1 / 66

Maximum flow problems

Maximum flow problems. Maximum flow problems. Introduction of: network, max-flow problem capacity, flow Ford-Fulkerson method pseudo code, residual networks, augmenting paths cuts of networks max-flow min-cut theorem example of an execution analysis, running time,

dana-mccall
Download Presentation

Maximum flow problems

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Maximum flow problems

  2. Maximum flow problems • Introduction of: • network, max-flow problem • capacity, flow • Ford-Fulkerson method • pseudo code, residual networks, augmenting paths • cuts of networks • max-flow min-cut theorem • example of an execution • analysis, running time, • variations of the max-flow problem

  3. Introduction – flow network • Practical examples of a flow network • - liquids flowing through pipes • parts through assembly lines • current through electrical network • information through communication network • goods transported on the road…

  4. Introduction – flow network Representation Example: oil pipeline Flow network: directed graph G=(V,E) v3 v3 v1 v1 S S t t source v2 v4 v2 v4 sink

  5. Introduction – max-flow problem Representation Example: oil pipeline Flow network: directed graph G=(V,E) v3 v3 v1 v1 S S t t source v2 v4 v2 v4 sink Informal definition of the max-flow problem: What is the greatest rate at which material can be shipped from the source to the sink without violating any capacity contraints?

  6. 3 v3 v1 6 8 3 S t 3 6 8 v2 v4 6 c(u,v)=12 6 c(u,v)=6 u u v v Introduction - capacity Representation Example: oil pipeline Flow network: directed graph G=(V,E) v3 v1 S t v2 v4 12 Big pipe Small pipe

  7. Introduction - capacity Representation Example: oil pipeline Flow network: directed graph G=(V,E) 3 v3 v3 v1 v1 6 8 3 S S t t 3 6 8 v2 v4 v2 v4 6 If (u,v)  E  c(u,v) = 0 6 v2 v4 0 0 v4 v3 0

  8. 3 v3 v1 6 8 3 S t 3 6 8 v2 v4 6 f(u,v)=6 f(u,v)=6 u u v v Introduction – flow Representation Example: oil pipeline Flow network: directed graph G=(V,E) v3 v1 S t v2 v4 6/12 Flow below capacity 6/6 Maximum flow

  9. 3 v3 v1 6 8 3 S t 3 6 8 v2 v4 6 Introduction – flow Representation Example: oil pipeline Flow network: directed graph G=(V,E) v3 v1 S t v2 v4

  10. 3 v3 v1 6 8 3 S t 3 6/6 6/8 v2 v4 6/6 Introduction – flow Representation Example: oil pipeline Flow network: directed graph G=(V,E) v3 v1 S t v2 v4

  11. 3 v3 v1 6 8 3 S t 3 6/6 6/8 v2 v4 6/6 Introduction – flow Representation Example: oil pipeline Flow network: directed graph G=(V,E) v3 v1 S t v2 v4

  12. 3/3 v3 v1 3/6 3/8 3 S t 3 6/6 6/8 v2 v4 6/6 Introduction – flow Representation Example: oil pipeline Flow network: directed graph G=(V,E) v3 v1 S t v2 v4

  13. 3/3 v3 v1 3/6 3/8 3 S t 3 6/6 6/8 v2 v4 6/6 Introduction – flow Representation Example: oil pipeline Flow network: directed graph G=(V,E) v3 v1 S t v2 v4

  14. 3/3 v3 v1 3/6 5/8 2/3 S t 3 6/6 8/8 v2 v4 6/6 Introduction – flow Representation Example: oil pipeline Flow network: directed graph G=(V,E) v3 v1 S t v2 v4

  15. 3/3 v3 v1 3/6 5/8 2/3 S t 3 6/6 8/8 v2 v4 6/6 Introduction – cancellation Representation Example: oil pipeline Flow network: directed graph G=(V,E) v3 v1 S t v2 v4

  16. 3/3 v3 v1 4/6 6/8 3/3 S t 1/3 6/6 8/8 v2 v4 5/6 Introduction – cancellation Representation Example: oil pipeline Flow network: directed graph G=(V,E) v3 v1 S t v2 v4 u u u u 8/10 3/4 5/10 4 10 4 8/10 4 v v v v

  17. Flow properties

  18. Flow properties Flow network G = (V,E) 12/12 v3 v1 Note: by skew symmetry f (v3,v1) = - 12 11/16 15/20 10 1/4 7/7 S t 4/9 4/4 8/13 v2 v4 11/14

  19. Net flow and value of a flow u u f(u,v) = 5 f(v,u) = -5 8/10 3/4 5/10 4 v v 3/3 v3 v1 4/6 6/8 3/3 S t 1/3 6/6 8/8 v2 v4 5/6

  20. The max-flow problem Informal definition of the max-flow problem: What is the greatest rate at which material can be shipped from the source to the sink without violating any capacity contraints? Formal definition of the max-flow problem: The max-flow problem is to find a valid flow for a given weighted directed graph G, that has the maximum value over all valid flows.

  21. The Ford-Fulkerson methoda way how to find the max-flow This method contains 3 important ideas:   1) residual networks     2) augmenting paths     3) cuts of flow networks

  22. Ford-Fulkerson – pseudo code 1 initialize flow f to 0 2 while there exits an augmenting path p 3do augment flow f along p 4 return f

  23. 5 11 5 8 cf(u,v) = c(u,v) – f(u,v) Ford Fulkerson – residual networks The residual network Gf of a given flow network G with a valid flow f consists of the same vertices v  V as in G which are linked with residual edges (u,v)  Ef that can admit more strictly positive net flow. The residual capacity cf represents the weight of each edge Ef and is the amount of additional net flow f(u,v) before exceeding the capacity c(u,v) Flow network G = (V,E) residual network Gf = (V,Ef) 12/12 v3 v3 v1 v1 11/16 15/20 10 1/4 7/7 S t S t 4/9 4/4 8/13 v2 v4 v2 v4 11/14

  24. cf(u,v) = c(u,v) – f(u,v) Ford Fulkerson – residual networks The residual network Gf of a given flow network G with a valid flow f consists of the same vertices v  V as in G which are linked with residual edges (u,v)  Ef that can admit more strictly positive net flow. The residual capacity cf represents the weight of each edge Ef and is the amount of additional net flow f(u,v) before exceeding the capacity c(u,v) Flow network G = (V,E) residual network Gf = (V,Ef) 12 12/12 v3 v3 v1 v1 5 5 11/16 15/20 4 11 10 1/4 11 7/7 7 15 3 S t S t 4/9 5 5 4/4 4 3 8/13 8 v2 v4 v2 v4 11/14 11

  25. Ford Fulkerson – augmenting paths Definition: An augmenting path p is a simple (free of any cycle) path from s to t in the residual network Gf Residual capacity of p cf(p) = min{cf (u,v): (u,v) is on p} Flow network G = (V,E) residual network Gf = (V,Ef) 12 12/12 v3 v3 v1 v1 5 5 11/16 15/20 4 11 10 1/4 11 7/7 7 15 3 S t S t 4/9 5 5 4/4 4 3 8/13 8 v2 v4 v2 v4 11/14 11

  26. Ford Fulkerson – augmenting paths Definition: An augmenting path p is a simple (free of any cycle) path from s to t in the residual network Gf Residual capacity of p cf(p) = min{cf (u,v): (u,v) is on p} Flow network G = (V,E) residual network Gf = (V,Ef) 12 12/12 v3 v3 v1 v1 5 5 11/16 15/20 4 11 10 1/4 11 7/7 7 15 3 S t S t 4/9 5 5 4/4 4 3 8/13 8 v2 v4 v2 v4 11/14 11 Augmenting path

  27. Ford Fulkerson – augmenting paths We define a flow: fp: V x V  R such as: cf(p) if (u,v) is on p fp(u,v) = - cf(p) if (v,u) is on p 0 otherwise Flow network G = (V,E) residual network Gf = (V,Ef) 12 12/12 v3 v3 v1 v1 5 5 11/16 15/20 4 11 10 1/4 11 7/7 7 15 3 S t S t 4/9 5 5 4/4 4 3 8/13 8 v2 v4 v2 v4 11/14 11

  28. cf(p) if (u,v) is on p fp(u,v) = - cf(p) if (v,u) is on p 0 otherwise Ford Fulkerson – augmenting paths We define a flow: fp: V x V  R such as: Flow network G = (V,E) residual network Gf = (V,Ef) 12 12/12 v3 v3 v1 v1 5 4/5 11/16 15/20 4/4 11 10 1/4 11 7/7 7 -4/15 3 S t S t 4/9 -4/5 4/5 4/4 4 3 8/13 -4/8 v2 v4 v2 v4 11/14 11 Our virtual flow fp along the augmenting path p in Gf

  29. cf(p) if (u,v) is on p fp(u,v) = - cf(p) if (v,u) is on p 0 otherwise Ford Fulkerson – augmenting the flow We define a flow: fp: V x V  R such as: Flow network G = (V,E) residual network Gf = (V,Ef) 12 12/12 v3 v3 v1 v1 5 4/5 11/16 19/20 4/4 11 10 1/4 11 7/7 7 15 3 S t S t 0/9 5 4/5 4/4 4 3 12/13 8 v2 v4 v2 v4 11/14 11 New flow: f´: V x V  R : f´=f + fp Our virtual flow fp along the augmenting path p in Gf

  30. cf(p) if (u,v) is on p fp(u,v) = - cf(p) if (v,u) is on p 0 otherwise Ford Fulkerson – new valid flowproof of capacity constraint Lemma: f´ : V x V  R : f´ = f + fp in G cf(p) = min{cf (u,v): (u,v) is on p} Capacity constraint: For all u,v  V, we require f(u,v) < c(u,v) cf(u,v) = c(u,v) – f(u,v) Proof: fp (u ,v) < cf (u ,v) = c (u ,v) – f (u ,v)  (f + fp) (u ,v) = f (u ,v) + fp (u ,v) < c (u ,v)

  31. Ford Fulkerson – new valid flowproof of Skew symmetry Lemma: f´ : V x V  R : f´ = f + fp in G Skew symmetry: For all u,v  V, we require f(u,v) = - f(v,u) Proof: (f + fp)(u ,v) = f (u ,v) + fp (u ,v) = - f (v ,u) – fp (v ,u) = - (f (v ,u) + fp (v ,u)) = - (f + fp) (v ,u)

  32. Ford Fulkerson – new valid flowproof of flow conservation Lemma: f´ : V x V  R : f´ = f + fp in G Flow conservation: For all u  V \ {s,t} :  f(u,v) = 0 v  V Proof: u  V – {s ,t}  (f + fp) (u ,v) =  (f(u ,v) + fp (u ,v)) =  f (u ,v) +  fp (u ,v) = 0 + 0 = 0 v  V v  V v  V v  V

  33. Ford Fulkerson – new valid flow Lemma: | (f + fp) | = | f | + | fp | Value of a Flow f:  Def: |f| =  f(s,v) v  V Proof: | (f + fp) | =  (f + fp) (s ,v) =  (f (s ,v) + fp (s ,v)) =  f (s ,v) +  fp (s ,v) = | f | + | fp | v  V v  V v  V v  V

  34. Ford Fulkerson – new valid flow Lemma: f´ : V x V  R : f´ = f + fp in G | (f + fp) | = | f | + | fp | > | f | Lemma shows: if an augmenting path can be found then the above flow augmentation will result in a flow improvement. Question: If we cannot find any more an augmenting path is our flow then maximum? Idea: The flow in G is maximum  the residual Gf contains no augmenting path.

  35. Practical example In the example: S = {s,v1,v2), T = {v3,v4,t} Net flow f(S ,T) = f(v1,v3) + f(v2,v4) + f(v2,v3) = 12 + 11 + (-0) = 23 Capacity c(S,T) = c(v1,v3) + c(v2,v4) = 12 + 14 = 26 12/12 v3 v1 11/16 19/20 10 1/4 7/7 S t 0/9 4/4 12/13 v2 v4 11/14 S T Implicit summation notation: f (S, T) =  f (u, v) u  S v  T Ford Fulkerson – cuts of flow networks New notion: cut (S,T) of a flow network A cut (S,T) of a flow network G=(V,E) is a partiton of V into S and T = V \ S such that s  S and t  T.

  36. Ford Fulkerson – cuts of flow networks Lemma: the value of a flow in a network is the net flow across any cut of the network f (S ,T) = | f | 12/12 v3 v1 11/16 19/20 10 1/4 7/7 S t 0/9 4/4 12/13 v2 v4 11/14 proof

  37. Ford Fulkerson – cuts of flow networks Assumption: The value of any flow f in a flow network G is bounded from above by the capacity of any cut of G Lemma: | f | <c (S, T) 12/12 | f | = f (S, T) =  f (u, v) < c (u, v) = c (S, T) v3 v1 11/16 19/20 10 1/4 7/7 S t 0/9 u S v T 4/4 u S v T 12/13 v2 v4 11/14

  38. F. Fulkerson: 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 Gf contains no augmenting paths. • | f | = c (S, T) for some cut (S, T) of G. • proof: •  (2): • We assume for the sake of contradiction that f is a maximum flow in G but that there still exists an augmenting path p in Gf. • Then as we know from above, we can augment the flow in G according to the formula: f´= f + fp. That would create a flow f´that is strictly greater than the former flow f which is in contradiction to our assumption that f is a maximum flow.

  39. residual network Gf 3/3 3 v3 v3 v1 v1 4/6 2 6/8 3/3 2 3 4 6 1 S S t t 1/3 2 6/6 6 5 8/8 8 v2 v4 v2 v4 5/6 1 F. Fulkerson: 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 Gf contains no augmenting paths. • | f | = c (S, T) for some cut (S, T) of G. proof: (2)  (3): original flow network G

  40. F. Fulkerson: 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 Gf contains no augmenting paths. • | f | = c (S, T) for some cut (S, T) of G. proof: (2)  (3): Define S = {v V |  path p from s to v in Gf } T = V \ S (note t  S according to (2)) residual network Gf 3 v3 v1 2 2 3 4 6 1 S t 2 6 5 8 v2 v4 1

  41. 3/3 v3 v1 4/6 6/8 3/3 S t 1/3 6/6 8/8 v2 v4 5/6 F. Fulkerson: 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 Gf contains no augmenting paths. • | f | = c (S, T) for some cut (S, T) of G. • proof: • (2)  (3): Define • S = {v V |  path p from s to v in Gf } • T = V \ S (note t  S according to (2)) • for  u  S, v  T: f (u, v) = c (u, v) (otherwise (u, v)  Ef and v  S) • | f | = f (S, T) = c (S, T) original network G 1

  42. F. Fulkerson: Max-flow min-cut theorem • If f is a fow 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 Gf contains no augmenting paths. • | f | = c (S, T) for some cut (S, T) of G. proof: (3)  (1): as proofed before | f | = f (S, T) < c (S, T) the statement of (3) : | f | = c (S, T) implies that f is a maximum flow

  43. The basic Ford Fulkerson algorithm

  44. The basic Ford Fulkerson algorithmexample of an execution (residual) network Gf 12 v3 v1 16 20 10 4 7 S t 9 4 13 v2 v4 14

  45. The basic Ford Fulkerson algorithmexample of an execution (residual) network Gf 0/12 v3 v1 0/16 0/20 0/10 0/4 7 S t 0/9 0/4 0/13 v2 v4 0/14

  46. The basic Ford Fulkerson algorithmexample of an execution (residual) network Gf 12 v3 v1 16 20 10 4 7 S t 9 4 13 v2 v4 14

  47. The basic Ford Fulkerson algorithmexample of an execution (residual) network Gf 12 v3 v1 16 20 10 4 7 S t 9 4 13 v2 v4 14 temporary variable: cf (p) = 12

  48. The basic Ford Fulkerson algorithmexample of an execution (residual) network Gf 12 v3 v1 16 20 10 4 7 S t 9 4 13 v2 v4 14 new flow network G 12/12 v3 v1 12/16 temporary variable: cf (p) = 12 12/20 10 4 7 S t 9 4 13 v2 v4 14

  49. The basic Ford Fulkerson algorithmexample of an execution (residual) network Gf 12 v3 v1 4 8 12 12 10 4 7 S t 9 4 13 v2 v4 14 new flow network G 12/12 v3 v1 12/16 12/20 10 4 7 S t 9 4 13 v2 v4 14

  50. The basic Ford Fulkerson algorithmexample of an execution (residual) network Gf 12 v3 v1 4 8 12 12 10 4 7 S t 9 4 13 v2 v4 14 new flow network G 12/12 v3 v1 12/16 12/20 10 4 7 S t 9 4 13 v2 v4 14

More Related