MIT 18 438 - Lecture 5 Notes (6 pages)

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Lecture 5 Notes



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Lecture 5 Notes

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Pages:
6
School:
Massachusetts Institute of Technology
Course:
18 438 - Advanced Combinatorial Optimization
Advanced Combinatorial Optimization Documents

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18 438 Advanced Combinatorial Optimization Updated February 18 2012 Lecture 5 Lecturer Michel X Goemans Scribe Yehua Wei 2009 In this lecture we establish the connection between nowhere zero k flows and nowhere zero Zk flows Then we present several theorems of relations between edge connectivity of a graph and the existence of nowhere zero flows 1 Nowhere zero k flow Let us first recall some definitions from the previous lecture Definition 1 Let G V E be a directed graph and be an abelian group A nowherezero flow is E 0 such that X X e e flow conservation e v e v If G is undirected then we say that it has a nowhere zero flow if the graph admits a nowhere zero flow after giving an orientation to all the edges As we saw if one orientation works then any does since inverses exist in abelian groups Definition 2 Let G be an undirected graph For integer k 2 a nowhere zero k flow is an assignment E 1 k 1 such that for some orientation of G flow conservation is achieved i e X X e e e v e v for all v V It is often convenient to fix an orientation and let take values in 1 k 1 Theorem 1 Tutte 1950 Let G be an undirected graph Then G has a nowhere zero k flow G has a nowhere zero Zk flow Proof By definition of nowhere zero flows P P Let be a nowhere zero Zk flow define e v e v e e v e for all v V under group operation in Z Observe that all e v s are multiples of Pk Without loss of generality we can assume is the nowhere zero Zk flow such that v V e v is minimized where P we minimize over all and all orientations of G Suppose v V e v 0 then we have obtained P a nowhere zero k flow POtherwise let S v e v 0 and T v e v 0 Since v V e v 0 and v V e v 0 we have that both S and Let U be the from S PT are nonempty P P set of vertices reachable If U T then 0 v U e v e U e e U e But U and 5 1 P P thus implies e U e e U e 0 which is a contradiction Thus we must have U T 6 which implies we can find a directed path P from some s S to t T Now we revert path P to create another nowhere zero



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