Graph_Theory_Notes5.pdf

# Lemma 1 for each s v such that x s and y s the net

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Lemma 1. For each S V such that x / S and y / S , the net flow out of S and the net flow into S both equal 0. That is, f + ( S ) = f - ( S ). Proof. By the conservation condition, 0 = u S { uv A f ( uv ) - wu A f ( wu ) } = uv A : u S f ( uv ) - wu A : u S f ( wu ) = (total flow on arcs with tails in S ) - (total flow on arcs with heads in S ) = uv A : u S,v V - S f ( uv ) - wu A : w V - S,u S f ( wu ) = f + ( S ) - f - ( S ) .

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Let N be a network with source x and sink y . Let f be a flow in N . Lemma 2. Net flow out of x equals net flow into y . f + ( x ) - f - ( x ) = f - ( y ) - f + ( y ) . Proof. Consider S = V - { x, y } . Assume first that there is no arc between x and y . Then f + ( S ) = f - ( x ) + f - ( y ) and f - ( S ) = f + ( x ) + f + ( y ) . By the previous lemma, f + ( S ) = f - ( S ), which implies the claim. Proof (continued). If xy is an arc but yx is not, then f + ( S ) = f - ( x ) + ( f - ( y ) - f ( xy )) f - ( S ) = ( f + ( x ) - f ( xy )) + f + ( y ) . Since f + ( S ) = f - ( S ), we get the same equality. Similarly, if yx is an arc but xy is not, or if both xy and yx are arcs, we still get the same equality. Definition 6. The value of the flow f is defined as val ( f ) := f + ( x ) - f - ( x ) = f - ( y ) - f + ( y ) . Maximum flow problem Definition 7. A flow f in N is a maximum flow if val ( f ) val ( g ) for every flow g in N . The value of a maximum flow is the maximum amount of the commodity that can be transported from x to y under the capacity constraints. Problem 1 (Maximum Flow Problem) . Find a maximum flow in a given network. Cuts Intuition: A set of arcs “separating” x and y limits flows from x to y . Definition 8. Let N = ( V, A ) be a network with source x and sink y . Let S be a set of vertices of N containing x but not containing y . The cut associated with S is the set of all arcs uv A , where u S and v V - S . This cut is denoted by ( S, V - S ). Each subset S V such that x S but y / S gives rise to a cut. Definition 9. The capacity of a cut ( S, V - S ) is the sum of the capacities of its arcs cap ( S, V - S ) = X uv A : u S,v V - S c ( uv ) . Lemma 3. For every flow f and cut ( S, V - S ) in N , val ( f ) = f + ( S ) - f - ( S ) . Proof. By the definition of val ( f ) and the conservation condition, val ( f ) = xv A f ( xv ) - wx A f ( wx ) = u S { uv A f ( uv ) - wu A f ( wu ) } = uv A : u S f ( uv ) - wu A : u S f ( wu ) = (total flow on arcs with tails in S ) - (total flow on arcs with heads in S ) = uv A : u S,v V - S f ( uv ) - wu A : w V - S,u S f ( wu ) = f + ( S ) - f - ( S ) .
Definition 10. An arc a is saturated by f if f ( a ) = c ( a ) and a is f -zero if f ( a ) = 0. Lemma 4. For every flow f and cut ( S, V - S ) of N , val ( f ) cap ( S, V - S ) . Furthermore, val ( f ) = cap ( S, V - S ) iff each arc in ( S, V - S ) is saturated by f and each arc from V - S to S is f -zero. Proof. By the previous lemma, val ( f ) = f + ( S ) - f - ( S ). Since f + ( S ) cap ( S, V - S ) and f - ( S ) 0, val ( f ) cap ( S, V - S ) . To prove the “furthermore” claim, note that val ( f ) = cap ( S, V - S ) iff f + ( S ) = cap ( S, V - S ) and f - ( S ) = 0. We have f + ( S ) = cap ( S, V - S ) iff each arc in ( S, V - S ) is saturated by f , and f - ( S ) = 0 iff each arc from V - S to S is f -zero.

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