1 Chapter 7 Network Flow Slides by Kevin Wayne. Copyright © 2005 Pearson-Addison Wesley. All rights reserved.

7.5 Bipartite Matching

3 Matching. n Input: undirected graph G = (V, E). n M  E is a matching if each node appears in at most one edge in M. n Max matching: find a max cardinality matching. Matching

4 Bipartite Matching Bipartite matching. n Input: undirected, bipartite graph G = (L  R, E). – bipartite graph (or bigraph) is a graph whose vertices can be divided into two disjoint sets (U and V) such that every edge connects a vertex in U to one in V. n M  E is a matching if each node appears in at most one edge in M. n Max matching: find a max cardinality matching ' 3' 5' 2 4 2' 4' matching 1-2', 3-1', 4-5' RL

5 Bipartite Matching Bipartite matching. n Input: undirected, bipartite graph G = (L  R, E). n M  E is a matching if each node appears in at most edge in M. n Max matching: find a max cardinality matching ' 3' 5' 2 4 2' 4' RL max matching 1-1', 2-2', 3-3' 4-4'

6 Max flow formulation. n Create digraph G' = (L  R  {s, t}, E' ). n Direct all edges from L to R, and assign infinite (or unit) capacity. n Add source s, and unit capacity edges from s to each node in L. n Add sink t, and unit capacity edges from each node in R to t. Bipartite Matching s ' 3' 5' t 2 4 2' 4' 1 1  RL G'

7 Theorem. Max cardinality matching in G = value of max flow in G'. Pf.  n Given max matching M of cardinality k. n Consider flow f that sends 1 unit along each of k paths. n f is a flow, and has cardinality k. ▪ Bipartite Matching: Proof of Correctness s ' 3' 5' t 2 4 2' 4' 11  ' 3' 5' 2 4 2' 4' G' G

8 Theorem. Max cardinality matching in G = value of max flow in G'. Pf.  n Let f be a max flow in G' of value k. n Integrality theorem  k is integral and can assume f is 0-1. n Consider M = set of edges from L to R with f(e) = 1. – each node in L and R participates in at most one edge in M – |M| = k: consider cut (L  s, R  t) ▪ Bipartite Matching: Proof of Correctness ' 3' 5' 2 4 2' 4' G s ' 3' 5' t 2 4 2' 4' 11  G'

9 Def. A matching M  E is perfect if each node appears in exactly one edge in M. Q. When does a bipartite graph have a perfect matching? Structure of bipartite graphs with perfect matchings. n Clearly we must have |L| = |R|. n What other conditions are necessary? n What conditions are sufficient? Perfect Matching

10 Notation. Let S be a subset of nodes, and let N(S) be the set of nodes adjacent to nodes in S. Observation. If a bipartite graph G = (L  R, E), has a perfect matching, then |N(S)|  |S| for all subsets S  L. Pf. Each node in S has to be matched to a different node in N(S). Perfect Matching No perfect matching: S = { 2, 4, 5 } N(S) = { 2', 5' } ' 3' 5' 2 4 2' 4' L R

11 Marriage Theorem. [Frobenius 1917, Hall 1935] Let G = (L  R, E) be a bipartite graph with |L| = |R|. Then, G has a perfect matching iff |N(S)|  |S| for all subsets S  L. Pf.  This was the previous observation. Marriage Theorem ' 3' 5' 2 4 2' 4' L R No perfect matching: S = { 2, 4, 5 } N(S) = { 2', 5' }.

12 Pf.  Suppose G does not have a perfect matching. n Formulate as a max flow problem and let (A, B) be min cut in G'. n By max-flow min-cut, cap(A, B) < | L |. n Define L A = L  A, L B = L  B, R A = R  A. n cap(A, B) = | L B | + | R A |. n Since min cut can't use  edges: N(L A )  R A. n |N(L A )|  | R A | = cap(A, B) - | L B | < | L | - | L B | = | L A |. – Choose S = L A, conflict with assumption ▪ Proof of Marriage Theorem L A = {2, 4, 5} L B = {1, 3} R A = {2', 5'} N(L A ) = {2', 5'} s ' 3' 5' t 2 4 4' 1  2' A  G' 

13 Which max flow algorithm to use for bipartite matching? n Generic augmenting path: O(m val(f*) ) = O(mn). n Capacity scaling: O(m 2 log C ) = O(m 2 ). n Shortest augmenting path: O(m n 1/2 ). Non-bipartite matching. n Structure of non-bipartite graphs is more complicated, but well-understood. [Tutte-Berge, Edmonds-Galai] n Blossom algorithm: O(n 4 ). [Edmonds 1965] n Best known: O(m n 1/2 ). [Micali-Vazirani 1980] Bipartite Matching: Running Time

7.6 Disjoint Paths

15 Disjoint path problem. Given a digraph G = (V, E) and two nodes s and t, find the max number of edge-disjoint s-t paths. Def. Two paths are edge-disjoint if they have no edge in common. Ex: communication networks. s Edge Disjoint Paths t

16 Disjoint path problem. Given a digraph G = (V, E) and two nodes s and t, find the max number of edge-disjoint s-t paths. Def. Two paths are edge-disjoint if they have no edge in common. Ex: communication networks. s Edge Disjoint Paths t

17 Max flow formulation: assign unit capacity to every edge. Theorem. Max number edge-disjoint s-t paths equals max flow value. Pf.  n Suppose there are k edge-disjoint paths P 1,..., P k. n Set f(e) = 1 if e participates in some path P i ; else set f(e) = 0. n Since paths are edge-disjoint, f is a flow of value k. ▪ Edge Disjoint Paths st

18 Max flow formulation: assign unit capacity to every edge. Theorem. Max number edge-disjoint s-t paths equals max flow value. Pf.  n Suppose max flow value is k. n Integrality theorem  there exists 0-1 flow f of value k. n Consider edge (s, u) with f(s, u) = 1. – by conservation, there exists an edge (u, v) with f(u, v) = 1 – continue until reach t, always choosing a new edge n Produces k (not necessarily simple) edge-disjoint paths. ▪ Edge Disjoint Paths st can eliminate cycles to get simple paths if desired

19 Network connectivity. Given a digraph G = (V, E) and two nodes s and t, find min number of edges whose removal disconnects t from s. Def. A set of edges F  E disconnects t from s if every s-t path uses at least one edge in F. Network Connectivity s t

20 Edge Disjoint Paths and Network Connectivity Theorem. [Menger 1927] The max number of edge-disjoint s-t paths is equal to the min number of edges whose removal disconnects t from s. Pf.  n Suppose the removal of F  E disconnects t from s, and |F| = k. n Every s-t path uses at least one edge in F. Hence, the number of edge-disjoint paths is at most k. ▪ s t s t

21 Disjoint Paths and Network Connectivity Theorem. [Menger 1927] The max number of edge-disjoint s-t paths is equal to the min number of edges whose removal disconnects t from s. Pf.  n Suppose max number of edge-disjoint paths is k. n Then max flow value is k. n Max-flow min-cut  cut (A, B) of capacity k. n Let F be set of edges going from A to B. n |F| = k and disconnects t from s. ▪ s t s t A

7.7 Extensions to Max Flow

23 Circulation with Demands Circulation with demands. n Directed graph G = (V, E). n Edge capacities c(e), e  E. n Node supply and demands d(v), v  V. Def. A circulation is a function that satisfies: n For each e  E:0  f(e)  c(e) (capacity) n For each v  V: (conservation) Circulation problem: given (V, E, c, d), does there exist a circulation? demand if d(v) > 0; supply if d(v) < 0; transshipment if d(v) = 0

24 Necessary condition: sum of supplies = sum of demands. Pf. Sum conservation constraints for every demand node v flow Circulation with Demands capacity demand supply

25 Circulation with Demands Max flow formulation. G: supply demand

26 Circulation with Demands Max flow formulation. n Add new source s and sink t. n For each v with d(v) < 0, add edge (s, v) with capacity -d(v). n For each v with d(v) > 0, add edge (v, t) with capacity d(v). n Claim: G has circulation iff G' has max flow of value D. G': supply s t saturates all edges leaving s and entering t demand

27 Circulation with Demands Integrality theorem. If all capacities and demands are integers, and there exists a circulation, then there exists one that is integer-valued. Pf. Follows from max flow formulation and integrality theorem for max flow. Characterization. Given (V, E, c, d), there does not exists a circulation iff there exists a node partition (A, B) such that  v  B d v > cap(A, B) Pf idea. Look at min cut in G'. demand by nodes in B exceeds supply of nodes in B plus max capacity of edges going from A to B

28 Circulation with Demands and Lower Bounds Feasible circulation. n Directed graph G = (V, E). n Edge capacities c(e) and lower bounds (e), e  E. n Node supply and demands d(v), v  V. Def. A circulation is a function that satisfies: n For each e  E: (e)  f(e)  c(e) (capacity) n For each v  V: (conservation) Circulation problem with lower bounds. Given (V, E,, c, d), does there exists a a circulation?

29 Circulation with Demands and Lower Bounds Idea. Model lower bounds with demands. n Send (e) units of flow along edge e. n Update demands of both endpoints. Theorem. There exists a circulation in G iff there exists a circulation in G'. If all demands, capacities, and lower bounds in G are integers, then there is a circulation in G that is integer-valued. Pf sketch. f(e) is a circulation in G iff f'(e) = f(e) - (e) is a circulation in G'. vw [2, 9] lower bound upper bound vw d(v)d(w) d(v) + 2 d(w) - 2 G G' 7 capacity

7.8 Survey Design

31 Survey Design Survey design. n Design survey asking n 1 consumers about n 2 products. n Can only survey consumer i about product j if they own it. n Ask consumer i between c i and c i ' questions. n Ask between p j and p j ' consumers about product j. Goal. Design a survey that meets these specs, if possible. Bipartite perfect matching. Special case when c i = c i ' = p i = p i ' = 1. one survey question per product

32 Survey Design Algorithm. Formulate as a circulation problem with lower bounds. n Include an edge (i, j) if consumer i owns product j. n Integer circulation  feasible survey design. – Flow on (t,s) is equal to overall # of questions asked – All nodes have demand of 0 s ' 3' 5' t 2 4 2' 4' [c 1, c 1 '] [0, 1] consumers [p 1, p 1 '] [0,  ] products [  ci,  ci’]

7.10 Image Segmentation

34 Image Segmentation Image segmentation. n Central problem in image processing. n Divide image into coherent regions. Ex: Three people standing in front of complex background scene. Identify each person as a coherent object.

35 Image Segmentation Foreground / background segmentation. n Label each pixel in picture as belonging to foreground or background. n V = set of pixels, E = pairs of neighboring pixels. n a i  0 is likelihood pixel i in foreground. n b i  0 is likelihood pixel i in background. n p ij  0 is separation penalty for labeling one of i and j as foreground, and the other as background. Goals. n Accuracy: if a i > b i in isolation, prefer to label i in foreground. n Smoothness: if many neighbors of i are labeled foreground, we should be inclined to label i as foreground. n Find partition (A, B) that maximizes: foreground background

36 Image Segmentation Formulate as min cut problem. n Maximization. n No source or sink. n Undirected graph. Turn into minimization problem. n Maximizing is equivalent to minimizing n or alternatively

37 Image Segmentation Formulate as min cut problem. n G' = (V', E'). n Add source to correspond to foreground; add sink to correspond to background n Use two anti-parallel edges instead of undirected edge. st p ij ij ajaj G' bibi

38 Image Segmentation Consider min cut (A, B) in G'. n A = foreground. n Precisely the quantity we want to minimize. G' st ij A if i and j on different sides, p ij counted exactly once p ij bibi ajaj

7.11 Project Selection

40 Project Selection Projects with prerequisites. n Set P of possible projects. Project v has associated revenue p v. – some projects generate money: create interactive e-commerce interface, redesign web page – others cost money: upgrade computers, get site license n Set of prerequisites E. If (v, w)  E, can't do project v and unless also do project w. n A subset of projects A  P is feasible if the prerequisite of every project in A also belongs to A. Project selection. Choose a feasible subset of projects to maximize revenue. can be positive or negative

41 Project Selection: Prerequisite Graph Prerequisite graph. n Include an edge from v to w if can't do v without also doing w. n {v, w, x} is feasible subset of projects. n {v, x} is infeasible subset of projects. n Easy to find feasible set, but what about max revenue? v w x v w x feasibleinfeasible

42 Min cut formulation. n Assign capacity  to all prerequisite edge. n Add edge (s, v) with capacity -p v if p v > 0. n Add edge (v, t) with capacity -p v if p v < 0. n For notational convenience, define p s = p t = 0. st -p w u v w x yz Project Selection: Min Cut Formulation  pvpv -p x      pypy pupu -p z 

43 Claim. (A, B) is min cut iff A  { s } is optimal set of projects. n Infinite capacity edges ensure A  { s } is feasible. n Max revenue because: s t -p w u v w x yz Project Selection: Min Cut Formulation pvpv -p x pypy pupu    A + and – all nodes in A that have positive p

44 Open-pit mining. (studied since early 1960s) n Blocks of earth are extracted from surface to retrieve ore. n Each block v has net value p v = value of ore - processing cost. n Can't remove block v before w or x. Project selection application#1: Open Pit Mining v xw

7.12 Baseball Elimination

46 Baseball Elimination Which teams have a chance of finishing the season with most wins? n Montreal eliminated since it can finish with at most 80 wins, but Atlanta already has 83. n w i + r i < w j  team i eliminated. n Only reason sports writers appear to be aware of. n Sufficient, but not necessary! Team i Against = r ij Wins w i To play r i Losses l i AtlPhiNYMon Montreal New York Philly Atlanta

47 Baseball Elimination Which teams have a chance of finishing the season with most wins? n Philly can win 83, but still eliminated... n If Atlanta loses a game, then some other team wins one. Remark. Answer depends not just on how many games already won and left to play, but also on whom they're against. Team i Against = r ij Wins w i To play r i Losses l i AtlPhiNYMon Montreal New York Philly Atlanta

48 Baseball Elimination Baseball elimination problem. n Set of teams S. n Distinguished team s  S. n Team x has won w x games already. n Teams x and y play each other r xy additional times. n Is there any outcome of the remaining games in which team s finishes with the most (or tied for the most) wins?

49 Can team 3 finish with most wins? n Assume team 3 wins all remaining games  w 3 + r 3 wins. n Find a solution of remaining games so that all teams have  w 3 + r 3 wins. Baseball Elimination: Max Flow Formulation s t r 24 = 7  w 3 + r 3 - w 4 team 4 can still win this many more games games left  game nodes team nodes

50 Theorem. Team 3 is not eliminated iff max flow saturates all edges leaving source s. n Integrality theorem  each remaining game between x and y added to number of wins for team x or team y. n Capacity on (x, t) edges ensure no team wins too many games. Baseball Elimination: Max Flow Formulation s t  team 4 can still win this many more games games left  game nodes team nodes r 24 = 7 w 3 + r 3 - w 4