Network Flow Lecturer: Shi Li Department of Computer Science and - PowerPoint PPT Presentation

CSE 632: Analysis of Algorithms II: Combinatorial Optimization and Linear Programming (Fall 2020) Network Flow Lecturer: Shi Li Department of Computer Science and Engineering University at Buffalo

Outline Network Flow 1 Ford-Fulkerson Method 2 Correctness of Ford-Fulkerson’s Method Running Time of Ford-Fulkerson-Type Algorithm Applications of Network-Flow 3 Single Commodity Flow with Multiple Sources and Sinks s - t Edge-Disjoint Paths Problem Bipartite Matching Problem 2/77

Flow Network Abstraction of fluid flowing through edges Digraph G = ( V, E ) with source s ∈ V and sink t ∈ V No edges enter s No edges leave t Edge capacity c ( e ) ∈ R > 0 for every e ∈ E 12 a c 20 16 s t 7 4 9 13 4 b d 14 3/77

Def. An s - t flow is a function f : E → R such that for every e ∈ E : 0 ≤ f ( e ) ≤ c ( e ) (capacity conditions) for every v ∈ V \ { s, t } : � � f ( e ) = f ( e ) . (conservation conditions) e into v e out of v The value of a flow f is � val ( f ) = f ( e ) . e out of s Maximum Flow Problem Input: directed network G = ( V, E ) , capacity function c : E → R > 0 , source s ∈ V and sink t ∈ V Output: an s - t flow f in G with the maximum val ( f ) 4/77

Maximum Flow Problem: Example 12/12 a c 19/20 12/16 0/4 7/7 s t 0/9 11/13 4/4 b d 11/14 5/77

Outline Network Flow 1 Ford-Fulkerson Method 2 Correctness of Ford-Fulkerson’s Method Running Time of Ford-Fulkerson-Type Algorithm Applications of Network-Flow 3 Single Commodity Flow with Multiple Sources and Sinks s - t Edge-Disjoint Paths Problem Bipartite Matching Problem 6/77

Greedy Algorithm Start with empty flow: f ( e ) = 0 for every e ∈ E Define the residual capacity of e to be c ( e ) − f ( e ) Find an augmenting path: a path from s to t , where all edges have positive residual capacity Augment flow along the path as much as possible Repeat until we got stuck 7/77

Greedy Algorithm: Example 12/12 a c 19/20 12/16 0/4 7/7 s t 0/9 11/13 4/4 b d 11/14 8/77

Greedy Algorithm Does Not Always Give a Optimum Solution a 0/1 0/1 1/1 1/1 s t 0/1 0/1 1/1 b 9/77

Fix the Issue: Allowing “Undo” Flow Sent a 1/1 1/1 0/1 s t 1/1 1/1 b 10/77

Assumption ( u, v ) and ( v, u ) can not both be in E Def. For a s - t flow f , the residual graph G f of G = ( V, E ) w.r.t f contains: the vertex set V , for every e = ( u, v ) ∈ E with f ( e ) < c ( e ) , a forward edge e = ( u, v ) , with residual capacity c f ( e ) = c ( e ) − f ( e ) , for every e = ( u, v ) ∈ E with f ( e ) > 0 , a backward edge e ′ = ( v, u ) , with residual capacity c f ( e ′ ) = f ( e ) . a a 0/1 0/1 1/1 1 1 1/1 s t s t 1 0/1 0/1 1 1 1/1 b b Residual Graph G f Original graph G and f 11/77

Residual Graph: One More Example G f G s s 4/16 0 / 12 13 1 3 4 0/4 4 a b a b 9 / 4/12 4/14 4 5 0 4 8 4 4 1 0/7 7 c d c d 0 4/4 / 20 4 2 0 t t 12/77

Agumenting Path Augmenting the flow along a path P from s to t in G f Augment ( P ) 1: b ← min e ∈ P c f ( e ) 2: for every ( u, v ) ∈ P do if ( u, v ) is a forward edge then 3: f ( u, v ) ← f ( u, v ) + b 4: ⊲ ( u, v ) is a backward edge else 5: f ( v, u ) ← f ( v, u ) − b 6: 7: return f 13/77

Example for Augmenting Along a Path a a 0/1 0/1 1/1 1 1 1 s t s t / 1 1 0/1 0/1 1/1 1 1 b b 14/77

Ford-Fulkerson’s Method Ford-Fulkerson ( G, s, t, c ) 1: let f ( e ) ← 0 for every e in G 2: while there is a path from s to t in G f do let P be any simple path from s to t in G f 3: f ← augment ( f, P ) 4: 5: return f 15/77

Ford-Fulkerson: Example G f G s s 1 8/16 1 / 2 8 1 3 8 11 0/4 4 a b a b 9 / 8/12 11/14 0 9 1 4 8 3 1 7/7 7 c d c d 1 15 5 4/4 / 4 2 5 0 t t 16/77

Outline Network Flow 1 Ford-Fulkerson Method 2 Correctness of Ford-Fulkerson’s Method Running Time of Ford-Fulkerson-Type Algorithm Applications of Network-Flow 3 Single Commodity Flow with Multiple Sources and Sinks s - t Edge-Disjoint Paths Problem Bipartite Matching Problem 17/77

Correctness of Ford-Fulkerson’s Method The procedure augment ( f, P ) maintains the two conditions: 1 for every e ∈ E : 0 ≤ f ( e ) ≤ c ( e ) (capacity conditions) for every v ∈ V \ { s, t } : � � f ( e ) = f ( e ) . (conservation conditions) e into v e out of v When Ford-Fulkerson’s Method terminates, val ( f ) is 2 maximized Ford-Fulkerson’s Method will terminate 3 18/77

Correctness of Ford-Fulkerson’s Method The procedure augment ( f, P ) maintains the two conditions: 1 for every e ∈ E : 0 ≤ f ( e ) ≤ c ( e ) (capacity conditions) for every v ∈ V \ { s, t } : � � f ( e ) = f ( e ) . (conservation conditions) e into v e out of v When Ford-Fulkerson’s Method terminates, val ( f ) is 2 maximized Ford-Fulkerson’s Method will terminate 3 19/77

for every e ∈ E : 0 ≤ f ( e ) ≤ c ( e ) (capacity conditions) for every v ∈ V \ { s, t } : � � f ( e ) = f ( e ) . (conservation conditions) e into v e out of v + b − b + b + b − b t s for an edge e correspondent to a forward edge : b ≤ c ( e ) − f ( e ) = ⇒ f ( e ) + b ≤ c ( e ) for an edge e correspondent to a backward edge : b ≤ f ( e ) = ⇒ f ( e ) − b ≥ 0 20/77

Correctness of Ford-Fulkerson’s Method The procedure augment ( f, P ) maintains the two conditions: 1 for every e ∈ E : 0 ≤ f ( e ) ≤ c ( e ) (capacity conditions) for every v ∈ V \ { s, t } : � � f ( e ) = f ( e ) . (conservation conditions) e into v e out of v When Ford-Fulkerson’s Method terminates, val ( f ) is 2 maximized Ford-Fulkerson’s Method will terminate 3 21/77

Def. An s - t cut of G = ( V, E ) is a pair ( S ⊆ V, T = V \ S ) such that s ∈ S and t ∈ T . Def. The cut value of an s - t cut is � c ( S, T ) := c ( e ) . e =( u,v ) ∈ E : u ∈ S,v ∈ T Def. Given an s - t flow f and an s - t cut ( S, T ) , the net flow sent from S to T is � � f ( e ) − f ( S, T ) := f ( e ) . e =( u,v ) ∈ E : u ∈ S,v ∈ T e =( u,v ) ∈ E : u ∈ T,v ∈ S 22/77

T S 9/14 b d 4/4 3 c ( S, T ) = 14 + 12 = 26 1 / 5 4 7 4 G s / / t / 0 9 5 6/16 f ( S, T ) = 9 + 6 − 4 = 11 7/20 a c 6/12 Obs. f ( S, T ) ≤ c ( S, T ) s - t cut ( S, T ) . Obs. f ( S, T ) = val ( f ) for any s - t flow f and any s - t cut ( S, T ) . Coro. val ( f ) ≤ s - t cut ( S,T ) c ( S, T ) for every s - t flow f. min 23/77

Coro. val ( f ) ≤ s - t cut ( S,T ) c ( S, T ) for every s - t flow f. min We will prove Main Lemma The flow f found by the Ford-Fulkerson’s Method satisfies val ( f ) = c ( S, T ) for some s - t cut ( S, T ) . Corollary and Main Lemma implies Maximum Flow Minimum Cut Theorem sup val ( f ) = s - t cut ( S,T ) c ( S, T ) . min s - t flow f 24/77

Maximum Flow Minimum Cut Theorem sup val ( f ) = s - t cut ( S,T ) c ( S, T ) . min s - t flow f 12/12 a c 19/20 12/16 0/4 7 s t 0/9 / 7 11/13 4/4 b d 11/14 25/77

Main Lemma The flow f found by the Ford-Fulkerson’s Method satisfies val ( f ) = c ( S, T ) for some s - t cut ( S, T ) . Proof of Main Lemma. When algorithm terminates, no path from s to t in G f , What can we say about G f ? There is a s - t cut ( S, T ) , such that there are no edges from S to T For every e = ( u, v ) ∈ E, u ∈ S, v ∈ T , we have f ( e ) = c ( e ) For every e = ( u, v ) ∈ E, u ∈ T, v ∈ S , we have f ( e ) = 0 � Thus, val ( f ) = f ( S, T ) = f ( e ) − e =( u,v ) ∈ E,u ∈ S,v ∈ T � � f ( e ) = c ( e ) = c ( S, T ) . e =( u,v ) ∈ E,u ∈ T,v ∈ S e =( u,v ) ∈ E,u ∈ S,v ∈ T 26/77

Correctness of Ford-Fulkerson’s Method The procedure augment ( f, P ) maintains the two conditions: 1 for every e ∈ E : 0 ≤ f ( e ) ≤ c ( e ) (capacity conditions) for every v ∈ V \ { s, t } : � � f ( e ) = f ( e ) . (conservation conditions) e into v e out of v When Ford-Fulkerson’s Method terminates, val ( f ) is 2 maximized Ford-Fulkerson’s Method will terminate 3 27/77

Ford-Fulkerson’s Method will Terminate Intuition: In every iteration, we increase the flow value by some amount There is a maximum flow value So the algorithm will finally reach the maximum value However, the algorithm may not terminate if some capacities are irrational numbers. (“Pathological cases”) 28/77

Lemma Ford-Fulkerson’s Method will terminate if all capacities are integers. Proof. The maximum flow value is finite (not ∞ ). In every iteration, we increase the flow value by at least 1. So the algorithm will terminate. Integers can be replaced by rational numbers. 29/77

Correctness of Ford-Fulkerson’s Method The procedure augment ( f, P ) maintains the two conditions: 1 for every e ∈ E : 0 ≤ f ( e ) ≤ c ( e ) (capacity conditions) for every v ∈ V \ { s, t } : � � f ( e ) = f ( e ) . (conservation conditions) e into v e out of v When Ford-Fulkerson’s Method terminates, val ( f ) is 2 maximized Ford-Fulkerson’s Method will terminate 3 30/77

Outline Network Flow 1 Ford-Fulkerson Method 2 Correctness of Ford-Fulkerson’s Method Running Time of Ford-Fulkerson-Type Algorithm Applications of Network-Flow 3 Single Commodity Flow with Multiple Sources and Sinks s - t Edge-Disjoint Paths Problem Bipartite Matching Problem 31/77