110 likes | 196 Views
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.
E N D
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