# The Matching Factory

Posted on September 26, 2019 by Spencer

Suppose I have a bipartite graph $$G = (L, R, E)$$, where $$L$$ and $$R$$ have the same number of vertices $$n$$, and nonnegative costs $$c(u, v)$$ for each edge $$(u, v)$$. I want to compute a minimum cost perfect matching; that is, a subset $$M$$ of $$E$$ such that every vertex of $$G$$ is contained in exactly one edge of $$M$$, with the minimum sum of edge costs among all such subsets.

One way to tackle this problem is linear programming duality. Chapter 3 of these excellent lecture notes by Robert Kleinberg describes a primal-dual algorithm for minimum cost bipartite perfect matching. Paraphrasing, a subset of the edge set $$E$$ of $$G$$ can be represented by a $$n \times n$$ matrix of zeros and ones $$X$$, where $$X_{ij} = 1$$ represents an edge from vertex $$i \in L$$ to vertex $$j \in R$$. The edge set represented by $$X$$ is a perfect matching iff the sum of each row and each column is $$1$$. We want to find the cheapest such edge set, that is to solve the optimization problem:

\begin{align} &\min \sum_{i, j} c(i, j) X_{ij} \quad \text{subject to}\\ &\forall j \quad \sum_i X_{ij} = 1 \\ &\forall i \quad \sum_j X_{ij} = 1 \\ &\forall i \forall j \quad X_{ij} \in \{0, 1\} \\ \end{align}

Relaxing this to a continuous optimization problem by letting the $$X_{ij}$$ take values in the interval $$[0, 1]$$ gives a linear program whose solutions are fractional perfect matchings; (see the lecture notes for more detail).

\begin{align} &\min \sum_{i, j} c(i, j) X_{ij} \quad \text{subject to}\\ &\forall j \quad \sum_i X_{ij} = 1 \\ &\forall i \quad \sum_j X_{ij} = 1 \\ &\forall i \forall j \quad X_{ij} \in [0, 1] \\ \end{align}

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We call the optimal objective value (in this case, the cost of the minimum cost fractional perfect matching) the value of this linear program. By combining the constraints of this linear program, we arrive at the dual program, a maximization problem whose value is equal to the value of the above program (called the primal program).

To be continued!