2. Flows sourcesink s t

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

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

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

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

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

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

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

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

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

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

13. 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

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

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

16. +1 +1 Augmenting path

17. +1 Augmenting path

18. +1 Augmenting path  can improve the flow F  zero-flow while  augmenting path p improve F along p Ford-Fulkerson algorithm

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

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

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

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

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

24. 9 4 4 2 3 4 7 2 2 2 2 4 2 1

25. 9 4 4 2 4 7 2 2 2 4 2 1 1 5

26. 9 4 8 4 2 7 1 1 5 4 7 6 Is there an augmenting path? Is there a path from s to t in the residual network?

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

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

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

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

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

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

33. 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 = ?

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

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

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

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

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

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

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

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

42. 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 * )

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

44. 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

45. 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 * )

46. Graph G flow f in G Residual network G f f’=flow in G f 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

47. Graph G flow f in G Residual network G f f’=flow in G f 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)

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

49. 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

50. 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 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 = flow f = p + (f-p)

51. 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 * ) Claim: any flow f in G is a sum of at most |E| path flows Graph G flow f in G Residual network G f f’=f * -f max-flow f * in G  path with bottleneck capacity  f’/|E|

52. 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 * ) Graph G flow f in G Residual network G f f’=f * -f max-flow f * in G  path with bottleneck capacity  f’/|E| max-flow in the residual graph decreases by a factor of at least (1-1/|E|)

53. 2. choose path p which has the fewest number of edges THEOREM: number of augmenting steps  O(E V)

54. 2. choose path p which has the fewest number of edges level graph = BFS in the residual network

55. 2. choose path p which has the fewest number of edges level graph = BFS in the residual network

56. 2. choose path p which has the fewest number of edges 1) one such edge saturated 2) no new introduced count edges in the residual graph which are on a path of length d from s to t

57. Plain Ford-Fulkerson O(E F * ) Maximum bottleneck capacity O( (E+V log V) E log F  ) Fewest edges O(E 2 V)

58. Max-matching in bipartite graphs

60. st solution using max-flow all edge weights =1

61. Max-matching in bipartite graphs solution using max-flow st all edge weights =1 Plain Ford-Fulkerson O(E F * ) Maximum bottleneck capacity O( (E+V log V) E log F  ) Fewest edges O(E 2 V) which one to use?