Theory of Computability Giorgi Japaridze Theory of Computability NP-completeness Section 7.4

NP-complete problems form a certain important subclass of NP. The Importance 7.4.a Giorgi Japaridze Theory of Computability NP-complete problems form a certain important subclass of NP. The phenomenon of NP-completeness was discovered in the early 1970s by Stephen Cook and Leonid Levin. If a polynomial time algorithm exists for any of the NP-complete problems, all problems in NP would be polynomial time solvable. To prove that P=NP, it would be sufficient to take any particular NP-complete problem A and show that AP. To prove that P≠NP, it would be sufficient to take any particular NP-complete problem A and show that AP. On the practical side, finding that a given problem A is NP-complete may prevent wasting time looking for a (probably nonexistent, or unlikely-to-be-found even if exists) polynomial time algorithm for A.

(1  (0  1))  (0  1) (1  (1  0))  (0  0) Boolean formulas 7.4.b Giorgi Japaridze Theory of Computability Boolean variables x,y,… take one of the two values 0 (false) or 1 (true). Boolean operations:  (NOT),  (AND),  (OR). We write A for A. Boolean formulas are constructed from variables and operations in the standard way. Once a truth assignment for variables is given, the value of a compound formula is calculated as follows: 0 = 1 0  0 = 0 0  0 = 0 1 = 0 0  1 = 0 0  1 = 1 1  0 = 0 1  0 = 1 1  1 = 1 1  1 = 1 If x=0 and y=1, what is the value of the following formula? (y  (x  y))  (x  y) (1  (0  1))  (0  1) (1  (1  0))  (0  0) (1  1 )  0 1  0 1

of 0s and 1s to its variables that makes the formula evaluate to 1. The SAT problem 7.4.c Giorgi Japaridze Theory of Computability We say that a Boolean formula is satisfiable iff there is an assignment of 0s and 1s to its variables that makes the formula evaluate to 1. Are the following formulas satisfiable? x(xy) x(xy) SAT = {<> |  is a satisfiable Boolean formula} SAT  P? SAT  NP?

Polynomial time reducibility Giorgi Japaridze Theory of Computability Definition 7.28 A polynomial time computable function is a function computed by some polynomial time TMO. Definition 7.29 Let A and B be be languages over an alphabet . We say that A is polynomial time mapping reducible, or simply polynomial time reducible, to B, written APB, if a polynomial time computable function f: *  * exists s.t. for every string w*, wA iff f(w)B. Such a function f is called a polynomial time reduction of A to B. Theorem 7.31 If APB and BP, then AP. Proof. Assume M is a polynomial time decider for B, and f is a polynomial time reduction from A to B. The following is a polynomial time algorithm deciding A: N = “On input w: 1. Compute f(w). 2. Run M on input f(w) and do (accept or reject) whatever M does.”

A literal is a Boolean variable x or a negated Boolean variable x. The 3SAT problem 7.4.e Giorgi Japaridze Theory of Computability A literal is a Boolean variable x or a negated Boolean variable x. A clause is several literals connected with s, as in (x  y  z  t). A Boolean formula is in conjunctive normal form, called a cnf-formula, if it comprises several clauses connected with s, as in (x  y  z  t)  (x  z)  (x  y t) A cnf-formula is a 3cnf-formula if all the clauses have 3 literals, as in (x  y  z)  (x  z  t)  (x  y  t)  (z  y  t) 3SAT = {<> |  is a satisfiable 3cnf-formula} 3SAT  P? 3SAT  NP?

Reducing 3SAT to CLIQUE (a) 7.4.f Giorgi Japaridze Theory of Computability Theorem 7.32 3SAT is polynomial time reducible to CLIQUE. Proof. Let  be a 3cnf-formula with k clauses such as  = (a1 b1 c1)  (a2 b2 c2)  …  (ak bk ck) Our reduction f is going to generate the string <G,k>, where G is an undirected graph defined as follows. The nodes of G are organized into k groups of three nodes each called the triples, t1,…,tk. Each triple corresponds to one of the clauses in , and each node in a triple corresponds to a literal in the associated clause. Label each node of G with its corresponding literal in . The edges of G connect all but two types of pairs of nodes: (1) no edge is present between two nodes in the same triple, and (2) No edge is present between nodes with contradictory labels, as in x and x.

Reducing 3SAT to CLIQUE (b) Giorgi Japaridze Theory of Computability  = (x  x  z)  (x  z  z)  (x  z  z) For instance, if  is as above, then G would be x z z x x x z z z Obviously transforming  into G takes polynomial time. Next we argue that (slide 7.4.h) if 3SAT, then <G,k>CLIQUE, and that (slide 7.4.i) if <G,k>CLIQUE, then 3SAT. So, we indeed have a polynomial time reduction.

Reducing 3SAT to CLIQUE (c) 7.4.h Giorgi Japaridze Theory of Computability x z z x x x z z z  = (x  x  z)  (x  z  z)  (x  z  z) Suppose  has a satisfying assignment. Then at least one literal should be true in each clause. Select one such literal in each clause, and select the corresponding nodes in the graph. Those nodes form a k-clique! Because there are k such nodes, and each pair is connected by an edge because they are in different triples and non-contradictory.

Reducing 3SAT to CLIQUE (d) Giorgi Japaridze Theory of Computability x z z x x x z x = z = 1 z z  = (x  x  z)  (x  z  z)  (x  z  z) Now suppose G has a k-clique. Each of its nodes should be in different triples as there are no edges within triples. So, each triple has exactly one node of the clique. Select the corresponding literals in , and select an assignment that makes each such literal true. This is possible (why?). Then the same assignment makes  true.

Definition of NP-completeness 7.4.j Giorgi Japaridze Theory of Computability Definition 7.34 A language B is NP-complete if it satisfies two conditions: 1. B is in NP, and 2. every language in NP is polynomial time reducible to B. Theorem 7.35 If a language B is NP-complete and BP, then P=NP. Proof. Immediately from the above clause 2 and Theorem 7.31. Theorem 7.36 If CNP, B is NP-complete and BPC, then C is NP-complete. Proof. We already know that CNP, so we only need to show that APC for every ANP. Consider any ANP. We must have APB, because B is NP-complete. Let f be a polynomial time reduction from A to B. Next, we know that BPC, so let g be a polynomial time reduction from B to C. Let now h be the composition of f and g, that is, h(w) = g(f(w)). It is easy to see that h is a polynomial time reduction from A to C. So, indeed A PC.

Cook-Levin theorem: Getting started Giorgi Japaridze Theory of Computability Theorem 7.37 (Cook-Levin Theorem) SAT is NP-complete. Proof. That SATNP is obvious (why?). So we only need to show that every language A from NP is polynomial time reducible to SAT. Pick an arbitrary ANP, and N be NTM deciding A. We assume that the running time of N is nk. We are going to show how to turn a string w into a Boolean formula  that “simulates” N on input w in the sense that  is satisfiable iff N accepts w.

Cook-Levin theorem: Tableaus Giorgi Japaridze Theory of Computability A tableau for N on w is an nknk table whose rows are the configurations of a(ny) computation branch of N on input w=w1…wn. The 1st and last columns contain #s. … - - … - # q0 w1 w2 wn # 1st (start) configuration # # 2nd configuration # # 3rd configuration nk window # # nkth configuration nk We say that a tableau is accepting if any of its rows is an accepting configuration. Every accepting tableau for N on w corresponds to an accepting computation branch of N on input w. Thus the problem of determining whether N accepts w is equivalent to determining whether there is an accepting tableau for N on w.

 = cell  start  move  accept Cook-Levin theorem:  7.4.m Giorgi Japaridze Theory of Computability … - - … - # q0 w1 w2 w3 wn # … - - … - # w1 q7 w2 w3 wn # … - - … - # q5 w1 $ w3 wn # We denote by C the set Q{#}, where Q is the set of states of N and  is the tape alphabet. C is thus the set of all possible contents of the cells of the tableau. The cell in row i and column j is called cell[i,j]. For each such cell and each sC, we create a Boolean variable xi,j,s. Its meaning is going to be “cell[i,j] contains symbol s”. Our formula  is going to be built from those variables, and we have  = cell  start  move  accept cell asserts that each cell contains exactly one symbol start asserts that the first row is the start configuration on input w move asserts that rows are related to each other in accordance with the transition function accept asserts that one of the cells contains the accept state  thus asserts that the tableau is for an accepting computation branch, i.e. that wA.

Cook-Levin theorem: cell Giorgi Japaridze Theory of Computability cell asserts that each cell contains exactly one symbol “cell[i,j] contains symbol s” = xi,j,s ( xi,j,s) sC “cell[i,j] contains at least one symbol” = (xi,j,s xi,j,t) “cell[i,j] does not contain both s and t” = (  (xi,j,s xi,j,t)) s,tC s≠t “cell[i,j] contains at most one symbol” = “each cell “cell[i,j] contains exactly one symbol” [ (xi,j,s ) sC  (  (xi,j,s xi,j,t) )] s,tC s≠t  1i,jnk cell

Cook-Levin theorem: start Giorgi Japaridze Theory of Computability start asserts that the first row is the start configuration on input w … - - … - # q0 w1 w2 wn # 1st (start) configuration “cell[1,1] contains #” = x1,1,# “cell[1,2] contains q0” = x1,2,q0 … x1,1,#  x1,2,q0  x1,3,w1  x1,4,w2  …  x1,n+2,wn  x1,n+3,-  … x1,nk-1,-  x1,nk,# start =

Cook-Levin theorem: accept Giorgi Japaridze Theory of Computability accept asserts that one of the cells contains the accept state “cell[i,j] contains qaccept” = xi,j,qaccept  xi,j,qaccept 1i,jnk accept =

Cook-Levin theorem: Windows 7.4.q Giorgi Japaridze Theory of Computability 1 2 3 … j … 1 2 3 . . . the (i,j) window i a1 a2 a3 a4 a5 a6 . . . We say that (the content of) a window is legal if it(s content) could appear in some (legal) tableau for N.

Cook-Levin theorem: Examples of legal and illegal windows Giorgi Japaridze Theory of Computability Where  is the transition function of N, assume we have (q1,a) = {(q1,b,R)} and (q1,b) = {(q2,c,L),(q2,a,R)}. Are the following windows legal or illegal? q1 q2 a b c q1 a b q2 a q1 b b # a b a q2 b c b a q1 a b q1 q2 b

Cook-Levin theorem: Claim about windows Giorgi Japaridze Theory of Computability Claim 7.41 If the top row of the tableau is the start configuration and every window in the tableau is legal, then each row in the tableau is a configuration that legally follows the preceding one. Proof. Consider any two adjacent rows (configurations). In the upper configuration, every cell that isn’t adjacent to a state symbol and doesn’t contain the boundary symbol #, is the center top cell in a window whose top row contains no states. Therefore that symbol, in a legal window, must appear unchanged in the center bottom of the window. Hence it appears (as it should) in the same position in the bottom configuration. b ? x y The window containing the state symbol in the center top cell guarantees that the corresponding three positions are updated consistently with the transition function. Therefore, if the upper configuration is a legal configuration, so is the lower configuration, and the lower one follows the upper one according to N’s rules. q1

Cook-Levin theorem: move Giorgi Japaridze Theory of Computability move asserts that rows are related to each other in accordance with the transition function Let us say that a 6-tuple (a1,a2,a3,a4,a5,a6) of symbols from C is legal if the window on the right is legal. Notice that the number of legal 6-tuples is fixed and it does not depend on w. BTW, at most how many legal 6-tuples could exist? a2 a1 a3 a4 a5 a6 |C|6 “the content of cell[i,j] is (a1,…,a6)” (xi,j-1,a1  xi,j,a2  xi,j+1,a3  xi+1,j-1,a4  xi+1,j,a5  xi+1,j+1,a6 )  (a1,…,a6) is legal “the (i,j) window is legal”  1 i nk-1 2 j nk-1 (the (i,j) window is legal) move =

Cook-Levin theorem: The complexity of the reduction Giorgi Japaridze Theory of Computability Our reduction does nothing but builds , so its time complexity is asymptotically the same as the size of . We want to see this size is polynomial in n. For this, in turn, it would be sufficient to verify the polynomiality (in n) of the four conjuncts of . What is the size of start? What is the size of accept? What is the size of cell? What is the size of move? The complexity of our reduction is thus O(n2k), i.e. polynomial.

Cook-Levin theorem: Wrapping it up Giorgi Japaridze Theory of Computability It remains to understand why our reduction is indeed a reduction, i.e., why it is the case that wA iff  is satisfiable. () Assume wA. Then N accepts w, i.e. there is an accepting computation branch of N on input w. Construct a tableau that describes such a branch. Then declare each variable xi,j,s to be true iff, in that tableau, cell[i,j] contains symbol s. Obviously this truth assignment satisfies . () Assume  is satisfiable, i.e. there is an assignment that makes  true. Fix it. Construct a tableau by putting symbol s in cell[i,j] iff xi,j,s is true (the truth of cell guarantees that such a construction is possible and unique). Obviously such a tableau describes a certain accepting computation branch of N on input w. Thus, N accepts w, meaning that wA.

The NP-completeness of 3SAT 7.4.w Giorgi Japaridze Theory of Computability Corollary 7.42 3SAT is NP-complete. Proof. From logic, a polynomial-time-computable function f: {<> |  is a(ny) Boolean formula}  {<> |  is a 3cnf-formula} is known such that, for any Boolean formula , we have  is satisfiable iff f() is satisfiable. Thus, f is a polynomial time reduction from SAT to 3SAT. Hence, in view of the already known NP-completeness of SAT together with Theorem 7.36, we find that 3SAT is NP-complete. The book gives a slightly different and full proof of this result.