Ahuja-Orlin: Difference between revisions
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# Let <math>v</math> be the top element of <math>S</math>, that is, the endnode of the corresponding path. | # Let <math>v</math> be the top element of <math>S</math>, that is, the endnode of the corresponding path. | ||
# If <math>v=t</math>: | # If <math>v=t</math>: | ||
## [[Basic flow definitions#Augmenting along a path|Augment]] the current flow along the path corresponding to <math>S</math>. | ## [[Basic flow definitions#Augmenting along a path|Augment]] the current flow along the path corresponding to <math>S</math> up to saturation. | ||
## Remove all nodes but <math>s</math> from <math>S</math>. | ## Remove all nodes but <math>s</math> from <math>S</math>. | ||
# Otherwise: | # Otherwise: |
Revision as of 09:13, 23 November 2015
General Information
Algorithmic problem: Max-flow problems (standard version)
Type of algorithm: loop
Abstract View
Auxiliary data:
- A nonnegative integral value [math]\displaystyle{ d(v) }[/math] for each node [math]\displaystyle{ v\in V }[/math].
- 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].
- A stack [math]\displaystyle{ S }[/math] of nodes.
Invariant: After [math]\displaystyle{ i \ge 0 }[/math] iterations:
- The flow [math]\displaystyle{ f }[/math] is a feasible flow.
- If all upper bounds are integral, [math]\displaystyle{ f }[/math] is integral as well.
- The values [math]\displaystyle{ d(\cdot) }[/math] are a valid distance labeling with respect to [math]\displaystyle{ f }[/math].
- 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].
- 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:
- An arc is appended to the current path (when going forward in the depth-first search for an augmenting path).
- At least one arc is saturated (when augmenting along the path up to saturation).
- The distance label of some node is increased (when going forward in the depth-first search for an augmenting path).
Break condition: It is [math]\displaystyle{ d(s)\geq n }[/math], where [math]\displaystyle{ n=|V| }[/math],
Induction basis
Abstract view:
- We start with some feasible flow, for example, the zero flow.
- Also, we start with some valid distance labeling with respect to the initial flow.
- Stack [math]\displaystyle{ S }[/math] is initialized so as to contain [math]\displaystyle{ s }[/math] and no other node.
- 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]; the resulting distances are the [math]\displaystyle{ d }[/math]-labels.
Proof of the implementation: 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:
- Let [math]\displaystyle{ v }[/math] be the top element of [math]\displaystyle{ S }[/math], that is, the endnode of the corresponding path.
- If [math]\displaystyle{ v=t }[/math]:
- Augment the current flow along the path corresponding to [math]\displaystyle{ S }[/math] up to saturation.
- Remove all nodes but [math]\displaystyle{ s }[/math] from [math]\displaystyle{ S }[/math].
- Otherwise:
- 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.
- If the current arc of [math]\displaystyle{ v }[/math] is now void:
- 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.
- Reset the current arc of [math]\displaystyle{ v }[/math] to the beginning of the list of outgoing arcs of [math]\displaystyle{ v }[/math].
- Remove [math]\displaystyle{ v }[/math] from [math]\displaystyle{ S }[/math].
- 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.