max flow min cut. theorem the maximum value of an st-flow in a digraph equals the minimum capacity...
TRANSCRIPT
Max Flow Min Cut
Theorem
The maximum value of an st-flow in a digraph equals the minimum capacity of an st-cut.
Theorem
If every arc has integer capacity, then in a maximum flow every arc has integer flow.
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Find a maximum st-flow and a minimum st-cut
s t
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2 3
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1 2
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Algorithm
s t1
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9
7
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2 3
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Algorithm
s t
}{sS
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8
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9
7
0f
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2 3
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Algorithm
s t
}{sS }4,{sS
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9
7
0f
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2 3
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1 2
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Algorithm
s t
}{sS }4,{sS
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3
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4
8
6
9
7
}7,4,{sS 0f
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2 3
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Algorithm
s t
}{sS }4,{sS
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4
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9
7
}7,4,{sS 0f
}9,7,4,{sS
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2 3
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1 2
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Algorithm
s t
}{sS }4,{sS
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4
8
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7
}7,4,{sS 0f
}9,7,4,{sS },9,7,4,{ tsS
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2 3
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1 2
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Algorithm
s t
}{sS }4,{sS
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3
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2
4
8
6
9
7
}7,4,{sS 0f
}9,7,4,{sS },9,7,4,{ tsS
STOP St
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2 3
33
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1 2
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Algorithm
s t
}{sS }4,{sS
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3
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2
4
8
6
9
7
}7,4,{sS 0f
}9,7,4,{sS },9,7,4,{ tsS
STOP St
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2
1
2 3
33
3
1
11
1 2
3
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Algorithm
s t
}{sS }4,{sS
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3
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2
4
8
6
9
7
}7,4,{sS
}9,7,4,{sS },9,7,4,{ tsS
STOP St 3),9()9,7(
)7,4()4,(
tff
fsf
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2
3
2
1
2 3
33
3
1
11
1 2
3
30
Algorithm
s t
}{sS }4,{sS
1
3
5
2
4
8
6
9
7
}7,4,{sS
}9,7,4,{sS },9,7,4,{ tsS
STOP St 3),9()9,7(
)7,4()4,(
tff
fsf
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2
1
2 3
33
3
1
11
1 2
3
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Algorithm
s t1
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2
4
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9
7
3),9()9,7(
)7,4()4,(
tff
fsf
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2
1
2 3
33
3
1
11
1 2
3
30
Algorithm
s t1
3
5
2
4
8
6
9
7
3),9()9,7(
)7,4()4,(
tff
fsf}{sS }1,{sS }2,1,{sS
},6,8,5,2,1,{ tsS
STOP St
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2
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2
1
2 3
33
3
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1 2
3
30
Algorithm
s t1
3
5
2
4
8
6
9
7
3),9()9,7(
)7,4()4,(
tff
fsf}{sS }1,{sS }2,1,{sS
},6,8,5,2,1,{ tsS
STOP St
1),6(...
)2,1()1,(
tf
fsf
1
21
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2
1
2 31
331
3
1
11
1 21
3
30
Algorithm
s t1
3
5
2
4
8
6
9
7
3),9()9,7(
)7,4()4,(
tff
fsf}{sS }1,{sS }2,1,{sS
},6,8,5,2,1,{ tsS
STOP St
1),6(...
)2,1()1,(
tf
fsf
1
21
3
2
1
2 31
331
3
1
11
1 21
3
30
Algorithm
s t1
3
5
2
4
8
6
9
7
3),9()9,7(
)7,4()4,(
tff
fsf
1),6(...
)2,1()1,(
tf
fsf
1
21
3
2
1
2 31
331
3
1
11
1 21
3
30
Algorithm
s t1
3
5
2
4
8
6
9
7
3),9()9,7(
)7,4()4,(
tff
fsf
1),6(...
)2,1()1,(
tf
fsf
}{sS
1
21
3
2
1
2 31
331
3
1
11
1 21
3
30
Algorithm
s t1
3
5
2
4
8
6
9
7
3),9()9,7(
)7,4()4,(
tff
fsf
1),6(...
)2,1()1,(
tf
fsf
}{sS }6,{sS
1
21
3
2
1
2 31
331
3
1
11
1 21
3
30
Algorithm
s t1
3
5
2
4
8
6
9
7
3),9()9,7(
)7,4()4,(
tff
fsf
1),6(...
)2,1()1,(
tf
fsf
}{sS }6,{sS
},6,{ tsS STOP St
1
21
3
2
1
2 31
331
3
1
11
1 21
3
30
Algorithm
s t1
3
5
2
4
8
6
9
7
3),9()9,7(
)7,4()4,(
tff
fsf
1),6(...
)2,1()1,(
tf
fsf
}{sS }6,{sS
},6,{ tsS STOP St
2),6(,1)6,( tfsf
1
21
3
2
1
2 31
332
3
1
11
1 21
3
30
Algorithm
s t1
3
5
2
4
8
6
9
7
3),9()9,7(
)7,4()4,(
tff
fsf
1),6(...
)2,1()1,(
tf
fsf
}{sS }6,{sS
},6,{ tsS STOP St
2),6(,1)6,( tfsf
1
21
3
2
1
2 31
332
3
1
11
1 21
3
30
Algorithm
s t1
3
5
2
4
8
6
9
7
3),9()9,7(
)7,4()4,(
tff
fsf
1),6(...
)2,1()1,(
tf
fsf
}{sS }6,{sS
},6,{ tsS STOP St
2),6(,1)6,( tfsf
5)( fv
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21
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2
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2 31
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Minimum Cut
s t1
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5
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9
7
5)( fv
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2 31
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Minimum Cut
s t1
3
5
2
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8
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9
7
5)( fv
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2 31
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1 21
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Minimum Cut
s t1
3
5
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4
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9
7
5)( fv5113),( SSc
1
1
31
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1
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s t1
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9
7
So f is a max flow
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2 3
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1 2
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Algorithm
s t
}{sS }4,{sS
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3
5
2
4
8
6
9
7
}7,4,{sS 0f
1
2
3
2
1
2 3
33
3
1
11
1 2
3
30
Algorithm
s t
}{sS }4,{sS
1
3
5
2
4
8
6
9
7
}7,4,{sS 0f
},6,8,7,4,{ tsS
1
2
3
2
1
2 3
33
3
1
11
1 2
3
30
Algorithm
s t
}{sS }4,{sS
1
3
5
2
4
8
6
9
7
}7,4,{sS
},6,8,7,4,{ tsS
3),6()6,8()8,9(
)9,7()7,4()4,(
tfff
ffsf
1
2
3
2
1
2 3
33
3
1
11
1 2
3
30
Algorithm
s t1
3
5
2
4
8
6
9
7
3),6()6,8()8,9(
)9,7()7,4()4,(
tfff
ffsf
1
2
3
2
1
2 3
33
3
1
11
1 2
3
30
Algorithm
s t1
3
5
2
4
8
6
9
7
3),6()6,8()8,9(
)9,7()7,4()4,(
tfff
ffsf}{sS }6,{sS
}8,6,{sS
},9,8,6,{ tsS
1
2
3
2
1
2 3
33
3
1
11
1 2
3
30
Algorithm
s t1
3
5
2
4
8
6
9
7
3),6()6,8()8,9(
)9,7()7,4()4,(
tfff
ffsf}{sS }6,{sS
}8,6,{sS
},9,8,6,{ tsS
2)9,8()8,6(
1),9()6,(
ff
tfsf
1
2
3
2
1
2 32
323
31
1
11
1 2
3
30
Algorithm
s t1
3
5
2
4
8
6
9
7
3),6()6,8()8,9(
)9,7()7,4()4,(
tfff
ffsf}{sS }6,{sS
}8,6,{sS
},9,8,6,{ tsS
2)9,8()8,6(
1),9()6,(
ff
tfsf
Hall’s Theorem from
Max Flow Min Cut
1 11
1
1
Direct all edges from s to t and assign all arcs unit capacity
Adds and t, adjacent to all of A and B respectively.
We have to show that Hall’s Condition XX )(gives a 1-factor.
A flow of value is enough to guarantee a 1-factor.
A
So all we have to do is show for each cut S.
ASSc ),(
So all we have to do is show for each cut S.
ASSc ),(
),()\,()\,(),( tSBeSBSAeSAseSSc )\,(\ SBSAeSBSA
)(\ SASA
ASASA \