1. CSCI 256 Data Structures and Algorithm Analysis Lecture 20 Some slides by Kevin Wayne copyright 2005, Pearson Addison Wesley all rights reserved, and some by Iker Gondra

2. Cuts Def: An s-t cut is a partition (A, B) of V with s  A and t  B Def: The capacity of a cut (A, B) is: s 2 3 4 5 6 7 t 15 5 30 15 10 8 15 9 6 10 15 4 4 Capacity = 10 + 5 + 15 = 30 A

3. Cuts Def: An s-t cut is a partition (A, B) of V with s  A and t  B Def: The capacity of a cut (A, B) is: s 2 3 4 5 6 7 t 15 5 30 15 10 8 15 9 6 10 15 4 4 A Capacity = 9 + 15 + 8 + 30 = 62

4. Minimum Cut Problem Min s-t cut problem: Find an s-t cut of minimum capacity s 2 3 4 5 6 7 t 15 5 30 15 10 8 15 9 6 10 15 4 4 A Capacity = 10 + 8 + 10 = 28

5. Find a Minimum Cut u st v 40 10

6. Find a Minimum Cut s t 6 6 10 7 3 5 36 2 4 5 8 5 4 8 a b c d e f g h

7. Interesting point We’ll see -- problem of finding min cut is solved using FF technique Definition: the net flow across the cut = sum of the flows out of A minus the sum of the flows into A ( written as f out (A) –f in (A) )

8. Flows and Cuts Flow value lemma: Let f be any flow, and let (A, B) be any s-t cut. Then, the net flow sent across the cut is equal to the amount leaving s: 10 6 6 11 1 10 3 8 8 0 0 0 11 s 2 3 4 5 6 7 t 15 5 30 15 10 8 15 9 6 10 15 4 4 0 Value = 24 4 A

9. Flows and Cuts Flow value lemma: Let f be any flow, and let (A, B) be any s-t cut. Then, the net flow sent across the cut is equal to the amount leaving s: 10 6 6 1 3 8 8 0 0 0 11 s 2 3 4 5 6 7 t 15 5 30 15 10 8 15 9 6 10 15 4 4 0 Value = 6 + 0 + 8 - 1 + 11 = 24 4 11 A

10. Flows and Cuts Flow value lemma: Let f be any flow, and let (A, B) be any s-t cut. Then, the net flow sent across the cut is equal to the amount leaving s: 10 6 6 11 1 10 3 8 8 0 0 0 11 s 2 3 4 5 6 7 t 15 5 30 15 10 8 15 9 6 10 15 4 4 0 Value = 10 - 4 + 8 - 0 + 10 = 24 4 A

11. Let us see how to prove the flow value lemma Proof: v(f) =  f(e) =  f(e) -  f(e) (def and s is source) e out of s e out of s e into s = f out (s) - f in (s) (definition) =  ( f out (v) - f in (v) ) conservation: nodes in A except s are internal v in A =  (  f(e) -  f(e) ) (definition) v in A e out of v e into v =  f(e) -  f(e) = f out (A) –f in (A) e out of A e into A (reason for second-last equality: if e has both ends in A, f(e) appears once with a + and once with a – and they cancel; if e has tail in A it appears with a +; if it has its head in A it appears with a minus).

12. Flows and Cuts Flow value lemma: Let f be any flow, and let (A, B) be any s-t cut. Then Proof: previous slide Thus by watching the amount of flow f sends across a cut we can measure the value of the flow – it is the amount out minus the amount in.

13. Relationship of Flows and Cuts We shall show: Let f be any flow, and let (A, B) be any s- t cut. Then the value of the flow is at most the capacity of the cut Cut capacity = 30  Flow value  30 s 2 3 4 5 6 7 t 15 5 30 15 10 8 15 9 6 10 15 4 4 Capacity = 30 A

14. Flows and Cuts Flow/Cut Lemma : Let f be any flow. Then, for any s-t cut (A, B) we have v(f)  cap(A, B) Proof: = v(f) = f out (A) - f in (A) flow value lemma (and definition) ≤ f out (A) =  f(e) definition e out of A ≤  c(e) capacity condition e out of A = c(A,B) definition The corollary below follows from the flow/cut lemma – do you see why??? (Focus on the two any’s in the statement of the lemma)

15. Certificate of Optimality Corollary: Let f be any flow, and let (A, B) be any cut. If v(f) = cap(A, B), then f is a max flow and (A, B) is a min cut Value of flow = 28 Cut capacity = 28  Flow value  28 10 9 9 14 4 10 4 8 9 1 0 0 0 14 s 2 3 4 5 6 7 t 15 5 30 15 10 8 15 9 6 10 15 4 4 0 A

16. The Ford-Fulkerson Algorithm terminates when the flow f has no s-t path in G f This is the property needed to prove its maximality

17. Max-Flow = Min Cut Theorem: If f is an s-t flow such that there is no s-t path in G f, then there is an s-t cut (A*,B*) for which v(f) = c(A*,B*). Consequently, f has the maximum value of any flow in G and (A*,B*) has the minimum capacity of any s-t cut. Proof: The second statement is the above corollary; so all we need to do is prove the first statement. Assume we have an s-t flow such that there is no s-t path in G f.Here is how to find (A*,B*): Let A* denote the set of nodes {v} such that there is an s-v path in G f. Let B* = V - A*.; then:

18. Proof of Max-Flow Min-Cut Theorem 1. (A*, B*) is a cut (Why?? --- well s is in A* and t must be in B* as there is no s-t path in G f ) 2. If e=(u,v) is an edge in G with u in A* and v in B*, then f(e) = c(e) (Why?? If not, e is a forward in edge in G f and thus there would be an s-v path in G f – impossible as v is in B*) 3. In a similar manner we can show that if e’ = (u’, v’) is an edge in G with u’ in B* and v’ in A*, then f(e’) = 0 (if not consider the backward edge (v’,u’) in G f ; it gives rise to an s-u’ path in G f (via s-v’-u’) – impossible as u’ is in B*) From 2 and 3 we conclude that all edges e out of A* are completely saturated with flow (i.e., f(e) = c(e)) while all edges e into A* are completely unused (i.e., f(e) = 0).

19. Proof of Max-Flow Min-Cut Theorem Thus: v(f) = f out (A*) – f in (A*) flow value lemma ((A*, B*) is a cut) = Σ f(e) - Σ f(e) definition e out of A* e into A* = Σ c(e) - 0 previous slide justifies this e out of A* = c(A*,B*) definition

20. Max-Flow Min-Cut Theorem Ford-Fulkerson algorithm finds a flow where the residual graph G f is disconnected, thus we can’t increase the flow any more; hence FF finds a maximum flow. If we want to find a minimum cut, we begin by finding a maximum flow. Then using the G f obtained by FF, perform a BFS to determine the set of nodes A* that s can reach. If B* = V–A*, (A*,B*) is a cut of minimum capacity.