Lecture 17: CookLevin Theorem, NPComplete Problems

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1 6.045 Lecture 17: CookLevin Theorem, NPComplete Problems 1
2 Is SAT solvable in O(n) time on a multitape TM? Logic circuits of 6n gates for SAT? If yes, then not only is P=NP, but there would be a dream machine that could crank out short proofs of theorems, quickly optimize all aspects of life recognizing quality work is all you need to produce THIS IS AN OPEN QUESTION! 2
3 Polynomial Time Reducibility f : Σ* Σ* is a polynomial time computable function if there is a polytime Turing machine M that on every input w, halts with just f(w) on its tape Language A is polytime reducible to language B, written as A P B, if there is a polytime computable f : Σ* Σ* so that: w A f(w) B f is a polynomial time reduction from A to B Note there is a k such that for all w, f(w) w k 3
4 A n w n f f B n c f(w) n c f converts any string w into a string f(w) such that w A f(w) B 4
5 Theorem: If A P B and B P C, then A P C A f B g C n n c n cd f g 5
6 Theorem: If A P B and B P, then A P Proof: Let M B be a polytime TM that decides B. Let f be a polytime reduction from A to B. We build a machine M A that decides A as follows: M A = On input w, 1. Compute f(w) 2. Run M B on f(w), output its answer w A f(w) B 6
7 Theorem: If A P B and B NP, then A NP Proof: Analogous 7
8 Theorem: If A P B and B 2 P, then A 2 P Theorem: If A P B and B 2 NP, then A 2 NP Corollary: If A P B and A P, then B P 8
9 Definition: A language B is NPcomplete if: 1. B NP 2. Every A in NP is polytime reducible to B That is, A P B When this is true, we say B is NPhard On homework, you showed A language L is recognizable iff L m A TM A TM is complete for recognizable languages : A TM is recognizable, and for all recognizable L, L m A TM 9
10 Suppose L is NPComplete P NP L If L P, then P = NP! If L P, then P NP! 10
11 Suppose L is NPComplete Then assuming the conjecture P NP, L is not decidable in n k time, for every k 11
12 There are thousands of NPcomplete problems! Your favorite topic certainly has an NPcomplete problem somewhere in it Even the other sciences are not safe: biology, chemistry, physics have NPcomplete problems too! 12
13 The CookLevin Theorem: SAT and 3SAT are NPcomplete 1. 3SAT NP A satisfying assignment is a proof that a 3cnf formula is satisfiable 2. 3SAT is NPhard Every language in NP can be polynomialtime reduced to 3SAT (complex logical formula) Corollary: 3SAT P if and only if P = NP 13
14 Theorem (CookLevin): 3SAT is NPcomplete Proof Idea: (1) 3SAT NP (done) (2) Every language A in NP is polynomial time reducible to 3SAT (this is the challenge) We give a polytime reduction from A to SAT The reduction converts a string w into a 3cnf formula such that w A iff 3SAT For any A NP, let N be a nondeterministic TM deciding A in n k time will simulate N on w 14
15 Deterministic Computation Nondeterministic Computation n k reject accept or reject accept exp(n k ) 15
16 Let L(N) NTIME(n k ). A tableau for N on w is an n k n k matrix whose rows are the configurations of some possible computation history of N on w # q 0 w 1 w 2 w n # # # n k Each cell contains a σ Q Γ { # } # # n k 16
17 A tableau is accepting if the last row of the tableau is an accepting configuration N accepts w if and only if there is an accepting tableau for N on w Given w, we ll construct a 3cnf formula with O( w 2k ) clauses, describing logical constraints that any accepting tableau for N on w must satisfy The 3cnf formula will be satisfiable if and only if there is an accepting tableau for N on w 17
18 Variables of formula will encode a tableau Let C = Q Γ { # } Each cell of a tableau contains a symbol from C cell[i,j] = symbol in the cell at row i and column j = the jth symbol in the ith configuration For every i and j (1 i, j n k ) and for every s C we make a Boolean variable x i,j,s in Total number of variables = C n 2k, which is O(n 2k ) The x i,j,s variables represent the cells of a tableau We will enforce the condition: for all i, j, s, x i,j,s = 1 cell[i,j] = s 18
19 Idea: Make so that every satisfying assignment to the variables x i,j,s corresponds to an accepting tableau for N on w (an assignment to all cell[i,j] s of the tableau) The formula will be the AND of four CNF formulas: = cell start accept move cell : for all i, j, there is a unique s C with x i,j,s = 1 start : the first row of the table equals the start configuration of N on w accept : the last row of the table has an accept state move : every row is a configuration that yields the configuration on the next row 19
20 start : the first row of the table equals the start configuration of N on w start = x 1,1,# x 1,2,q 0 x 1,3,w x 1,4,w x 1,n+2,w 1 2 n x 1,n+3, x 1,n 1, x k 1,n k,# # q 0 w 1 w 2 w n # # # O(n k ) clauses 20
21 accept : the last row of the table has an accept state accept = x k n,j, q accept 1 j n k # q 0 w 1 w 2 w n # # # # q accept # 21
22 accept : the last row of the table has an accept state accept = x k n,j, q accept 1 j n k How can we convert accept into a 3cnf formula? The clause (a 1 a 2 a t ) is equivalent to (a 1 a 2 z 1 ) ( z 1 a 3 z 2 ) ( z t3 a t1 a t ) where z i are new variables. This produces O(t) new 3cnf clauses. O(n k ) clauses 22
23 cell : for all i, j, there is a unique s C with x i,j,s = 1 cell = 1 i, j n k s C x i,j,s ( x i,j,s x i,j,t ) s,t C s t for all i,j at least one x i,j,s is set to 1 at most one x i,j,s is set to 1 O(n 2k ) clauses 23
24 move : every row is a configuration that yields the configuration on the next row Key Question: If one row yields the next row, how many cells can be different between the two rows? Answer: AT MOST THREE CELLS! # b a # b a a q 1 b q 2 a c c b # c b # 24
25 move : every row is a configuration that yields the configuration on the next row Key Question: If one row yields the next row, how many cells can be different between the two rows? Answer: AT MOST THREE CELLS! # b a # b a a q 1 b q 2 a c c b # c b # 25
26 move : every row is a configuration that yields the configuration on the next row Idea: check that every 2 3 window of cells is legal (consistent with the transition function of N) # q 0 w 1 w 2 w n # # # j i the (i,j) window # # 26
27 move = 1 i, j n k ( the (i, j) window is legal ) the (i, j) window is legal = (a 1,, a 6 ) ISN T legal ( x i,j,a x i,j+1,a x i,j+2,a x i+1,j,a x i+1,j+1,a x i+1,j+2,a ) O(n 2k ) clauses 31
28 Summary. We wanted to prove: Every A in NP has a polynomial time reduction to 3SAT For every A in NP, we know A is decided by some nondeterministic n k time Turing machine N We gave a generic method to reduce a string w to a 3CNF formula of O( w 2k ) clauses such that satisfying assignments to the variables of directly correspond to accepting computation histories of N on w The formula is the AND of four 3CNF formulas: = cell start accept move 32
29 Theorem (CookLevin): SAT and 3SAT are NPcomplete Corollary: SAT P if and only if P = NP Given a favorite problem 2 NP, how can we prove it is NPhard? Generic Recipe: 1. Take a problem that you know to be NPhard (3SAT) 2. Prove that P Then for all A 2 NP, A P and P We conclude that A P, and is NPhard 33
30 is NPComplete P NP 34
31 Reading Assignment Read Luca Trevisan s notes for an alternative proof of the CookLevin Theorem! Sketch: 1. Define CIRCUITSAT: Given a logical circuit C(y), is there an input a such that C(a)=1? 2. Show that CIRCUITSAT is NPhard: The n k x n k tableau for N on w can be simulated using a logical circuit of O(n 2k ) gates 3. Reduce CIRCUITSAT to 3SAT in polytime 4. Conclude 3SAT is also NPhard 35
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