# Is the previous shop from which it is cheapest to go

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] is the previous shop from which it is cheapest to go shop i . Q5 : a) Yes, it will. The algorithm works by finding a path from s to t , and send- ing as much flow along it as possible. If all edges have even capacity, then in the first iteration, clearly the flow will be even. We then modify the edges by subtracting this flow, and reversing edges, and so in the new graph, again all edges have even capacity. Thus in each step, the amount of flow on each edge is even. b) Suppose a node n in the original graph has a capacity c . Replace it with two nodes, called n 1 , n 2 . Add an edge from n 1 to n 2 , with capacity c . For all edges incoming to node n in the original graph, replace them with an edge incoming to n 1 in the new graph. For all edges outgoing from node n in the original graph, replace them with an edge outgoing from n 2 in the new graph. In other words, make a new graph G 0 to replace the original graph G . G 0 has twice as many nodes as G . For every node n in G , make nodes n 1 and n 2 in G 0 . Add an edge ( n 1 , n 2 ) in G 0 , with capacity corresponding to the capacity of node n in G . For an edge ( n, m ) in G , add an edge ( n 2 , m 1 ) in G 0 . If there is f units of flow going through node n in G , there are f units of flow along the edge ( n 1 , n 2 ) in G 0 . 2
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