Chapter 10 Part 1: Reduction

Transcription

1 //06 Polynomial-Time Reduction Suppose we could solve Y in polynomial-time. What else could we solve in polynomial time? don't confuse with reduces from Chapter 0 Part : Reduction Reduction. Problem X polynomial reduces to problem Y if arbitrary instances of problem X can be solved using: Polynomial number of standard computational steps, plus Polynomial number of calls to the blackbox that solves problem Y. Notation. X P Y. Examples: Max-Bipartite-Matching P Min-Bipartite-Vertex-Cover Sudoku P SAT N-queen problem P SAT Knight-tour P SAT SAT P CSP SAT P Integer Linear Programming Polynomial-Time Reduction Independent Set If X P Y, and the code of Y is B, we may obtain the code A which uses B to solve X (the time spent by B is not cared). Design algorithms. If X P Y and Y can be solved in polynomial-time, then X can also be solved in polynomial time. That is, if Y is easy, so is X. Establish equivalence. If X P Y and Y P X, we use notation X P Y. INDEPENDENT SET: Given a graph G = (V, E) and an integer k, is there a subset of vertices S V such that S k, and for each edge at most one of its endpoints is in S? Ex. Is there an independent set of size 6? Yes. Ex. Is there an independent set of size 7? No. up to cost of reduction E.g., Max-Bipartite-Matching P Min-Bipartite-Vertex-Cover independent set Clique Vertex Cover Clique: Given a graph G = (V, E) and an integer k, is there a subset of vertices S V such that S k, and for each pair (x, y) of points in S, (x, y) is an edge of E? Claim. CLIQUE P INDEPENDENT-SET. Pf. We show S is an independent set of G iff S is a clique of G, where G is the complement of G: G = (V, V E). VERTEX COVER: Given a graph G = (V, E) and an integer k, is there a subset of vertices S V such that S k, and for each edge, at least one of its endpoints is in S? Ex. Is there a vertex cover of size? Yes. Ex. Is there a vertex cover of size? No. vertex cover 6

2 //06 Vertex Cover and Independent Set Vertex Cover and Independent Set Claim. VERTEX-COVER P INDEPENDENT-SET. Pf. We show S is an independent set iff V S is a vertex cover. Consequently, S is a maximum independent set iff V S is a minimum vertex cover. If we have an efficient algorithm to solve one, we will have efficient algorithm to solve the other. Claim. VERTEX-COVER P INDEPENDENT-SET. Pf. We show S is an independent set iff V S is a vertex cover. Let S be any independent set. Consider an arbitrary edge (u, v). S independent u S or v S u V S or v V S. Thus, V S covers (u, v). independent set vertex cover Let V S be any vertex cover. Consider two nodes u S and v S. Observe that (u, v) E since V S is a vertex cover. Thus, no two nodes in S are joined by an edge S independent set. 8 Set Cover SET COVER: Given a set U of elements, a collection S, S,..., S m of subsets of U, and an integer k, does there exist a collection of k of these sets whose union is equal to U? Sample application. m available pieces of software. Set U of n capabilities that we would like our system to have. The i th piece of software provides the set S i U of capabilities. Goal: achieve all n capabilities using fewest pieces of software. Vertex Cover Reduces to Set Cover Claim. VERTEX-COVER P SET-COVER. Pf. Given a VERTEX-COVER instance G = (V, E), k, we construct a set cover instance whose size equals the size of the vertex cover instance. Construction. Create SET-COVER instance: k = k, U = E, S v = {e E : e incident to v } Set-cover of size k iff vertex cover of size k. Ex: U = {,,,,, 6, 7 } k = S = {, 7} S = {, } S = {,,, 6} S = {} S = {} S 6 = {,, 6, 7} VERTEX COVER a b e 7 e e f e 6 e k = e d e e c SET COVER U = {,,,,, 6, 7 } k = S a = {, 7} S b = {, } S c = {,,, 6} S d = {} S e = {} S f = {,, 6, 7} 9 0 SAT: Given formula, does it have a satisfying truth assignment? Literal: A Boolean variable or its negation. Clause: A disjunction of literals. -Satisfiability Conjunctive normal form: A propositional formula that is the conjunction of clauses. x i or x i : x x x -SAT: A SAT formula in conjunctive formula where each clause contains exactly literals. C j : C C C C -SAT is as hard as SAT Claim. SAT P -SAT Pf. Let K be any circuit representing the formula of SAT. Create a -SAT variable x i for each circuit element i. Make circuit compute correct values at each node: x x add clauses: x x, x x x x x add clauses: x x, x x, x x x x 0 x x add clauses: x 0 x, x 0 x, x 0 x x Hard-coded input values and output value. x = 0 add clause: x 0 = add clause: x x 0 output x 0 Ex: x x x x x x Yes: x = true, x = true x = false. each corresponds to a different variable x x x x x Final step: turn clauses of length < into clauses of length exactly by introducing new variables. x x x x x 0??

3 //06 Basic reduction strategies. Summary Simple equivalence: INDEPENDENT-SET P VERTEX-COVER. Special case to general case: VERTEX-COVER P SET-COVER, or SAT P -SAT. Transitivity. If X P Y and Y P Z, then X P Z. Pf idea. Compose the two algorithms. Ex: INDEPENDENT-SET P VERTEX-COVER P SET-COVER. So INDEPENDENT-SET P SET-COVER Decision vs Optimization For the clique problem: Decision Problem: Input: a graph G = (V, E), integer k. Question: Does G have a clique of size >= k? Optimization problem: Find a clique of maximum cardinality. In general, Decision problem is easier than Optimization problem. However, the difference is by a polynomial reduction. Decision vs Optimization For the clique problem, suppose we have an algorithm A to answer the decision problem: A(G, k) = yes iff G has a clique of size k. How can we find the maximum clique of G? Step : Decide k, the size of the maximum clique: for i from n downto, if A(G, i) = yes and A(G, i+) = no, then return i; Step : Decide the actual max clique: for each vertex v in G if (A(G - v, k) = yes) G = G - v; // G v means v is deleted from G. Suppose we have an algorithm B which returns the maximum clique of G. How can we solve the decision version of the clique problem? Vertex Cover Problem Decision Problem: Input: a graph G = (V, E), integer k. Question: Does G have a vertex cover of size k? Optimization problem. Find vertex cover of minimum cardinality of G. Self-reducibility: Decision and Optimization Problems are all equivalent (= P ) Applies to all (NP-complete) problems. Justifies our focus on decision problems. 6 Vertex Cover Problem Ex: Reduce Optimization to Decision To find min cardinality vertex cover. (Binary) search for cardinality k* of min vertex cover. Find a vertex v such that G v has a vertex cover of size k* -. delete v and all incident edges any vertex in any min vertex cover will have this property Include v in the vertex cover. Recursively find a min vertex cover in G v. What is Linear Programming? Toy manufacturer can produce skateboards and dolls. Both require the precious resource of plastic, of which there are 60 units available. Skateboards take five units of plastic and make $ profit. Dolls take two units of plastic and make $0. profit. What is the number of dolls and skateboards the company can produce to maximize profit?

4 //06 What is Linear Programming? A mathematical tool for maximizing or minimizing a quantity (usually profit or cost of production), subject to certain constraints. Of all computations and decisions made by management in business, 0 90% of those involve linear programming. Setting Up Problems First identify components of the problem:. Resources Plastic (60 available). Products Skateboards & Dolls. Recipes Skateboards ( units), Dolls ( units). Profits Skateboards ($.00), Dolls $0.). Objective Maximize profit Second, make a mixture chart: Resources Products Skateboards (x units) Dolls (y units) Plastic (60) Profit $.00 $0. Translate Mixture Chart into Formulas Products Skateboards (x units) Dolls (y units) Groups of Equations: Resources Plastic (60) Profit $.00 $0. Objective Equation (profit equation) Constraints (minimum constraints, resource constraints ) Objective Equation total profit given number of units produced P = x + 0.y Constraints usually inequalities x + y 60 Linear Programming Feasible Region region which consists of all possible solution choices for a particular problem Using the constraint equation we get the following graph: Constraints: x + y 60 With these, create Feasible Region Corner Point Principle Corner Point Principle Which point is optimal? Any point in feasible region will satisfy constraint equation, but which will maximize profit equation? ( 0, 0 ) Plug in corner points to profit formula: Point Calculation of Profit Formula $.00x + $0.y = P (0, 0) $.00 (0) + $0. (0) = $0.00 (0, 0) $.00 (0) + $0. (0) = $6.0 ( 0, 0 ) Corner Point Principle In LP problem, if a maximal value exists, one of corner points on feasible region should have it. ( 0, 0 ) (, 0 ) (, 0) $.00 () + $0. (0) = $.00 Corner point (0,0) is the optimal point Therefore the optimal solution would be to produce 0 skateboards and 0 dolls. In general, corner points are not always integers. If only integer solutions are allowed, ( 0, 0 ) (, 0 )

5 //06 Types of Integer Programming Models A linear program in which all the variables are restricted to be integers is called an integer linear program (ILP). If only a subset of the variables are restricted to be integers, the problem is called a mixed integer linear program (MILP). Binary variables are variables whose values are restricted to be 0 or. If all variables are restricted to be 0 or, the problem is called a 0- or binary integer program. Consider the following all-integer linear program: Max x + x (e.g., profit) s.t. x + x < 9 (resource, e.g., plastic) x + x < 7 (resource, e.g., workers) -x + x < (x < x + ) x, x > 0 and integer LP Relaxation x -x + x < x + x < 9 Max z = x + x LP Optimal (.,.) x + x < 7 LP Relaxation Solving the problem as a linear program ignoring the integer constraints, the optimal solution to the linear program gives fractional values for both x and x. From the graph on the previous slide, we see that the optimal solution to the linear program is: x =., x =., z = 0. x 6 7 Rounding Up If we round up the fractional solution (x =., x =.) to the LP relaxation problem, we get x = and x =. From the graph on the next page, we see that this point lies outside the feasible region, making this solution infeasible. Rounded Up Solution x -x + x < x + x < 9 Max z = x + x ILP Infeasible (, ) LP Optimal (.,.) x + x < 7 x 6 7

6 //06 Rounding Down By rounding the optimal solution down to x =, x =, we see that this solution indeed is an integer solution within the feasible region, and substituting in the objective function, it gives z = 8. We have found a feasible all-integer solution, but have we found the optimal all-integer solution? The answer is NO! The optimal solution is x = and x = 0 giving z = 9, as evidenced in the next two slides. x -x + x < x + x < 9 Max z = x + x ILP Optimal (, 0) x + x < 7 x 6 7 Decision Problem: 0- Programming Complete Enumeration of Feasible ILP Solutions There are eight feasible integer solutions to this problem: x x z optimal solution PROGRAMMING. Given a n by m matrix A, a vector B of m numbers, a vector X of n variables, is there a binary solution of X such that AX B? Claim. -SAT P 0- PROGRAMMING. Pf. 6