Minimum cut problem definition 11 a cut s v s in n is

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Minimum cut problem Definition 11. A cut ( S, V - S ) in N is a minimum cut if cap ( V, S ) cap ( S 0 , V - S 0 ) for all S 0 V with x S 0 and y 6∈ S 0 . Problem 2 (Minimum Cut Problem) . Find a minimum cut in a network. Lemma 5. If val ( f ) = cap ( S, V - S ) for a flow f and a cut ( S, V - S ), then f is a maximum flow and ( S, V - S ) is a minimum cut. Proof. By the lemma above, for every flow f 1 and cut ( S 1 , V - S 1 ), val ( f 1 ) cap ( S 1 , V - S 1 ) . Let f 0 be an arbitrary flow and ( S 0 , V - S 0 ) be an arbitrary cut. Applying the inequality above to f 0 , ( S, V - S ) and f, ( S 0 , V - S 0 ) respectively, val ( f 0 ) cap ( S, V - S ) = val ( f ) cap ( S 0 , V - S 0 ) . Hence f is a maximum flow and ( S, V - S ) is a minimum cut. Summary so far val ( f ) cap ( S, V - S ) for every flow f and cut ( S, V - S ), and if val ( f ) = cap ( S, V - S ) then f is a maximum flow and ( S, V - S ) is a minimum cut. Is the maximum flow always a flow with val ( f ) = cap ( S, V - S ) for some cut ( S, V - S )? Yes! Our statement can be extended: a flow f cannot be maximum unless val ( f ) = cap ( S, V - S ) for some cut. Augmenting paths Definition 12. For a path P from x to some vertex in a network, define I ( P ) = min a A ( P ) ( c ( a ) - f ( a ) if a is forward in P f ( a ) if a is backward in P. P is f -unsaturated if I ( P ) > 0. In other words, a path starting from x is f -unsaturated iff every forward arc is unsaturated by f and every backward arc is not f -zero. An f -unsaturated path from x to y is called an f -augmenting path .
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Augmentation For an f -augmenting path P , we can increase the flow on each forward arc by I ( P ) and decrease the flow on each backward arc by I ( P ). In this way we obtain a new flow with larger value. This is called augmentation along P . Lemma 6. Let P be an f -augmenting path in N . Define ˆ f ( a ) = f ( a ) + I ( P ) if a is a forward arc of P f ( a ) - I ( P ) if a is a backward arc of P f ( a ) if a 6∈ A ( P ). Then ˆ f is a flow in N with value val ( ˆ f ) = val ( f ) + I ( P ). Proof. (i) Capacity constraint If an arc e is not in P , then 0 ˆ f ( e ) = f ( e ) c ( e ). Suppose e is an arc in P . If e is a forward arc of P , then by the definition of I ( P ), ˆ f ( e ) = f ( e ) + I ( P ) > f ( e ) 0 and ˆ f ( e ) = f ( e ) + I ( P ) f ( e ) + ( c ( e ) - f ( e )) = c ( e ). If e is a backward arc of P , then ˆ f ( e ) = f ( e ) - I ( P ) 0 and ˆ f ( e ) = f ( e ) - I ( P ) f ( e ) c ( e ). Hence ˆ f satisfies the capacity constraint. Proof (continued). (ii) Conservation condition If a vertex u is not in P , then ˆ f + ( u ) = f + ( u ) = f - ( u ) = ˆ f - ( u ). Suppose u 6∈ { x, y } is a vertex in P . Let e and e 0 denote the arcs of P incident with u in order of occurrence in P . If both e and e 0 are forward, then ˆ f + ( u ) = f + ( u )+ I ( P ) and ˆ f - ( u ) = f - ( u )+ I ( P ). If e is forward and e 0 is backward, or e is backward and e 0 is forward, then ˆ f + ( u ) = f + ( u ) and ˆ f - ( u ) = f ( u ). If both e and e 0 are backward, then ˆ f + ( u ) = f + ( u ) - I ( P ) and ˆ f - ( u ) = f - ( u ) - I ( P ). Since f + ( u ) = f - ( u ), in each case we have ˆ f + ( u ) = ˆ f - ( u ). Proof (continued). (iii) Value of ˆ f Let e 1 be the first arc of P (i.e. the one incident with x ). If e 1 is forward, then ˆ f + ( x ) = f + ( x ) + I ( P ) and ˆ f - ( x ) = f - ( x ). Hence val ( ˆ f ) = ˆ f + ( x ) - ˆ f - ( x ) = ( f + ( x ) - f - ( x ))+ I ( P ) = val ( f )+ I ( P ). If e 1 is backward, then ˆ f + ( x ) = f + ( x ) and ˆ f - ( x ) = f - ( x ) - I ( P ). Hence val ( ˆ f ) = ˆ f + ( x ) - ˆ f - ( x ) = f + ( x ) - ( f - ( x ) - I ( P )) = val ( f ) + I ( P ).
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