1. Max Flow / Min Cut

2. Max Flow
Given a directed graph G with a special source node s, a special sink node t
s has no inbound edges, and t has no outbound edges
For each edge e in the graph, c(e) is the given “capacity” of the edge
The capacity must be greater than 0
For simplicity, assume the capacity is an integer or ∞
Define f(e) to be the “flow” along an edge
The flow must be non-negative
The flow must also no greater than the capacity for a given edge
For any given node, the sum of flows of inbound edges must equal the sum of flows of outbound edges (“conservation of flow”)
Exceptions: s may have any amount of outbound flow, and t may have any amount of inbound flow
Question: what is the maximum amount of flow that can be sent from s to t?

3. Max Flow d a e s t g b h 0/10 0/∞ 0/4 0/3 0/∞ 0/10 0/2 0/2 0/3 0/15

4. Max Flow d a e s t g b h 0/10 0/∞ 0/4 0/3 0/∞ 0/10 0/2 0/2 0/3 0/15

5. Max Flow d a e s t g b h 0/10 0/∞ 0/4 0/3 0/∞ 0/10 0/2 0/2 0/3 15/15

6. Max Flow d a e s t g b h 0/10 0/∞ 0/4 0/3 0/∞ 0/10 0/2 0/2 0/3 15/15

7. Max Flow d a e s t g b h 0/10 0/∞ 0/4 0/3 0/∞ 0/10 2/2 0/2 3/3 6/15
1/1 h

8. Max Flow d a e s t g b h 0/10 0/∞ 0/4 0/3 0/∞ 0/10 2/2 0/2 3/3 6/15
1/1 h

9. Max Flow d a e s t g b h 0/10 0/∞ 0/4 0/3 0/∞ 0/10 0/2 0/2 0/3 0/15

10. Max Flow d a e s t g b h 0/10 0/∞ 0/4 0/3 0/∞ 0/10 0/2 0/2 0/3 0/15

11. Max Flow d a e s t g b h 0/10 0/∞ 0/4 0/3 0/∞ 0/10 0/2 0/2 0/3 0/15

12. Max Flow d a e s t g b h 0/10 0/∞ 0/4 0/3 0/∞ 0/10 0/2 2/2 2/3 2/15

13. Max Flow d a e s t g b h 0/10 0/∞ 0/4 0/3 0/∞ 0/10 0/2 2/2 2/3 2/15

14. Max Flow d a e s t g b h 0/10 0/∞ 0/4 0/3 0/∞ 0/10 0/2 2/2 2/3 2/15

15. Max Flow d a e s t g b h 0/10 3/∞ 3/4 3/3 0/∞ 3/10 0/2 2/2 2/3 2/15

16. Max Flow d a e s t g b h 0/10 3/∞ 3/4 3/3 0/∞ 3/10 0/2 2/2 2/3 2/15

17. Max Flow d a e s t g b h 0/10 3/∞ 3/4 3/3 0/∞ 3/10 0/2 2/2 2/3 2/15

18. Max Flow d a e s t g b h 0/10 3/∞ 3/4 3/3 0/∞ 3/10 0/2 2/2 2/3 2/15

19. Max Flow d a e s t g b h 0/10 4/∞ 4/4 3/3 0/∞ 4/10 1/2 2/2 2/3 3/15

20. Max Flow d a e s t g b h 0/10 4/∞ 4/4 3/3 0/∞ 4/10 1/2 2/2 2/3 3/15

21. Max Flow d a e s t g b h 1/10 5/∞ 4/4 3/3 0/∞ 4/10 2/2 2/2 2/3 4/15
0/1
0/1 h

22. Max Flow d a e s t g b h 1/10 5/∞ 4/4 3/3 0/∞ 4/10 2/2 2/2 2/3 4/15
0/1
0/1 h

23. Max Flow d a e s t g b h 1/10 5/∞ 4/4 3/3 0/∞ 4/10 2/2 2/2 2/3 5/15

24. Max Flow d a e s t g b h 1/10 5/∞ 4/4 3/3 0/∞ 4/10 2/2 2/2 2/3 5/15

25. Max Flow d a e s t g b h 1/10 5/∞ 4/4 3/3 0/∞ 4/10 2/2 2/2 2/3 5/15

26. Max Flow d a e s t g b h 1/10 5/∞ 4/4 3/3 0/∞ 4/10 2/2 2/2 2/3 5/15

27. Max Flow d a e s t g b h 1/10 5/∞ 4/4 3/3 0/∞ 4/10 2/2 2/2 2/3 5/15

28. Max Flow d a e s t g b h 1/10 5/∞ 4/4 3/3 0/∞ 4/10 2/2 2/2 2/3 5/15

29. Max Flow / Min Cut
Define a “cut” to be a partition of the nodes into two sets. An s-t cut is one where one set contains s and the other contains t
Define the capacity of an s-t cut to be the sum of the capacities of the edges that “cross the partition boundary” from the s set to the t set
In other words, if the nodes are partitioned into sets A and B with s ∈ A and t ∈ B, the edges from u to v where u ∈ A and v ∈ B
The Min Cut problem is to find the s-t cut with the minimum capacity
The Max Flow / Min Cut Theorem says the answers are the same!
Essentially formalizes the notion of “bottlenecks”
...but does our bottleneck finding/saturating algorithm always work?

30. Max Flow v
0/3
0/2 s t
0/5
0/2
0/3 w

31. Max Flow v
0/3
0/2 s t
0/5
0/2
0/3 w

32. Max Flow v
3/3
0/2 s t
3/5
0/2
3/3 w

33. Max Flow v
3/3
0/2 s t
3/5
0/2
3/3 w
Max flow = 3 ?

34. Max Flow v
3/3
0/2 s t
3/5
0/2
3/3 w
Max flow = 3 ?

35. Max Flow v
3/3
0/2 s t
3/5
0/2
3/3 w
Max flow = 3 ?

36. Max Flow v
3/3
0/2 s t
3/5
0/2
3/3 w
Max flow = 3 ?

37. Max Flow v
3/3
0/2 s t
3/5
0/2
3/3 w
Max flow = 3 ?

38. Max Flow v
0/3
0/2 s t
0/5
0/2
0/3 w

39. Max Flow v
0/3
0/2 s t
0/5
0/2
0/3 w

40. Max Flow v
2/3
2/2 s t
0/5
0/2
0/3 w

41. Max Flow v
2/3
2/2 s t
0/5
0/2
0/3 w

42. Max Flow v
2/3
2/2 s t
0/5
0/2
0/3 w

43. Max Flow v
2/3
2/2 s t
0/5
2/2
2/3 w

44. Max Flow v
2/3
2/2 s t
0/5
2/2
2/3 w

45. Max Flow v
2/3
2/2 s t
0/5
2/2
2/3 w

46. Max Flow v
3/3
2/2 s t
1/5
2/2
3/3 w

47. Max Flow v
3/3
2/2 s t
1/5
2/2
3/3 w
Max flow = 5 !

48. Max Flow Being greedy isn’t quite good enough
At least, the order you pick the paths seems to matter
Key idea: just because we *can* saturate a path doesn’t necessarily mean we *should*
We want to be able to “undo” choices if it turns out they boxed us in a corner
Solution: “residual graphs”
Augment the graph with information that allows algorithm to “undo” or “push flow back”

49. Max Flow v
0/3
0/2 s t
0/5
0/2
0/3 w

50. Max Flow v v s t s t w w Original Graph Residual Graph 3 2 0/3 0/2 5
0/5 2 3
0/2
0/3 w w

51. Max Flow v v s t s t w w Original Graph Residual Graph 3 2 0/3 0/2 5
0/5 2 3
0/2
0/3 w w

52. Max Flow v v s t s t w w Original Graph Residual Graph 3 2 3/3 0/2 2 3
3/5 2 3
0/2
3/3 w w

53. Max Flow v v s t s t w w Original Graph Residual Graph 3 2 3/3 0/2 2 3
3/5 2 3
0/2
3/3 w w

54. Max Flow v v s t s t w w Original Graph Residual Graph 3 2 3/3 0/2 2 3
3/5 2 3
0/2
3/3 w w

55. Max Flow v v s t s t w w Original Graph Residual Graph 3 2 3/3 2/2 4 1
1/5 2 3
2/2
3/3 w w

56. Max Flow v v s t s t w w Original Graph Residual Graph 3 2 3/3 2/2 4 1
1/5 2 3
2/2
3/3 w w

57. Max Flow: Ford-Fulkerson Method
Given graph G, define GR (the residual graph) to be the same graph but with only capacity (no flow) and capacities denoted by cR(e).
Maintain that for any edge (u, v) in G that cR(u, v) + cR(v, u) = c(u, v)
Called “skew symmetry”
While there’s a path P from s to t in GR:
Find the minimum capacity cR among all edges in P, call it m
For each edge (u, v) in P:
If (u, v) in G, update f(u, v) += m
Otherwise, (v, u) is in G so update f(v, u) -= m
Update cR(u, v) and cR(v, u) accordingly to match remaining capacity in G and maintain skew symmetry
Called a “method” because the path finding mechanism is not explicitly defined
If you use Breadth-First search, it’s called the Edmonds-Karp algorithm

58. Max Flow: Applications
Many seemingly unrelated problems map nicely into a network flow equivalent
Useful fact: If all edge capacities are integers, the max flow will also be an integer (and the flow along any given edge will also be an integer)
Known as the Integral Flow Theorem

59. Max Flow: Network Connectivity
You have two computers that are indirectly connected through a network of other computer systems. How many internal network disconnections is your connection resilient to? s t
0/1

60. Max Flow: Network Connectivity
You have two computers that are indirectly connected through a network of other computer systems. How many internal network disconnections is your connection resilient to?
1/1
1/1
1/1
0/1
1/1
1/1 s t
0/1
0/1
0/1
0/1

61. Max Flow: School Dance
Boys and girls need to be paired up for the school dance, but the kids only want to be paired with someone that they know. Is such a pairing possible? And if so, what’s the pairing? s b1 b2 b3 b4 g1 g2 g3 g4 t
(all capacities = 1)

62. Min Cut: Project Selection
You have a set of projects pi which will each net a revenue of r(pi). Each project will require purchasing one or more machines qj each of which costs c(qj). Machines can be shared by multiple projects. The goal is to maximize profit.
Let P̄ be the set of projects NOT taken, and Q be the set of machines purchased. Then we want:
max [ ∑i r(pi) - ∑pi∈ P̄ r(pi) - ∑pj∈ Q c(qj) ]
Which can be reformulated as:
∑i r(pi) - min [ ∑pi∈ P̄ r(pi) + ∑pj∈ Q c(qj) ]
OPTIONAL MATERIAL

63. Min Cut: Project Selection
∑i r(pi) - min [ ∑pi∈ P̄ r(pi) + ∑pj∈ Q c(qj) ] s p1 p2 p3 q1 q2 q3 t ∞
r(p1)
r(p2)
r(p3)
c(q1)
c(q2)
c(q3)
OPTIONAL MATERIAL