Max-flow min-cut: Difference between revisions
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# <math>f</math> is maximum. | # <math>f</math> is maximum. | ||
# There is a [[Basic flow definitions#Cuts and saturated cuts|saturated cut]]. | # There is a [[Basic flow definitions#Cuts and saturated cuts|saturated cut]]. | ||
# There is no [[Basic flow definitions#Flow-augmenting path|flow-augmenting path]]. | # There is no [[Basic flow definitions#Flow-augmenting path|flow-augmenting path]] from <math>s</math> to <math>t</math>. | ||
== Proof == | |||
First suppose there is no flow-augmenting path. Let <math>S</math> denote the set of all nodes reachable from <math>s</math> via flow-augmenting paths, and let <math>T:=V\setminus S</math>. Then we have <math>s�\in S</math> and <math>t\in T</math>. Obviously, <math>(S,T)</math> is saturated. | |||
Next suppose there is a saturated cut. Then no flow <math>f'</math> can carry more units of flow from <math>S</math> to <math>T</math> than <math>f</math>. Consequently, <math>f</math> is maximum. | |||
Finally suppose <math>f</math> is maximum. Clearly, a flow-augmenting path cannot exist. | |||
Revision as of 19:23, 19 October 2014
Max-flow min-cut theorem
Let [math]\displaystyle{ G=(V,A) }[/math], [math]\displaystyle{ s,t\in V }[/math], and for [math]\displaystyle{ a\in A }[/math] let [math]\displaystyle{ u(a) }[/math] and [math]\displaystyle{ f(a) }[/math] be real values such that [math]\displaystyle{ 0\leq f(a)\leq u(a) }[/math] and [math]\displaystyle{ f }[/math] is an [math]\displaystyle{ (s,t) }[/math]-flow. Then the following three statements are equivalent:
- [math]\displaystyle{ f }[/math] is maximum.
- There is a saturated cut.
- There is no flow-augmenting path from [math]\displaystyle{ s }[/math] to [math]\displaystyle{ t }[/math].
Proof
First suppose there is no flow-augmenting path. Let [math]\displaystyle{ S }[/math] denote the set of all nodes reachable from [math]\displaystyle{ s }[/math] via flow-augmenting paths, and let [math]\displaystyle{ T:=V\setminus S }[/math]. Then we have [math]\displaystyle{ s�\in S }[/math] and [math]\displaystyle{ t\in T }[/math]. Obviously, [math]\displaystyle{ (S,T) }[/math] is saturated.
Next suppose there is a saturated cut. Then no flow [math]\displaystyle{ f' }[/math] can carry more units of flow from [math]\displaystyle{ S }[/math] to [math]\displaystyle{ T }[/math] than [math]\displaystyle{ f }[/math]. Consequently, [math]\displaystyle{ f }[/math] is maximum.
Finally suppose [math]\displaystyle{ f }[/math] is maximum. Clearly, a flow-augmenting path cannot exist.