L is in NP means: There is a language L ’ in P and a polynomial p so that

1. L is in NP means: There is a language L’ in P and a polynomial p so that • L1·L2 means: For some polynomial time computable map r : 8 x: x 2 L1 iff r(x) 2L2 • L is NP-hard means: 8L’ 2NP: L’ ·L • L is in NPC means: L2NP and L is NP-hard

2. Polynomial time computable maps f: {0,1}* ! {0,1}* is called polynomial time computable if for some polynomial p, - For all x, |f(x)| ·p(|x|). - Lf2P.

3. Equivalent definition • A map is polynomial time computable if and only if there is a Turing machine that on every input x accepts after at most a polynomial number of steps and leaves f(x) on its tape when terminating.

4. How to establish NP-hardness • Lemma: If L1 is NP-hard and L1· L2 then L2 is NP-hard.

5. SAT • SAT: Given a Boolean function in CNF representation, is there a way to assign truth values to the variables so that the function evaluates to true? • SAT: Given a CNF, is it true that it does not represent the constant-0 function? • Input: (: x1Ç: x2) Æ (x1Ç x2) • Output: Yes. • Input: (: x1Ç: x2) Æ (x1Ç x2) Æ (x1Ç: x2 ) Æ (: x1Ç x2) • Output: No.

6. SAT • SAT is in NP. • Cook’s theorem (1971): SAT is NP-hard.

7. TSP HAMILTONIAN CYCLE MIN VERTEX COLORING SAT MAX INDEPENDENT SET SET COVER ILP MILP KNAPSACK TRIPARTITE MATCHING BINPACKING

8. Usefulness of NPC • Languages in NPC are the least likely problems in NP to be in P. • Suppose we would like to find algorithm for L 2NPC. • If we believe that P is not NP, we know that no worst case efficient algorithm exists. • If we have no opinion about P vs. NP, we know that if we find an efficient algorithm for L, we’ll earn $1,000,000.

9. Cooks’ theorem: SAT is NP-hard • Proof of Cook’s theorem: • CIRCUIT SAT is NP-hard. • CIRCUIT SAT reduces to SAT. • Hence, SAT is NP-hard.

10. Boolean Circuits ⋀ ⋁ ⋀ ⌐ X3 X2 X1

11. Circuits vs. Turing Machines • Circuits can be given as inputs to algorithms but they can also be seen as computational devices themselves! • Like Turing Machines, circuits C: {0,1}n! {0,1} solve decision problems on {0,1}n. • Unlike Turing machines, circuits takes inputs of afixedinput length n only.

12. Theorem • Given Turing Machine M running in time at most p(n) on inputs of length n, where p is a polynomial. • For every n, there is a circuit Cn with at most O(p(n)2) gates so that 8x2 {0,1}n: Cn(x)=1 iff M accepts x. • The map 1n!Cn is polynomial time computable.

13. Remark • This is really just like the Polynomial Turing’s Thesis, only “in reverse”: • We show that “a reasonable sequential model of computation” (computation by “uniform” families of circuits) has at least as much power as Turing Machines.

14. Intuition behind proof

15. Problem: Cycles! Flip-Flop, stores one bit.

16. The Tableau Method Time t … Time 1 Can be replaced by acyclic Boolean circuit of size ≈ s Time 0

17. Cell state vectors • Given a Turing Machine computation, an integer t and an integer i let cti2 {0,1}s be a Boolean representation of the following information, a cell state vector: • The symbol in cell i at time t • Whether or not the head is pointing to cell i at time t • If the head is pointing to cell i, what is the state of the finite control of the Turing machine at time t? • The integer s depends only on the Turing machine (not the input to the computation, nor t,i). • To make cti defined for all t, we let c(t+1)i = ct i if the computation has already terminated at time t.

18. Crucial Observation • If we know the Turing machine and ct-1,i-1, ct-1,i, ct-1, i+1, we also can determine ct,i. • In other words, there is a Boolean function h depending only on the Turing machine so that ct,i = h(ct-1,i-1, ct-1,i, ct-1,i+1). • A circuit D for h is the central building block in a circuit computing all cell state vectors for all times for a given input.

19. t(n) x1 xn 2 t(n)

20. Cooks’ theorem: SAT is NP-hard • Proof of Cook’s theorem: • CIRCUIT SAT is NP-hard. • CIRCUIT SAT reduces to SAT. • Hence, SAT is NP-hard.

21. CIRCUIT SAT • CIRCUIT SAT: Given a Boolean circuit, is there a way to assign truth values to the input gates, so that the output gate evaluates to true? • Generalizes SAT, as CNFs are formulas and formulas are circuits.

22. Example ⋁ Input: ⋀ Output: Yes ⋀ ⌐ X3 X2 X1

23. Example ⋀ Input: ⋀ Output: No ⋀ ⌐ X3 X2 X1

24. CIRCUIT SAT is NP-hard • Given an arbitrary language L in NP we must show that L reduces to CIRCUIT SAT. • This means: We must construct a polynomial time computable map r mapping instances of L to circuits, so that 8 x: x 2 L , r(x) 2 CIRCUIT SAT • The only thing we know about L is that there is a language L’ in P and a polynomial p, so that:

25. Theorem • Given Turing Machine M for L’ running in time at most q(n) on inputs of length n, where q is a polynomial. • For every n, there is a circuit Cn with at most O(q(n)2) gates so that 8x2 {0,1}n: Cn(x)=1 iff M accepts x. • The map 1n!Cn is polynomial time computable.

26. Cooks’ theorem: SAT is NP-hard • Proof of Cook’s theorem: • CIRCUIT SAT is NP-hard. • CIRCUIT SAT reduces to SAT. • Hence, SAT is NP-hard.

27. TSP HAMILTONIAN CYCLE MIN VERTEX COLORING SAT MAX INDEPENDENT SET SET COVER ILP MILP KNAPSACK TRIPARTITE MATCHING BINPACKING

28. Remarks on Papadimitriou’s terminology • When Papadimitriou writes “log space reduction”, just substitute “polynomial time reduction”. • When Papadimitriou writes NL, just substitute P. • Papadimtriou’s concepts are more restrictive, but the more restrictive definitions will play no role in this course.