Negative cycle-canceling: Difference between revisions
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'''Proof:''' | '''Proof:''' | ||
The invariant and the variant are obviously fulfilled. | The invariant and the variant are obviously fulfilled. | ||
== Correctness == | |||
We have to show that a feasible flow is minimum if it admits no negative cycle. So assume a feasible flow <math>f</math> is ''not'' minimum. Then there is |
Revision as of 17:02, 22 October 2014
Abstract view
Algorithmic problem: Min-cost flow problem
Type of algorithm: loop
Invariant: The flow is feasible.
Variant: The cost of the flow decreases.
Definition: A cycle is negative if the sum of the cost values on all of its arcs is negative.
Break condition: There is no more negative cycle in the residual network.
Induction basis
Abstract view: Start with an arbitrary feasible flow, for example, the zero flow.
Induction step
Abstract view:
- Find a cycle of negative cost in the residual network.
- Augment the flow on all arcs of this cycle by the minmum residual capacity of all arcs on this cycle.
Implementation:
- Run an all-pairs shortest-paths algorithm on the residual network, where the length of an arc [math]\displaystyle{ (v,w)in A }[/math] is defined as the minimum of [math]\displaystyle{ c(v,w) }[/math] and [math]\displaystyle{ -c(w,v) }[/math]. The number of all
- If the distance [math]\displaystyle{ v\rightarrow v }[/math] is 0 for all [math]\displaystyle{ v\in V }[/math], the break condition applies.
- Otherwise (that is, the distances [math]\displaystyle{ v\rightarrow v }[/math] of some nodes [math]\displaystyle{ v\in V }[/math] are negative):
- Reconstruct one negative cycle [math]\displaystyle{ C }[/math] from the output of step 1.
- Augment the flow on all arcs of [math]\displaystyle{ C }[/math] (in the [Basic flow definitions#Residual network|residual network]]) by the minimum residual capacity of all arcs of [math]\displaystyle{ C }[/math].
Proof: The invariant and the variant are obviously fulfilled.
Correctness
We have to show that a feasible flow is minimum if it admits no negative cycle. So assume a feasible flow [math]\displaystyle{ f }[/math] is not minimum. Then there is