lecture_21

lecture_21 - Introduction to Algorithms 6.046J/18.401J...

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Introduction to Algorithms 6.046J/18.401J L ECTURE 21 Network Flow II Max-flow, min-cut theorem Ford-Fulkerson algorithm and analysis Edmonds-Karp algorithm and analysis Best algorithms to date Prof. Charles E. Leiserson
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Recall from Lecture 22 Flow value: | f | = f ( s , V ) . Cut: Any partition ( S , T ) of V such that s S and t T . Lemma. | f | = f ( S , T ) for any cut ( S , T ) . Corollary. | f | c ( S , T ) for any cut ( S , T ) . Residual graph: The graph G f = ( V , E f ) with strictly positive residual capacities c f ( u , v ) = c ( u , v ) – f ( u , v ) > 0 . Augmenting path: Any path from s to t in G f . Residual capacity of an augmenting path: , c f ( p ) = min { c f ( v u )} . , ( v u ) p © 2001–4 by Charles E. Leiserson Introduction to Algorithms November 29, 2004 L21.2
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Max-flow, min-cut theorem Theorem. The following are equivalent: 1. | f | = c ( S , T ) for some cut ( S , T ) . 2. f is a maximum flow. 3. f admits no augmenting paths. © 2001–4 by Charles E. Leiserson Introduction to Algorithms November 29, 2004 L21.3
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Max-flow, min-cut theorem Theorem. The following are equivalent: 1. | f | = c ( S , T ) for some cut ( S , T ) . 2. f is a maximum flow. 3. f admits no augmenting paths. Proof. ( 1 ) ( 2 ): Since | f | c ( S , T ) for any cut ( S , T ) (by the corollary from Lecture 22), the assumption that | f | = c ( S , T ) implies that f is a maximum flow. © 2001–4 by Charles E. Leiserson Introduction to Algorithms November 29, 2004 L21.4
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Max-flow, min-cut theorem Theorem. The following are equivalent: 1. | f | = c ( S , T ) for some cut ( S , T ) . 2. f is a maximum flow. 3. f admits no augmenting paths. Proof. ( 1 ) ( 2 ): Since | f | c ( S , T ) for any cut ( S , T ) (by the corollary from Lecture 22), the assumption that | f | = c ( S , T ) implies that f is a maximum flow. ( 2 ) ( 3 ): If there were an augmenting path, the flow value could be increased, contradicting the maximality of f . © 2001–4 by Charles E. Leiserson Introduction to Algorithms November 29, 2004 L21.5
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( 3 ) ( 1 ): Suppose that f admits no augmenting paths. Define S = { v V : there exists a path in G f from s to v } , and let T = V S . Observe that s S and t T , and thus ( S , T ) is a cut. Consider any vertices u S and v T . s s u u v v S T path in G f We must have c f ( u , v ) = 0 , since if c f ( u , v ) > 0 , then v S , not v T as assumed. Thus, f ( u , v ) = c ( u , v ) , since c f ( u , v ) = c ( u , v ) – S , T ) = c f ( u , v ) . Summing over all u S and v T yields f ( ( S , T ) , and since | f | = f ( S , T ) , the theorem follows. © 2001–4 by Charles E. Leiserson
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This note was uploaded on 04/10/2008 for the course CSE 6.046J/18. taught by Professor Piotrindykandcharlese.leiserson during the Fall '04 term at MIT.

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lecture_21 - Introduction to Algorithms 6.046J/18.401J...

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