Post on 31-Dec-2015
description
Lecture 16
Max Flows 3
Preflow-Push Algorithms
Review
The Ford Fulkerson Maximum Flow Algorithm
Begin x := 0;
create the residual network G(x);
while there is some directed path from s to t in G(x) do
begin
let P be a path from s to t in G(x);
∆:= δ(P);
send ∆ units of flow along P;
update the r's;
End
end {the flow x is now maximum}.
e(3)=1
Exact distance: d(i) equals the length of the shortest path from node i to node t in G(x)
The Goldberg-Tarjan Preflow Push Algorithm for the
Maximum Flow Problem
Preflow Push
4
11
4
21
2
3
3
1
s
2
4
5
3
t
This is the original network, and the original residual network.
Initialize Distances
4
11
4
21
2
3
3
1
s
2
4
5
3
t
The node label henceforth will be the distance label.
0
5
4
3
2
1
t
4 53
s2
d(j) is at most the distance of j to t in G(x)
02
2
1
1
1
Saturate Arcs out of node s
4
11
4
21
2
3
3
1
s
2
4
5
3
t
Saturate arcs out of node s.
0
5
4
3
2
1
t
4 53
s2
Move excess to the adjacent arcs
02
2
1
1
1
3
2
12
13
6s
s
Relabel node s after all incident arcs have been saturated.
2
3 4
Select, then relabel/push
4
11
4
21
2
3
3
1
s
2
4
5
3
t
Select an active node, that is, one with excess
0
5
4
3
2
1
t
4 53
s2
Push/Relabel
02
2
1
1
1
3
2
12
13
6s
s
Update excess after a push
11
Select, then relabel/push
4
11
4
21
2
3
3
1
s
2
4
5
3
t
Select an active node, that is, one with excess
0
5
4
3
2
1
t
4 53
s202
2
1
1
1
3
2
12
1
6s
s
No arc incident to the selected node is admissible. So relabel.
11
23
3
Select, then relabel/push
4
11
4
21
2
3
3
1
s
2
4
5
t
Select an active node, that is, one with excess
0
5
4
3
2
1
t
4 5
s202
2
1
1
3
2
26
s
s
Push/Relabel
1
2
3
21
31
2
3
Select, then relabel/push
4
11
4
21
2
3
3
1
s
2
4
5
t
Select an active node.
0
5
4
3
2
1
t
4 5
s202
2
1
1
3
2
26
s
s
Push/Relabel
1
2
3
21
31
2
3 31
32
2
1
5
Select, then relabel/push
4
11
4
21
2
3
3
1
s
2
4
5
t
Select an active node.
0
5
4
3
2
1
t
4 5
s202
2
1
12
26
s
s
Push/Relabel
1
2
3
21
31
2
31
3
2
1
Select, then relabel/push
4
11
4
21
2
3
3
1
s
2
4
5
t
Select an active node.
0
5
4
3
2
1
t
4 5
s202
2
1
12
26
s
s
There is no incident admissible arc. So Relabel.
1
2
3
21
31
2
31
3
2
12
5
5
Select, then relabel/push
4
11
4
21
2
3
3
1
s
2
4
5
t
Select an active node.
0
5
4
3
2
1
t
4
5s202
2
1
12
26
s
s
Push/Relabel
1
2
3
21
3
2
31
3
2
22
1
41
Select, then relabel/push
4
11
4
21
2
3
3
1
s
2
4
5
t
Select an active node.
0
5
4
3
2
1
t
4
5s202
2
1
12
26
s
s
There is no incident admissible arc. So relabel.
1
2
3
21
3
2
31
3
2
22
1
41
3
5
5
Select, then relabel/push
4
11
4
21
2
3
3
1
s
2
4
5
t
Select an active node.
0
5
4
3
2
1
t
4 5
s202
2
1
12
26
s
s
Push/Relabel
1
2
3
21
3
2
31
3
2
22
1
41
3
5
5
3
Select, then relabel/push
4
11
4
21
2
3
3
1
s
2
4
5
t
Select an active node.
0
5
4
3
2
1
t
4 5
s202
2
1
12
26
s
s
Push/Relabel
1
2
3
21
3
22
222
41
3
5
5
3
1
141
4
Select, then relabel/push
4
11
4
21
2
3
3
1
s
2
4
5
t
Select an active node.
0
5
4
3
2
1
t
4 5
s202
2
1
12
26
s
s
Push/Relabel
1
2
3
21
3
22
222
41
3
5
5
3
1
4
24
2
2
Select, then relabel/push
4
11
4
21
2
3
3
1
s
2
4
5
t
Select an active node.
0
5
4
3
2
1
t
4 5
s02
2
1
12
26
s
s
Push/Relabel
1
2
3
21
3
22
222
41
3
5
5
3
1
4
4
2
4
Select, then relabel/push
4
11
4
21
2
3
3
1
s
2
4
5
t
Select an active node.
0
5
4
3
2
1
t
4 5
s02
2
1
12
26
s
s
Push/Relabel
1
2
3
21
3
21
322
41
3
5
5
3
1
4
4
2
4 35
5
5
Select, then relabel/push
4
11
4
21
2
3
3
1
s
2
4
5
t
One can keep pushing flow between nodes 2 and 5 until eventually all flow returns to node s.
0
5
4
3
2
1
t
4 5
s02
2
1
12
26
s
s
There are no paths from nodes 2 and 5 to t, and there are ways to speed up the last iterations.
1
2
3
21
3
21
322
41
3
5
5
3
1
4
4
2
4 35
5
5
Assignments
Read sections 7.3 and 7.6
Exercises
7.4