Classical bipartite cardinality matching

Abstract view

Algorithmic problem: Cardinality-maximal matching in bipartite graphs.

Type of algorithm: loop.

Invariant: [math]\displaystyle{ M }[/math] is a matching in [math]\displaystyle{ G }[/math].

Variant: [math]\displaystyle{ |M| }[/math] is increased by one.

Break condition: There is no more augmenting path.

Induction basis

Abstract view: Initialize [math]\displaystyle{ M }[/math] to be a feasible matching, for example, the empty matching.

Induction step:

Abstract view:

1. Search for an augmenting path.
2. If there is none, the loop terminates.
3. Let [math]\displaystyle{ p }[/math] denote the augmenting path found in step 1.
4. Augment [math]\displaystyle{ M }[/math] along [math]\displaystyle{ p }[/math].

Implementation of step 1: In a loop over all exposed nodes [math]\displaystyle{ v }[/math], a graph traversal algorithm is started at [math]\displaystyle{ v }[/math]. An algorithm must be chosen that generates an arborescence rooted at the start node [math]\displaystyle{ v }[/math], for example, a DFS or a BFS. This terminates once an augmenting path has been found. To find an augmenting path, the graph search algorithm is modified as follows:

1. Whenever the current node [math]\displaystyle{ v }[/math] has been reached via an edge in [math]\displaystyle{ M }[/math], only incident edges in [math]\displaystyle{ E\setminus M }[/math] are considered for seeing new nodes.
2. Mirror-symmetrically, whenever the current node [math]\displaystyle{ v }[/math] has been reached via an edge in [math]\displaystyle{ E\setminus M }[/math], only the (unique) incident edge in [math]\displaystyle{ M }[/math], if existing, is considered for seeing a new node.

Proof: Basically, we have to show that there is no more augmenting path if this repeated graph search does not find one. So suppose for a contradiction that a search from an exposed node [math]\displaystyle{ u }[/math] fails although there is an augmenting path [math]\displaystyle{ p }[/math] from [math]\displaystyle{ u }[/math] to some other exposed node [math]\displaystyle{ v }[/math]. Then [math]\displaystyle{ v }[/math] is not seen by this graph search. Let [math]\displaystyle{ x }[/math] be the last node on [math]\displaystyle{ p }[/math] seen by this graph search, let [math]\displaystyle{ e }[/math] denote the edge over which [math]\displaystyle{ x }[/math] was seen, and let [math]\displaystyle{ e' }[/math] be the next edge on [math]\displaystyle{ p }[/math] (as seen in the direction from [math]\displaystyle{ u }[/math] to [math]\displaystyle{ v }[/math]). Since [math]\displaystyle{ e }[/math] was considered by the graph search and [math]\displaystyle{ e' }[/math] was not, we may conclude [math]\displaystyle{ e,e'\not\in M }[/math]. Let [math]\displaystyle{ p' }[/math] be the subpath of [math]\displaystyle{ p }[/math] from [math]\displaystyle{ u }[/math] to [math]\displaystyle{ x }[/math] from [math]\displaystyle{ u }[/math] to [math]\displaystyle{ x }[/math] and [math]\displaystyle{ p'' }[/math] the path from [math]\displaystyle{ u }[/math] to [math]\displaystyle{ x }[/math] in the arborescence. Note that the last edge of [math]\displaystyle{ p' }[/math] is in [math]\displaystyle{ M }[/math], so [math]\displaystyle{ p' }[/math] has even length. Since [math]\displaystyle{ e\not\in M }[/math], [math]\displaystyle{ p'' }[/math] has odd legngth. In summary, the concatenation [math]\displaystyle{ p+p' }[/math] is a (usually non-simple) cycle of odd length. In particular, [math]\displaystyle{ G }[/math] is not bipartite.

Remark: This modified graph traversal in an undirected graph could be implemented as a regular graph traversal in a directed graph:

1. Duplicate each matched node [math]\displaystyle{ v }[/math] giving [math]\displaystyle{ v_1 }[/math] and [math]\displaystyle{ v_2 }[/math].
2. Replace each edge [math]\displaystyle{ \{v,w\} }[/math] by two arcs, [math]\displaystyle{ (v,w) }[/math] and [math]\displaystyle{ (w,v) }[/math].
3. For each matched node [math]\displaystyle{ v }[/math]:
   1. Let all incoming arcs of [math]\displaystyle{ v }[/math] in [math]\displaystyle{ M }[/math] point to [math]\displaystyle{ v_1 }[/math] and all outgoing arcs in [math]\displaystyle{ E\setminus M }[/math] leave [math]\displaystyle{ v_1 }[/math].
   2. Mirror-symmetrically, let all incoming arcs of [math]\displaystyle{ v }[/math] in [math]\displaystyle{ E\setminus M }[/math] point to [math]\displaystyle{ v_2 }[/math] and all outgoing arcs of [math]\displaystyle{ v }[/math] in [math]\displaystyle{ M }[/math] leave [math]\displaystyle{ v_2 }[/math].