hw6-sol - Assignment #6, CS4/531 Solution Due Date: Friday....

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Unformatted text preview: Assignment #6, CS4/531 Solution 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 its 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 ).) False . The following Fig shows a counter example. s t 1/2 1/1 Figure 1: A Counter example for Problem 1. 2. (4 points) Describe if the following statement is true or false. If it is true, give a short explanation. If its 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 ( ). False . The following Fig shows a counter example. In this Figure, the integer next to each edge is the capacity of the edge. The min-cut of G is: S = { s,a } and T = { b,c,d,t } . The capacity of the cut is c ( S,T ) = 3. After the capacity of each edge is increased by one, the new min-cut is: S = { s } and T = { a,b,c,d,t } with capacity c ( S ,T ) = 5. 1 s t a b c d 1 1 1 1 1 1 4 Figure 2: A Counter example for Problem 2. 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). Answer: Let G = ( V,E ) be a directed acyclic graph. Assume that V = { 1 , 2 ,... ,n } , construct the graph G = ( V ,E ), where V = { x ,x 1 ,... ,x n } { y ,y 1 ,... ,y n } E = { ( x ,x i ) : i V } { ( y i ,y ) : i V } { ( x i ,y j ) : ( i,j ) E } ....
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hw6-sol - Assignment #6, CS4/531 Solution Due Date: Friday....

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