Hungarian method: Difference between revisions

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## Initialize two [[Sets and sequences|sets]], <math>S:=\{v\}</math>, <math>T:=\emptyset</math>.
## Initialize two [[Sets and sequences|sets]], <math>S:=\{v\}</math>, <math>T:=\emptyset</math>.
## While <math>T\subsetneq N(S)</math>:
## While <math>T\subsetneq N(S)</math>:
### Let <math>w\in N(s)\setminus T</math>..
### Let <math>w\in N(s)\setminus T</math>.
### Let <math>\{u,w\}</math> be the unique edge incident to <math>w</math> that is in <math>M</math>.
### Let <math>\{u,w\}</math> be the unique edge incident to <math>w</math> that is in <math>M</math>.
### Insert <math>w</math> in <math>T</math> and <math>u</math> in <math>S</math>.
### Insert <math>w</math> in <math>T</math> and <math>u</math> in <math>S</math>.

Revision as of 08:25, 23 November 2014

Abstract view

Algorithmic problem: Maximum-weight matching in complete bipartite graphs [math]\displaystyle{ G=(V_1\dot\cup V_2,E) }[/math] with [math]\displaystyle{ |V_1|=|V_2| }[/math].

Type of algorithm: loop.

Auxiliary data:

  1. A real number [math]\displaystyle{ x(v) }[/math] for each node [math]\displaystyle{ v\in V_1 }[/math].
  2. A real number [math]\displaystyle{ y(w) }[/math] for each node [math]\displaystyle{ w\in V_2 }[/math].

Invariant:

  1. [math]\displaystyle{ M }[/math] is a matching in [math]\displaystyle{ G }[/math].
  2. For each edge [math]\displaystyle{ e=\{v,w\}\in M }[/math], where [math]\displaystyle{ v\in V_1 }[/math] and [math]\displaystyle{ w\in V_2 }[/math], it is [math]\displaystyle{ c(e)\leq x(v)+y(w) }[/math].

Variant:

Break condition: For each edge [math]\displaystyle{ e=\{v,w\} in M }[/math], where [math]\displaystyle{ v\in V_1 }[/math] and [math]\displaystyle{ w\in V_2 }[/math], it is [math]\displaystyle{ c(e)=x(v)+y(w) }[/math].

Induction basis

  1. initialize [math]\displaystyle{ M }[/math] to be a feasible matching, for example, the empty matching.
  2. Initialize all [math]\displaystyle{ x }[/math] and [math]\displaystyle{ y }[/math] values such that the invariant is fulfilled, for example:
    1. [math]\displaystyle{ x(v):=\max\{c(\{v,w\})|w\in V_2\} }[/math] for all [math]\displaystyle{ v\in V_1 }[/math].
    2. [math]\displaystyle{ y(w):=0 }[/math] for all [math]\displaystyle{ w\in V_2 }[/math],

Induction step

Notation:

  1. For a node [math]\displaystyle{ v\in V_1 }[/math]: [math]\displaystyle{ N(v):=\{w\in V_2|c(\{v,w\})=x(v)+y(w)\} }[/math], the "neighbors" of [math]\displaystyle{ v }[/math].
  2. For a set [math]\displaystyle{ S\subseteq V_1 }[/math]: [math]\displaystyle{ N(S):=\bigcup_{v\in S}N(v) }[/math].

Abstract view:

  1. Let [math]\displaystyle{ G'=(V,E') }[/math] be the subgraph of [math]\displaystyle{ G }[/math] where [math]\displaystyle{ E' }[/math] comprises all edges [math]\displaystyle{ e=\{v,w\}\in E }[/math] ([math]\displaystyle{ v\in V_1 }[/math], [math]\displaystyle{ w\in V_2 }[/math]) such that [math]\displaystyle{ c(e)=x(v)+y(w) }[/math].
  2. Try to find an augmenting path in [math]\displaystyle{ G' }[/math] with respect to [math]\displaystyle{ M }[/math].
  3. If step 2 succeeds, augment [math]\displaystyle{ M }[/math] along this path.
  4. Otherwise:
    1. Choose some exposed node [math]\displaystyle{ v\in V_1 }[/math].
    2. Initialize two sets, [math]\displaystyle{ S:=\{v\} }[/math], [math]\displaystyle{ T:=\emptyset }[/math].
    3. While [math]\displaystyle{ T\subsetneq N(S) }[/math]:
      1. Let [math]\displaystyle{ w\in N(s)\setminus T }[/math].
      2. Let [math]\displaystyle{ \{u,w\} }[/math] be the unique edge incident to [math]\displaystyle{ w }[/math] that is in [math]\displaystyle{ M }[/math].
      3. Insert [math]\displaystyle{ w }[/math] in [math]\displaystyle{ T }[/math] and [math]\displaystyle{ u }[/math] in [math]\displaystyle{ S }[/math].
    4. Set [math]\displaystyle{ \delta:=\min\{c(\{u,w\})-x(u)-y(w)|u\in S,w\in V_2\setminus T\} }[/math].
    5. For all [math]\displaystyle{ v\in V_1 }[/math], decrease [math]\displaystyle{ x(v) }[/math] by [math]\displaystyle{ \delta }[/math].
    6. For all [math]\displaystyle{ w\in V_2 }[/math], decrease [math]\displaystyle{ x(v) }[/math] by [math]\displaystyle{ \delta }[/math].

Remark: Steps 2 and 4 can be folded into one loop, which breaks if [math]\displaystyle{ N(S)=T }[/math] or an exposed node (and thus an augmenting path) is found.

Proof: Note that [math]\displaystyle{ w }[/math] is indeed matched because, otherwise, step ?? had found an augmenting path, which had rather been found in step 2.

Correctness

Termination of the main loop will follow from the complexity considerations below. Let [math]\displaystyle{ M }[/math] be a perfect matching and [math]\displaystyle{ x }[/math] and [math]\displaystyle{ y }[/math] be given such that [math]\displaystyle{ c(e)=x(v)+y(w) }[/math] for all [math]\displaystyle{ e=\{v,w\}\in M }[/math], where [math]\displaystyle{ v\in V_1 }[/math] and [math]\displaystyle{ w\in V_2 }[/math]. Let [math]\displaystyle{ M' }[/math] be another matching. Due to the invariant, it is

[math]\displaystyle{ c(M)=\sum_{v\in V_1,w\in V_2\atop e=\{v,w\}\in M}c(e)=\sum_{v\in V_1}x(v)+\sum_{w\in V_2}y(w)\geq\sum_{v\in V_1,w\in V_2\atop e=\{v,w\}\in M'}c(e) }[/math].

Therefore, [math]\displaystyle{ M }[/math] is maximal.

Remark: In particular, the following equivalence is proved: A perfect matching [math]\displaystyle{ M }[/math] has maximum weight if, and only if, there are [math]\displaystyle{ x }[/math] and [math]\displaystyle{ y }[/math] such that [math]\displaystyle{ c(e)=x(v)+y(w) }[/math] for all [math]\displaystyle{ e=\{v,w\}\in M }[/math].