Types of Networks. Internet Telephone Cell Highways Rail Electrical Power Water Sewer Gas

Transcription

1 Flow Networks

2 Network Flows 2

3 Types of Networks Internet Telephone Cell Highways Rail Electrical Power Water Sewer Gas 3

4 Maximum Flow Problem How can we maximize the flow in a network from a source or set of sources to a destination or set of destinations? The problem reportedly rose to prominence in relation to the rail networks of the Soviet Union, during the 1950's. The US wanted to know how quickly the Soviet Union could get supplies through its rail network to its satellite states in Eastern Europe. In addition, the US wanted to know which rails it could destroy most easily to cut off the satellite states from the rest of the Soviet Union. It turned out that these two problems were closely related, and that solving the max flow problem also solves the min cut problem of figuring out the cheapest way to cut off the Soviet Union from its satellites. Source: lbackstrom, The Importance 4 of Algorithms, at

5 Instance: Network Flow A Network is a directed graph G Edges represent pipes that carry flow Each edge (u,v) has a maximum capacity c(u,v) A source node s in which flow arrives A sink node t out which flow leaves Goal: Max Flow 5

6 The Problem Use a graph to model material that flows through conduits. Each edge represents one conduit, and has a capacity, which is an upper bound on the flow rate, in units/time. Can think of edges as pipes of different sizes. Want to compute max rate that we can ship material from a designated source to a designated sink. 6

7 What is a Flow Network? Each edge (u,v) has a nonnegative capacity c(u,v). If (u,v) is not in E, assume c(u,v)=0. We have a source s, and a sink t. Assume that every vertex v in V is on some path from s to t. e.g., c(s,v 1 )=16; c(v 1,s)=0; c(v 2,v 3 )=0 7

8 What is a Flow in a Network? For each edge (u,v), the flow f(u,v) is a real-valued function that must satisfy 3 conditions: Capacity constraint: u, v V, f ( u, v ) c( u, v ) Skew symmetry: u, v V, f ( u, v ) f ( v, u) Flow conservation : u V { s, t }, f ( u, v ) 0 Notes: The skew symmetry condition implies that f(u,u)=0. We show only the positive flows in the flow network. v V 8

9 capacity Example of a Flow: flow capacity f(v 2, v 1 ) = 1, c(v 2, v 1 ) = 4. f(v 1, v 2 ) = -1, c(v 1, v 2 ) = 10. f(v 3, s) + f(v 3, v 1 ) + f(v 3, v 2 ) + f(v 3, v 4 ) + f(v 3, t) = 0 + (-12) (-7) + 15 = 0 9

10 The Value of a flow The value of a flow is given by 10 V v V v t v f v s f f ), ( ), ( This is the total flow leaving s = the total flow arriving in t.

11 Example: f = f(s, v 1 ) + f(s, v 2 ) + f(s, v 3 ) + f(s, v 4 ) + f(s, t) = = 19 f = f(s, t) + f(v 1, t) + f(v 2, t) + f(v 3, t) + f(v 4, t) = = 19 11

12 A flow in a network We assume that there is only flow in one direction at a time. Sending 7 trucks from Edmonton to Calgary and 3 trucks from Calgary to Edmonton has the same net effect as sending 4 trucks from Edmonton to Calgary. 12

13 Multiple Sources Network We have several sources and several targets. Want to maximize the total flow from all sources to all targets. Reduce to max-flow by creating a supersource and a supersink: 13

14 Residual Networks The residual capacity of an edge (u,v) in a network with a flow f is given by: c f ( u, v) c( u, v) f ( u, v) The residual network of a graph G induced by a flow f is the graph including only the edges with positive residual capacity, i.e., G ( V, E ), where E {( u, v) V V : c ( u, v) 0} f f f f 14

15 Example of Residual Network Flow Network: Residual Network: 15

16 Augmenting Paths An augmenting path p is a simple path from s to t on the residual network. We can put more flow from s to t through p. We call the maximum capacity by which we can increase the flow on p the residual capacity of p. c f ( p) min{ c ( u, v) :( u, v) is on p} f 16

17 Augmenting Paths Network: Residual Network: Augmenting path The residual capacity of this augmenting path is 4. 17

18 Computing Max Flow Classic Method: Identify augmenting path Increase flow along that path Repeat 18

19 Ford-Fulkerson Method 19

20 Example Flow(1) No more augmenting paths max flow attained. Residual(1) Flow(2) Residual(2) Cut 20

21 Cuts of Flow Networks A cut ( S, T ) of a flow network is a partition of V into S and T V S such that s S and t T. 21

22 The Net Flow through a Cut (S,T) f ( S, T) f u S, v T ( u, v) f(s,t) = = 19 22

23 The Capacity of a Cut (S,T) c( S, T) u S, v T c( u, v) c(s,t)= = 26 23

24 Augmenting Paths example Capacity of the cut = maximum possible flow through the cut = = 23 Flow(2) The network has a capacity of at most 23. In this case, the network does have a capacity of 23, because this is a minimum cut. 24 cut

25 Net Flow of a Network The net flow across any cut is the same and equal to the flow of the network f. 25

26 Bounding the Network Flow The value of any flow f in a flow network G is bounded from above by the capacity of any cut of G. 26

27 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: 1. f is a maximum flow in G. 2. The residual network G f contains no augmented paths. 3. f = c(s,t) for some cut (S,T) (a min-cut). 27

28 The Basic Ford-Fulkerson Algorithm 28

29 Example augmenting path Original Network Flow Network Resulting Flow = 4 29

30 Example Flow Network Resulting Flow = 4 Residual Network augmenting path Flow Network Resulting Flow = 11 30

31 Example Flow Network Resulting Flow = 11 Residual Network augmenting path Flow Network Resulting Flow = 19 31

32 Example Flow Network Resulting Flow = 19 augmenting path Residual Network Flow Network Resulting Flow = 23 32

33 Example Resulting Flow = 23 No augmenting path: Maxflow=23 Residual Network 33

34 Analysis O(E)? O(E) 34

35 Analysis If capacities are all integer, then each augmenting path raises f by 1. If max flow is f*, then need f* iterations time is O(E f* ). Note that this running time is not polynomial in input size. It depends on f*, which is not a function of V or E. If capacities are rational, can scale them to integers. If irrational, FORD-FULKERSON might never terminate! 35

36 The Basic Ford-Fulkerson Algorithm With time O ( E f* ), the algorithm is not polynomial. This problem is real: Ford-Fulkerson may perform very badly if we are unlucky: f* =2,000,000 36

37 Run Ford-Fulkerson on this example Augmenting Path Residual Network 37

38 Run Ford-Fulkerson on this example Augmenting Path Residual Network 38

39 Run Ford-Fulkerson on this example Repeat 999,999 more times Can we do better than this? 39

40 The Edmonds-Karp Algorithm A small fix to the Ford-Fulkerson algorithm makes it work in polynomial time. Select the augmenting path using breadth-first search on residual network. The augmenting path p is the shortest path from s to t in the residual network (treating all edge weights as 1). Runs in time O(V E 2 ). 40

41 The Edmonds-Karp Algorithm - example The Edmonds-Karp algorithm halts in only 2 iterations on this graph. 41

42 An Application of Max Flow: Maximum Bipartite Matching

43 Maximum Bipartite Matching A bipartite graph is a graph G=(V,E) in which V can be divided into two parts L and R such that every edge in E is between a vertex in L and a vertex in R. e.g. vertices in L represent skilled workers and vertices in R represent jobs. An edge connects workers to jobs they can perform. 43

44 A matching in a graph is a subset M of E, such that for all vertices v in V, at most one edge of M is incident on v. 44

45 A maximum matching is a matching of maximum cardinality (maximum number of edges). not maximum maximum 45

46 A Maximum Matching No matching of cardinality 4, because only one of v and u can be matched. In the workers-jobs example a max-matching provides work for as many people as possible. v u 46

47 Solving the Maximum Bipartite Matching Problem Reduce the maximum bipartite matching problem on graph G to the max-flow problem on a corresponding flow network G. Solve using Ford-Fulkerson method. 47

48 Corresponding Flow Network To form the corresponding flow network G' of the bipartite graph G: Add a source vertex s and edges from s to L. Direct the edges in E from L to R. Add a sink vertex t and edges from R to t. Assign a capacity of 1 to all edges. Claim: max-flow in G corresponds to a max-bipartite-matching on G. s G G t L 48 R

49 Solving Bipartite Matching as Max Flow L et G ( V, E ) be a bipartite graph with vertex partition V L R. Let G ( V, E ) be its corresponding flow netwok r. If M is a matching in G, then there is an integer-valued flow f in G with value f M. Convers ey l, if f is an integer-valued flow in G, then there is a matching M in G with cardinality M f. Thus max M max(intege r f ) 49

50 Does this mean that max f = max M? Problem: we haven t shown that the max flow f(u,v) is necessarily integer-valued. 50

51 Integrality Theorem If the capacity function c takes on only integral values, then: 1. The maximum flow f produced by the Ford-Fulkerson method has the property that f is integer-valued. 2. For all vertices u and v the value f(u,v) of the flow is an integer. So max M = max f 51

52 Example min cut M = 3 max flow = f = 3 52

53 Conclusion Network flow algorithms allow us to find the maximum bipartite matching fairly easily. Similar techniques are applicable in other combinatorial design problems. 53

54 Example In a department there are n courses and m instructors. Every instructor has a list of courses he or she can teach. Every instructor can teach at most 3 courses during a year. The goal: find an allocation of courses to the instructors subject to these constraints. 54

55 Network flows on directed graphs Flow network: 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 flow network, source s and sink t. The source produces the material at a steady rate. The sink consumes the material at a steady rate. Other nodes pass any material through them. Objective: What is the max amount of material can be imposed through the network subject to the capacity of every edge?

56 Consider the network G=(V,E) shown in the figure below. Each edge (u,v) E in the network is labeled with its capacity c(u,v). 2 a 4 s 1 t 5 b 3 2 a 4 Flow: 1 s b t

57 Flow: A flow in G is a real-valued function f : V V R that satisfies three properties: 1. Capacity constraint : For all u,v V, we require f(u,v) c(u,v). The net flow from one vertex to another must not exceed the given capacity. 2. Skew symmetry : For all u,v V, we require f(u,v) = -f(v,u). The net flow from a vertex u to a vertex v is the negative of the net flow in the 3. Flow conservation : For all u V - {s,t}, we require f(u,v) = 0. v V

58 The quantity f(u,v),which can be positive or negative, is called the net flow from vertex u to v. The value of the flow is defined as f = f(s,v) that is the totat net flow out of s. v V f(u,v)/c(u,v) 2/2 a 1/4 s 1/1 t 0/5 b 1/3 f = f(s,a) + f(s,b) = = 2

59 Residual Capacity Given a flow f in network G = (V, E). Consider a pair of vertices u, v V. Residual capacity: The amount of additional flow we can push directly from u to v. c f (u, v) = c(u, v) f (u, v) 0 (since f (u, v) c(u, v)) c f (u, v) = f (v, u) if f(v, u) 0 Example: c(u,v) = 16, f(u,v) = 5 c f (u, v) = 11, c f (v, u) = 5

60 Residual Network Residual network: G f = (V, E f ), E f = {(u, v) V V : c f (u, v) > 0}. Each edge of the residual network can admit a positive flow. Given flows f 1 and f 2, the flow sum f 1 + f 2 is (f 1 + f 2 )(u, v) = f 1 (u, v) + f 2 (u, v) for all u, v V.

61 Consider the network G=(V,E) shown in the figure below. Each edge (u,v) E in the network is labeled with its capacity c(u,v). G 0 2 a 4 s 1 t 5 b 3 1/2 a 4 s 1/1 t Flow: f 1 = 1 5 b 1/3

62 Residual graph with respect to f 1 1/2 a 4 1 a 4 s 1/1 t s 1 1 t 5 Flow: f 1 = 1 b 1/3 5 b 1 2 2/2 a 1/4 s 1/1 t Flow: f 2 = 1 5 b 1/3

63 Residual graph with respect to f 2 2/2 s 5 Flow: f 2 = 2 a b 1/1 1/4 1/3 t s 2 5 a b t 2/2 a 2/4 s 0/1 t Flow: f 3 = 3 1/5 b 1/3

64 Residual graph with respect to f 3 2/2 a 2/4 2 a 2 2 s 0/1 t s t 1/5 Flow: f 3 = 3 b 1/3 4 b 2 2/2 a 2/4 s 1 t 3/5 3/3 Flow: f 4 = 5 b

65 Residual graph with respect to f 4 s 2/2 a 1 2/4 t s 2 2 a t 3/5 Flow: f 4 = 5 b 3/3 3 b 3 Max Flow : f* = 5

66 Ford-Fulkerson Algorithm The Ford-Fulkerson method is iterative: Starts with f(u,v) for (u,v) V, initial flow of value 0. The method is based on the augmenting path which is defined as a path from s to t in the residual network G f along which we can push more flow and then augment flow along this path. If no augmenting path exists in the residual network, then f is the max flow.

67 Ford-Fulkerson Algorithm Ford-Fulkerson(G, s, t) // G = (V, E) 1 for each edge (u,v) in E 2 f(u,v) = f(v,u) = 0 3 while exists path p from s to t in residual network G f 4 c f (p) = min{c f (u, v) : (u, v) is in p} 5 for each edge (u,v) on p 6 f(u,v) = f(u,v) + c f (p) 7 f(v,u) = -f(u,v)

68 Running time for this algorithms is O(E * f* ) where f* is the maximum flow found by algorithm. First two lines take time θ(e). The condition of the while loop takes time θ(e). The body of the while loop is executed at most f* times, since the flow value increases by at least one unit in each iteration.

69 Example: 16 v1 12 v2 20 s t 13 v3 14 v4 4 4/12 4/16 v1 v2 20 s /9 7 t 13 v3 4/14 v4 4/4

70 8 s v v t 13 v3 10 v /12 v1 v2 11/16 7/20 s 7/10 4 4/9 7/7 t 13 v3 11/14 v4 4/4

71 8 s v v t 13 v3 3 v /12 v1 v2 11/16 15/20 s 10 1/4 4/9 7/7 t 8/13 v3 11/14 v4 4/4

72 5 v1 12 v2 5 s t 5 v3 3 v /12 11/16 v1 v2 19/20 s 10 1/4 9 7/7 t 12/13 v3 11/14 v4 4/4

73 5 v1 12 v2 1 s t 1 v3 3 4 v4 11