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- 1. Network flow problems
- 2. Network 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 – network Practical examples of a network - liquids flowing through pipes - parts through assembly lines - current through electrical network - information through communication network - goods transported on the road…
- 4. Introduction - network Example: oil pipelineRepresentation Flow network: directed graph G=(V,E) S t v1 v2 v3 v4 S t v1 v2 v3 v4source sink
- 5. Introduction – max-flow problemRepresentation Flow network: directed graph G=(V,E) S t v1 v2 v3 v4 S t v1 v2 v3 v4source 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? Example: oil pipeline
- 6. Introduction - capacity S t v1 v2 v3 v4 Representation Flow network: directed graph G=(V,E) S t v1 v2 v3 v4 u v 12 u v 6 c(u,v)=12 c(u,v)=6 Big pipe Small pipe 8 3 6 8 6 6 3 3 Example: oil pipeline
- 7. Introduction - capacity S t v1 v2 v3 v4 Representation Flow network: directed graph G=(V,E) S t v1 v2 v3 v4 8 3 6 8 6 6 3 3 Example: oil pipeline v4 v2 v3 v4 If (u,v) E c(u,v) = 0 0 0 0 6
- 8. Introduction – flow S t v1 v2 v3 v4 Representation Flow network: directed graph G=(V,E) S t v1 v2 v3 v4 u v 6/12 u v 6/6 f(u,v)=6 f(u,v)=6 Flow below capacity Maximum flow 8 3 6 8 6 6 3 3 Example: oil pipeline
- 9. Introduction – flow S t v1 v2 v3 v4 Representation Flow network: directed graph G=(V,E) S t v1 v2 v3 v4 8 3 6 8 6 6 3 3 Example: oil pipeline
- 10. Introduction – flow S t v1 v2 v3 v4 Representation Flow network: directed graph G=(V,E) S t v1 v2 v3 v4 8 3 6 6/8 6/6 6/6 3 3 Example: oil pipeline
- 11. Introduction – flow S t v1 v2 v4 v3 Representation Flow network: directed graph G=(V,E) S t v1 v2 v3 v4 8 3 6 6/8 6/6 6/6 3 3 Example: oil pipeline
- 12. Introduction – flow S t v1 v2 v4 v3 Representation Flow network: directed graph G=(V,E) S t v1 v2 v3 v4 3/8 3/3 3/6 6/8 6/6 6/6 3 3 Example: oil pipeline
- 13. Introduction – flow S t v1 v2 v4 v3 Representation Flow network: directed graph G=(V,E) S t v1 v2 v3 v4 3/8 3/3 3/6 6/8 6/6 6/6 3 3 Example: oil pipeline
- 14. Introduction – flow S t v1 v2 v4 v3 Representation Flow network: directed graph G=(V,E) S t v1 v2 v3 v4 5/8 3/3 3/6 8/8 6/6 6/6 2/3 3 Example: oil pipeline
- 15. Introduction – cancellation S t v1 v2 v4 v3 Representation Flow network: directed graph G=(V,E) S t v1 v2 v3 v4 5/8 3/3 3/6 8/8 6/6 6/6 2/3 3 Example: oil pipeline
- 16. Introduction – cancellation S t v1 v2 v4 v3 Representation Flow network: directed graph G=(V,E) S t v1 v2 v3 v4 6/8 3/3 4/6 8/8 5/6 6/6 3/3 1/3 u v 10 4 u v 8/10 4 u v 8/10 3/4 u v 5/10 4 Example: oil pipeline
- 17. Maximum Flow Problem: definitions D M K S B P 8 14 146 12 10 10 7 17 6 D M K S B P 8 14 146 12 10 10 7 17 6 SOURCE: Node with net outflow: Production point SINK: Node with net inflow; Consumption point CAPACITY: Maximum flow on an edge Efficient method to solve such problems: Ford-Fulkerson Method
- 18. Flow properties Flow in G = (V,E): f: V x V R with 3 properties: 1) Capacity constraint: For all u,v V : f(u,v) < c(u,v) 2) Skew symmetry: For all u,v V : f(u,v) = - f(v,u) 3) Flow conservation: For all u V {s,t} : f(u,v) = 0 v V
- 19. Flow properties Flow in G = (V,E): f: V x V R with 3 properties: 1) Capacity constraint: For all u,v V : f(u,v) < c(u,v) 2) Skew symmetry: For all u,v V : f(u,v) = - f(v,u) 3) Flow conservation: For all u V {s,t} : f(u,v) = 0 v V S t v1 v2 v3 v4 12/12 11/14 Flow network G = (V,E) Note: by skew symmetry f (v3,v1) = - 12
- 20. Net flow and value of a flow Net Flow: positive or negative value of f(u,v) Value of a Flow f: Def: |f| = f(s,v)v V S t v1 v2 v3 v4 6/8 3/3 4/6 8/8 5/6 6/6 3/3 1/3 u v 8/10 3/4 u v 5/10 4 f(u,v) = 5 f(v,u) = -5
- 21. 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.
- 22. The Ford-Fulkerson method a way how to find the max-flow This method contains 3 important ideas: 1) residual networks 2) augmenting paths 3) cuts of flow networks
- 23. Ford-Fulkerson – pseudo code 1 initialize flow f to 0 2 while there exits an augmenting path p 3 do augment flow f along p 4 return f
- 24. Ford Fulkerson – residual networks S t v1 v2 v3 v4 12/12 11/14 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) cf(u,v) = c(u,v) – f(u,v) S t v1 v2 v3 v4 residual network Gf = (V,Ef)
- 25. Ford Fulkerson – residual networks S t v1 v2 v3 v4 12/12 11/14 Flow network G = (V,E) cf(u,v) = c(u,v) – f(u,v) S t v1 v2 v3 v4 residual network Gf = (V,Ef) 11 12 3 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)
- 26. Ford Fulkerson – augmenting paths S t v1 v2 v3 v4 12/12 11/14 cf(p) = min{cf (u,v): (u,v) is on p} Flow network G = (V,E) residual network Gf = (V,Ef) 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 S t v1 v2 v3 v4 12 3 11
- 27. Ford Fulkerson – augmenting paths S t v1 v2 v3 v4 12/12 11/14 cf(p) = min{cf (u,v): (u,v) is on p} Flow network G = (V,E) S t v1 v2 v3 v4 12 3 residual network Gf = (V,Ef) 11 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 Augmenting path
- 28. Ford Fulkerson – augmenting paths S t v1 v2 v3 v4 12/12 11/14 Flow network G = (V,E) S t v1 v2 v3 v4 12 3 residual network Gf = (V,Ef) 11 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
- 29. Ford Fulkerson – augmenting paths S t v1 v2 v3 v4 12/12 11/14 Flow network G = (V,E) S t v1 v2 v3 v4 12 3 residual network Gf = (V,Ef) 11 Our virtual flow fp along the augmenting path p in Gf cf(p) if (u,v) is on p fp(u,v) = - cf(p) if (v,u) is on p 0 otherwise We define a flow: fp: V x V R such as:
- 30. Ford Fulkerson – augmenting the flow S t v1 v2 v3 v4 12/12 11/14 Flow network G = (V,E) S t v1 v2 v3 v4 12 3 residual network Gf = (V,Ef) 11 Our virtual flow fp along the augmenting path p in Gf New flow: f´: V x V R : f´=f + fp cf(p) if (u,v) is on p fp(u,v) = - cf(p) if (v,u) is on p 0 otherwise We define a flow: fp: V x V R such as:
- 31. Ford Fulkerson – new valid flow proof of capacity constraint 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) Lemma: f : V x V R : f = f + fp in G Capacity constraint: For all u,v V, we require f(u,v) < c(u,v) cf(p) if (u,v) is on p fp(u,v) = - cf(p) if (v,u) is on p 0 otherwise cf(p) = min{cf (u,v): (u,v) is on p} cf(u,v) = c(u,v) – f(u,v)
- 32. Ford Fulkerson – new valid flow proof of Skew symmetry 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) Skew symmetry: For all u,v V, we require f(u,v) = - f(v,u) Lemma: f : V x V R : f = f + fp in G
- 33. Ford Fulkerson – new valid flow proof of flow conservation Proof: u V – {s ,t} (f + fp) (u ,v) = (f(u ,v) + fp (u ,v)) = f (u ,v) + fp (u ,v) = 0 + 0 = 0 Flow conservation: For all u V {s,t} : f(u,v) = 0 v V v Vv V v Vv V Lemma: f : V x V R : f = f + fp in G
- 34. Ford Fulkerson – new valid flow Proof: | (f + fp) | = (f + fp) (s ,v) = (f (s ,v) + fp (s ,v)) = f (s ,v) + fp (s ,v) = | f | + | fp | Lemma: | (f + fp) | = | f | + | fp | v Vv V v Vv V Value of a Flow f: Def: |f| = f(s,v) v V
- 35. Ford Fulkerson – new valid flow Lemma: | (f + fp) | = | f | + | fp | > | f | f : V x V R : f = f + fp in G 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.
- 36. 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. Implicit summation notation: f (S, T) = f (u, v) S t v1 v2 v3 v4 12/1 2 11/1 4 TS 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 Practical example u S v T
- 37. 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 | S t v1 v2 v3 v4 12/1 2 11/1 4 proof
- 38. 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) | f | = f (S, T) = f (u, v) < c (u, v) = c (S, T) v Tu S v Tu S S t v1 v2 v3 v4 12/1 2 11/1 4
- 39. 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: 1. f is a maximum flow in G. 2. The residual network Gf contains no augmenting paths. 3. | f | = c (S, T) for some cut (S, T) of G. proof: (1) (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.
- 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: 1. f is a maximum flow in G. 2. The residual network Gf contains no augmenting paths. 3. | f | = c (S, T) for some cut (S, T) of G. proof: (2) (3): S t v1 v2 v3 v4 6/8 3/3 4/6 8/8 5/6 6/6 3/3 1/3 S t v1 v2 v3 v4 2 3 2 8 1 6 3 2 5 6 1 4 residual network Gforiginal flow network G
- 41. 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: 1. f is a maximum flow in G. 2. The residual network Gf contains no augmenting paths. 3. | 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)) S t v1 v2 v3 v4 2 3 2 8 1 6 3 2 5 6 1 4 residual network Gf
- 42. 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: 1. f is a maximum flow in G. 2. The residual network Gf contains no augmenting paths. 3. | 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) 1 S t v1 v2 v3 v4 6/8 3/3 4/6 8/8 5/6 6/6 3/3 1/3 original network G
- 43. 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: 1. f is a maximum flow in G. 2. The residual network Gf contains no augmenting paths. 3. | 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
- 44. 1 for each edge (u, v) E [G] 2 do f [u, v] = 0 3 f [v, u] = 0 4 while there exists a path p from s to t in the residual network Gf 5 do cf (p) = min {cf (u, v) | (u, v) is in p} 6 for each edge (u, v) in p 7 do f [u, v] = f [u, v] + cf (p) 8 f [v, u] = - f [u, v] The basic Ford Fulkerson algorithm
- 45. The basic Ford Fulkerson algorithm example of an execution S t v1 v2 v3 v4 12 14 1 for each edge (u, v) E [G] 2 do f [u, v] = 0 3 f [v, u] = 0 4 while there exists a path p from s to t in the residual network Gf 5 do cf(p) = min{cf(u, v) | (u, v) p} 6 for each edge (u, v) in p 7 do f [u, v] = f [u, v] + cf(p) 8 f [v, u] = - f [u, v] (residual) network Gf
- 46. The basic Ford Fulkerson algorithm example of an execution S t v1 v2 v3 v4 0/12 0/14 (residual) network Gf 1 for each edge (u, v) E [G] 2 do f [u, v] = 0 3 f [v, u] = 0 4 while there exists a path p from s to t in the residual network Gf 5 do cf(p) = min{cf(u, v) | (u, v) p} 6 for each edge (u, v) in p 7 do f [u, v] = f [u, v] + cf(p) 8 f [v, u] = - f [u, v]
- 47. The basic Ford Fulkerson algorithm example of an execution S t v1 v2 v3 v4 12 14 (residual) network Gf 1 for each edge (u, v) E [G] 2 do f [u, v] = 0 3 f [v, u] = 0 4 while there exists a path p from s to t in the residual network Gf 5 do cf(p) = min{cf(u, v) | (u, v) p} 6 for each edge (u, v) in p 7 do f [u, v] = f [u, v] + cf(p) 8 f [v, u] = - f [u, v]
- 48. The basic Ford Fulkerson algorithm example of an execution S t v1 v2 v3 v4 12 14 (residual) network Gf temporary variable: cf (p) = 12 1 for each edge (u, v) E [G] 2 do f [u, v] = 0 3 f [v, u] = 0 4 while there exists a path p from s to t in the residual network Gf 5 do cf(p) = min{cf(u, v) | (u, v) p} 6 for each edge (u, v) in p 7 do f [u, v] = f [u, v] + cf(p) 8 f [v, u] = - f [u, v]
- 49. The basic Ford Fulkerson algorithm example of an execution S t v1 v2 v3 v4 12 14 (residual) network Gf temporary variable: cf (p) = 12 S t v1 v2 v3 v4 12/1 2 14 new flow network G 1 for each edge (u, v) E [G] 2 do f [u, v] = 0 3 f [v, u] = 0 4 while there exists a path p from s to t in the residual network Gf 5 do cf(p) = min{cf(u, v) | (u, v) p} 6 for each edge (u, v) in p 7 do f [u, v] = f [u, v] + cf(p) 8 f [v, u] = - f [u, v]
- 50. The basic Ford Fulkerson algorithm example of an execution S t v1 v2 v3 v4 12 14 (residual) network Gf S t v1 v2 v3 v4 12/1 2 14 new flow network G 1 for each edge (u, v) E [G] 2 do f [u, v] = 0 3 f [v, u] = 0 4 while there exists a path p from s to t in the residual network Gf 5 do cf(p) = min{cf(u, v) | (u, v) p} 6 for each edge (u, v) in p 7 do f [u, v] = f [u, v] + cf(p) 8 f [v, u] = - f [u, v]
- 51. The basic Ford Fulkerson algorithm example of an execution S t v1 v2 v3 v4 12 14 (residual) network Gf S t v1 v2 v3 v4 12/1 2 14 new flow network G 1 for each edge (u, v) E [G] 2 do f [u, v] = 0 3 f [v, u] = 0 4 while there exists a path p from s to t in the residual network Gf 5 do cf(p) = min{cf(u, v) | (u, v) p} 6 for each edge (u, v) in p 7 do f [u, v] = f [u, v] + cf(p) 8 f [v, u] = - f [u, v]
- 52. The basic Ford Fulkerson algorithm example of an execution S t v1 v2 v3 v4 12 14 (residual) network Gf S t v1 v2 v3 v4 12/1 2 14 new flow network G 1 for each edge (u, v) E [G] 2 do f [u, v] = 0 3 f [v, u] = 0 4 while there exists a path p from s to t in the residual network Gf 5 do cf(p) = min{cf(u, v) | (u, v) p} 6 for each edge (u, v) in p 7 do f [u, v] = f [u, v] + cf(p) 8 f [v, u] = - f [u, v]
- 53. The basic Ford Fulkerson algorithm example of an execution (residual) network Gf S t v1 v2 v3 v4 12/1 2 14 new flow network G temporary variable: cf (p) = 4 S t v1 v2 v3 v4 12 14 1 for each edge (u, v) E [G] 2 do f [u, v] = 0 3 f [v, u] = 0 4 while there exists a path p from s to t in the residual network Gf 5 do cf(p) = min{cf(u, v) | (u, v) p} 6 for each edge (u, v) in p 7 do f [u, v] = f [u, v] + cf(p) 8 f [v, u] = - f [u, v]
- 54. The basic Ford Fulkerson algorithm example of an execution S t v1 v2 v3 v4 12 14 (residual) network Gf S t v1 v2 v3 v4 12/1 2 new flow network G temporary variable: cf (p) = 4 1 for each edge (u, v) E [G] 2 do f [u, v] = 0 3 f [v, u] = 0 4 while there exists a path p from s to t in the residual network Gf 5 do cf(p) = min{cf(u, v) | (u, v) p} 6 for each edge (u, v) in p 7 do f [u, v] = f [u, v] + cf(p) 8 f [v, u] = - f [u, v]
- 55. The basic Ford Fulkerson algorithm example of an execution S t v1 v2 v3 v4 12 10 (residual) network Gf S t v1 v2 v3 v4 12/1 2 new flow network G 4 1 for each edge (u, v) E [G] 2 do f [u, v] = 0 3 f [v, u] = 0 4 while there exists a path p from s to t in the residual network Gf 5 do cf(p) = min{cf(u, v) | (u, v) p} 6 for each edge (u, v) in p 7 do f [u, v] = f [u, v] + cf(p) 8 f [v, u] = - f [u, v]
- 56. The basic Ford Fulkerson algorithm example of an execution S t v1 v2 v3 v4 12 10 (residual) network Gf S t v1 v2 v3 v4 12/1 2 new flow network G 4 1 for each edge (u, v) E [G] 2 do f [u, v] = 0 3 f [v, u] = 0 4 while there exists a path p from s to t in the residual network Gf 5 do cf(p) = min{cf(u, v) | (u, v) p} 6 for each edge (u, v) in p 7 do f [u, v] = f [u, v] + cf(p) 8 f [v, u] = - f [u, v]
- 57. The basic Ford Fulkerson algorithm example of an execution S t v1 v2 v3 v4 12 10 (residual) network Gf S t v1 v2 v3 v4 12/1 2 new flow network G 4 1 for each edge (u, v) E [G] 2 do f [u, v] = 0 3 f [v, u] = 0 4 while there exists a path p from s to t in the residual network Gf 5 do cf(p) = min{cf(u, v) | (u, v) p} 6 for each edge (u, v) in p 7 do f [u, v] = f [u, v] + cf(p) 8 f [v, u] = - f [u, v]
- 58. The basic Ford Fulkerson algorithm example of an execution S t v1 v2 v3 v4 12 10 (residual) network Gf S t v1 v2 v3 v4 12/1 2 new flow network G 4 temporary variable: cf (p) = 7 1 for each edge (u, v) E [G] 2 do f [u, v] = 0 3 f [v, u] = 0 4 while there exists a path p from s to t in the residual network Gf 5 do cf(p) = min{cf(u, v) | (u, v) p} 6 for each edge (u, v) in p 7 do f [u, v] = f [u, v] + cf(p) 8 f [v, u] = - f [u, v]
- 59. The basic Ford Fulkerson algorithm example of an execution S t v1 v2 v3 v4 12 10 (residual) network Gf S t v1 v2 v3 v4 12/1 2 new flow network G 4 temporary variable: cf (p) = 7 1 for each edge (u, v) E [G] 2 do f [u, v] = 0 3 f [v, u] = 0 4 while there exists a path p from s to t in the residual network Gf 5 do cf(p) = min{cf(u, v) | (u, v) p} 6 for each edge (u, v) in p 7 do f [u, v] = f [u, v] + cf(p) 8 f [v, u] = - f [u, v]
- 60. The basic Ford Fulkerson algorithm example of an execution (residual) network Gf S t v1 v2 v3 v4 12/1 2 new flow network G S t v1 v2 v3 v4 12 3 11 1 for eachedge(u,v) E[G] 2 do f[u,v]=0 3 f [v, u] = 0 4 while thereexistsapathpfromstot intheresidualnetworkG f 5 do cf(p)=min{c f(u,v)|(u,v) p} 6 for eachedge(u,v)inp 7 do f[u,v]=f[u,v]+c f(p) 8 f [v, u] = - f[u,v]
- 61. The basic Ford Fulkerson algorithm example of an execution (residual) network Gf S t v1 v2 v3 v4 12/1 2 new flow network G S t v1 v2 v3 v4 12 3 11 Finally we have: | f | = f (s, V) = 23 1 for each edge (u, v) E [G] 2 do f [u, v] = 0 3 f [v, u] = 0 4 while there exists a path p from s to t in the residual network Gf 5 do cf(p) = min{cf(u, v) | (u, v) p} 6 for each edge (u, v) in p 7 do f [u, v] = f [u, v] + cf(p) 8 f [v, u] = - f [u, v]
- 62. Analysis of the Ford Fulkerson algorithm Running time The running time depends on how the augmenting path p in line 4 is determined. 1 for each edge (u, v) E [G] 2 do f [u, v] = 0 3 f [v, u] = 0 4 while there exists a path p from s to t in the residual network Gf 5 do cf(p) = min{cf(u, v) | (u, v) p} 6 for each edge (u, v) in p 7 do f [u, v] = f [u, v] + cf(p) 8 f [v, u] = - f [u, v]
- 63. Analysis of the Ford Fulkerson algorithm Running time (arbitrary choice of p) O(|E|) O(|E| ) O(|E| |fmax|) running time: O ( |E| |fmax| ) with fmax as maximum flow (1) The augmenting path is chosen arbitrarily and all capacities are integers 1 for each edge (u, v) E [G] 2 do f [u, v] = 0 3 f [v, u] = 0 4 while there exists a path p from s to t in the residual network Gf 5 do cf(p) = min{cf(u, v) | (u, v) p} 6 for each edge (u, v) in p 7 do f [u, v] = f [u, v] + cf(p) 8 f [v, u] = - f [u, v] O(|E| )
- 64. Analysis of the Ford Fulkerson algorithm Running time (arbitrary choice of p) (1) The augmenting path is chosen arbitrarily and all capacities are integers Consequencies of an arbitrarily choice: Example if |f*| is large: t v2 v1 s 1 running time: O ( |E| |fmax| ) with fmax as maximum flow
- 65. Analysis of the Ford Fulkerson algorithm Running time (arbitrary choice of p) (1) The augmenting path is chosen arbitrarily and all capacities are integers Consequencies of an arbitrarily choice: Example if |f*| is large: t v2 v1 s 1/1 t v2 v1 s 1 residual network Gf running time: O ( |E| |fmax| ) with fmax as maximum flow
- 66. Analysis of the Ford Fulkerson algorithm Running time (arbitrary choice of p) (1) The augmenting path is chosen arbitrarily and all capacities are integers Consequencies of an arbitrarily choice: Example if |f*| is large: t v2 v1 s 1 t v2 v1 s 1 residual network Gf running time: O ( |E| |fmax| ) with fmax as maximum flow
- 67. Analysis of the Ford Fulkerson algorithm Running time with Edmonds-Karp algorithm (2) Edmonds-Karp algorithm augmenting path is found by breath-first search and has to be a shortest path from p to t running time: O ( |V| |E|² ) S t u1 u2 v3 v4 4 4 Informal idea of the proof: (1) for all vertices v V{s,t}: the shortest path distance f(s,v) in Gf increases monotonically with each flow augmentation f (s,v) < f (s,v)
- 68. Analysis of the Ford Fulkerson algorithm Running time with Edmonds-Karp algorithm running time: O ( |V| |E|² ) S t u1 u2 v3 v4 4 4 Informal idea of the proof: (1) for all vertices v V{s,t}: the shortest path distance f(s,v) in Gf increases monotonically with each flow augmentation . f(s,v) < f (s,v) f(s,u2) = 1 (2) Edmonds-Karp algorithm augmenting path is found by breath-first search and has to be a shortest path from p to t
- 69. Analysis of the Ford Fulkerson algorithm Running time with Edmonds-Karp algorithm running time: O ( |V| |E|² ) t u1 u2 v3 v4 4 4 f(s,u2) = 3 Two flow augmentations later S Informal idea of the proof: (1) for all vertices v V{s,t}: the shortest path distance f(s,v) in Gf increases monotonically with each flow augmentation . f(s,v) < f (s,v) (2) Edmonds-Karp algorithm augmenting path is found by breath-first search and has to be a shortest path from p to t
- 70. Analysis of the Ford Fulkerson algorithm Running time with Edmonds-Karp algorithm running time: O ( |V| |E|² ) S t u1 u2 v3 v4 4 4 The total number of flow augmentations performed by the algorithm is at most O(|V| |E|) (2) Edmonds-Karp algorithm augmenting path is found by breath-first search and has to be a shortest path from p to t
- 71. Analysis of the Ford Fulkerson algorithm Running time with Edmonds-Karp algorithm running time: O ( |V| |E|² ) S t u1 u2 v3 v4 4 4 Critical edges on the path p Def: critical edge (u,v) if cf(p) = cf(u,v) At least one edge on p must be critical. How many times can (u,v) be critical during an execution of Edmonds-Karp algorithm? (2) Edmonds-Karp algorithm augmenting path is found by breath-first search and has to be a shortest path from p to t
- 72. Analysis of the Ford Fulkerson algorithm Running time with Edmonds-Karp algorithm running time: O ( |V| |E|² ) S t u1 u2 v3 v4 4 4 our example: edge (u2,v4) Idea: (u,v) can become critical at most O(V) times (After beeing critical it dissappears in Gf and can only reappear after net flow is decreased) (2) Edmonds-Karp algorithm augmenting path is found by breath-first search and has to be a shortest path from p to t
- 73. Analysis of the Ford Fulkerson algorithm Running time with Edmonds-Karp algorithm running time: O ( |V| |E|² ) our example: edge (u2,v4) has disappeared Idea: (u,v) can become critical at most O(V) times (After beeing critical it dissappears in Gf and can only reappear after net flow is decreased) t u1 u2 v3 v4 4 4 Two flow augmentations later S (2) Edmonds-Karp algorithm augmenting path is found by breath-first search and has to be a shortest path from p to t
- 74. Analysis of the Ford Fulkerson algorithm Running time with Edmonds-Karp algorithm running time: O ( |V| |E|² ) our example: edge (u2,v4) reappeared but is now unreachable from s Idea: (u,v) can become critical at most O(V) times (After beeing critical it disappears in Gf and can only reappear after net flow is decreased) t u1 u2 v3 v4 4 2 After the third flow augmentation S 2 (2) Edmonds-Karp algorithm augmenting path is found by breath-first search and has to be a shortest path from p to t
- 75. Analysis of the Ford Fulkerson algorithm Running time with Edmonds-Karp algorithm running time: O ( |V| |E|² ) Idea: (u,v) can become critical at most O(|V|) times O(|E|) pairs of vertices exist in Gf All critical edges in O(|V| |E|) O(|E|) for breath first search for p total running time O(|V| |E|²) (2) Edmonds-Karp algorithm augmenting path is found by breath-first search and has to be a shortest path from p to t
- 76. Different variants of the max-flow problem - The algorithm of Dinic (different implementations reaps better running times than Edmonds-Karp): e.g. simultanous saturation of augmenting paths of the same length O(|V|² |E|) - A second capacity function lim for a lower limit of the flow function f with lim(u,v) < f(u,v) < c(u,v) - A cost function cost where each edge (u,v) has got a second weight cost(u,v) and the min-cost max-flow problem - Networks with multiple sources and sinks
- 77. Networks with multiple sources and sinks s1 s3 s2 t2 t1 s4 S´ T´supersource supersink
- 78. Ford Fulkerson – cuts of flow networks Assumption: the value of a flow in a network is the net flow across any cut of the network Lemma: f (S ,T) = | f | f (S, T) = f (S, VS) = f (S, V) – f (S, S) = f (S, V) = f (s [Ss], V) = f (s, V) + f (Ss, V) = f (s, V) = | f | Working with flows: X,Y,Z V and X Y = f (X, Y) = f (x, y) f (X, X) = 0 f (X, Y) = - f (Y, X) f (u, V) = 0 for all u V {s, t} f (X Y, Z) = f (X, Z) + f (Y, Z) f (Z, X Y) = f (Z, X) + f (Z, Y) x X y Y Proof:

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