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HW6

Course: CSE 431/531, Fall 2011
School: SUNY Buffalo
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#6, Assignment CS4/531 Due Date: Friday. Dec. 9, 2011 Total points: 52 You MUST turn in your HW by 2:10pm on Dec 9. After that, I will NOT accept your HW. This rule will be STRICTLY ENFORCED. Please PRINT YOUR LAST NAME, FIRST NAME and UB number on the first page. Write solution of each problem on a separate sheet. Staple them in the order of problem numbers. If your homework solution deviates significantly...

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#6, Assignment CS4/531 Due Date: Friday. Dec. 9, 2011 Total points: 52 You MUST turn in your HW by 2:10pm on Dec 9. After that, I will NOT accept your HW. This rule will be STRICTLY ENFORCED. Please PRINT YOUR LAST NAME, FIRST NAME and UB number on the first page. Write solution of each problem on a separate sheet. Staple them in the order of problem numbers. If your homework solution deviates significantly from these guidelines, TA may deduct up to 20% of the points. 1. (4 points) Describe if the following statement is true or false. If it is true, give a short explanation. If it's false, give a counter example. Let G be a (basic) flow-network with source s and sink t. Each directed edge e of G has a positive integer capacity c(e). If f is a max-flow of G, then f saturates every edge out of s. (Namely, for every edge e = s v, f (e) = c(e).) 2. (4 points) Describe if the following statement is true or false. If it is true, give a short explanation. If it's false, give a counter example. Let G be a (basic) flow-network with source s and sink t. Each directed edge e of G has a positive integer capacity c(e). Let (S, T ) be a minimum capacity st-cut with respect to the capacity function c(). Now suppose that we add 1 to the capacity of every edge e. Namely, we define a new capacity function c (e) = c(e) + 1 for every edge e in G. Then (S, T ) is still a minimum capacity st-cut of G with respect to the new capacity function c (). 3. (7 points) Page 761, Problem 26-2 (a). Note 1: First try to solve the problem on a small graph using the method described in the problem definition. This will help you to guess the correct solution. Note 2: The Hamiltonian Path (HP) problem is a special case of the Minimum Path Cover problem. Can you see it? The HP problem for general directed graphs is NP complete. So the answer to the part (b) of this problem is "no". Although not required, you should think about why the method in part (a) does not work for part (b). 4. (2+3+2+3=10 points). The Graph Edge Coloring (GEC) problem is defined as follows. The input is an undirected graph G = (V, E) with n vertices and m edges. An edge coloring of G is an assignment of colors to the edges of G so that for any two edges e1 , e2 , if e1 and e2 share a common end vertex, then e1 and e2 must have different colors. The problem is: Given G, determine the smallest integer k such that G has an edge coloring using k colors. (a) Describe the decision version of the GEC problem. 1 (b) Suppose that we have an algorithm A for solving the decision version of the GEC problem with run time T (n, m). Describe an algorithm B for solving the optimization version of the GEC problem. Analyze the run time of B (in terms of T (n, m).) (c) What is the "certificate" of the GEC problem? (d) Consider the following Time Table (TT) problem. In a school, there p are teachers X1 , X2 , . . . , Xp and q classes Y1 , Y2 , . . . , Yq . For each i (1 i p) and j (1 j q), the teacher Xi meets the class Yj tij times. Obviously, during any period, each teacher can meet only one class, and each class can meet only one teacher. The problem is to schedule a complete timetable using the minimum possible number of periods. Describe how to convert the TT problem to the GEC problem. 5. (6 points) An eye-glass graph of order 2n is a graph consisting of two cycles (each with n vertices) connected by an edge. The graph shown in Figure 1 is an eye-glass graph of 8 vertices. Figure 1: The eye-glass graph with 8 vertices Consider the following problem: Given an undirected graph G = (V, E), decide if G has a figure-8 graph as a subgraph. Show this problem is NP-complete. 6. (7 points) Let G = (V, E) be an undirected graph. A Hamiltonian Path (HP) of G is a path that visits each vertex of G exactly once. A Hamiltonian Cycle (HC) of G is a cycle that visits each vertex of G exactly once. The Hamiltonian Cycle (HC) problem is: Given an undirected graph G, decide if G has a Hamiltonian Cycle or not. The Hamiltonian Path (HP) problem is: Given an undirected graph G, decide if G has a Hamiltonian path or not. We know that HC problem is NP-complete. Show HP-problem is NP-complete, by showing that HC P HP . 7. (7 points) The Traveling Salesman Problem (TSP) has been defined in class. The TSP problem with triangle inequality, denoted by TSP, is a restricted version of the TSP problem: it requires that the edge weight function w satisfies the triangle inequality. (Namely, for any three vertices u, v, w of G, w(u, v) + w(v, w) w(u, w).) Show that TSP P TSP. (Since TSP is NP-complete, this reduction implies that TSP is also NP-complete). Hint: You need to show that in polynomial time we can transform an instance G of the TSP problem into another instance G that satisfies the triangle inequality. This can be done by adding a large number K to the weight of every edge of G. 8. (7 points) The SAT problem is defined as follows: 2 Input: A CNF Boolean formula F with variables x1 , . . . , xk . Output: "yes" if F is satisfiable; "no" otherwise. The Constructive-SAT problem is defined as follows: Input: A CNF Boolean formula F with variables x1 , . . . , xk . Output: If F is satisfiable, find a boolean assignment of x1 , . . . , xk (namely, specifically assign 0 or 1 to each variable xi ) so that F evaluates 1. If F is not satisfiable, output "no". Note that the Constructive-SAT problem is not a decision problem. Assume that we have an algorithm A for solving the SAT problem. Describe an algorithm B for solving the Constructive-SAT problem by calling A polynomially many times. Hint: The algorithm B goes through a loop. Each iteration i try to find a correct 0/1 value for variable xi . 3
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