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# sol4 - Massachusetts Institute of Technology 6.854J/18.415J...

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Massachusetts Institute of Technology Handout 10 6.854J/18.415J: Advanced Algorithms Friday, October 9, 2009 David Karger Problem Set 4 Solutions Friday, October 9, 2009 Problem 1. First, we go through the elements in Y . For all d ij given in Y , we reduce r i by d ij and c j by d ij . In other words, we set r i = r i j d ij and c j = c j i d ij , where d ij is given in Y . Next, we construct a graph. We create vertices for each of the rows R i and columns C j along with a source s and sink t . We draw edges from the source to each of the rows R i with capacity equal to the adjusted row sum. That is to say, we draw the edge ( s, R i ), with u ( s, R i ) = r i . Then we draw edges ( C j , t ) from each of the column vertices C j to the sink t with capacities u ( C j , t ) = c j . Finally, we draw edges from all rows R i to columns C j with u ( R i , C j ) = provided that d ij wasn’t given in Y . If d ij was given, we do not draw the edge ( R i , C j ). For example, suppose Y is empty. Then we can draw the graph as follows: R 1 R 2 R p middledotmiddledotmiddledot s C 1 C 2 C q middledotmiddledotmiddledot t r 1 r 2 r p c 1 c 2 c q Claim 0.1 Consider a flow f on the graph. This flow corresponds to a solution to the matrix and all the constraints if and only if f = i r i and is feasible. Proof. ( ) Suppose there is a solution { d ij } to the matrix. Then we claim that f ( s, R i ) = r i , f ( R i , C j ) = d ij , and f ( C j , t ) = c j provides a feasible flow. It is fairly obvious that f = i r i , as we saturated all the edges leaving s . Next, we just need to show that the flow is feasible. Consider the vertex representing the row R i . We know that j d ij = r i as we satisfy our matrix constraints. Thus, if we send r i flow to R i , we can send (exactly) the d ij flow necessary along the edge ( R i , C j ) (these edges have infinite capacity). Thus, we are in accord with the capacity and conservation conditions. Similarly for C j .

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2 Handout 10: Problem Set 4 Solutions ( ) This direction is also trivial. Just reverse argument from above. If we’re given a flow, we let d ij = f ( R i , C j ), then we argue that that is a matrix solution. If we have a flow with f = i r i , then we must be saturating all edges ( s, R i ) and ( C j , t ). Therefore, we must have j d ij = j ( R i , C j ) = r i for all i . Similarly, we have i d ij = c j for all j , and the solution is valid. Thus, we can just find a solution to the matrix by finding a max flow on the graph. If the max flow f is not equal to i r i , then there is no solution. If it is, then we proceed to verify each d ij as being protected or unprotected. To verify whether d ij is protected, we just need to ask the question, “can any other value work for d ij ?” We can break this into two smaller questions: “Does there exists a valid solution with d ij < d ij ?” and “Does there exist a valid solution with d ij > d ij ?” If the answer to either of these questions is “yes,” then there is not a unique answer for d ij . Thus, d ij is protected.
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