Network Flow Algorithms Lecture 14 October 12, 2016 Chandra & - PowerPoint PPT Presentation

CS 473: Algorithms, Fall 2016 Network Flow Algorithms Lecture 14 October 12, 2016 Chandra & Ruta (UIUC) CS473 1 Fall 2016 1 / 42

Part I Algorithm(s) for Maximum Flow Chandra & Ruta (UIUC) CS473 2 Fall 2016 2 / 42

Greedy Approach Begin with f ( e ) = 0 for each edge. u 1 /20 / Find a s - t path P with f ( e ) < c ( e ) 2 1 0 for every edge e ∈ P . s / t 3 Augment flow along this path. 0 3 /20 /10 Repeat augmentation for as long as 4 possible. v Chandra & Ruta (UIUC) CS473 3 Fall 2016 3 / 42

Greedy Approach Begin with f ( e ) = 0 for each edge. u 1 /20 / Find a s - t path P with f ( e ) < c ( e ) 2 1 0 for every edge e ∈ P . s / t 3 Augment flow along this path. 10 0 3 10 /20 /10 Repeat augmentation for as long as 4 10 possible. v Chandra & Ruta (UIUC) CS473 3 Fall 2016 3 / 42

Greedy Approach Begin with f ( e ) = 0 for each edge. u 1 /20 1 0 / Find a s - t path P with f ( e ) < c ( e ) 2 10 1 0 0 1 for every edge e ∈ P . s / t 3 Augment flow along this path. 0 3 10 /20 /10 Repeat augmentation for as long as 4 10 possible. v Chandra & Ruta (UIUC) CS473 3 Fall 2016 3 / 42

Greedy Approach Begin with f ( e ) = 0 for each edge. u 1 /20 1 0 / Find a s - t path P with f ( e ) < c ( e ) 2 20 1 0 1 for every edge e ∈ P . 0 s / t 10 3 Augment flow along this path. 0 3 10 /20 /10 Repeat augmentation for as long as 4 20 possible. v Chandra & Ruta (UIUC) CS473 3 Fall 2016 3 / 42

Greedy Approach: Issues Issues = What is this nonsense? u Begin with f ( e ) = 0 for each edge /20 1 / 1 0 Find a s - t path P with f ( e ) < c ( e ) 2 for every edge e ∈ P s / t 3 0 Augment flow along this path 3 /20 /10 Repeat augmentation for as long as 4 v possible. Chandra & Ruta (UIUC) CS473 4 Fall 2016 4 / 42

Greedy Approach: Issues Issues = What is this nonsense? u Begin with f ( e ) = 0 for each edge /20 1 / 1 20 0 Find a s - t path P with f ( e ) < c ( e ) 2 2 0 for every edge e ∈ P s / 20 t 3 0 Augment flow along this path 3 /20 /10 20 Repeat augmentation for as long as 4 v possible. Chandra & Ruta (UIUC) CS473 4 Fall 2016 4 / 42

Greedy Approach: Issues Issues = What is this nonsense? u Begin with f ( e ) = 0 for each edge /20 /20 1 / / 1 1 20 0 0 Find a s - t path P with f ( e ) < c ( e ) 2 2 0 for every edge e ∈ P s / / t 3 3 0 0 Augment flow along this path 3 /20 /20 /10 /10 20 Repeat augmentation for as long as 4 v possible. Greedy can get stuck in sub-optimal flow! Chandra & Ruta (UIUC) CS473 4 Fall 2016 4 / 42

Greedy Approach: Issues Issues = What is this nonsense? Begin with f ( e ) = 0 for each edge u 1 /20 /20 / / Find a s - t path P with f ( e ) < c ( e ) 2 20 1 1 0 0 2 2 for every edge e ∈ P 0 0 s / / t 3 3 Augment flow along this path 10 0 0 3 10 /20 /20 /10 /10 Repeat augmentation for as long as 4 20 possible. v Greedy can get stuck in sub-optimal flow! Chandra & Ruta (UIUC) CS473 4 Fall 2016 4 / 42

Greedy Approach: Issues Issues = What is this nonsense? Begin with f ( e ) = 0 for each edge u 1 /20 / Find a s - t path P with f ( e ) < c ( e ) 2 20 1 � �� � 2 0 0 =10 − for every edge e ∈ P 1 0 s / t 3 Augment flow along this path 10 0 3 10 /20 /10 Repeat augmentation for as long as 4 20 possible. v Greedy can get stuck in sub-optimal flow! Need to “push-back” flow along edge ( u , v ) . Chandra & Ruta (UIUC) CS473 4 Fall 2016 4 / 42

Greedy Approach: Issues Issues = What is this nonsense? Begin with f ( e ) = 0 for each edge u 1 /20 1 0 / Find a s - t path P with f ( e ) < c ( e ) 2 20 1 0 1 for every edge e ∈ P 0 s / t 3 Augment flow along this path 10 0 3 10 /20 /10 Repeat augmentation for as long as 4 20 possible. v Greedy can get stuck in sub-optimal flow! Need to “push-back” flow along edge ( u , v ) . Chandra & Ruta (UIUC) CS473 4 Fall 2016 4 / 42

Residual Graph The “leftover” graph Definition For a network G = ( V , E ) and flow f , the residual graph G f = ( V ′ , E ′ ) of G with respect to f is V ′ = V , 1 Forward Edges : For each edge e ∈ E with f ( e ) < c ( e ) , we 2 add e ∈ E ′ with capacity c ( e ) − f ( e ) . Backward Edges : For each edge e = ( u , v ) ∈ E with 3 f ( e ) > 0 , we add ( v , u ) ∈ E ′ with capacity f ( e ) . Chandra & Ruta (UIUC) CS473 5 Fall 2016 5 / 42

Residual Graph Example u u /20 /20 /20 / / 20 1 1 / 0 0 1 0 2 0 0 1 s / / t / / 3 3 s t 2 0 0 0 /20 /20 /10 /10 /20 /10 20 v v Figure: Flow on edges is indicated in Figure: Residual Graph red Chandra & Ruta (UIUC) CS473 6 Fall 2016 6 / 42

Residual graph has... Given a network with n vertices and m edges, and a valid flow f in it, the residual network G f , has (A) m edges. (B) ≤ 2 m edges. (C) ≤ 2 m + n edges. (D) 4 m + 2 n edges. (E) nm edges. (F) just the right number of edges - not too many, not too few. Chandra & Ruta (UIUC) CS473 7 Fall 2016 7 / 42

Residual Graph Property Observation: Residual graph captures the “residual” problem exactly. Chandra & Ruta (UIUC) CS473 8 Fall 2016 8 / 42

Residual Graph Property Observation: Residual graph captures the “residual” problem exactly. Lemma Let f be a flow in G and G f be the residual graph. If f ′ is a flow in G f then f + f ′ is a flow in G of value v ( f ) + v ( f ′ ) . Chandra & Ruta (UIUC) CS473 8 Fall 2016 8 / 42

Residual Graph Property Observation: Residual graph captures the “residual” problem exactly. Lemma Let f be a flow in G and G f be the residual graph. If f ′ is a flow in G f then f + f ′ is a flow in G of value v ( f ) + v ( f ′ ) . Lemma Let f and f ′ be two flows in G with v ( f ′ ) ≥ v ( f ) . Then there is a flow f ′′ of value v ( f ′ ) − v ( f ) in G f . Chandra & Ruta (UIUC) CS473 8 Fall 2016 8 / 42

Residual Graph Property Observation: Residual graph captures the “residual” problem exactly. Lemma Let f be a flow in G and G f be the residual graph. If f ′ is a flow in G f then f + f ′ is a flow in G of value v ( f ) + v ( f ′ ) . Lemma Let f and f ′ be two flows in G with v ( f ′ ) ≥ v ( f ) . Then there is a flow f ′′ of value v ( f ′ ) − v ( f ) in G f . Definition of + and - for flows is intuitive and the above lemmas are easy in some sense but a bit messy to formally prove. Chandra & Ruta (UIUC) CS473 8 Fall 2016 8 / 42

Residual Graph Property: Implication Recursive algorithm for finding a maximum flow: MaxFlow ( G , s , t ): if the flow from s to t is 0 then return 0 Find any flow f with v ( f ) > 0 in G Recursively compute a maximum flow f ′ in G f Output the flow f + f ′ Chandra & Ruta (UIUC) CS473 9 Fall 2016 9 / 42

Residual Graph Property: Implication Recursive algorithm for finding a maximum flow: MaxFlow ( G , s , t ): if the flow from s to t is 0 then return 0 Find any flow f with v ( f ) > 0 in G Recursively compute a maximum flow f ′ in G f Output the flow f + f ′ Iterative algorithm for finding a maximum flow: MaxFlow ( G , s , t ): Start with flow f that is 0 on all edges while there is a flow f ′ in G f with v ( f ′ ) > 0 do f = f + f ′ Update G f Output f Chandra & Ruta (UIUC) CS473 9 Fall 2016 9 / 42

Ford-Fulkerson Algorithm algFordFulkerson for every edge e , f ( e ) = 0 G f is residual graph of G with respect to f while G f has a simple s - t path do let P be simple s - t path in G f f = augment( f , P ) Construct new residual graph G f . Chandra & Ruta (UIUC) CS473 10 Fall 2016 10 / 42

Ford-Fulkerson Algorithm algFordFulkerson for every edge e , f ( e ) = 0 G f is residual graph of G with respect to f while G f has a simple s - t path do let P be simple s - t path in G f f = augment( f , P ) Construct new residual graph G f . augment ( f , P ) let b be bottleneck capacity, i.e., min capacity of edges in P (in G f ) for each edge ( u , v ) in P do if e = ( u , v ) is a forward edge then f ( e ) = f ( e ) + b else (* ( u , v ) is a backward edge *) let e = ( v , u ) (* ( v , u ) is in G *) f ( e ) = f ( e ) − b return f Chandra & Ruta (UIUC) CS473 10 Fall 2016 10 / 42

Example u u 10 30 10 s s t t 15 20 v v u u 10 30 10 s t s t 15 20 v v Chandra & Ruta (UIUC) CS473 11 Fall 2016 11 / 42

Example continued u u 10 30 10 s s t t 15 20 v v u u 10 30 10 s t s t 15 20 v v Chandra & Ruta (UIUC) CS473 12 Fall 2016 12 / 42

Example continued u u 10 30 10 s s t t 15 20 v v u u 10 30 10 s t s t 15 20 v v Chandra & Ruta (UIUC) CS473 13 Fall 2016 13 / 42

Example continued u u 10 30 10 s s t t 15 20 v v u u 10 30 10 s t s t 15 20 v v Chandra & Ruta (UIUC) CS473 14 Fall 2016 14 / 42

Properties about Augmentation: Flow Lemma If f is a flow and P is a simple s - t path in G f , then f ′ = augment ( f , P ) is also a flow. Chandra & Ruta (UIUC) CS473 15 Fall 2016 15 / 42