1. Chapter 27 Maximum Flow Maximum Flow Problem The Ford-Fulkerson method m bipartite matching

2. Background: Just as we can model a road map as a directed graph in order to find the shortest path from one point to another, we can also interpret a directed graph as a “flow network” and use it to answer questions about material coursing through a system from a source,where the material is produced, to a sink, where it is consumed.

3. Flow networks: A flow network G=(V,E) is a directed graph in which each edge (u,v)  E has a nonnegative capacity c(u,v)>=0.If (u,v)  E, we assume that c(u,v)=0.We distinguish two vertices in a flow network:a source s and a sink t. st 16 12 20 10 4 9 7 4 13 14

4. Flow: Let G=(V,E) be a flow network (with capacity function c). Let s be the source of the network, and let t be the sink.A flow in G is a real-valued function f:V*V R that satisfies the following three properties: Capacity constraint: For all u,v  V, we require f(u,v) <= c( u,v). Skew symmetry: For all u,v  V,we require f(u,v)=-f(v,u). Flow conservation: For all u  V-{s,t}, we require  f(u,v) =0 v  V

5. Net flow and value of a flow f: The quantity f (u,v), which can be positive or negative, is called the net flow from vertex u to vertex v. The value of a flow is defined as

6. st 11/16 12/12 15/20 10 1/44/97/7 4/48/13 11/14 A flow f in G with value.

7. v1 v2 104 v1 v2 8/104 v1 v2 8/103/4 v1 v2 5/104 v1 v2 102/4 (a)(b)(c)(d)(e) Cancellation.(a) Vertices v1 and v2,with c(v1,v2)=10 and c(v2,v1) =4.(b)How we indicate the net flow 8 from v1 to v2. (c) An additional shipment of 3 is made from v2 to v1. (d) By cancelling flow going in opposite directions,we can represent the situation in (c) with positive net flow in one direction only.(e) Another 7 is shipped from v2 to v1.

8. Maximum-flow problem: Given a flow network G with source s and sink t Find a flow of maximum value from s to t. How to solve it efficiently?

9. The Ford-Fulkerson method: This section presents the Ford-Fulkerson method for solving the maximum-flow problem.We call it a “method” rather than an “algorithm” because it encompasses several implementations with different running times.The Ford- Fulkerson method depends on three important ideas that transcend the method and are relevant to many flow algorithms and problems: residual networks,augmenting paths,and cuts. These ideas are essential to the important max-flow min-cut theorem,which characterizes the value of maximum flow in terms of cuts of the flow network.

10. Continue: FORD-FULKERSON-METHOD(G,s,t) initialize flow f to 0 while there exists an augmenting path p do augment flow f along p return f

11. Residual networks: Intuitively,given a flow network and a flow, the residual network consists of edges that can admit more net flow.More formally,suppose that we have a flow network G=(V,E) with source s and sink t.Let f be a flow in G, and consider a pair of vertices u,v  V.The amount of additional net flow we can push from u to v before exceeding the capacity c(u,v) is the residual capacity of (u,v), given by: c f (u,v)=c(u,v)-f(u,v) Given a flow network G=(V,E) and a flow f,the residual network of G induced by f is G f =(V,E f ),where E f ={(u,v)  V*V: c f (u,v) > 0}

12. Continue: Lemma 27.2 Let G=(V,E) be a flow network with source s and sink t, and let f be a flow in G.Let G f be the residual network of G induced by f,and let f’ be a flow in G f.Then, the flow sum f+f’ is a flow in G with value.

13. Augmenting paths: Given a flow network G=(V,E) and a flow f, an augmenting path is a simple path from s to t in the residual network G f. We call the maximum amount of net flow that we can ship along the edges of an augmenting path p the residual capacity of p, given by c f (p)=min{c f (u,v):(u,v) is on p}.

14. Example: (a)The flow network G and flow f. (b)The residual network G f with augmenting path p shaded; its residual capacity is c f (p)=c(v 2,v 3 )=4. (c) The flow in G that results from augmenting along path p by its residual capacity 4. (d) The residual network induced by the flow in (c).

15. 4/9 s v2v2 v4v4 t v3v3 v1v1 11/16 8/13 12/12 101/4 15/20 4/4 7/7 11/14 5 s v2v2 v4v4 t v3v3 v1v1 5 8 12 11 3 5 4 7 4 3 5 15 9 s v2v2 v4v4 t v3v3 v1v1 11/16 12/13 12/12 101/4 19/20 4/4 7/7 11/14 11 7 s v2v2 v4v4 t v3v3 v1v1 5 12 11 3 1 4 9 3 1 19 (a) (b) (c) (d)

16. Continue: Lemma 27.3 Let G=(V,E) be a flow network, let f be a flow in G,and let p be an augmenting path in G f.Define a function f p :V*V R by if (u,v) is on p, f p (u,v)= if (v,u) is on p, otherwise. Then, f p is a flow in G f with value.

17. Continue: Corollary 27.4 Let G=(V,E) be a flow network, let f be a flow in G,and let p be an augmenting path in G f. Let f p be defined as in Lemma 27.3.Define a function f’:V*V R by f’=f+f p. Then, f’ is a flow in G’ with value.

18. Cuts of flow networks: The Ford-Fulkerson method repeatedly augments the flow along augmenting paths until a maximum flow has been found.The max-flow min-cut theorem,which we shall prove shortly, tells us that a flow is maximum if and only if its residual network contains no augmenting path. A cut (S,T) of flow network G=(V,E) is a partition of V into S and T=V-S such that s  S and t  T.If f is a flow, then the net flow across the cut (S,T) is defined to be f(S,T).The capacity of the cut (S,T) is c(S,T).

19. s v2 v1 t v4 v3 11/16 8/13 101/4 12/12 11/14 4/9 15/20 4/4 7/7 S T A cut (S,T) in the flow network G,where S={s,v1,v2} and T={v3,v4,t}.The vertices in S are black,and the vertices in T are white.The net flow across (S,T) is f(S,T)=19,and the capacity is c(S,T)=26.

20. Continue: Lemma 27.5 Let f be a flow in a flow network G with source s and sink t,and let (S,T) be a cut of G.Then, the net flow across (S,T) is f(S,T)=.

21. Continue: Corollary 27.6 The value of any flow f in a flow network G is bounded from above by the capacity of any cut of G.

22. Continue: Theorem 27.7(Max-flow min-cut theorem) If f is a flow in a flow network G=(V,E) with source s and sink t,then the following conditions are equivalent: f is a maximum flow in G; The residual network G f contains no augmenting paths; =c(S,T) for some cut (S,T) of G.

23. The basic Ford-Fulkerson algorithm: FORD-FULKERSON(G,s,t) for each edge (u,v)  E[G] do f[u,v] 0 f[v,u] 0 while there exists a path p from s to t in the residual network G f do c f (p) min{c f (u,v): (u,v) is in p} for each edge (u,v) in p do f[u,v] f[u,v]+c f (p) f[v,u] -f[u,v]

24. Example: The execution of the basic Ford-Fulkerson algorithm.(a)-(d) Successive iterations of the while loop. The left side of each part shows the residual network G f from line 4 with a shaded augmenting path p.The right side of each part shows the new flow f that results from adding f p to f.The residual network in (a) is the input network G.(e) The residual network at the last while loop test.It has no augmenting paths,and the flow f shown in (d) is therefore a maximum flow.

25. s 20 7 9 v2v2 v4v4 t v3v3 v1v1 16 13 12 10 4 4 14 4/9 s v2v2 v4v4 t v3v3 v1v1 4/16 13 4/12 10 4 20 4/4 7 4/14 (a)

26. 5 s v2v2 v4v4 t v3v3 v1v1 12 138 10 4 20 4 7 4 4 4 4 10 4/9 s v2v2 v4v4 t v3v3 v1v1 11/16 13 4/12 7/10 4 7/20 4/4 7/7 11/14 (b)

27. 5 s v2v2 v4v4 t v3v3 v1v1 5 13 8 311 4 7 4 4 3 13 7 4/9 s v2v2 v4v4 t v3v3 v1v1 11/16 8/13 12/12 101/4 15/20 4/4 7/7 11/14 (c)

28. 5 s v2v2 v4v4 t v3v3 v1v1 5 8 113 4 7 12 4 3 5 15 5 9 s v2v2 v4v4 t v3v3 v1v1 11/16 12/13 12/12 101/4 19/20 4/4 7/7 11/14 (d)

29. s v2v2 v4v4 t v3v3 v1v1 5 12 11 3 1 4 9 3 1 19 7 (e)

30. Time complexity: If each c(e) is an integer, then time complexity is O(|E|f*), where f* is the maximum flow. Reason: each time the flow is increased by at least one. This might not be a polynomial time algorithm since f* can be represented by log (f*) bits. So, the input size might be log(f*).

31. The Edmonds-Karp algorithm Find the augmenting path using breadth-first search. Breadth-first search gives the shortest path for graphs with edge length 1. Time complexity of Edmonds-Karp algorithm is O(VE 2 ).

32. Lemma: With each flow augmentation, the shortest-path distance between s and u for any u increases monotonically. Proof: (Fun part, not required) Prove it by contradiction. Let f’ is obtained from f by an augmentation and d(f’,s,v)<d(f, s, v). Assume that v is a node such that d(f’, s, v) is the minimum satisfying d(f’,s,v)<d(f, s, v).

33. Consider u with d(f’, s, u)=d(f’, s, v)-1. Based on the choice of v, we know that d(f’, s, y)>=d(f, s, u). Then (u, v) is not in e(

34. Maximum bipartite matching: Given an undirected graph G=(V,E), a matching is a subset of edges M  E such that for all vertices v  V,at most one edge of M is incident on v.We say that a vertex v  V is matched by matching M if some edge in M is incident on v;otherwise, v is unmatched. A maximum matching is a matching of maximum cardinality,that is, a matching M such that for any matching M’, we have.

35. L R A bipartite graph G=(V,E) with vertex partition V=L  R.(a)A matching with cardinality 2.(b) A maximum matching with cardinality 3. (a) LR (b)

36. Finding a maximum bipartite matching: We define the corresponding flow network G’=(V’,E’) for the bipartite graph G as follows. Let the source s and sink t be new vertices not in V, and let V’=V  {s,t}.If the vertex partition of G is V=L  R, the directed edges of G’ are given by E’={(s,u):u  L}  {(u,v):u  L,v  R,and (u,v)  E}  {(v,t):v  R}.Finally, we assign unit capacity to each edge in E’. We will show that a matching in G corresponds directly to a flow in G’s corresponding flow network G’. We say that a flow f on a flow network G=(V,E) is integer-valued if f(u,v) is an integer for all (u,v)  V*V.

37. LR (a) LR (b) s t (a)The bipartite graph G=(V,E) with vertex partition V=L  R. A maximum matching is shown by shaded edges.(b) The corresponding flow network shown.Each edge has unit capacity.Shaded edges have a flow of 1,and all other edges carry no flow.

38. Continue: Lemma 27.10 Let G=(V,E) be a bipartite graph with vertex partition V=L  R,and let G’=(V’,E’) be its corresponding flow network.If M is a matching in G, then there is an integer- valued flow f in G’ with value.Conversely, if f is an integer-valued flow in G’,then there is a matching M in G with cardinality.

39. The End: