Hopcroft-Karp: Difference between revisions
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'''Statement:''' | '''Statement:''' | ||
The asymptotic complexity is in <math>\mathcal{O}(m\!\cdot\!\sqrt{n})</math>, where <math>n=|V|</math> and <math>m= | The asymptotic complexity is in <math>\mathcal{O}(m\!\cdot\!\sqrt{n})</math>, where <math>n=|V|</math> and <math>m=E|</math>. | ||
'''Proof:''' | '''Proof:''' | ||
Each edge is touched at most twice in each iteration | |||
After <math>\lfloor\sqrt{n}\rfloor</math> iterations, any augmenting path has a length of at least <math>\lfloor\sqrt{n}\rfloor</math>. It suffices to show that all |
Revision as of 16:55, 22 November 2014
Abstract view
Algorithmic problem: Cardinality-maximal matching in bipartite graphs.
Type of algorithm: loop.
Auxiliary data structures: a set [math]\displaystyle{ S }[/math] of BFS runs as iterators (cf. the corresponding remark). The modification to find augmenting paths is applied.
Invariant: Before and after each iteration:
- [math]\displaystyle{ M }[/math] is a matching in [math]\displaystyle{ G }[/math].
- All exposed nodes that are end nodes of augmenting paths, are in [math]\displaystyle{ S }[/math].
Variant: The shortest length of an augmenting path in terms of the number of edges is increased.
Break condition: [math]\displaystyle{ S=\emptyset }[/math].
Induction basis
Abstract view:
- Start with an arbitrary matching, for example, the empty matching.
- Create and initialize one BFS run for each exposed node.
Induction step
Abstract view: In the [math]\displaystyle{ i }[/math]-th iteration:
- Process all nodes of distance [math]\displaystyle{ i-1 }[/math] in the queues of all BFS runs.
- For each BFS run: If it had no nodes to process in step 1, remove it from [math]\displaystyle{ S }[/math].
- For each edge [math]\displaystyle{ \{v,w\} }[/math] that is seen by more than one:
- Choose two such BFS runs and such an edge [math]\displaystyle{ \{v,w\} }[/math], say.
- Augment [math]\displaystyle{ M }[/math] along the concatenation of the paths in the arborescences of these two BFS runs from the start nodes to [math]\displaystyle{ v }[/math] and [math]\displaystyle{ w }[/math], respectively, and of [math]\displaystyle{ \{v,w\} }[/math].
- Remove these two BFS runs from the set of all BFS runs.
Proof: The variant is obviously maintained. To prove the variant, consider an execution of step 3.2. Let [math]\displaystyle{ p }[/math] denote the path along which the augmentation takes place. Let [math]\displaystyle{ p' }[/math] and [math]\displaystyle{ p'' }[/math] the paths in he arborescences from the start nodes to [math]\displaystyle{ v }[/math] and [math]\displaystyle{ w }[/math], respectively.
For a contradiction, suppose augmenting along [math]\displaystyle{ p }[/math] has created a a new augmenting path [math]\displaystyle{ q }[/math] that is not longer than [math]\displaystyle{ p }[/math]. In either orientation of [math]\displaystyle{ p }[/math] and [math]\displaystyle{ q }[/math], let [math]\displaystyle{ v' }[/math] be the first node of [math]\displaystyle{ p }[/math] that also belongs to [math]\displaystyle{ q }[/math], and let [math]\displaystyle{ w' }[/math] be the last node of [math]\displaystyle{ p }[/math] that also belongs to [math]\displaystyle{ q }[/math]. Without loss of generality, let [math]\displaystyle{ v }[/math] be the last node of [math]\displaystyle{ p' }[/math] and let [math]\displaystyle{ v }[/math] be on [math]\displaystyle{ p' }[/math]. Note that each edge of [math]\displaystyle{ q }[/math] that is incident to exactly one node on [math]\displaystyle{ p }[/math] must be unmatched.
First, [math]\displaystyle{ v'=v }[/math] must hold because, otherwise, one subpath of [math]\displaystyle{ p }[/math] up to [math]\displaystyle{ v' }[/math] and one of [math]\displaystyle{ q }[/math] up to [math]\displaystyle{ v' }[/math] would be a shorter augmenting path, which contradicts the induction hypothesis. Moreover, [math]\displaystyle{ v'=v }[/math] implies that [math]\displaystyle{ w }[/math] must belong to [math]\displaystyle{ p'' }[/math] and, by the same argument, [math]\displaystyle{ w'=w }[/math] must hold. So, the intersection of [math]\displaystyle{ p }[/math] and [math]\displaystyle{ q }[/math] is exactly [math]\displaystyle{ \{v,w\} }[/math]. Therefore, after the augmentation along [math]\displaystyle{ p }[/math], [math]\displaystyle{ q }[/math] becomes a candidate. Augmenting along [math]\displaystyle{ q }[/math] does not generate new augmenting paths because these new paths would intersect with [math]\displaystyle{ q }[/math] exactly on [math]\displaystyle{ \{v,w\} }[/math] for the same reason, but [math]\displaystyle{ \{v,w\} }[/math] is not matched anymore, and the two incident edges on this assumed new path are not matched, either, so this assumed new path is not alternating.
Remark: The union of all arborescences at termination is called a Hungarian forest in the literature. The algorithm proves, constructively, that a matching in a bipartite graph is cardinality-maximal if, and only if, a Hungarian forest exists.
Correctness
Due to the invariant, the matching is cardinality-maximal if the main loop terminates. Termination of the main loop results from the following complexity considerations.
Complexity
Statement: The asymptotic complexity is in [math]\displaystyle{ \mathcal{O}(m\!\cdot\!\sqrt{n}) }[/math], where [math]\displaystyle{ n=|V| }[/math] and [math]\displaystyle{ m=E| }[/math].
Proof: Each edge is touched at most twice in each iteration After [math]\displaystyle{ \lfloor\sqrt{n}\rfloor }[/math] iterations, any augmenting path has a length of at least [math]\displaystyle{ \lfloor\sqrt{n}\rfloor }[/math]. It suffices to show that all