously shown any non prime does not satisfy both conditions

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Complexity 49 Q.E.D! This finally proves the validity of our alternative characterization of primes, which implies that PRIMES is in NP.

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Complexity 50 Place PRIMES P PRIMES NPcoNP

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Complexity 51 Summary We’ve studied the complexity class coNP, and explored the relations between coNP and other classes, such as P and NP. We’ve introduced PRIMES and showed it’s in NP coNP, though it’s believed not to be in P.

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