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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 1. Make vertex a: (0, ). 2. Scan the first vertex and check each incoming and outgoing edge and label if they are unlabeled. 3. Choose another labeled vertex to scan and label. 4. Find an a-z chain K of slack edges by backtracking. 5. Increase the flow in the edges of K by ( z ) units. b 10 12 d 8 Example: Find the maximal a-z flow in the network. 5 L1 a b d z a L2 a c e z z 4 10 L3 a b e z 18 12 c (a ,3) b 12,9 e d (b ,3) 10,5 8,0 ( , ) b 5 d 3 8 5,5 Slack a a z z 4 4,4 10,10 18,14 c 12,10 e 5 L1 10 L 2 4 L3 2 c e P {a, b, d } 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 ( P , P ) then is a minimal a-z cut set. Proof of Theorem 3 • If is the current flow and k is the augmenting a-z unit flow chain, we must show the new flow + m k is a legal flow. • and k satisfy flow conditions: (e) 0 if e In(a ) or e Out( z ) For x a or z , eIn(x ) and are integer valued. (e) eOut(x ) ( e) Proof of Theorem 3 (cont.) • Where m and ( z ) is the amount of additional flow that can be sent from a-z. m ( z ) is the minimum slack, so m k satisfies the flow condition. 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 ( P, P ) 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 k ( P, P) is maximal. The value of the final flow equals k ( P, P) and is maximal. ( P, P) is a minimal a z cut. 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 10 30 a 5 10 5 20 20 d 5 20 c 10 L1 a b e z e 10 30 10 f 30 10 g 20 z L2 a c g z L3 a g f z L4 a b d e f z L5 a b d f z 10 L1 20 L 2 10 L3 5 L 4 5 L5 (a ,10) (d ,5) b 30,20 ( , ) a e 10,10 10,10 10,5 20,0 20,20 5,5 5,5 d (c ,10)5,0 20,0 c (b ,10) 30,20 10,10 f 30,20 z 10,10 20,20 g (c ,10) 10,10 Max Flow=50 P { f , z} 50