Hopcroft-Karp

Abstract view

Algorithmic problem: Cardinality-maximal matching in bipartite graphs.

Type of algorithm: loop.

Auxiliary data: a set [math]\displaystyle{ S }[/math] of BFS runs as iterators (cf. the remark on iterators). The modification to find augmenting paths is applied.

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

1. [math]\displaystyle{ M }[/math] is a matching in [math]\displaystyle{ G }[/math].
2. All exposed nodes that are end nodes of augmenting paths, are start nodes of BFS iterators in [math]\displaystyle{ S }[/math] (there may be start nodes in [math]\displaystyle{ S }[/math] that are not end nodes of augmenting paths).
3. Each node in [math]\displaystyle{ S }[/math] is the root of an arborescence:
   1. These arborescences are pairwise node-disjoint.
   2. Each leaf of each arborescence has distance [math]\displaystyle{ i }[/math] form the root of this arborescence (both in [math]\displaystyle{ G }[/math] and in the arborescence).

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:

1. Start with an arbitrary matching, for example, the empty matching.
2. Create and initialize one BFS run for each exposed node, and insert it in [math]\displaystyle{ S }[/math].

Induction step

Abstract view: In the [math]\displaystyle{ i }[/math]-th iteration:

1. Process all nodes of distance [math]\displaystyle{ i-1 }[/math] in the queues of all BFS runs.
2. For each BFS run: If it had no nodes to process in step 1, remove it from [math]\displaystyle{ S }[/math].
3. For each edge [math]\displaystyle{ \{v,w\} }[/math] that is seen by one BFS run from [math]\displaystyle{ v }[/math] and by one BFS run from [math]\displaystyle{ w }[/math]:
   1. Choose two such BFS runs and such an edge [math]\displaystyle{ \{v,w\} }[/math], say.
   2. 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].
   3. Remove these two BFS runs from the set of all BFS runs.

Proof: The invariant 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_1 }[/math] and [math]\displaystyle{ p_2 }[/math] be the paths in the two arborescences from the two start nodes to [math]\displaystyle{ v }[/math] and [math]\displaystyle{ w }[/math], respectively. So, it is [math]\displaystyle{ p=p_1+\{v,w\}+p_2 }[/math]. Suppose augmenting along [math]\displaystyle{ p }[/math] has created a new augmenting path [math]\displaystyle{ q }[/math]. We have to show that [math]\displaystyle{ q }[/math] is longer than [math]\displaystyle{ p }[/math].

In the orientation of [math]\displaystyle{ p }[/math] from the start node of [math]\displaystyle{ p_1 }[/math] to the start node of [math]\displaystyle{ p_2 }[/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]. Since [math]\displaystyle{ q }[/math] has become augmenting by the augmentation along [math]\displaystyle{ p }[/math],[math]\displaystyle{ q }[/math] must share at least one edge with [math]\displaystyle{ p }[/math], which means [math]\displaystyle{ v'\neq w' }[/math]. Let [math]\displaystyle{ q_1 }[/math] and [math]\displaystyle{ q_2 }[/math] denote the loose ends of [math]\displaystyle{ q }[/math] up to [math]\displaystyle{ v }[/math] and [math]\displaystyle{ w }[/math], respectively. If [math]\displaystyle{ v' }[/math] belongs to [math]\displaystyle{ p_1 }[/math], it must be [math]\displaystyle{ v'=v }[/math] because, otherwise, the concatenation [math]\displaystyle{ p_1+q_1 }[/math] would yield an augmenting path that is shorter than [math]\displaystyle{ p }[/math], which contradicts the induction hypothesis. Mirror-symmetrically, if [math]\displaystyle{ w' }[/math] belongs to [math]\displaystyle{ p_2 }[/math], it must be [math]\displaystyle{ w'=w }[/math]. In summary, [math]\displaystyle{ v'\neq w' }[/math] implies that [math]\displaystyle{ v'=v }[/math] and [math]\displaystyle{ w'=w }[/math] is the only possible scenario.

Note that each edge of [math]\displaystyle{ q }[/math] that is incident to exactly one node on [math]\displaystyle{ p }[/math] must be unmatched. Therefore, the parities of the lengths of [math]\displaystyle{ p_1 }[/math] and [math]\displaystyle{ q_1 }[/math] are different. In particular, the lengths themselves are different. Again, if [math]\displaystyle{ q_1 }[/math] were shorter than [math]\displaystyle{ p_1 }[/math], [math]\displaystyle{ p_1+q_1 }[/math] were an augmenting path shorter than [math]\displaystyle{ p }[/math]. For the same reason, [math]\displaystyle{ q_2 }[/math] is longer than [math]\displaystyle{ p_2 }[/math]. In summary, [math]\displaystyle{ q }[/math] is longer than [math]\displaystyle{ p }[/math].

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, once from either incident node. Hence, it suffices to prove that the number of iterations is in [math]\displaystyle{ \mathcal{O}(\sqrt{n}) }[/math].

After [math]\displaystyle{ \lceil\sqrt{n}\rceil }[/math] iterations, any augmenting path has a length of at least [math]\displaystyle{ \lceil\sqrt{n}\rceil }[/math]. Consequently, there cannot be more than [math]\displaystyle{ \lfloor\sqrt{n}\rfloor }[/math] node-disjoint augmenting paths. The remark on the proof of Berge's theorem implies that the maximal cardinality of a matching is at most [math]\displaystyle{ |M|+\lfloor\sqrt{n}\rfloor }[/math]. Therefore, [math]\displaystyle{ |M| }[/math] is maximal after another [math]\displaystyle{ \lfloor\sqrt{n}\rfloor }[/math] iterations. The claim follows.