1. Maximum flow: The preflow/push method of Goldberg and Tarjan (87)

2. Definitions G=(V,E) is a directed graph
capacity c(v,w) for every v,w V: If (v,w)  E then c(v,w) = 0
Two distinguished vertices s and t. s 4 3 1 a b 3 3 2 c d 4 3 t

3. Definitions (cont)
A flow is a function on the edges which satisfies the following requirements
f(v,w) = -f(w,v) skew symmetry
f(v,w)  c(v,w)
For every v except s and t wf(v,w) = 0
The value of the flow |f| = wf(s,w)
The maxflow problem is to find f with maximum value

4. Flows and s-t cuts Let (X,X’) be a cut such that s X, t X’.
Flow is the same across any cut:
f(X,X’) = f(v,w) = f(v,w) - f(v,w) = |f| - 0 = |f|
v X, w  X’
v X, w  V
v X, w  X so
|f|  cap(X,X’) = c(v,w)
The value of the maximum flow is smaller than the minimum capacity of a cut.

5. More definitions
The residual capacity of a flow is a function r on the edges such that
r(v,w) = c(v,w) - f(v,w) a
2, 1 d
Interpretation: We can push r(v,w) more flow from v to w by increasing f(v,w) and decreasing f(w,v)

6. More definitions (cont)
We define the residual graph R on V such that there is an arc from v to w with capacity r(v,w) for every v and w such that r(v,w) > 0
An augmenting path p  R is a path from s to t in R
r(p) = min r(v,w)
(v,w)  p
We can increase the flow by r(p)

7. Example 1 3 3 4 1 3 3 1 1 3 1 3 1 2 2 1 2 2 1 2 1 3 4 1 3 3 3
The residual network
A flow

8. Basic theorem (1) f is max flow <==>
(2) There is no augmenting path in R <==>
(3) |f| = cap(X,X’) for some X
Proof. (3) ==> (1), (1) ==> (2) obvious
To prove (2) ==>(3) let X be all vertices reachable from s in R. By assumption t  X. So (X,X’) is an s-t cut. Since there is no edge from X to X’ in R
|f| = f(X,X’) = f(v,w) = c(v,w) = cap(X,X’)

9. Augmenting path methods
Repeat the following step:
Find an augmenting path in R, increase the flow, update R
Stop when s and t are disconnected in R.
Need to be careful about how you choose those augmenting paths !
The best algorithm in this family is Dinic’s algorithm, that can be implemented in O(nmlog(n)) time

10. But we’ll go for the preflow/push method

11. Distance labels Defined with respect to residual capacities d(t) = 0
d(v) ≤ d(w) + 1 if r(v,w) > 0

12. Example (distance labels)1 3 3 4 1 3 3 1 1 3 1 3 1 2 2 1 2 2 1 2 1 3 4 1 3 3 3
The residual network
A flow

13. Example (distance labels)1 3 3 3 4 1 3 3 1 1 2 2 3 1 3 1 2 2 1 2 2 1 2 1 1 1 4 1 3 3 3 3
The residual network
A flow

14. Example (distance labels)1 3 3 3 4 1 3 3 1 1 2 2 3 1 3 1 2 2 1 2 2 1 2 1 1 2 4 1 3 3 3 3
The residual network
A flow

15. Example (distance labels)1 4 3 3 4 1 3 3 1 1 2 3 3 1 3 1 2 2 1 2 2 1 2 1 1 3 4 1 3 3 3 3
The residual network
A flow

16. Distance labels – basic lemma
Lemma: d(v) is a lower bound on the length of the shortest path from v to the sink
Proof: Let the s.p. to the sink be: v v1 v2 t
d(v) ≤ d(v1) + 1 ≤ d(v2) ≤ d(t) + k = k

17. Preflow (definition)
A preflow is a function on the edges which satisfies the following requirements
f(v,w) = -f(w,v) skew symmetry
f(v,w)  c(v,w)
For every v, except s and t, vf(v,w) ≥ 0
Let e(w) = vf(v,w) be the excess at the node v
(we’ll also have e(t) ≥ 0, and e(s) ≤ 0)

18. Example (preflow) 2 2 2 3 1 3 2 s t 3 2 1
Nodes with positive excess are called active.
The preflow push algorithm will try to push flow from active nodes towards the sink, relying on d( ).

19. Initialization (preflow)3 4 2 1 3 3 4 4 3 4 1 3 3 2 4 3

20. Initialization (distance labels)6 3 3 4 4 3 4 3 4 1 1 3 3 3 3 2 2 4 3 4 3
Note: s must be disconnected from t when d(s) = n, and the labeling is valid…

21. Admissible arc in the residual graphv w
d(v) = d(w) + 1

22. The preflow push algorithm
While there is an active node {
pick an active node v and push/relabel(v) }
Push/relabel(v) {
If there is an admissible arc (v,w) then {
push  = min {e(v) , r(v,w)} flow from v to w
} else {
d(v) := min{d(w) + 1 | r(v,w) > 0} (relabel) }

23. Example (preflow-push)2 4 1 1 4

24. 4 4 2 4 1 4 1 1 4 1

25. 4 4 2 4 1 4 1 1 4 1
relabel 4 2 1 4 1 4

26. 4 4 2 1 4 1 4 1 1 4 1
push 4 2 1 1 4 1 4

27. 2 4 2 1 2 4 1 4 1 1 4 1 4 2 1
push 1 4 1 4

28. relabel 1 4 2 1 2 4 1 1 4 1 1 4 2 4 2 1 1 4 1 4

29. 1 4 2 5 2 4 1 1 4 1 1 4 2
push 4 2 5 1 4 1 4

30. 3 2 5 2 4 1 1 4 1 1 4 2 2 5 3 1 1 4 1 4
relabel

31. 3 2 5 2 4 1 1 4 1 1 4 1 2 2 5 3 1 1 4 1 4
push 1

32. 3 2 5 2 4 1 1 4 1 2 1 4 1 2 5 3 1 1 4 1 2 1 2

33. Correctness
Lemma 1: The source is reachable from every active vertex in the residual network
Proof:
Assume that’s not the case: v s S
Which means that no flow enters S..

34. Correctness (cont)
Corollary: There is an outgoing arc incident with every active vertex
so assuming distance labels are valid, we can always either push or relabel an active node.

35. Correctness (cont) Lemma 1: Distance labels remain valid at all times
Proof:
By induction on the number of pushing and relabeling operations.
For relabel this is clear by the definition of relabel
For push: w v
d(v) = d(w) + 1 so even if we add (w,v) to the residual network then it is still a valid labeling

36. Correctness (cont)
Corollary: So we can push-relabel as long as there is an active vertex
Lemma: When (and if) the algorithm stops the preflow is a maximum flow
Proof:
It is a flow since there is no active vertex.
It is maximum since the sink is not reachable from the source in the residual network. (d(s) = n, and the labeling is valid)

37. Complexity analysis

38. Well, lets do another example first2 4 2 2 2 1 4

39. 4 2 4 2 2 6 2 2 1 4

40. relabel 4 2 4 2 2 6 2 2 1 4 2 4 2 2 6 2 1 4

41. push 4 2 1 4 2 2 6 2 2 1
push 4 2 1 4 2 2 6 2 1 4

42. 4 2 2 1 4 2 2 2 6 2 2 1
relabel 4 2 1 4 2 2 6 2 1 4

43. 4 2 2 1 4 2 2 2 2 6 2 2
push 1 4 2 1 4 2 2 2 6 2 1 4

44. relabel 4 2 2 1 4 2 2 2 2 2 6 2 2 1 4 2 1 4 2 2 2 6 2 1 4

45. 4 2 2 3 4 2 2 2 2 2 6 2 2
push 1 4 2 3 4 2 2 2 6 2 1 4

46. 4 2 2 3 4 2 2 2 2 6 2 2 1 4 2 3 4 2 2 2 6 2 1 4

47. 4 2 2 7 4 2 2 2 2 6 6 2 2 1 4
Skipping some steps… 2 7 4 2 2 6 6 2 1 4

48. Complexity analysis Observation: d(v) increases when we relabel v !
Lemma: d(v) ≤ 2n-1
Proof: v v1 v2 s
d(v) ≤ d(v1) + 1 ≤ d(v2) ≤ d(s) + (n-1) = 2n-1

49. Complexity analysis (cont)
Lemma: The # of relabelings is (2n-1)(n-2) < 2n2
Proof:
At most 2n-1 per each node other than s and t

50. Complexity analysis (cont)
Def: Call a push saturating if min{e(v), r(v,w)} = r(v,w)
Lemma: The # of saturating pushes is at most 2nm
Proof: Before another saturating push on (v,w), we must push from w to v.
d(w) must increase by at least 2
Since d(w) ≤ 2n-1, this can happen at most n times

51. Nonsaturating pushes
Lemma: The # of nonsaturating pushes is at most 4n2m
Proof:
Let Φ = Σv active d(v)
Decreases (by at least one) by every nonsaturating push
Increases by at most 2n-1 by a saturating push : total increase (2n-1)2nm
Increases by each relabeling: total increase < (2n-1)(n-2)

52. Implementation
Maintain a list of active nodes, so finding an active node is easy
Given an active node v, we need to decide if there is an admissible arc (v,w) to push on ? v
All edges, not only those in R
current edge

53. Current edge v
current edge
If the current edge (v,w) is admissible, push on it (updating the list of active vertices)
Otherwise, advance the current edge pointer
if you are on the last edge, relabel v and set the current edge to be the first one.

54. Is this implementation correct?
Lemma: When we relabel v there is no admissible arc (v,w)
Proof: After we scanned (v,w) either (v,w) dropped off the residual network or d(v) ≤ d(w)
If d(v) ≤ d(w) then this must be the case now since v has not been relabeled.
If (v,w) got back on the list of v since it was scanned then when that happened d(w) = d(v) + 1  d(v) ≤ d(w) and this must be the case now

55. Analysis
Lemma: The total time spent at v between two relabelings of v is Δv plus O(1) per push out of v
Summary: Since we relabel v at most (2n-1) times we get that the total work at v is O(nΔv) + O(1) per push out of v.
Summing over all vertices we get that the total time is O(nm) + #of pushes
 O(n2m)

56. Reducing the # of nonsaturating pushes
Maintain the list of active vertices as a FIFO queue (Q)
Discharge the first vertex of the queue:
Discharge(v) {
While v is active and hasn’t been relabeled then push/relabel(v).
(If the loop stops because v is relabeled then add v to the end of Q) }

57. Example (FIFO order) 2 4 2 2 2 1 4

58. 4 2 4 2 2 6 2 2 1 4

59. relabel 4 4 2 4 z 2 2 2 6 2 2 1 y w x 4 u v
Q: z y 2 4 2 2 6 2 1 4

60. 4 4 2 1 4 z 2 2 2 6 2 2 1 y w x 4
relabel u v
Q: y z 2 1 4 2 2 6 2 1 4

61. 4
push 4 2 1 4 z 2 2 2 2 6
push 2 2 1 y w x 4 u v
Q: z y 2 1 4 2 2 2 6 2 1 4

62. 4 2 2 1 4 z 2 2 2 2 2 6 2 2
push 1 y w x 4 2 u v
Q: y u 2 1 4 2 2 2 6 2 1 4

63. 2 4 2 2 1 4 z 2 2 2 2 2 6 2 2 1 y w x 4 2 u v
Q: u z
relabel 2 1 4 2 2 2 6 2 1 4

64. relabel 2 4 2 2 1 4 z 2 2 2 2 2 6 2 2 1 y w x 4 2 1 u v
Q: z u 2 1 4 2 2 2 6 2 1 4 1

65. 4 2 2 3 4 z 2 2 2 2 2 6 2 2 1 y w x 4 1 u v
Q: u z 2 3 4 2 2 2 6 2 1 4

66. Passes
Pass 1: Until you finish discharging all vertices initially in Q
Pass i: Until you finish discharging all vertices added to Q in pass (i-1)

67. relabel 4 4 2 4 z 2 2 2 6 2 2 1 y w x 4 u v
Q: z y 2 4 2 2 6 2 1 4

68. 4 4 2 1 4 z 2 2 2 6 2 2 1 y w x 4
relabel u v
Q: y z 2 1 4 2 2 6 2 1 4

69. End of pass 1 4 4 2 1 4 z 2 2 2 2 6 2 2 1 y w x 4 u v
Q: z y 2 1 4 2 2 2 6 2 1 4

70. 4
push 4 2 1 4 z 2 2 2 2 6
push 2 2 1 y w x 4 u v
Q: z y 2 1 4 2 2 2 6 2 1 4

71. 4 2 2 1 4 z 2 2 2 2 2 6 2 2
push 1 y w x 4 2 u v
Q: y u 2 1 4 2 2 2 6 2 1 4

72. End of pass 2 2 4 2 2 1 4 z 2 2 2 2 2 6 2 2 1 y w x 4 2 u v
Q: u z 2 1 4 2 2 2 6 2 1 4

73. 2 4 2 2 1 4 z 2 2 2 2 2 6 2 2 1 y w x 4 2 u v
Q: u z
relabel 2 1 4 2 2 2 6 2 1 4

74. relabel 2 4 2 2 1 4 z 2 2 2 2 2 6 2 2 1 y w x 4 2 1 u v
Q: z u 2 1 4 2 2 2 6 2 1 4 1

75. End of pass 3 4 2 2 3 4 z 2 2 2 2 2 6 2 2 1 y w x 4 1 u v
Q: u z 2 3 4 2 2 2 6 2 1 4

76. Analysis Note that we still have the O(n2m) bound
How many passes are there ?
Let Φ = maxactive vd(v)
1) If the algorithm does not relabel during a pass then Φ decreases by at least 1 (each active node at the beginning of a pass moved its excess to a vertex with lower label)
2) If we relabel then Φ may increase by at most the maximum increase of a distance label
There are at most O(n2) passes of the second kind.
These passes increase Φ by at most O(n2) 
There are at most O(n2) passes of the first kind

77. Analysis (Cont) So we have O(n2) passes
In each pass we have at most one nonsaturating push per vertex
 O(n3) nonsaturating pushes
 O(n3) total running time

78. A faster implementation
Maintain a (dynamic) forest of some of the admissible current edges

79. Reminder: Admissible arc in the residual graphv w
d(v) = d(w) + 1

80. A faster implementation
Maintain a (dynamic) forest of some of the admissible current edges

81. A faster implementation
Maintain a (dynamic) forest of some of the admissible current edges
Active guys are among the roots

82. At a high level the algorithm is almost the same
While there is an active node in Q {
Let v be the first in Q
discharge(v) }
discharge(v) {
While v is active and hasn’t been relabeled then Treepush/relabel(v).
(If the loop stops because v is relabeled then add v to the end of Q) }

83. A faster implementation
Q: v…..  discharge(v)  Treepush/relabel(v) v w

84. Case 1: (v,w) is admissible
link(v,w,rf(v,w)),
(v,c) = findmin(v), addcost(v,-c)
Let (u,c) = findmin(v)
If c=0 cut(u) and repeat
If e(v) > 0 and v is not a root then repeat

85. Case 2: (v,w) is not admissible
If (v,w) is not the last edge then advance the current edge
If (v,w) is the last edge we relabel v and perform cut(u) for every child u of v

86. Analysis O(1) work + tree operations in Treepush/relabel +
O(1) work + tree operations per cut
How many cuts do we do ?
O(mn)
How many times do we call Treepush/relabel ?
O(mn) + # of times we add a vertex to Q

87. Analysis (Cont) How many times a vertex can become active ?
We can charge each new active vertex to a link or a cut
 A vertex becomes active O(mn) times
Summary: we do O(mn) dynamic tree operations
We’ll see how to do those in O(log n) each so we get running time of O(mnlog n)

88. Can we improve on that ?
Notice that we have not really used the fact that Q is a queue, any list would do !

89. Idea: Don’t get the trees to grow too large
Case 1: (v,w) is admissible v
We won’t do the link if a too large tree is created (say larger than k) w
link(v,w,rf(v,w)),
(v,c) = findmin(v), addcost(v,-c)
Let (u,c) = findmin(v)
If c=0 cut(u) and repeat
If e(v) > 0 and v is not a root then repeat

90. If we are about to create a tree with at least k vertices
Push from v to w min{e(v),rf(v,w)} flow
(w,c) = findmin(w), addcost(w,-c)
Let (u,c) = findmin(w)
If c=0 cut(u) and repeat
If e(w) > 0 and w is not a root then repeat

91. What collapses in our analysis ?
Note r may become active and we cannot charge it to a link or a cut ! v w r
So we cannot say that the # of insertions into Q is O(mn)

92. How do we recover ? Either Tv or Tr is large ≥ k/2
May assume that the push from v is not saturating..(there are only O(nm) saturating ones)
There cannot be another such push involving v as the pushing root and r as the “becoming active” node in a phase
Either Tv or Tr is large ≥ k/2

93. Tv Tr v w r
Charge it to the large tree if it existed at the beginning of the phase
Charge it to the link or cut creating it if it did not exist at the beginning of the phase.

94. Each link is charged once, a cut is charged twice O(mn) such charges over all phases.
At the beginning of a phase we have O(n/k) large trees, each charged once  O(n3/k)
So we get O(mn + n3/k) dynamic tree operations each costs O(log k)
For k=n2/m we get O(mnlog(n2/m))