Network Flow II Lecture 13 October 10, 2013 - PowerPoint PPT Presentation

CS 573: Algorithms, Fall 2013 Accountability Network Flow II Lecture 13 October 10, 2013 http://www.cs.berkeley.edu/˜jrs/ Calvin Sariel (UIUC) CS573 1 Fall 2013 1 / 24 Sariel (UIUC) CS573 2 Fall 2013 2 / 24 Accountability Ford Fulkerson People that do not know maximum flows: essentially everybody. 1 algFordFulkerson ( G , s , t ) Initialize flow f to zero Average salary on earth ¡ $ 5 , 000 2 while ∃ path π from s to t in G f do People that know maximum flow – most of them work in 3 � � � c f ( π ) ← min c f ( u , v ) � ( u → v ) ∈ π � programming related jobs and make at least $ 10 , 000 a year. for ∀ ( u → v ) ∈ π do Salary of people that learned maximum flows: > $ 10 , 000 f ( u , v ) ← f ( u , v ) + c f ( π ) 4 f ( v , u ) ← f ( v , u ) − c f ( π ) Salary of people that did not learn maximum flows: < $ 5 , 000 . 5 Salary of people that know Latin: 0 (unemployed). 6 Lemma Conclusion If the capacities on the edges of G are integers, then algFordFulkerson runs in O ( m | f ∗ | ) time, where | f ∗ | is the amount Thus, by just learning maximum flows (and not knowing of flow in the maximum flow and m = | E ( G ) | . Latin) you can double your future salary! Sariel (UIUC) CS573 3 Fall 2013 3 / 24 Sariel (UIUC) CS573 4 Fall 2013 4 / 24

Proof of Lemma... Integrality theorem Proof. Observation (Integrality theorem) Observe that the algFordFulkerson method performs only If the capacity function c takes on only integral values, then the subtraction, addition and min operations. Thus, if it finds an maximum flow f produced by the algFordFulkerson method has the augmenting path π , then c f ( π ) must be a positive integer number. property that | f | is integer-valued. Moreover, for all vertices u and Namely, c f ( π ) ≥ 1 . Thus, | f ∗ | must be an integer number (by v , the value of f ( u , v ) is also an integer. induction), and each iteration of the algorithm improves the flow by at least 1 . It follows that after | f ∗ | iterations the algorithm stops. Each iteration takes O ( m + n ) = O ( m ) time, as can be easily verified. Sariel (UIUC) CS573 5 Fall 2013 5 / 24 Sariel (UIUC) CS573 6 Fall 2013 6 / 24 Edmonds-Karp algorithm The disappearing/reappearing lemma Edmonds-Karp : modify algFordFulkerson so it always returns the Lemma shortest augmenting path in G f . During execution Edmonds-Karp , edge ( u → v ) might Definition disappear/reappear from G f at most n / 2 times, n = | V ( G ) | . For a flow f , let δ f ( v ) be the length of the shortest path from the Proof. source s to v in the residual graph G f . Each edge is considered to be iteration when edge ( u → v ) disappears. of length 1 . 1 ( u → v ) appeared in augmenting path π . 2 Assume the following key lemma: Fully utilized: c f ( π ) = c f ( uv ) . f flow in beginning of iter. 3 Lemma till ( u → v ) “magically” reappears. 4 ∀ v ∈ V \ { s , t } the function δ f ( v ) increases. ... augmenting path σ that contained the edge ( v → u ) . 5 g : flow used to compute σ . 6 We have: δ g ( u ) = δ g ( v ) + 1 ≥ δ f ( v ) + 1 = δ f ( u ) + 2 7 distance of s to u had increased by 2 . QED. 8 Sariel (UIUC) CS573 7 Fall 2013 7 / 24 Sariel (UIUC) CS573 8 Fall 2013 8 / 24

Comments... Edmonds-Karp # of iterations δ ? ( u ) might become infinity. 1 Lemma u is no longer reachable from s . 2 Edmonds-Karp handles O ( nm ) augmenting paths before it stops. By monotonicity, the edge ( u → v ) would never appear again. � nm 2 � 3 Its running time is O , where n = | V ( G ) | and m = | E ( G ) | . Observation Proof. For every iteration/augmenting path of Edmonds-Karp algorithm, Every edge might disappear at most n / 2 times. 1 at least one edge disappears from the residual graph G ? . At most nm / 2 edge disappearances during execution 2 Edmonds-Karp . In each iteration, by path augmentation, at least one edge 3 disappears. Edmonds-Karp algorithm perform at most O ( mn ) iterations. 4 Computing augmenting path takes O ( m ) time. 5 � nm 2 � Overall running time is O . 6 Sariel (UIUC) CS573 9 Fall 2013 9 / 24 Sariel (UIUC) CS573 10 Fall 2013 10 / 24 Shortest distance increases during Edmonds-Karp Proof continued... execution π = s → · · · → u → v : shortest path in G g from s to v . 1 ( u → v ) ∈ E ( G g ) , and thus δ g ( u ) = δ g ( v ) − 1 . 2 Lemma By choice of v : δ g ( u ) ≥ δ f ( u ) . 3 Edmonds-Karp run on G = ( V , E ) , s , t , then ∀ v ∈ V \ { s , t } , (i) If ( u → v ) ∈ E ( G f ) then the distance δ f ( v ) in G f increases monotonically. δ f ( v ) ≤ δ f ( u ) + 1 ≤ δ g ( u ) + 1 = δ g ( v ) − 1 + 1 = δ g ( v ) . Proof This contradicts our assumptions that δ f ( v ) > δ g ( v ) . By Contradiction. f : flow before (first fatal) iteration. 1 g : flow after. 2 v : vertex s.t. δ g ( v ) is minimal, among all counter example 3 vertices. v : δ g ( v ) is minimal and δ g ( v ) < δ f ( v ) . 4 Sariel (UIUC) CS573 11 Fall 2013 11 / 24 Sariel (UIUC) CS573 12 Fall 2013 12 / 24

Proof continued II Bipartite Matching (ii) f ( u → v ) / ∈ E ( G f ) : 1 π used in computing g from f contains ( v → u ) . 1 s t ( u → v ) reappeared in the residual graph G g (while not being 2 present in G f ). 1 = ⇒ π pushed a flow in the other direction on the edge 3 ( u → v ) . Namely, ( v → u ) ∈ π . 1 Algorithm always augment along the shortest path. By 4 assumption δ g ( v ) < δ f ( v ) , and definition of u : δ f ( u ) = δ f ( v ) + 1 > δ g ( v ) = δ g ( u ) + 1 , = ⇒ δ f ( u ) > δ g ( u ) 5 = ⇒ monotonicity property fails for u . But: δ g ( u ) < δ g ( v ) . A contradiction. Sariel (UIUC) CS573 13 Fall 2013 13 / 24 Sariel (UIUC) CS573 14 Fall 2013 14 / 24 Bipartite matching Computing bipartite matching Theorem Definition Compute maximum bipartite matching in O ( nm ) time. G = ( V , E ) : undirected graph. M ⊆ E : matching if all vertices v ∈ V , at most one edge of M is Proof. incident on v . G : bipartite graph G . ( n vertices and m edges) 1 M is maximum matching if for any matching M ′ : | M | ≥ | M ′ | . Create new graph H with source on left and sink right. 2 M is perfect if it involves all vertices. Direct all edges from left to right. Set all capacities to one. 3 By Integrality theorem, flow in H is 0 / 1 on edges. 4 A flow of value k in H = ⇒ a collection of k vertex disjoint 5 s − t paths = ⇒ matching in G of size k . M : matching of k edge in G , = ⇒ flow of value k in H . 6 Running time of the algorithm is O ( nm ) . Max flow is n , and as 7 such, at most n augmenting paths. Sariel (UIUC) CS573 15 Fall 2013 15 / 24 Sariel (UIUC) CS573 16 Fall 2013 16 / 24

Extension: Multiple Sources and Sinks Proof by figures Question Given a flow network with several sources and sinks, how can we s 1 s 1 t 1 t 1 ∞ compute maximum flow on such a network? s ∞ t ∞ s 2 s 2 ∞ Solution t 2 t 2 The idea is to create a super source, that send all its flow to the old sources and similarly create a super sink that receives all the flow. Clearly, computing flow in both networks in equivalent. Sariel (UIUC) CS573 17 Fall 2013 17 / 24 Sariel (UIUC) CS573 18 Fall 2013 18 / 24