CS154, Lecture 13: P vs NP
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1 CS154, Lecture 13: P vs NP
2 The EXTENDED Church-Turing Thesis Everyone s Intuitive Notion of Efficient Algorithms Polynomial-Time Turing Machines More generally: TM can simulate every reasonable model of computation with only polynomial increase in time A controversial thesis! Potential counterexamples: quantum algorithms
3 Nondeterministic Turing Machines are just like standard TMs, except: 1. The machine may proceed according to several possible transitions (like an NFA) 2. The machine accepts an input string if there exists an accepting computation history for the machine on the string
4
5 Definition: A nondeterministic TM is a 7-tuple T = (Q, Σ, Γ,, q 0, q accept, q reject ), where: Q is a finite set of states Σ is the input alphabet, where Σ Γ is the tape alphabet, where Γ and Σ Γ : Q Γ 2 (Q Γ {L,R}) q 0 Q is the start state q accept Q is the accept state q reject Q is the reject state, and q reject q accept
6 Defining Acceptance for NTMs Let N be a nondeterministic Turing machine An accepting computation history for N on w is a sequence of configurations C 0,C 1,,C t where 1. C 0 is the start configuration q 0 w, 2. C t is an accepting configuration, 3. Each configuration C i yields C i+1 Def. N(w) accepts in t time Such a history exists N has time complexity T(n) if for all n, for all inputs of length n and for all histories, N halts in T(n) time
7 Definition: NTIME(t(n)) = { L L is decided by a O(t(n)) time nondeterministic Turing machine } TIME(t(n)) NTIME(t(n)) Is TIME(t(n)) = NTIME(t(n)) for all t(n)?
8 What problems can we efficiently solved nondeterministically, but not deterministically?
9 The Clique Problem a d f c b e g k-clique = complete subgraph on k nodes
10 The Clique Problem Find a clique of 1 million nodes?
11 Assume a reasonable encoding of graphs (example: the adjacency matrix is reasonable) CLIQUE = { (G,k) G is an undirected graph with a k-clique } Theorem: CLIQUE NTIME(n c ) for some c > 1 N((V,E),k): Nondeterministically guess a subset S of V with S = k For all u, v in S, if (u,v) is not in E then reject Accept
12 The Hamiltonian Path Problem b f e g a d i c h A Hamiltonian path traverses through each node exactly once
13 HAMPATH = { (G,s,t) G is a directed graph with a Hamiltonian path from s to t } Theorem: HAMPATH NTIME(n c ) for some c > 1 N((V,E),s,t): Nondeterministically guess a sequence v 1,, v V of vertices If v i = v j for some i j, reject For all i = 1,, V -1, if (v i,v i+1 ) is not in E then reject If (v 1 = s & v n = t) then accept else reject
14 Nondeterministic Polynomial Time NP = NTIME(n k ) k N
15 Theorem: L NP There is a constant k and polynomial-time TM V such that L = { x y ϵ Σ* [ y x k and V(x,y) accepts ] } Proof: 1. If L = { x y y x k and V(x,y) accepts } then L NP Define the NTM N(x): Guess y of length at most x k Run V(x,y) and output answer Then, L(N) is the set of x s.t. [ y x k & V(x,y) accepts] (2) If L NP then L = { x y y x k and V(x,y) accepts } Suppose N is a poly-time NTM that decides L. Define V(x,y) to accept iff y encodes an accepting computation history of N on x
16 A language L is in NP if and only if there are polynomial-length proofs (aka. certificates or witnesses) for membership in L CLIQUE = { (G,k) subset of nodes S such that S is a k-clique in G } HAMPATH = { (G,s,t) Hamiltonian path in graph G from node s to node t }
17 Boolean Formula Satisfiability logical operations parentheses recedes precedes = ( x y) z Boolean variables (0 or 1)
18 Boolean Formula Satisfiability = ( x y) z A satisfying assignment is a setting of the variables that makes the formula true x = 1, y = 1, z = 1 is a satisfying assignment for (in fact, any assignment with z = 1 is satisfying) = (x y) (z x)
19 A Boolean formula is satisfiable if there is a true/false setting to the variables that makes the formula true YES NO a b c d (x y) x SAT = { is a satisfiable Boolean formula }
20 A 3cnf-formula has the form: (x 1 x 2 x 3 ) (x 4 x 2 x 5 ) (x 3 x 2 x 1 ) literals clauses 3SAT = { is a satisfiable 3cnf-formula }
21 3SAT = { is a satisfiable 3cnf-formula } Theorem: 3SAT NP We can express 3SAT as 3SAT = { is in 3cnf and string y that encodes a satisfying assignment to } The number of variables of is at most, so y. Then, argue that the language 3SAT-CHECK = {(,y) is in 3cnf and y is a satisfying assignment to } is in P. (Similarly, SAT NP)
22 NP = Problems with the property that, once you have the solution, it is easy to verify the solution When ϵ SAT, or (G, k) ϵ CLIQUE, or (G,s,t) ϵ HAMPATH, Can prove that with a short proof that can easily been verified What if SAT? (G, k) CLIQUE? Or (G,s,t) HAMPATH?
23 P = the problems that can be efficiently solved NP = the problems where proposed solutions can be efficiently verified Is P = NP? can problem solving be automated? Clay Math Institute in the year 2000: millennium problems
24 If P = NP: Mathematicians may be out of a job Cryptography as we know it may be impossible In principle, every aspect of life could be efficiently and globally optimized life as we know it would be different! Conjecture: P NP
25 Polynomial Time Reducibility f : Σ* Σ* is a polynomial time computable function if there is a poly-time Turing machine M that on every input w, halts with just f(w) on its tape Language A is poly-time reducible to language B, written as A P B, if there is a poly-time 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
26 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
27 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
28 Theorem: If A P B and B P, then A P Proof: Let M B be a poly-time TM that decides B. Let f be a poly-time 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
29 Theorem: If A P B and B NP, then A NP Proof: Analogous
30 Theorem: If A P B and B P, then A P Theorem: If A P B and B NP, then A NP Corollary: If A P B and A P, then B P
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