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Section 4.2 Network Flows. By Christina Touhey. Recall from last class. The flow out of a equals the flow into z . Algorithm Make vertex a : (0, ). Scan the first vertex and check each incoming and outgoing edge and label if they are unlabeled.

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## Section 4.2 Network Flows

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**Section 4.2 Network Flows**By Christina Touhey**Recall from last class**• The flow out of a equals the flow into z. • Algorithm • Make vertex a: (0, ). • Scan the first vertex and check each incoming and outgoing edge and label if they are unlabeled. • Choose another labeled vertex to scan and label. • Find an a-z chain K of slack edges by backtracking. • Increase the flow in the edges of K by units.**Example: Find the maximal a-z**flow in the network. b 10 d 12 8 5 a z 4 10 18 12 c e b 5 d b 10,5 d 3 Slack 8 12,9 8,0 5,5 a z a z 4 4,4 2 10,10 18,14 c e 12,10 c e Max Flow=19**Theorem 3**• For any given a-z flow , a finite number of applications of the augmenting flow algorithm yields a maximum flow. • If P is the set of vertices labeled during the final application of the algorithm then is a minimal a-z cut set.**Proof of Theorem 3**• If is the current flow and is the augmenting a-z unit flow chain, we must show the new flow + m is a legal flow. • and satisfy flow conditions: and are integer valued.**Proof of Theorem 3 (cont.)**• Where m and is the amount of additional flow that can be sent from a-z.**Proof of Theorem 3 (cont.)**• Since m is a positive integer, each new flow is increased. • The capacities and the number of edges are finite, so eventually z is not labeled. • Let P be the set of labeled vertices when z is not labeled. • Clearly is an a-z cut since a is labeled and z is not.**Proof of Theorem 3 (cont.)**• There is no unsaturated edge from labeled vertex p to unlabeled vertex q or else it would have been previously labeled. • Therefore there is no flow between q and p because**Max Flow-Min Cut Theorem**• In a directed flow network, the value of a maximal a-z flow is equal to the capacity of a minimal a-z cut.**For Class to try**• Find a maximal flow from a to z in the network. Give a minimal capacity a-z cut. b e 10 10 10 5 30 10 z 5 30 a 20 f 20 d 5 10 20 20 c 30 g 10**b**e 10,10 10,10 30,20 10,10 5,5 10,5 z 5,5 30,20 a 20,0 f d 5,0 10,10 20,20 20,0 20,20 c 30,20 g 10,10 Max Flow=50

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