Maximum Flow The Maximum Flow Problem Given Directed graph G = ( V - PowerPoint PPT Presentation

Maximum Flow

The Maximum Flow Problem Given ◮ Directed graph G = ( V , E ) ◮ A capacity function c : E → R + ◮ Two vertices s and t . s , t -Flow ◮ Function f : E → R + such that ◮ f ( e ) ≤ c ( e ) , � � ◮ for all v ∈ V \ { s , t } , f ( uv ) = f ( vw ) , uv ∈ E vw ∈ E � � f ( vs ) = 0 , and f ( tv ) = 0 . ◮ vs ∈ E tv ∈ E � The value | f | of an s , t -flow f is f ( s , v ) sv ∈ E 2 / 12

The Maximum Flow Problem Maximum Flow Problem For a given directed graph and two given vertices s and t , find an s , t -flow f with maximum value. Multi-Source and Multi-Sink Flow ◮ Sources: s 1 , s 2 , . . . , s k ◮ Sinks: t 1 , t 2 , . . . , t l ◮ Add vertex s ′ and edges s ′ s i with c ( s ′ s i ) = ∞ ( 1 ≤ i ≤ k ). ◮ Add vertex t ′ and edges ( t i t ′ ) with c ( t i t ′ ) = ∞ ( 1 ≤ i ≤ l ). ◮ Find maximum s ′ , t ′ -flow. 3 / 12

Max Flow Algorithm “Naive” Approach ◮ Find some path P from s to t where each edge e has f ( e ) < c ( e ) . ◮ Add increase f ( e ) as much as possible for each edge of P . ◮ Repeat. This approach will not always create an optimal solution 1 | 1 0 | 1 1 | 1 1 | 1 s s 1 | 1 t 0 | 1 t 0 | 1 1 | 1 1 | 1 1 | 1 algorithm output optimal solution 4 / 12

Residual Graph Observation ◮ If f ( uv ) = k , we can assume there is an augmented edge vu with c ( vu ) ≥ k , even if vu / ∈ E . Residual Graph G f = ( V , E f ) ◮ E f = { uv | uv ∈ E , f ( uv ) < c ( uv ) } ∪ { vu | uv ∈ E , f ( uv ) > 0 } � c ( uv ) − f ( uv ) if uv ∈ E ◮ c f ( uv ) = f ( vu ) if vu ∈ E f ′ is an s , t -flow for G f if and only if f + f ′ is an s , t -flow for G . 5 / 12

Ford-Fulkerson Algorithm Ford-Fulkerson Algorithm ◮ Start with f ( e ) = 0 for all edges e ∈ E . ◮ Find a simple s , t -path P in G f . ◮ Update f with maximum flow for P . ◮ Update G f . ◮ Repeat. What is the worst case runtime for this algorithm? 6 / 12

Ford-Fulkerson Algorithm – Runtime Consider the following case: u 1 , 000 1 , 000 s 1 t 1 , 000 1 , 000 v If the algorithm picks uv or vu in each iteration, it takes 2 , 000 iterations to find a maximum flow. Worst case runtime ◮ If c : E → N , O ( | E || f max | ) ◮ If c : E → R , algorithm may not terminate. 7 / 12

Edmonds-Karp Algorithm Edmonds-Karp Algorithm ◮ Start with f ( e ) = 0 for all edges e ∈ E . ◮ Find a shortest s , t -path P in G f . ◮ Update f with maximum flow for P . ◮ Update G f . ◮ Repeat. Ford-Fulkerson ◮ Finds some s , t -path in G f . ◮ Runtime: O ( | E || f max | ) (integer capacities) Edmonds-Karp Algorithm ◮ Finds a shortest s , t -path in G f . ◮ Runtime: O ( | V || E | 2 ) Best known result today: O ( | V || E | ) (complicated) 8 / 12

Max-Flow Min-Cut Theorem

Cut s , t -Cut An s , t -cut ( S , T ) in a graph G = ( V , E ) is a partition of V into S and T such that s ∈ S and t ∈ T . The flow f ( S , T ) is defined as � � f ( S , T ) = f ( xy ) − f ( yx ) . x ∈ S , y ∈ T , xy ∈ E x ∈ S , y ∈ T , yx ∈ E The capacity c ( S , T ) is defined as � c ( S , T ) = c ( xy ) . x ∈ S , y ∈ T , xy ∈ E 10 / 12

Flows and Cuts Lemma Maximum s , t -flow ≤ minimum s , t -cut, i. e., max f | f | ≤ min ( S , T ) c ( S , T ) . Proof. Follows from definition. � Lemma For each s , t -cut ( S , T ) , f ( S , T ) = | f | . Proof. ◮ Let V = { s = v 0 , v 1 , v 2 , . . . , v n − 1 , v n = t } and V i = { v i , v i + 1 , . . . , v n } . � � � ◮ | f | = f ( s , v ) and, for all v ∈ V \ { s , t } , f ( uv ) = f ( vw ) . sv ∈ E uv ∈ E vw ∈ E ◮ Thus, f ( { s } , V 1 ) = f ( { s , v 1 } , V 2 ) = . . . = f ( S , T ) . � 11 / 12

Max-Flow Min-Cut Theorem Theorem For any s , t -flow f , the following three conditions are equivalent. ( i ) f is a maximum s , t -flow in G . ( ii ) There is no augmenting s , t -path in G f . ( iii ) There is an s , t -cut ( S , T ) in G with c ( S , T ) = | f | . ( i ) → ( ii ) : If there would be an s , t -path P in G f , | f | could be increased by the capacity of P . ( ii ) → ( iii ) : There is no augmenting s , t -path in G f . Let S be the set of vertices reachable from s in G f and T = V \ S . If c ( S , T ) > f ( S , T ) , there is an edge xy in G f with x ∈ S and y ∈ T . Thus, y is reachable from s . Contradiction. ( iii ) → ( i ) : There is an s , t -cut ( S , T ) in G with c ( S , T ) = | f | . Because max f | f | ≤ min ( S , T ) c ( S , T ) , f is a maximum s , t -flow in G . � 12 / 12