Classical bipartite cardinality matching: Difference between revisions
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This modified | This modified graph traversal in an undirected graph could be implemented as a regular graph traversal in a directed graph: | ||
# Duplicate each matched node <math>v</math> giving <math>v_1</math> and <math>v_2</math>. | # Duplicate each matched node <math>v</math> giving <math>v_1</math> and <math>v_2</math>. | ||
# Replace each edge <math>\{v,w\}</math> by two arcs, <math>(v,w)</math> and <math>(w,v)</math>. | # Replace each edge <math>\{v,w\}</math> by two arcs, <math>(v,w)</math> and <math>(w,v)</math>. |
Revision as of 07:45, 18 November 2014
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:
- Search for an augmenting path.
- If there is none, the loop terminates.
- Let [math]\displaystyle{ p }[/math] denote the augmenting path found in step 1.
- Augment [math]\displaystyle{ M }[/math] along [math]\displaystyle{ p }[/math].
Implementation of step 1: From every exposed node, a graph traversal algorithm is started that generates an arborescence from the start node (for example, a DFS or a BFS). This repeated graph search is finished once an augmenting path has been found. However, the graph search algorithm is modified as follows:
- 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.
- 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 and let [math]\displaystyle{ \{x,y\} }[/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{ \{x,y\} }[/math] was not considered by the graph search, it is [math]\displaystyle{ \{x,y\}\in E\setminus }[/math] and the edge over which [math]\displaystyle{ x }[/math] was seen is not in [math]\displaystyle{ M }[/math], either. Let [math]\displaystyle{ p' }[/math] be the path from [math]\displaystyle{ u }[/math] to [math]\displaystyle{ x }[/math] in the arborescence. 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:
- Duplicate each matched node [math]\displaystyle{ v }[/math] giving [math]\displaystyle{ v_1 }[/math] and [math]\displaystyle{ v_2 }[/math].
- Replace each edge [math]\displaystyle{ \{v,w\} }[/math] by two arcs, [math]\displaystyle{ (v,w) }[/math] and [math]\displaystyle{ (w,v) }[/math].
- For each matched node [math]\displaystyle{ v }[/math]:
- 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].
- 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].