Successive shortest paths

Abstract view

Definition:

1. For a node [math]\displaystyle{ v\in V }[/math], let [math]\displaystyle{ \Delta f(v):=\sum_{w:(v,w)\in A}f(v,w)-\sum_{w:(w,v)\in A}f(w,v) }[/math].
2. The imbalance of a node [math]\displaystyle{ v\in V }[/math] is defined as [math]\displaystyle{ I_f(v):=\Delta f(v)-b(v) }[/math].
3. The imbalance of a node [math]\displaystyle{ v\in V }[/math] is underestimating if [math]\displaystyle{ 0\leq \Delta f(v)\leq b(v) }[/math] or [math]\displaystyle{ 0\geq\Delta f(v)\geq b(v) }[/math].
4. The total imbalance of [math]\displaystyle{ f }[/math] is the defined as [math]\displaystyle{ \sum_{v\in V}|I_f(v)| }[/math].

Invariant:

1. The capacity constraints are fulfilled, that is, [math]\displaystyle{ 0\leq f(a)\leq u(a) }[/math] for all [math]\displaystyle{ a\in A }[/math].
2. There is no negative cycle in the residual network of [math]\displaystyle{ f }[/math].
3. The imbalance of every node is underestimating.

Variant: The total imbalance strictly decreases.

Break condition: The imbalances of all nodes are zero.

Induction basis

Abstract view: Start with the zero flow.

Proof: Obvious.

Induction step

1. In the residual network of [math]\displaystyle{ f }[/math], find a shortest path [math]\displaystyle{ p }[/math] from the set of nodes [math]\displaystyle{ v\in V }[/math] with [math]\displaystyle{ I_f(v)\lt 0 }[/math] to the set of nodes [math]\displaystyle{ w\in V }[/math] with [math]\displaystyle{ I_f(v)\gt 0 }[/math].
2. Let [math]\displaystyle{ v_0 }[/math] be the node where [math]\displaystyle{ p }[/math] actually starts and [math]\displaystyle{ w_0 }[/math] the node where [math]\displaystyle{ p }[/math] actually ends.
3. Increase the flow on all arcs of [math]\displaystyle{ p }[/math] by the minimum of [math]\displaystyle{ |I_f(v_0)| }[/math], of [math]\displaystyle{ I_f(w_0)\gt 0 }[/math], and of the residual capacities of all arcs on [math]\displaystyle{ p }[/math].

Correctness

The variant follows from the fact that the amount by which the flow is increased in step 3 is strictly positive. The first and third points of the invariant result from the choice of this amount as the minimum of imbalances and residual capacities. Thus, it remains to show point 2 of the invariant.

By induction hypothesis, there were nonegative cycles immediately before the iteration. So, any negative c<cle must have been introduced by the flow augmentation along [math]\displaystyle{ p }[/math]. This can only happen as follows: There is at least one arc [math]\displaystyle{ (v,w) }[/math] on [math]\displaystyle{ p }[/math]such that the residual capacity [math]\displaystyle{ w\rightarrow v }[/math] was zero immediately before the iteration (and is positive afterwards, of course). Let [math]\displaystyle{ p' }[/math]