We claim further that every arc from v s to s is f

Info icon This preview shows pages 6–9. Sign up to view the full content.

View Full Document Right Arrow Icon
We claim further that every arc from V - S to S is f -zero. Suppose to the contrary that uv is an arc from V - S to S such that f ( uv ) > 0, where
Image of page 6

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
u V - S and v S . Since v S , v is labelled. Since f ( uv ) > 0, by rule 2(ii) we should label u from v . This contradicts the assumption u V - S . We have proved that for the final flow f , every arc from S to V - S is saturated, and every arc from V - S to S is f -zero. Thus f + ( S ) = cap ( S, V - S ) and f - ( S ) = 0 . Thus val ( f ) = cap ( S, V - S ) . Hence, f is a maximum flow and ( S, V - S ) is a minimum cut by an earlier theorem, Remarks The labelling algorithm is essentially due to Ford and Fulkerson (and two other groups), who proposed the idea of augmentation. The version above omits a number of implementation issues, e.g. rules for choosing a vertex to label in Step 2. There is a refined ‘augmenting’ algorithm with polynomial time com- plexity. 4 Connectivity revisited Menger’s Theorem – Vertex version Theorem 3 (Menger’s Theorem – Vertex version) . Let s and t be non- adjacent vertices of a graph G . Then the maximum number of internally vertex-disjoint ( s, t )-paths in G is equal to the minimum number of vertices whose deletion destroys all ( s, t )-paths in G . In other words, p ( s, t ) = c ( s, t ) . Menger’s Theorem – Edge version Theorem 4 (Menger’s Theorem – Edge version) . Let s and t be distinct vertices of a graph G . Then the maximum number of edge-disjoint ( s, t )- paths in G is equal to the minimum number of edges whose deletion destroys all ( s, t )-paths in G . In other words, p 0 ( s, t ) = c 0 ( s, t ) . Menger’s Theorem – Arc version for digraphs Both vertex and edge versions of Menger’s Theorem can be proved by using the following: Theorem 5 (Menger’s Theorem – Arc version for digraphs) . Let s and t be distinct vertices of a directed graph G . Then the maximum number of arc-disjoint directed ( s, t )-paths in G is equal to the minimum number of arcs whose deletion destroys all directed ( s, t )-paths in G . A lemma Lemma 7. If the capacity function of a network is integral (that is, every arc has an integer capacity), then the network has a maximum flow which is integral (that is, the flow value on every arc is an integer). Proof. Take an integral flow f * with maximum possible value. We claim that f * must be a maximum flow. Suppose otherwise. Then there exists an augmenting path P with respect to f * . Since both f * and the capacity function are integral, I ( P ) is an integer. So augmentation along P yields an integral flow whose value is greater than f * , contradicting the choice of f * . Therefore, f * must be a maximum flow of the network.
Image of page 7
A key lemma A set of arcs of G whose deletion destroys all directed ( s, t )-paths is called an ( s, t ) -disconnecting set . Lemma 8. Let G be a network with source s and sink t in which each arc has unit capacity. Then (a) the value of a maximum flow in G is equal to the maximum number p of arc-disjoint directed ( s, t )-paths in G ; and (b) the capacity of a minimum ( s, t )-cut in G is equal to the minimum number c of arcs whose deletion destroys all directed ( s, t )-paths in G .
Image of page 8

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 9
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern