1. Max Flow Min Cut

2. Theorem The maximum value of an st-flow in a digraph equals the minimum capacity of an st-cut. Theorem If every arc has integer capacity, then in a maximum flow every arc has integer flow.

3. 1 2 3 2 1 2 3 3 3 3 1 1 1 1 2 3 30

4. 1 2 3 2 1 2 3 3 3 3 1 1 1 1 2 3 30 Find a maximum st-flow and a minimum st-cut s t

5. 1 2 3 2 1 2 3 3 3 3 1 1 1 1 2 3 30 Algorithm s t 1 3 5 2 4 8 6 9 7

6. 1 2 3 2 1 2 3 3 3 3 1 1 1 1 2 3 30 s t 1 3 5 2 4 8 6 9 7

7. 1 2 3 2 1 2 3 3 3 3 1 1 1 1 2 3 30 s t 1 3 5 2 4 8 6 9 7

8. 1 2 3 2 1 2 3 3 3 3 1 1 1 1 2 3 30 s t 1 3 5 2 4 8 6 9 7

9. 1 2 3 2 1 2 3 3 3 3 1 1 1 1 2 3 30 s t 1 3 5 2 4 8 6 9 7

10. 1 2 3 2 1 2 3 3 3 3 1 1 1 1 2 3 30 s t 1 3 5 2 4 8 6 9 7

11. 1 2 3 2 1 2 3 3 3 3 1 1 1 1 2 3 30 s t 1 3 5 2 4 8 6 9 7 STOP

12. 1 2 3 2 1 2 3 3 3 3 1 1 1 1 2 3 30 Algorithm s t 1 3 5 2 4 8 6 9 7 STOP

13. 1 2 3 2 1 2 3 3 3 3 1 1 1 1 2 3 30 Algorithm s t 1 3 5 2 4 8 6 9 7 STOP

14. 1 2 3 2 1 2 3 3 3 3 1 1 1 1 2 3 30 Algorithm s t 1 3 5 2 4 8 6 9 7 STOP

15. 1 2 3 2 1 2 3 3 3 3 1 1 1 1 2 3 30 Algorithm s t 1 3 5 2 4 8 6 9 7

16. 1 2 3 2 1 2 3 3 3 3 1 1 1 1 2 3 30 s t 1 3 5 2 4 8 6 9 7 STOP

17. 1 2 3 2 1 2 3 3 3 3 1 1 1 1 2 3 30 Algorithm s t 1 3 5 2 4 8 6 9 7 STOP

18. 1 2121 3 2 1 2 3131 3 3131 3 1 1 1 1 2121 3 30 Algorithm s t 1 3 5 2 4 8 6 9 7 STOP

19. 1 2121 3 2 1 2 3131 3 3131 3 1 1 1 1 2121 3 30 Algorithm s t 1 3 5 2 4 8 6 9 7

20. 1 2121 3 2 1 2 3131 3 3131 3 1 1 1 1 2121 3 30 s t 1 3 5 2 4 8 6 9 7

21. 1 2121 3 2 1 2 3131 3 3131 3 1 1 1 1 2121 3 30 s t 1 3 5 2 4 8 6 9 7

22. 1 2121 3 2 1 2 3131 3 3131 3 1 1 1 1 2121 3 30 s t 1 3 5 2 4 8 6 9 7 STOP

23. 1 2121 3 2 1 2 3131 3 3131 3 1 1 1 1 2121 3 30 Algorithm s t 1 3 5 2 4 8 6 9 7 STOP

24. 1 2121 3 2 1 2 3131 3 3232 3 1 1 1 1 2121 3 30 Algorithm s t 1 3 5 2 4 8 6 9 7 STOP

25. 1 2121 3 2 1 2 3131 3 3232 3 1 1 1 1 2121 3 30 Algorithm s t 1 3 5 2 4 8 6 9 7 STOP

26. 1 2121 3 2 1 2 3131 3 3232 3 1 1 1 1 2121 3 30 Minimum Cut s t 1 3 5 2 4 8 6 9 7

27. 1 2121 3 2 1 2 3131 3 3232 3 1 1 1 1 2121 3 30 s t 1 3 5 2 4 8 6 9 7

28. 1 2121 3 2 1 2 3131 3 3232 3 1 1 1 1 2121 3 30 s t 1 3 5 2 4 8 6 9 7

29. 1 1 3 1 1 2 3 1 1 3 3 s t 1 3 5 2 4 8 6 9 7 So f is a max flow

30. 1 2 3 2 1 2 3 3 3 3 1 1 1 1 2 3 30 Algorithm s t 1 3 5 2 4 8 6 9 7

31. 1 2 3 2 1 2 3 3 3 3 1 1 1 1 2 3 30 s t 1 3 5 2 4 8 6 9 7

32. 1 2 3 2 1 2 3 3 3 3 1 1 1 1 2 3 30 s t 1 3 5 2 4 8 6 9 7

33. 1 2 3 2 1 2 3 3 3 3 1 1 1 1 2 3 30 s t 1 3 5 2 4 8 6 9 7

34. 1 2 3 2 1 2 3 3 3 3 1 1 1 1 2 3 30 s t 1 3 5 2 4 8 6 9 7

35. 1 2 3 2 1 2 3 3 3 3 1 1 1 1 2 3 30 s t 1 3 5 2 4 8 6 9 7

36. 1 2 3 2 1 2 3232 3232 3 3131 1 1 1 1 2 3 30 s t 1 3 5 2 4 8 6 9 7

37. Hall’s Theorem from Max Flow Min Cut

39. 11111 Direct all edges from s to t and assign all arcs unit capacity Adds and t, adjacent to all of A and B respectively.

40. We have to show that Hall’s Condition gives a 1-factor.

41. A flow of value is enough to guarantee a 1-factor.

42. So all we have to do is show for each cut S.