Chapter_26_MaxFlow

16
v1
10 4
12
v3
7
20
s
13
9 14
t
4
v2
v4
Residual network
Example
Residual Networks
16
Flows
20 4/16
v1
10 4
12
v3
7
v1
0/10 0/4
4/12
v3
0/7
0/20
s
13
9 14
8 4 4 5 10
t
4
s
4/9 4/14 4/12
u
f(u,v) > 0 c(u,v)
v
u
v
Residual Network
u
f(u,v) > 0 c(u,v)
v
u
v
Residual Network Example
Short Test in Class
• Give the residual network of the next graph
v1
Max-Flow, Min-Cut Theorem
Max-Flow, Min-Cut Theorem
Max-Flow, Min-Cut Theorem
Max-Flow, Min-Cut Theorem
(S,T) is a cut! Why?
Max-Flow, Min-Cut Theorem
No augmenting path
Problem：Time Complexity
Problem：Time Complexity
Problem：Time Complexity
Problem：Time Complexity
Problem：Time Complexity
Problem：Time Complexity
(Proof). Exercise
Simple Properties of Net Flow
Net Flow into Sink
Cut
Flow of A Cut
Capacity of A Cut
Upper Bound on Flow Value
Residual Network
Residual Network
v T (u ,v ) E f cf(u ,v ) 0
f (u, v) c(u, v) c f (u, v) c(u, v) f (u, v)
f (u, v) c(u, v)
uS vT uS vT
f (S, T ) c(S, T ) f
Maximum flow!
Ford-Fulkerson Algorithm
A Story
Rough Idea
Algorithm
Example—Basic Implementation
0/16
v1
0/10 0/4
0/12
v3
0/7
0/20
s
0/9 0/14
t
0/4
0/13
v2
v4
Flow initialization
T&R Team of Algorithm Design College of Computer Science and Engineering, CQU
Algorithm Analysis & Design Introduction to Algorithm
Chapter 26: Maximum Flow
reachable vertices.
v.
– breadth-first tree = a tree with root s that contains all
Definitions on BSF
• Path between vertices u and v: vertices (v1, v2, …, vk) such that u=v1 and v =vk, (vi,vi+1) E, for all 1 i k-1. • Length of the path: Number of edges in the path. • Path is simple if no vertex is repeated.
Capacity and Flow
Capacity and Flow
Flow Value
The Maximum-Flow Problem
Flow Cancellation
Net Flow Definitions
Net Flow Value
Simple Properties of Net Flow
– V ={vV : [v] NIL} + {s} – E ={([v],v)E : v V - {s}}
• G is a breadth-first tree if:
– V consists of the vertices reachable from s – for all vV , there is a unique simple path from s to v in G – the path is also a shortest path from s to v in G.
Example (BFS)
r
s
0
t
u
v
w
Q: s 0
x
frontier
y
Example (BFS)
r
s
0
t
u
v
w
Q: s 0
x
frontier
y
Example (BFS)
r 1
s
0
t
u
v
1 w
x Q: w r 1 1
y
Example (BFS)
r 1
s
0
t 2
u
v
1 w
2
x Q: r t x 1 2 2
y
Example (BFS)
r 1
s
0
t 2
u
2 v
1 w
2
x Q: t x v 2 2 2
y
Example (BFS)
r 1
s
0
t 2
u 3
2 v
1 w
2
x Q: x v u 2 2 3
the maximum flow | f | can increased by increasing the flow on each edge in p
Augmenting Path Example
11/16
v1
12/12
v3
7/7
15/20
s
0/10 1/4 8/13
4/9 11/14
t
4/4 5
y
Example (BFS)
r 1
s
0
t 2
u 3
2 v
1 w
2
x Q: v u y 2 3 3
3 y
Example (BFS)
r 1
s
0
t 2
u 3
2 v
1 w Q: u y 3 3
2
x
3 y
Example (BFS)
r 1
s
0
t 2
u 3
2 v
1 w Q: y 3
2
x
3 y
Example (BFS)
augmenting path p
v2
v4
v1
12
v3
7
5
s
11 11 3 5 8
4
5 3 11
15 4
t
v2
v4
11/16
v1
0/9 11/14
v3
7/7
19/20
4
s
0/10 1/4 12/13
t
4/4
v2
v4
Maximum flow increased by 4!
Maximum Flow Theorem
Example
Residual Networks
11/16
Flows
v1
1/4 12/12
v3
7/7
19/20
s
0/10
12/13
0/9 11/14
t
4/4
v2
v4
12 5
v1
9 3
11
v3
7
1 19 4
Maximum flow |f| = 11+12 = 23
s
11 11 3 1 12
t
v2
v4
Outlines
• Flow networks • Ford-Fulkerson method
Flow Networks
The Tao of Flow
Flow Network
- two distinct vertices
- each vertex on some path from source to sink
Problem：Time Complexity
Time Complexity
Edmonds & Karp Algorithm
Breadth-First Search
• Input: - Graph G = (V, E), either directed or undirected, - source vertex s V. • Output: for all v V