1. Ford-Fulkerson Recap

2. Residual Flow Network s w v t 5 (3) 2 (0) 3 (3) G Gf v s t w
c(e) (f(e))

3. Residual Flow Network s w v t 5 (3) 2 (0) 3 (3) G Gf v 2 3 s t w
Forward edge.
capacity: c(e)-f(e) w
Backward edge.
capacity: f(e)

4. Residual Flow Network s w v t
5 (3)
2 (0)
3 (3) G Gf v s 2 3 t w

5. Residual Flow Network s w v t
5 (3)
2 (0)
3 (3) G Gf 3 v s 2 3 t w

6. Residual Flow Network s w v t 5 (3) 2 (0) 3 (3) G Gf 3 v 2 s 2 3 t 3 22 s 2 3 t 3 2 w

7. Delete all edges with residual capacity 0!
Residual Flow Network s w v t
5 (3)
2 (0)
3 (3) G Gf s w v t 2 3
Delete all edges with residual capacity 0!

8. Gf G u v 2 4 u v
6 (4)
saturated u v
6 (6) u v 6
empty u v 6 u v
6 (0)

9. Gf u v 2 1 4 G
6 (4) v u
3 (2)

10. Ford-Fulkerson Algorithm
Start: f(e)=0 for all edges e
Compute Gf.
While Gf contains an s-t path:
Let P be any simple s-t path in Gf.
Let Δ(P)=mine Pcf(e).
For each edge e=(u,v) on P:
If e is a forward edge, increase f(e) by Δ
Otherwise, let e'=(v,u) be the edge of G corresponding to e. Reduce f(e’) by Δ.
Recompute Gf.
Iterate:
improve the flow
re-compute Gf

11. Ford-Fulkerson Algorithm
Start: f(e)=0 for all edges e
Compute Gf.
While Gf contains an s-t path:
Let P be any simple s-t path in Gf.
Let Δ(P)=mine Pcf(e).
For each edge e=(u,v) on P:
If e is a forward edge, increase f(e) by Δ
Otherwise, let e'=(v,u) be the edge of G corresponding to e. Reduce f(e’) by Δ.
Recompute Gf.
Flow Update: push Δ(P) flow units along P

12. Ford-Fulkerson Algorithm
Start: f(e)=0 for all edges e
Compute Gf.
While Gf contains an s-t path:
Let P be any simple s-t path in Gf.
Let Δ(P)=mine Pcf(e).
For each edge e=(u,v) on P:
If e is a forward edge, increase f(e) by Δ
Otherwise, let e'=(v,u) be the edge of G corresponding to e. Reduce f(e’) by Δ.
Recompute Gf.

13. We Saw: Correctness proof: optimality is missing
The flow is always feasible and integral
The algorithm terminates
Each iteration takes O(m) time
Flow value increases by at least 1 in each iteration.
Number of iterations at most sum of capacities.
Correctness proof: optimality is missing
Running time: want faster

14. Is the Flow Optimal? s w v t
5 (1)
2 (2)
3 (3)
2(2)

15. Is the Flow Optimal? s w v t 1
100 s w v t
1 (1)
100 (2)

16. Is the Flow Optimal? v 100 s 1 t v 1 (1) 100 (2) w s 1 (1) t 100 (2)

17. Ford-Fulkerson Algorithm
Start: f(e)=0 for all edges e
Compute Gf.
While Gf contains an s-t path:
Let P be any simple s-t path in Gf.
Let Δ(P)=mine Pcf(e).
For each edge e=(u,v) on P:
If e is a forward edge, increase f(e) by Δ
Otherwise, let e'=(v,u) be the edge of G corresponding to e. Reduce f(e’) by Δ.
Recompute Gf.

18. Ford-Fulkerson Algorithm
Start: f(e)=0 for all edges e
Compute Gf.
While Gf contains an s-t path:
Let P be any simple s-t path in Gf.
Let Δ(P)=mine Pcf(e).
For each edge e=(u,v) on P:
If e is a forward edge, increase f(e) by Δ
Otherwise, let e'=(v,u) be the edge of G corresponding to e. Reduce f(e’) by Δ.
Recompute Gf.

19. Edmonds-Karp Algorithm
Start: with f(e)=0 for all edges e
Compute Gf.
While Gf contains an s-t path:
Let P be the shortest simple s-t path in Gf.
Let Δ(P)=mine Pcf(e).
For each edge e=(u,v) on P:
If e is a forward edge, increase f(e) by Δ
Otherwise, let e'=(v,u) be the edge of G corresponding to e. Reduce f(e’) by Δ.
Recompute Gf.

20. Changes to Gf Gf s t P

21. Changes to Gf Gf’ s t Claim: Only edges of P may disappear
At least one of them has to disappear
The only new edges are anti-parallel to edges of P

22. Claim: distance from s to t in the residual graph never decreases

23. Claim: distance from s to t in the residual graph never decreases

24. d is length of augmenting path
BFS from s in Gf t s
d is length of augmenting path
d+1

25. forward-looking edges
BFS from s in Gf
shortcut edge
forward-looking edges t s
d+1

26. backward-looking edges
BFS from s in Gf
backward-looking edges t s
d+1

27. sideways-looking edges
BFS from s in Gf
sideways-looking edges t s
d+1

28. BFS from s in Gf t s
d+1

29. BFS from s in Gf t s
d+1

30. BFS from s in Gf t s d+1 We do not create shortcut edges!
The s-t distance cannot decrease!
d+1

31. Claim: a phase has no more than m iterations.

32. BFS from s in Gf at the Start of the Phase
d+1

33. BFS from s in Gf at the Start of the Phase
d+1

34. BFS from s in Gf at the Start of the Phase
d+1

35. BFS from s in Gf at the Start of the Phase
d+1

36. BFS from s in Gf at the Start of the Phase
d+1

37. BFS from s in Gf at the Start of the Phase
d+1

38. BFS from s in Gf at the Start of the Phase
d+1

39. BFS from s in Gf at the Start of the Phase
We never create shortcut edges
As long as augmenting path length is d, it has to visit every layer in turn
Induction
Conclusion: at most m iterations to a phase! t s
We do not create new forward-looking edges in a phase.
d+1
At least one forward-looking edge deleted in each iteration