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Lecture 21 NP-complete problems Why do we care about NP-complete problems? Because if we wish to solve the P=NP problem, we need to deal with the hardest problems in NP. Why do we want to solve the P=NP problem? Because it will solve other 3000 NPC problems. It is one of the fundamental mathematical problems: Millennium Prize (CMI) -- $1M for each problem: P versus NP The Hodge conjecture The Poincaré conjecture - solved, by Grigori Perelman The Poincaré conjectureGrigori Perelman The Riemann hypothesis Yang-Mills existence and mass gap Navier-Stokes existence and smoothness The Birch and Swinnerton-Dyer conjecture

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Theorem. If there is a polynomial-time algorithm for any one NP-complete problem, then there is a polynomial- time algorithm for every problem in NP. (Proved in previous lecture.) Corollary. If no polynomial-time algorithm exists for some problem in NP, then there is no polynomial-time algorithm for any NP-complete problem. NP-completeness

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Easy vs Hard Here are some examples of easy (P) and hard problems (NP-hard). Easy Hard 2SAT 3SAT Minimum spanning tree Traveling salesman Shortest path Longest path linear programming integer linear programming Eulerian cycle Hamiltonian cycle

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“First Natural” NP-hard problem Circuit Satisfiability problem. An instance of the problem is a Boolean circuit (using AND, OR, and NOT gates connected by wires) that has n Boolean inputs and a single Boolean output. The circuit has no cycles. The size of the circuit is defined to be the total number of gates and wires. CIRCUIT-SAT = the decision problem, given a Boolean circuit C, is it satisfiable? CIRCUIT-SAT can be solved by simply trying all possible 2 n assignments, not polynomial time. By S.A. Cook, 1971 STOC

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Circuit SAT AND OR AND OR ANDNOTOR NOT x1x1 x2x2 x3x3 x4x4 x5x5 AND NOT AND OR NOT AND OR AND Find a satisfying assignment

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Thm. CIRCUIT-SAT is NP-complete Proof by hand-waving. CIRCUIT-SAT is clearly in NP. The "certificate” is a satisfying assignment. Given this, we can easily verify in polynomial time that the circuit outputs a 1. We now give a polynomial-time transformation from every problem L in NP to CIRCUIT-SAT. If L is in NP, then there is a verifier A(x,y) that runs in time T(|x|) = O(|x| k ) for some k. We now construct a single boolean circuit M that maps one "configuration" of a machine that carries out the computation of A(x,y) (recording such things as the memory state, program counter, etc.) to the next "configuration". We now hook together T(|x|) of these circuits together, making the inputs to the circuit at the top the value of y, and the output the single bit that reflects the value of A(x,y). This big circuit C(x) is satisfiable (by a value of y) if and only if x was a "yes" instance of L. The size of this circuit is polynomial in x, and the transformation can be done in polynomial time. QED

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4 step routine for proving NPC Four-step routine for proving NP-completeness of a decision problem A: 1. Prove A is in NP by describing the polynomial-time verifier V that verifies "yes" instances of A. What is the certificate? How is it verified? 2. Select a problem B that you already know to be NP- complete. 3. Design a function f that maps "yes" instances of B to "yes" instances of A, and “no” instances of B to “no” instances of A, and justify that. 4. Show that f can be computed in polynomial time.

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Proving other NPC problems The following is the scheme of reductions we will use to prove some problems to be NP-complete (there are over 3000 of them – actually many more now): CIRCUIT-SAT | SAT | 3-CNF-SAT / \ CLIQUE SUBSET-SUM / VERTEX-COVER

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SAT problem. SAT. A boolean formula consists of boolean (1/0) variables joined by connectives: AND, OR, IMPLIES, IFF, NOT. Also, parentheses may be used. An example, F = ((x 1 IMPLIES x 2 ) OR NOT ((NOT x 1 IFF x 3 ) OR x 4 )) AND NOT x 2. An assignment is a specification of truth values of the various variables. For example, (x 1,x 2,x 3,x 4 )=(0,0,0,1) is an assignment that makes F true (equal to 1). Such an assignment is called a satisfying assignment. If a formula F has a satisfying assignment, it is called satisfiable.

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Theorem. SAT is NP-complete ( Given a boolean formula F, is it satisfiable?) Proof. First, we need to show that SAT is in NP. Clearly there is a polynomial-time algorithm A(x,y) to verify that x is a "yes" instance of SAT, using y as the purported satisfying assignment for the boolean formula x. Second, we choose CIRCUIT-SAT (we do not have other choices at this moment) to reduce it to SAT. Third, we design a function that maps circuits to Boolean formulas. For each gate, label each wire coming out of a gate with a new variable name. Thus, for example, if an AND gate has inputs x 6, x 7, and x 8, and output x 9, introduce a clause x 9 IFF (x 6 AND x 7 AND x 8 ) Add a clause that just consists of x output, the label for output wire. Take the AND of all these clauses.

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NPC of SAT, proof continues Clearly, C is a satisfiable circuit if and only if the formula f(C) is a satisfiable boolean formula. If C is satisfiable, then there exists some assignment to the input wires that results in the output being 1. Therefore, the exists some assignment to the variables input wires, plus the additional variables representing the output wires, that results in the formula f(C) evaluating to 1. Similarly, if the formula f(C) is satisfiable, then it corresponds to the circuit being satisfiable, and each output wire being correctly computed in terms of the gate type. Step 4. Clearly, f can be computed in polynomial time. We can simply do a breadth-first search on the graph representation of the circuit, outputting a formula for each gate. This completes our proof that SAT is NP-complete. QED

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