CSE 202 Network flow II

1. CSE 202 Network flow II Fan Chung Graham UC San Diego

2. Flow Network: Oil Through Pipelines A D How much oil can be shipped from S to T ? 2 5 3 2 2 T S B 1 10 1 4 6 C E • Directed graph G = (V,E) • Identified sourceS and sinkT • Edge capacitiesce

3. A Feasible Flow in the Network A D How much oil can be shipped from S to T ? 2/2 4/5 2/3 0/2 T 2/2 S B 0/1 2/10 1/1 5 units of flow – is this the maximum possible? 1/4 1/6 C E • Directed graph G = (V,E) • Identified sourceS and sinkT • Edge capacitiesce • The flow along an edge is ≤capacity

4. Formal Definition of Flow Flow never exceeds capacity Conservation of flow at all nodes except source, sink • A flow on a graph G is a function f : E   such that: • 0 ≤ f(e) ≤ ce for all edges e  E • Flow into a node = flow out of that node For all u  s, t : (u,w)  E f(u,w) = (z,u)  E f(z,u) • Size of flow: size(f) = (s,u)  E f(s,u) Amount of flow leaving the source 2 units of s-t flow A D 2/2 4/5 2/3 0/2 T 2/2 2 units of s-t flow S B 0/1 2/10 1/1 1 unit of s-t flow 1/4 1/6 C E

5. Flows and Cuts • The size of a flow can be measured across any cut • A cut (L,R) satisfies: • V = L  R, with L R =  disjoint partition of V • S  L, T  R source is in L, sink is in R • Flow across an (L,R) cut:  f(u,w) – f(w,z) u L, w R (u,w)  E w R, z L (w,z)  E From R to L From L to R A D 2/2 4/5 2/3 0/2 T 2/2 S B 0/1 2/10 1/1 1/4 1/6 C E

6. Observations • The flow across an (L,R) cut cannot exceed the capacity of the cut capacity(L,R) =  c(u,w) •  For any flow f and any cut C, size(f) ≤ capacity(C) “maximum flow ≤ minimum cut” • Previous example had size(f) = 5 and capacity(C) = 5 • The cut is a certificate of optimality (maximality) of the flow • The flow is a certificate optimality (minimality) of the cut u L, w R

7. Ford-Fulkerson Algorithm – Basic Idea • Start with zero flow • REPEAT: • Find a path from S to T along which flow can be increased • Increase the flow along this path

9. Residual Graph • We have some flow, and want to improve it • Two ways to “push” or “advance” flow in G • Find some unused capacity on an edge • Find some cancelable flow on an edge •  residual graph Gf= “what’s unused or cancelable” Flow f Residual Graph Gf A T A T 1 1 1 1 1 1 1 S B S B 1 REPEAT: Find an S-T path in Gf ; Increase f along this path as much as possible

10. Recipe for Constructing Gf reversed edges • Gf = (V, Ef) • Ef  E  ER • For any (u,w) in E or ER capacity cf(u,w) = c(u,w) – f(u,w) + f(w,u) • Note 1: Can ignore edges with cf( , ) = 0 • Note 2: If (u,w)  E, write c(u,w) = 0, f(u,w) = 0 remaining capacity cancelable flow REPEAT: Find an S-T path in Gf ; Increase f along this path as much as possible

11. A D 2 2 2 T S Worked Example A D 5 2 3 2 2 T S B 1 10 1 • Initial Gf • Augment flow along a path New Gf 4 6 C E 2 A D 2 2 1 3 2 2 T S B 1 10 1 4 6 C E

12. Worked Example 2 A D 2 2 1 3 2 2 T S B 1 10 1 4 6 C E T S 1 D 1 1 C E 2 2 T S B 4 A D 2 2 2 1 1 2 2 T 2 S B 1 Final Gf 8 1 1 What is the significance of this cut? 3 5 C E 1

13. Ford-Fulkerson Algorithm – Summary • Initialize f = 0 • REPEAT: • Construct the residual graph Gf • Find a path P from s to t in Gf • If there is no such path, HALT • cp = minimum cf-capacity edge on path P • Increase f by cp units along path P • Flow always increases  the algorithm terminates • But, if capacities are bounded by f*, can take up to |E|  f* iterations in worst case. running time is |E|2 f* 106 A T 106 1 106 S B 106

14. The Max-Flow Min-Cut Theorem • Ford-Fulkerson constructively proves that the maximum flow equals the minimum cut • Define an (L,R) cut as follows: • L = nodes reachable from s in final residual graph Gf • R = rest of nodes = V – L Note that t cannot be in L  T must be in R • Consider edges between L and R in G • Edges e going from L  R : must have flow = capacity (by def. of L, R) • Edges e’ going from R  L : must have flow = 0 (by def. of L, R)  The flow across this cut = LR edges e c(e) = the capacity of the cut  Since any flow is ≤ any cut, we have max flow = min cut e L R S T e’

15. Integrality theorem: If all capacities in the flow network are integers, then there is a maximum flow for which all flow values are integers. Proof: Can choose augumenting paths with integer value flows. The flow network can be maintained to be integer valued.

20. Ford-Fulkerson Algorithm – Summary • Initialize f = 0 • REPEAT: • Construct the residual graph Gf • Find a path P from s to t in Gf • If there is no such path, HALT • cp = minimum cf-capacity edge on path P • Increase f by cp units along path P • Flow always increases  the algorithm terminates • But, if capacities are bounded by f*, can take up to |E|  f* iterations in worst case. The running time is |E|2 f* 106 A T 106 1 106 S B 106

21. Improving Ford-Fulkerson Algorithm • Find better augumenting paths 106 A T 106 1 106 S B 106

22. Edmond Karp Algorithm– an improvement of Ford-Fulkerson Algorithm • Initialize f = 0 • REPEAT: • Construct the residual graph Gf • Find a shortest saturating path P using BFS from s to t in Gf • If there is no such path, HALT • cp = minimum cf-capacity edge on path P • Increase f by cp units along path P • Flow always increases  the algorithm terminates • Can take up to |V| |E| iterations in worst case. The running time is |V| |E|2 106 A T 106 1 106 S B 106

23. Edmond Karp Algorithm– an improvement of Ford-Fulkerson Algorithm Analysis: Observation 1: The length of shortest paths in residual graph can not decrease. Observation 2: Choose the augumenting path so that it contains at least one saturated edge. Can take up to |V| |E| iterations in worst case. The running time is |V| |E|2 saturated T S A B augumenting T S A B Length increased by 2.