1. Maximum Flow Problem flow capacity Actual flow  capacity
The source node can produce flow as much as possible
For every node other than s and t, flow out  flow in
Source s
actual flow 3/ 4 2/ 2 1/ 1 a b 2/ 3 2/ 2 1/ 2
our textbook says = d c e e
The target node’s flow-in is called the flow of the graph f g
A node may have different ways to dispatch its flow-out . 1/ 2 2/ 3 t
What is the maximum flow of the graph?
Target
(sink)
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2. What makes the maximum flow problem more difficult than the shortest path problem?
Source s
The shortest path of t is based on the shortest path of g. 3/ 4 2/ 2
once a node’s shortest path from s is determined, we will never have to revise the decision 1/ 1 a b 2/ 3 2/ 2 1/ 2 d c e e
The maximum flow to t is not based on the maximum flows to f and g. f g 1/ 2 2/ 3 t
Target
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3. local optimum does not imply global optimum
Source
Source
Source s s s
5/5
2/2
4/5
2/2
5/5
2/2
1/1
0/1 a
1/1 a b a b b
4/4
3/3
1/4
2/3
0/3
3/4
3/3
3/3
1/3
0/3 d
3/3 d c
1/3 d c c
Target
Target
2/2
5/5
sc
max flow = 6
sd
max flow = 7 t
Target
st
max flow = 7
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4. A naïve approach: adding all possible paths together
Source
Source
Source s s s 1 2/ 2 1/ 1 2/ 3 2 1 1 1 a a a b b b 1/ 4 4 4 1 2/ 2 1 2/ 3 2 d d d c c c 2/ 3 1/ 1 2/ 2 3 t t t
Target
Target
Target
Residual Graph
Residual Graph
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5. Not always work. 3/3 2/2 1 1 3 1/4 1 3 2/2 2/ 2 3/3 We are just lucky
Source
Source s s
3/3
2/2 1 a 1 b a b 3
1/4 1 3
2/2 d c d c 2/ 2
3/3 t t
Target
Not always work.
Target
We are just lucky
Residual Graph
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6. A wrong choice: 3/ 3 2 2 1 1 3/ 4 1 3 2 3 2 2 3/ 3 2 s s a b a b d c d
Source
Source s s 3/ 3 2 2 1 a 1 b a b 3/ 4 1 3 2 3 2 d c d c 2 3/ 3 2 t t
Target
Target
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7. Keep a path for changing decision:
Source
Source s s 3/ 3 2 2 3 1 a 1 b a b 3/ 4 1 3 2 3 2 3 d c d c 2 3/ 3 2 3 t t
Target
Target
Augmented Graph
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8. Find another path in the augmented graph
No more path from s to t
Source s s s 2/ 2 3 2 2 3 3 1 a 1 1 b a a b b 2 1 2/ 3 2/ 2 3 1 2 1 2 1 2/ 3 2 2 d 1 1 c d d c c 2/ 2 3 2 2 3 3 t t t
Target
Augmented Graph
new Augmented Graph
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9. + = Add flows together 3 3 2 2 3 2 2 2 2 2 1 3 3 2 2 Maximum Flow s s
Source + = s s 3 s 3 2 2 a b a b a b 3 2 2 2 2 2 1 d c d c d c 3 3 2 2 t t
Target t
Target
Target
Maximum Flow
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10. An algorithm for finding the maximum flow of graphs
Input: G = (V, E), and s,t  V
Let Gmax = (, ), Gaug = G
Find a path p from s to t in Gaug
If no such p exists, output Gmax and stop the program
Add p into Gmax and update Gaug
Repeat 1, 2, and, if possible, 3
Complexity:
O(f |E|)
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11. Claim: This algorithm always terminates with a maximum flow of the input graph
Our textbook says: The proof is somewhat difficult and beyond the scope of this text.
Input: G = (V, E), and s,t  V
Let Gmax = (, ), Gaug = G
Find a path p from s to t in Gaug
If no such p exists, output Gmax and stop the program
Add p into Gmax and update Gaug
Repeat 1, 2, and 3 if possible
I say: No, it is not difficult and every one here should know.
Since any given graph has a finite maximum flow, and every path from s to t, if any, provides a positive flow, the program therefore cannot run forever.
By contradiction, suppose the algorithm terminates with a flow in Gmax. that is not maximum. Then, there must be a flow not included in Gmax. But, if this is the case, there must be a path from s to t to carries this flow, and hence the program should not terminate at this moment. A contradiction.
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12. An inefficient situation
Source s 1/ 1/
1000 1/
1000 1/ 1 a b 1/ 1 1/
1000 1/
1000 t
Target
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13. An inefficient situation
Source s 2/ 1/
1000 2/ 1/
1000 1/ 1 a b 2/ 1/
1000 2/ 1/
1000 t
Target
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IT 279

14. Always chose the maximum next edge
Solution
Always chose the maximum next edge
Source s
1000/
1000
1000 a b
1000/
1000 1/
1000 t
Target
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