Ahuja-Orlin

General Information

Algorithmic problem: Max-flow problems (standard version)

Type of algorithm: loop

Abstract View

Auxiliary data:

1. A nonnegative integral value [math]\displaystyle{ d(v) }[/math] for each node [math]\displaystyle{ v\in V }[/math].
2. Each node [math]\displaystyle{ v\in V\setminus\{t\} }[/math] has a current arc, which may be implemented as an iterator on the list of outgoing arcs of [math]\displaystyle{ v }[/math].
3. A stack [math]\displaystyle{ S }[/math] of nodes.

Invariant: After [math]\displaystyle{ i \ge 0 }[/math] iterations:

1. The flow [math]\displaystyle{ f }[/math] is a fleasible flow.
2. If all upper bounds are integral, [math]\displaystyle{ f }[/math] is integral as well.
3. The values [math]\displaystyle{ d(\cdot) }[/math] are a valid distance labeling with respect to [math]\displaystyle{ f }[/math].
4. The current arc of a node [math]\displaystyle{ v\in V\setminus\{s,t\} }[/math] is either void or one of the outgoing arcs of [math]\displaystyle{ v }[/math]. No admissible outgoing arc of [math]\displaystyle{ v }[/math] precedes the current arc of [math]\displaystyle{ v }[/math].
5. Stack [math]\displaystyle{ S }[/math] contains a cycle-free flow-augmenting path with respect to [math]\displaystyle{ f }[/math]. This path starts with [math]\displaystyle{ s }[/math] and ends at an arbitrary node of [math]\displaystyle{ G }[/math]. Each arc on this path is admissible and the current arc of its tail node.

Variant: One of the following actions will take place in the current iteration:

1. An arc is appended to the current path.
2. At least one arc is saturated.
3. The distance label of some node is increased.

Break condition: It is [math]\displaystyle{ d(s)\geq n }[/math], where [math]\displaystyle{ n=|V| }[/math],

Induction basis

Abstract view:

1. We start with some feasible flow, for example, the zero flow.
2. Also, we start with some valid distance labeling.
3. Stack [math]\displaystyle{ S }[/math] is initialized so as to contain [math]\displaystyle{ s }[/math] and no other node.
4. The current arc of each node [math]\displaystyle{ v\in V }[/math] is reset to be the very first outgoing arc of [math]\displaystyle{ v }[/math].

Implementation of step 2 in case [math]\displaystyle{ f }[/math] is initialized as the zero flow: A BFS is run on the transpose of [math]\displaystyle{ G }[/math] with start node [math]\displaystyle{ t }[/math] and unit arc lengths; the resulting distances are the [math]\displaystyle{ d }[/math]-labels.

Proof: Clearly, it is [math]\displaystyle{ d(t)=0 }[/math], so it remains to show [math]\displaystyle{ d(v)\leq d(w)+1 }[/math] for each arc [math]\displaystyle{ a\in A }[/math]. If [math]\displaystyle{ f }[/math] is the zero flow, the arcs of the residual network [math]\displaystyle{ G_f }[/math] are exactly the arcs of [math]\displaystyle{ G }[/math]. So, the claim follows immediately from the valid distance property.

Induction step

Abstract view:

1. Let [math]\displaystyle{ v }[/math] be the top element of [math]\displaystyle{ S }[/math], that is, the endnode of the corresponding path.
2. If [math]\displaystyle{ v=t }[/math]:
   1. Augment the current flow along the path corresponding to [math]\displaystyle{ S }[/math].
   2. Remove all nodes but [math]\displaystyle{ s }[/math] from [math]\displaystyle{ S }[/math].
3. Otherwise:
   1. While the current arc of [math]\displaystyle{ v }[/math] is not void and not admissible or points to [math]\displaystyle{ s }[/math]: Move the current arc one step forward.
   2. If the current arc of [math]\displaystyle{ v }[/math] is now void:
      1. Set [math]\displaystyle{ d(v):=\tilde{d}+1 }[/math], where [math]\displaystyle{ \tilde{d} }[/math] denotes the minimum value [math]\displaystyle{ d(w) }[/math] over all outgoing arcs [math]\displaystyle{ (v,w) }[/math] of [math]\displaystyle{ v }[/math] in the residual network.
      2. Reset the current arc of [math]\displaystyle{ v }[/math] to the beginning of the list of outgoing arcs of [math]\displaystyle{ v }[/math].
   3. Otherwise, that is, if the current arc is some arc [math]\displaystyle{ (v,w) }[/math]: Push [math]\displaystyle{ w }[/math] on [math]\displaystyle{ S }[/math].

Proof: The variant, points 1, 2, and 5 of the invariant, and the first sentence of point 4 of the invariant are obvious. For the second sentence of point 4, note that no admissible arc of [math]\displaystyle{ v }[/math] is skipped when the current arc is moved forward, and the current arc is reset whenever outgoing arcs may become admissible due to an increase of [math]\displaystyle{ d(v) }[/math]. Finally, point 3 of the invariant follows from the conservative increase of [math]\displaystyle{ d(v) }[/math] in step 3.2.

Correctness

First we show that the flow is maximum when the break condition applies. Due to the max-flow min-cut theorem, it suffices to show that, in this moment, there is no more flow-augmenting [math]\displaystyle{ (s,t) }[/math]-path. However, such a path would have at most [math]\displaystyle{ n-1 }[/math] arcs, and for each arc [math]\displaystyle{ (v,w) }[/math] on this path, it is [math]\displaystyle{ d(v)\leq d(w)+1 }[/math]. This implies [math]\displaystyle{ d(s)-d(t)=d(s)\leq n-1 }[/math], which contradicts the break condition.

It remains to show that the algorithm indeed terminates. This is proved by the following complexity considerations.

Complexity

Statement: The asymptotic complexity of the algorithm is in [math]\displaystyle{ \mathcal{O}(n^2m) }[/math], where [math]\displaystyle{ n=|V| }[/math] and [math]\displaystyle{ m=|A| }[/math].

Proof: First we show that no distance label [math]\displaystyle{ d(v) }[/math] may become higher than [math]\displaystyle{ n }[/math] (in particular, the total number of executions of step 3.2 over all iterations is in [math]\displaystyle{ \mathcal{O}(n^2) }[/math]).

To see that, recall that [math]\displaystyle{ d(s)\lt n }[/math] as long as the break condition is not fulfilled. Whenever the distance label [math]\displaystyle{ d(v) }[/math] of a node [math]\displaystyle{ v\in V }[/math] is increased, [math]\displaystyle{ v }[/math] is the endnode of the current path, which consists solely of admissible arcs. Due to the definition of admissible arcs, it is [math]\displaystyle{ d(v)\leq d(s) }[/math] immediately before the current iteration (equality only if [math]\displaystyle{ v=s }[/math]). Let [math]\displaystyle{ (v,w) }[/math] be an arc such that [math]\displaystyle{ \tilde{d}=d(w) }[/math], in particular, it is [math]\displaystyle{ w\neq s }[/math]. For the same reason, it is [math]\displaystyle{ d(w)\lt d(s) }[/math], which implies [math]\displaystyle{ d(v)\leq d(s) }[/math] immediately after the current iteration.

Now we know [math]\displaystyle{ d(u)\leq n }[/math] for all nodes [math]\displaystyle{ u\in V }[/math]. Next we consider an arbitrary but fixed arc [math]\displaystyle{ (v,w)\in A }[/math]. Suppose [math]\displaystyle{ (v,w) }[/math] has ben saturated at least twice during the algorithm. In both iterations, [math]\displaystyle{ (v,w) }[/math] is admissible, which means [math]\displaystyle{ d(v)\gt d(w) }[/math]. However, in the meantime, some flow must have been sent back from [math]\displaystyle{ w }[/math] to [math]\displaystyle{ v }[/math], which implies [math]\displaystyle{ d(w)\gt d(v) }[/math] in that intermediate iteration. As distance labels are non-decreasing, [math]\displaystyle{ d(v) }[/math] must have been increased between any two iterations in which [math]\displaystyle{ (v,w) }[/math] is saturated. In summary, each arc is saturated only [math]\displaystyle{ \mathcal{O}(n) }[/math] times.

In each execution of step 2.1 of the induction step, at least one arc is saturated, so the total number of executions of step 2.1+2.2 is in [math]\displaystyle{ \mathcal{O}(nm) }[/math]. The path contains at most [math]\displaystyle{ n-1 }[/math] arcs, so the total asymptotic complexity for all executions of step 2.1+2 is in [math]\displaystyle{ \mathcal{O}(n^2m) }[/math].

Finally, the sequence of outgoing arcs is passed once for each node in step 3.1 between two increases of [math]\displaystyle{ d(v) }[/math], so the total asymptotic complexity of all executions of step 3.1 over all iterations is in [math]\displaystyle{ \mathcal{O}(n\cdot m) }[/math].

Remark

This algorithm may be seen as a "lazy" variant on Edmonds-Karp. In fact, the most expensive step there is the computation of a shortest flow-augmenting [math]\displaystyle{ (s,t) }[/math]-path. This task amounts to computing the true distance from every node to [math]\displaystyle{ t }[/math]. A valid distance labeling may be seen as "lazily evaluated" true distances.