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Flows

Understand and implement the Ford-Fulkerson algorithm for maximum flow problems on directed graphs, with skew symmetry, capacity constraints, and flow conservation. Learn how to find augmenting paths and improve flow efficiently. Explore runtime complexities and optimization strategies for solving max-flow challenges effectively.

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Flows

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  1. Flows t s source sink

  2. Flows edge-weights = capacities 2 3 5 7 4 4 2 2 source sink

  3. Flows edge-weights = capacities 2 2 2 3 2 5 7 2 4 4 2 2 2 2 4 units of flow source sink

  4. Flows edge-weights = capacities 2 2 2 3 4 5 7 2 2 4 4 4 2 2 2 6 units of flow source sink

  5. Flows edge-weights = capacities 2 2 3 3 5 5 7 3 2 1 4 4 4 2 1 2 7 units of flow source sink

  6. Optimal? 2 2 3 3 5 5 7 3 2 1 4 4 4 2 1 2 a larger flow?

  7. Cuts capacity of a cut cap(C)= c(u,v) C s uC vV-C C  V s  C t  V-C for any cut and any flow flow  cut  max-flow  min-cut

  8. Flows edge-weights = capacities 2 2 5 5 3 4 FLOW CONSERVATION

  9. Skew symmetry edge-weights = capacities 2 2 5 5 3 4 x -x FLOW CONSERVATION

  10. Flows edge-weights = capacities 2 2 -5 5 3  f(u,v) = 0 vV 4 x -x FLOW CONSERVATION

  11. Flow – formal definition FLOW CONSERVATION CAPACITY CONSTRAINTS  f(u,v) = 0 vV f(u,v)  c(u,v) SKEW SYMMETRY f(u,v) = - f(v,u)

  12. Max-flow problem INPUT: directed graph G=(V,E), source sV, sink tV a capacity c(e) for each edge eE OUTPUT: maximum flow from s to t

  13. Max-flow problem F  zero-flow while can improve F improve F

  14. Max-flow problem 2 2 2 2 3 2 3 3 4 5 5 5 7 7 3 2 2 2 1 4 4 4 4 4 4 2 2 2 1 2 2 improving can decrease flow on some edges

  15. Augmenting path +1 +1 +1 +1 -1

  16. Augmenting path +1 +1 +1 +1 +1

  17. Augmenting path +1 +1 +1 +1 +1  can improve the flow Ford-Fulkerson algorithm F  zero-flow while  augmenting path p improve F along p

  18. Residual capacity f(u,v) c(u,v) r(u,v) = c(u,v) – f(u,v) makes sense for negative f(u,v)

  19. Residual network 2 2 5 2 3 7 2 4 2 4 4 4 2 2 2

  20. Residual network 2 2 2 3 9 7 2 1 2 4 4 4 2 2 2

  21. Residual network 2 2 2 3 9 7 2 1 2 4 4 4 2 2 2

  22. Residual network 4 2 3 9 7 2 1 2 4 4 4 2 2 2

  23. Residual network 4 2 3 9 7 2 1 2 4 4 4 2 2 2

  24. Residual network 4 5 9 7 2 1 1 2 4 4 4 2 2 2

  25. Residual network 4 5 6 9 7 1 1 7 2 8 4 4 Is there an augmenting path? Is there a path from s to t in the residual network?

  26. Correct ? F  zero-flow while  augmenting path p improve F along p

  27. Correct ? F  zero-flow while  augmenting path p improve F along p YES Theorem: no augmenting path  max-flow

  28. Theorem: no augmenting path  max-flow Proof: no augmenting path  no path from s to t in the residual network s t

  29. Theorem: no augmenting path  max-flow Proof: no augmenting path  no path from s to t in the residual network s t vertices to which we can get from s

  30. Theorem: no augmenting path  max-flow Proof: no augmenting path  no path from s to t in the residual network all edges from C to V-C saturated C s t vertices to which we can get from s

  31. Theorem: no augmenting path  max-flow Theorem 2: max-flow = min-cut

  32. Is there an augmenting path? Is there a path from s to t in the residual network? time O(E) F  zero-flow while  augmenting path p improve F along p time = ?

  33. F  zero-flow while  augmenting path p improve F along p 100 100 0 0 1 0 0 0 100 100

  34. F  zero-flow while  augmenting path p improve F along p 100 100 0 0 1 0 0 0 100 100

  35. F  zero-flow while  augmenting path p improve F along p 100 100 1 0 1 1 0 1 100 100

  36. F  zero-flow while  augmenting path p improve F along p 100 100 1 0 1 1 0 1 100 100

  37. F  zero-flow while  augmenting path p improve F along p 100 100 1 0 1 1 0 1 100 100

  38. F  zero-flow while  augmenting path p improve F along p 100 100 1 2 1 -1 2 1 100 100

  39. F  zero-flow while  augmenting path p improve F along p 100 100 1 2 1 -1 2 1 100 100

  40. F  zero-flow while  augmenting path p improve F along p 100 100 3 2 1 1 2 3 100 100

  41. F  zero-flow while  augmenting path p improve F along p Assume that the capacities are integers. Then the number of augementations is bounded by F* (=max-flow) running time O(E F*)

  42. F  zero-flow while  augmenting path p improve F along p can a good choice of p improve the algorithm? running time O(E F*)

  43. F  zero-flow while  augmenting path p improve F along p 1. choose path p which increases the flow the most (i.e., p has the maximum bottleneck capacity) 2. choose path p which has the fewest number of edges

  44. 1. choose path p which increases the flow the most (i.e., p has the maximum bottleneck capacity) THEOREM: number of augmenting steps O(E log F*)

  45. Graph G flow f in G Residual network Gf f’=flow in Gf Claim: h=f+f’ is a flow in G SKEW SYMMETRY: h(u,v)=f(u,v)+f’(u,v)=-f(v,u)-f’(v,u) = - (f(v,u) + f’(v,u)) = - h(v,u) FLOW CONSERVATION: u h(u,v) = u (f(u,v)+f’(u,v))= u f(u,v) + u f’(u,v) = 0

  46. Graph G flow f in G Residual network Gf f’=flow in Gf Claim: h=f+f’ is a flow in G CAPACITY CONSTRAINTS: f’(u,v) c(u,v)-f(u,v) h(u,v)=f(u,v)+f’(u,v)  c(u,v)

  47. Graph G flow f in G Residual network Gf f’=flow in Gf flow h in G Claim: f’=h-g is a flow in Gf CAPACITY CONSTRAINTS: f’(u,v)= h(u,v)-f(u,v)c(u,v)-f(u,v)

  48. Claim: any flow f in G is a sum of at most |E| path flows Induction on the number of non-zero edges in f

  49. Claim: any flow f in G is a sum of at most |{e E;f(e)0}| path flows Induction on the number of non-zero edges in f f = flow G’ with capacities c’(u,v)=max{0,f(u,v)} p = augmenting path from zero flow in G’ p has bottleneck edge  f-p has less non-zero edges than f f = p + (f-p)

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