The Cook-Levin Theorem Matvey Soloviev (Cornell University) CS 4820, - PowerPoint PPT Presentation

The Cook-Levin Theorem Matvey Soloviev (Cornell University) CS 4820, Summer 2020 1

Today By using our formal definition of nondeterministic Turing machines, we will prove the existence of a universal polynomial time. This problem will be provably hard under the assumption that any problem in NP is. 2 subproblem which every problem in NP can be reduced to in

Hardness from reductions, made concrete (1) Before, we talked about how a reduction from P to Q lets us infer hardness of Q from hardness of P conceptually. Let’s make a concrete instance of the statement. Proposition Suppose P is a problem for which there exists no polynomial-time algorithm, and there is a polynomial- time reduction from P to Q . Then there is no polynomial-time algorithm for Q . 3

Hardness from reductions, made concrete (1) Before, we talked about how a reduction from P to Q lets us infer hardness of Q from hardness of P conceptually. Let’s make a concrete instance of the statement. Proposition Suppose P is a problem for which there exists no polynomial-time algorithm, and there is a polynomial- time reduction from P to Q . Then there is no polynomial-time algorithm for Q . 3

Hardness from reductions, made concrete (1) Before, we talked about how a reduction from P to Q lets us infer hardness of Q from hardness of P conceptually. Let’s make a concrete instance of the statement. Proposition Suppose P is a problem for which there exists no polynomial-time algorithm, and there is a polynomial- time reduction from P to Q . Then there is no polynomial-time algorithm for Q . 3

Hardness from reductions, made concrete (2) Proof. Suppose not, and there is in fact a polynomial-time Plug this algorithm into the reduction. By definition of a polynomial-time reduction, we obtain a correct algorithm that solves size- n instances of P in time h n g n O p f n . Since all of f , g and h are polynomials, this is again a polynomial (check this!). But then we actually do have a polynomial-time algorithm for P , contradicting the assumption. 4 algorithm for Q that runs in time O ( p ( n )) .

Hardness from reductions, made concrete (2) Proof. Suppose not, and there is in fact a polynomial-time Plug this algorithm into the reduction. By definition of a polynomial-time reduction, we obtain a correct algorithm that Since all of f , g and h are polynomials, this is again a polynomial (check this!). But then we actually do have a polynomial-time algorithm for P , contradicting the assumption. 4 algorithm for Q that runs in time O ( p ( n )) . solves size- n instances of P in time h ( n ) + g ( n ) · O ( p ( f ( n ))) .

Hardness from reductions, made concrete (2) Proof. Suppose not, and there is in fact a polynomial-time Plug this algorithm into the reduction. By definition of a polynomial-time reduction, we obtain a correct algorithm that Since all of f , g and h are polynomials, this is again a polynomial (check this!). But then we actually do have a polynomial-time algorithm for P , contradicting the assumption. 4 algorithm for Q that runs in time O ( p ( n )) . solves size- n instances of P in time h ( n ) + g ( n ) · O ( p ( f ( n ))) .

Hardness from reductions, made concrete (2) Proof. Suppose not, and there is in fact a polynomial-time Plug this algorithm into the reduction. By definition of a polynomial-time reduction, we obtain a correct algorithm that Since all of f , g and h are polynomials, this is again a polynomial (check this!). But then we actually do have a polynomial-time algorithm for 4 algorithm for Q that runs in time O ( p ( n )) . solves size- n instances of P in time h ( n ) + g ( n ) · O ( p ( f ( n ))) . P , contradicting the assumption. ✷

Boolean formulae A Boolean formula is an expression generated recursively by the grammar assignment is itself a function from V to truth values. 5 φ, ψ ::= φ ∧ ψ | φ ∨ ψ | ¬ φ | x , y , z , . . . | ( φ ) , where x , y , z , . . . ∈ V are variable names, which represents a function from assignments to truth values { T , F } . An For instance, φ = x ∧ ( y ∨ z ) → ( ¬ v ) .

i n i is called a clause. Conjunctive normal form A Boolean formula is said to be in conjunctive normal form (CNF) if it is of the form Each i 1 6 ( ℓ 1 , 1 ∨ . . . ℓ 1 , n 1 ) ∧ . . . ∧ ( ℓ m , 1 ∨ . . . ∨ ℓ m , n m ) , where each ℓ i , j is a literal: that is, either a variable v ∈ V or the negation ¬ v of one. For example, ( x ∨ ¬ z ) ∧ ( y ∨ z ∨ x ∨ v ) ∧ ( x ) .

Conjunctive normal form A Boolean formula is said to be in conjunctive normal form (CNF) if it is of the form 6 ( ℓ 1 , 1 ∨ . . . ℓ 1 , n 1 ) ∧ . . . ∧ ( ℓ m , 1 ∨ . . . ∨ ℓ m , n m ) , where each ℓ i , j is a literal: that is, either a variable v ∈ V or the negation ¬ v of one. For example, ( x ∨ ¬ z ) ∧ ( y ∨ z ∨ x ∨ v ) ∧ ( x ) . Each ( ℓ i , 1 ∨ . . . ∨ ℓ i , n i ) is called a clause.

The CNF-SAT problem The conjunctive normal form Boolean satisfiability problem (CNF-SAT) is: Given a Boolean formula in CNF, does there exist any assignment to its variables that makes this formula true? This problem is in NP: We can write out the assignment (as a string of the form x T y F ) and use it as a certificate. 7

The CNF-SAT problem The conjunctive normal form Boolean satisfiability problem (CNF-SAT) is: Given a Boolean formula in CNF, does there exist any assignment to its variables that makes this formula true? This problem is in NP: We can write out the assignment (as a 7 string of the form x �→ T , y �→ F , . . . ) and use it as a certificate.

The Cook-Levin Theorem Theorem (Cook-Levin) The CNF-SAT problem is NP-complete. 8

That is, we need to show that there exists a polynomial-time reduction from every problem in NP to CNF-SAT. Proof of Cook-Levin: High-level overview (1) We already know that the problem is in NP, so it just remains to establish that it is NP-hard. 9

Proof of Cook-Levin: High-level overview (1) We already know that the problem is in NP, so it just remains to establish that it is NP-hard. reduction from every problem in NP to CNF-SAT. 9 That is, we need to show that there exists a polynomial-time

if and only if M P accepts the string polynomial-time Turing machine M P which decides P . Proof of Cook-Levin: High-level overview (2) Boolean formula rejected. 4. Accept if the algorithm accepted. Reject if the algorithm 3. Run our algorithm for CNF-SAT on this formula. . which has a satisfying assignment M P , write out a specially prepared CNF We will establish this as follows: 2. In time polynomial in to P . 1. Read the input following reduction: Using our knowledge of this machine, we can write the Since P is in NP, there must exist a nondeterministic Let P be an arbitrary problem in NP. 10

if and only if M P accepts the string polynomial-time Turing machine M P which decides P . Proof of Cook-Levin: High-level overview (2) Boolean formula rejected. 4. Accept if the algorithm accepted. Reject if the algorithm 3. Run our algorithm for CNF-SAT on this formula. . which has a satisfying assignment M P , write out a specially prepared CNF We will establish this as follows: 2. In time polynomial in to P . 1. Read the input following reduction: Using our knowledge of this machine, we can write the Since P is in NP, there must exist a nondeterministic Let P be an arbitrary problem in NP. 10

if and only if M P accepts the string Proof of Cook-Levin: High-level overview (2) Boolean formula rejected. 4. Accept if the algorithm accepted. Reject if the algorithm 3. Run our algorithm for CNF-SAT on this formula. . which has a satisfying assignment M P , write out a specially prepared CNF We will establish this as follows: 2. In time polynomial in to P . 1. Read the input following reduction: Using our knowledge of this machine, we can write the Since P is in NP, there must exist a nondeterministic Let P be an arbitrary problem in NP. 10 polynomial-time Turing machine M P which decides P .

if and only if M P accepts the string Proof of Cook-Levin: High-level overview (2) Boolean formula rejected. 4. Accept if the algorithm accepted. Reject if the algorithm 3. Run our algorithm for CNF-SAT on this formula. . which has a satisfying assignment M P , write out a specially prepared CNF We will establish this as follows: 2. In time polynomial in to P . 1. Read the input following reduction: Using our knowledge of this machine, we can write the Since P is in NP, there must exist a nondeterministic Let P be an arbitrary problem in NP. 10 polynomial-time Turing machine M P which decides P .

if and only if M P accepts the string Proof of Cook-Levin: High-level overview (2) Boolean formula rejected. 4. Accept if the algorithm accepted. Reject if the algorithm 3. Run our algorithm for CNF-SAT on this formula. . which has a satisfying assignment M P , write out a specially prepared CNF We will establish this as follows: 2. In time polynomial in following reduction: Using our knowledge of this machine, we can write the Since P is in NP, there must exist a nondeterministic Let P be an arbitrary problem in NP. 10 polynomial-time Turing machine M P which decides P . 1. Read the input α to P .