二分图完全匹配 不完全匹配 / linear_sum_assignment 详解

https://jack.valmadre.net/notes/2020/12/08/non-perfect-linear-assignment/

\(G = (U,V,E)\)

• \(|U| = r\)
• \(|V| = n\)
• without loss of generality, assume \(r \leq n\)

\[\begin{bmatrix} \infty & 3 & -1 \\ \infty & 5 & \infty \\ 2 & -3 & 0 \\ \end{bmatrix}\]

• \(E(i,j) = \infty\) means no connection
  • Missing edges
• \(E(i,j) = 0\) means cost is zero.
  • however, given that cost can be either positive or negative, zero can be chosen.

In a matching problem, there will be \(\nu\) chosen edges.

\[0<\nu<=r \]

Balanced

Balanced means \(r = n\)

two subsets have equal cardinality

Perfect/Complete Matching

note that the problem is not actuallysolved using a general-purpose ILP(integer linear programming) solver, it is just a convenient framework in which to express the problem

Perfect/Complete matching = every vertex has a match

The constraint that the sum of each row and column is equal to one ensures that each element has exactly one match.

Unbalanced

assume \(r < n\)

an unbalanced probem cannot have a perfect matching, since there will be at least \(n-r\) unmatched elements in the larger set.

One-sided Perfect Matching

One-sided Perfect Matching = every vertex in the smaller set has a match.

In one-sided perfect matching, \(\nu = r < n\)

What if there is no perfect matching?

Imperfect Matching(with given edge number)

- when there does not exist a (one-sided)perfect matching.

when \(\nu < r\), i.e. one-sided perfect matching cannot be achieved due to the lack of edges, all possible matchings are imperfect/incomplete.

[Note💡]: The algorithm need to achieve CERTAIN NUMBER(\(\nu\)) OF MATCHING

The constraints means that unmatched nodes on both part are allowed.

If \(\nu = r = n\), then perfect and imperfect matching coincide.

Minimum-Weight Matching(merely consider min cost)

no constraint on the size of the matching.

[Note💡]: It puts no constraint on the size of the matching.

• 

Meeting the given \(\nu\) may force the algorithm to give up limiting the cost as small as possible, and choose a SUBOPTIMAL, yet bigger matching.

Tension between max matching-num and min cost

Add edges(imperfect -> perfect matching) may lead to bigger cost.

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Transformations

Transform imperfect and minimum-weight matching problems into perfect matching problems while preserving sparsity

Unbalanced/Imperfect to Balanced/Perfect

Adding $n-r$ rows of zeros

The resulting balanced graph has a perfect matching if and only if the original unbalanced graph had a matching of size \(r\)(in which every vertex in the smaller setis matched).

How to solve a Imperfect matching problem

replace the infinite edge weights with a large,finite cost

The resulting graph must contain a perfect matching since it is a completebipartite graph, and each high-cost edge is worth more than all original edges combined.

The high-cost edges(modified from nonexistent edges) can be excluded at the end.

A complete bipartite graph \(\mathbf{K}_{m,n}\) has a maximum matching of size \(\min(m,n)\).

wiki

Solve minimum-weight matching

replace all positive edges (including infinite edges) with zero

Minimum-weight matching will never select an edge with positive cost: it is better to simply leave it unselected.

Edges with zero cost have noimpact.

Behavior of scipy.optimize.linear_sum_assignment

It solved a one-sized perfect matching problem. Row/col with higher cardinality will be left partially matched.