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**Having Proofs for Incorrectness**

coNP Having Proofs for Incorrectness Complexity

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**Introduction Objectives: Overview:**

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 Complexity

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

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**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: Complexity

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**Indeed it doesn’t satisfy x!**

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

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**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... Complexity

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**P coNP: As coP = P, and P NP**

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

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**NP-Complete & coNP-Complete**

L NP-Complete Lc coNP-Complete. AcNP R LNP-Complete AcoNP R LccoNP -Complete Complexity

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**Does the opposite direction also hold?**

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? Complexity

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**coNP=NP? & Completeness in coNP**

Claim: If a coNP-Complete problem L is in NP, under Karp reduction, then coNP=NP. Proof: in that case, any AcoNP, must be in NP AcoNP ANP R LcoNP-Complete LNP Complexity

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**What’s coNP’s Proper Position?**

Complexity

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

Complexity

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**PRIMES Instance: A number in binary representation.**

Problem: To decide if this number is prime. Yes instance: 10111 No instance: 10110 Complexity

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**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 Complexity

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**Don’t forget to make sure this takes**

PRIMES is in coNP Don’t forget to make sure this takes polynomial time Given a number N Guess two numbers i and j Verify i*j=N . . . 1 # . . . 1 # Complexity

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**5 is prime. What are its primitive roots?**

PAP Is PRIMES in NP? Claim: A number p > 2 is prime iff a number 1<r<p (called primitive root) s.t 1) rp-1 = 1 (mod p) 2) prime divisor q of p-1: r(p-1)/q 1 (mod p) 5 is prime. What are its primitive roots? Complexity

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**Pratt’s Theorem Pratt’s Theorem: PRIMES is in NPcoNP.**

Proof: Assuming the above claim we need to find some type of a guess that can be easily verify... Complexity

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**What Can We Get By Guessing r?**

We first need to verify rp-1=1 (mod p) BUT rp-1 mod p requires only poly-space rp-1 can be super-exponential! Complexity

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**What Can We Get By Guessing r?**

We first need to verify rp-1=1 (mod p) Performing p-1 multiplications is not polynomial! But you can start with r and square log(p-1) times! Complexity

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**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 klogn prime divisors. Proof: Denote the prime divisors of n by q1,...,qk. Note that nq1·... ·qk and all qi2. Thus n2k, i.e - klogn. Complexity

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**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! Complexity

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**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! Complexity

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**Make sure it’s succinct**

Claim Theorem The certificate that a natural p is a prime is the following: p=2 C(p)=() p>2 C(p)=(r,q1,C(q1),...,qk,C(qk)) Make sure it’s succinct Complexity

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**Make sure it takes poly-time**

The Verification 1. If p=2, accept 2. Otherwise, verify rp-1=1 (mod p). 3. Check that p can be reduced to 1 by repeated divisions by the qi’s. 4. Check r(p-1)/qi1 (mod p) for all the qi’s. 5. Recursively apply this algorithm upon every qi,C(qi) Make sure it takes poly-time Complexity

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

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**Observe: For any prime p, (p)={1,...,p-1}**

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

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**Fermat’s Little Theorem**

Fermat’s Little Theorem: Let p be a prime number 0 < a < p, ap-1 =1 (mod p) p=5; a=2 25-1 mod 5 = 16 mod 5 = 1 Example: Complexity

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**Observation 0<a<p, a·(p):={a·m (mod p) | m(p)} = (p)**

Example: 1 2 4 3 (5) ·2 (mod 5) 2 4 1 3 Complexity

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**Fermat’s Theorem: Proof**

Therefore, for any 0<a<p: 0 (mod p) Complexity

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**Generalization Claim: For all a(n) , a(n)=1 (mod n). Example:**

Complexity

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**Generalization: Proof**

Again: For any a(n), a·(n)=(n) Again: m(n)m 0 (mod n) 1 3 5 7 (8) Example: * (mod 8) And the claim follows. Complexity

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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 an-1-(n)=1 (mod n) Complexity

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

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**All Residues Have Exponents**

Let s (p). j > i N that satisfy si=sj (mod p). si is indivisible by p. sj-i=1 (mod p). Complexity

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

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**Non Primes Must Fail For a non prime n: It must be that (p) < p-1.**

Assume there is r s.t rp-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! Complexity

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**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: 2 6 4 1 7 . . . 3 5 n-1 all the residues modulo n are candidates for (n) Complexity

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**Corollaries Corollary: If gcd(m,n)=1, (mn)=(m)(n). Proof: **

(6)=|{1,5}|=2 (2)=|{1}|=1 (3)=|{1,2}|=2 Complexity

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**The Chinese Remainder Theorem**

The Chinese Remainder Theorem: If n is the product of distinct primes p1,...,pk, for each k-tuple of residues (r1,...,rk), where ri(pi), there is a unique r(n), where ri=r mod pi for every 1ik. 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} Complexity

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**The Chinese Remainder Theorem**

Proof: If n is the product of distinct primes p1,...,pk, then (n)=1ik(pi-1). This means |(n)|=|(p1)...(pk)|. The following is a 1-1 correspondence between the two sets: r (r mod p1,...,r mod pk) Complexity

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**Another Property of the Euler Function**

Claim: m|n(m)=n. Example: m|12(m)= (1) + (2) + (3) + (4) + (6) + (12)= |{1}| + |{1}| + |{1,2}| + |{1,3}| + |{1,5}| + |{1,5,7,11}|= = 12 Complexity

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**Another Property of the Euler Function**

Claim: m|n(m)=n. Proof: Let 1ilpiki be the prime factorization of n. (n)=np|n(1-1/p) m|n(m)= Since (ab)=(a)(b) telescopic sum Complexity

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**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)? Complexity

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**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 x1,...,xk+1 are roots of (x)=akxk+...+a0. ’(x)= (x)-ak1ik(x-xi) is of degree k-1 and not identically zero. x1,...,xk are its roots - Contradiction! Complexity

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**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,s2,…,sk-1) are k distinct solutions of xk=1 (mod p) (why?) If sl has exponent k, l(k) (otherwise its exponent is lower). Complexity

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**All p-1 residues have exponents**

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

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**Summing Up R(k)=(k) for all divisors of p-1 R(p-1) = (p-1) > 0**

p has at least one primitive root Complexity

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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 Complexity

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

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Place PRIMES PRIMES P NP coNP Complexity

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** 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 NPcoNP, though it’s believed not to be in P. Complexity

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