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

2. 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|). - L f 2 P.

3. 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. 4 How to establish NP-hardness Lemma: If L 1 is NP-hard and L 1 · L 2 then L 2 is NP-hard.

5. 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: ( : x 1 Ç : x 2 ) Æ (x 1 Ç x 2 ) Output: Yes. Input: ( : x 1 Ç : x 2 ) Æ (x 1 Ç x 2 ) Æ (x 1 Ç : x 2 ) Æ ( : x 1 Ç x 2 ) Output: No.

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

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

8. 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 2 NPC. 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. 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. 10 ⋀ Boolean Circuits ⋁ ⌐ X1X1 X2X2 X3X3 ⋀

11. 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 a fixed input length n only.

12. 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 C n with at most O(p(n) 2 ) gates so that 8 x 2 {0,1} n : C n (x)=1 iff M accepts x. The map 1 n ! C n is polynomial time computable.

13. 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. 14 Intuition behind proof

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

16. 16 The Tableau Method Time 0 Time 1 Time t … 12 s Can be replaced by acyclic Boolean circuit of size ≈ s

17. 17 Cell state vectors Given a Turing Machine computation, an integer t and an integer i let c ti 2 {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 c ti defined for all t, we let c (t+1)i = c t i if the computation has already terminated at time t.

18. 18 Crucial Observation If we know the Turing machine and c t-1,i-1, c t-1,i, c t- 1, i+1, we also can determine c t,i. In other words, there is a Boolean function h depending only on the Turing machine so that c t,i = h(c t-1,i-1, c t-1,i, c t-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. 19 2 t(n) t(n) x1x1 xnxn

20. 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. 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. 22 Example ⋀ ⋁ ⌐ X1X1 X2X2 X3X3 ⋀ Input: Output: Yes

23. 23 Example ⋀ ⌐ X1X1 X2X2 X3X3 ⋀ Input: Output: No ⋀

24. 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. 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 C n with at most O(q(n) 2 ) gates so that 8 x 2 {0,1} n : C n (x)=1 iff M accepts x. The map 1 n ! C n is polynomial time computable.

26. 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. 27 SAT ILP MILP MAX INDEPENDENT SET MIN VERTEX COLORING HAMILTONIAN CYCLE TSP TRIPARTITE MATCHING SET COVER KNAPSACK BINPACKING

28. 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.