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] is the previous shop from which it is cheapest to go shopi.Q5:a) Yes, it will. The algorithm works by finding a path fromstot, and send-ing as much flow along it as possible. If all edges have even capacity, then inthe first iteration, clearly the flow will be even. We then modify the edgesby subtracting this flow, and reversing edges, and so in the new graph, againall edges have even capacity. Thus in each step, the amount of flow on eachedge is even.b) Suppose a nodenin the original graph has a capacityc. Replace it withtwo nodes, calledn1, n2. Add an edge fromn1ton2, with capacityc. For alledges incoming to nodenin the original graph, replace them with an edgeincoming ton1in the new graph. For all edges outgoing from nodenin theoriginal graph, replace them with an edge outgoing fromn2in the new graph.In other words, make a new graphG0to replace the original graphG.G0hastwice as many nodes asG. For every nodeninG, make nodesn1andn2inG0. Add an edge (n1, n2) inG0, with capacity corresponding to the capacityof nodeninG. For an edge (n, m) inG, add an edge (n2, m1) inG0. If thereisfunits of flow going through nodeninG, there arefunits of flow alongthe edge (n1, n2) inG0.2
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