Linear Assignment Problem (or Weighted Bipartite Matching Problem)

3 posts in this topic

In this topic I would like to discuss everything about Linear Assignment Problems (LAPs). Another important class of assignment problems, that is Quadratic Assignment Problems, will be discussed later in another topic.

Let's begin with the classic LAP, defined as follows (Wikipedia):


The problem instance has a number of agents and a number of tasks. Any agent can be assigned to perform any task, incurring some cost that may vary depending on the agent-task assignment. It is required to perform all tasks by assigning exactly one agent to each task and exactly one task to each agent in such a way that the total cost of the assignment is minimized.

This problem is also called Linear Sum Assignment Problem (LSAP), since the objective function is a sum, to be differentiated with other types of assignment problems such as the Linear Bottleneck Assignment Problem (LBAP). These problems will also be discussed later in this topic. When we say LAP without specification, we refer to the LSAP.

The formal mathematical definition of the LAP is the following.

Problem. (LAP) Given a matrix $C = (c_{ij})_{1\le i,j \le n} \in \mathbb{R}^{n\times n}$, called cost matrix, where $n$ is a given positive integer.
\begin{align} \mbox{minimize}\quad & \sum_{i=1}^n \sum_{j=1}^n c_{ij}x_{ij} \\ \mbox{subject to}\quad & \sum_{i=1}^n x_{ij} = 1 \quad \forall j, \\ & \sum_{j=1}^n x_{ij} = 1 \quad \forall i,\\ & x_{ij} \in \set{0,1} \quad\forall i,j. \end{align}

It is straightforward to see that this problem is equivalent to finding a perfect matching of a bipartite graph, thus it is also known as the bipartite minimum cost perfect matching problem.

It is well-known that the LP relaxation of the above integer program (i.e. the integral constraints $x_{ij} \in \set{0,1}$ are relaxed to $x_{ij} \ge 0$) has optimal solutions at extreme points, thus the LP relaxed problem and the original integer program are equivalent.

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The Hungarian Method

This famous method is the first to solve the LAP in polynomial time. It is recognized as the predecessor of the primal-dual method for linear programming. Primal-dual methods are also a very important topic, which I will discuss later in a separate forum topic (as they deserve it).

It is easy to derive the following dual of the LP relaxation of the LAP: \begin{align} \mbox{maximize}\quad & \sum_{i=1}^n u_i+ \sum_{j=1}^n v_j \\ \mbox{subject to}\quad & u_i+v_j \le c_{ij} \quad\forall i,j. \end{align}

If a primal feasible solution $x^*$ and a dual feasible solution $(u^*,v^*)$ satisfy the complementary slackness conditions: 
$$x_{ij}(c_{ij} -u_i-v_j) \quad \forall ij,$$
then $x^*$ is an optimal solution to the primal and $(u^*,v^*)$ is an optimal solution to the dual. 

For a given dual feasible solution, let $J=\set{ij\in E: u_i+v_j=c_{ij}}$. The restricted primal:
\begin{align*} \mbox{minimize}\quad & \sum_{i} s_i \sum_{j}s_j \\
\mbox{subject to}\quad & \sum_{i=1}^n x_{ij} +s_j= 1 \quad \forall j, \\
& \sum_{j=1}^n x_{ij} +s_i = 1 \quad \forall i,\\
& x_{ij}=0\quad\forall ij\not\in J, \\
& x_{ij}\ge 0\quad\forall ij\in J, \\
&s\ge 0. \end{align*}

Since $$\sum_{i} s_i +  \sum_{j}s_j =  \sum_{i} \left(1-\sum_{j}x_{ij}\right) + \sum_{j} \left(1-\sum_{i}x_{ij}\right) = 2n -2\sum_{i}\sum_{j}x_{ij},$$ it is straightforward to see that the above RP is equivalent to the following problem:
\begin{align*} \mbox{maximize}\quad & \sum_{i}\sum_{j}x_{ij} \\
\mbox{subject to}\quad & \sum_{i=1}^n x_{ij} \le 1 \quad \forall j, \\
& \sum_{j=1}^n x_{ij} \le 1 \quad \forall i,\\
& x_{ij}=0\quad\forall ij\not\in J, \\
& x_{ij}\ge 0\quad\forall ij\in J. \end{align*}
Clearly, this is the Maximum Bipartite Matching Problem for the graph $G'=(A\cup B,J)$. One can solve it efficiently using one of the methods presented in that topic.

If the maximum cardinality of this matching is $n=|A|=|B|$ then the obtained matching is also the optimal solution to the original assignment problem. Otherwise, we can use this matching to find an optimal solution to the following dual of the RP, called the DRP:
\begin{align*} \mbox{maximize}\quad & \sum_{i}u_i' + \sum_{j}v_j' \\
\mbox{subject to}\quad & u_i' \le 1 \quad \forall i, \\
&v_j' \le 1 \quad \forall j,\\
& u_i'+v_j'\le 0 \quad\forall ij\in J. \end{align*}

Deriving an optimal solution to the DRP from an optimal solution to the RP

This is an important step. Let $x$ be a maximum matching of RP. The corresponding optimal $s$ can be trivially obtained: $s_i=0$ if $i$ is matched and $s_i=1$ otherwise, for any $i\in V=A\cup B$. Now, a solution $(u',v')$ to the DRP is optimal if and only if it is dual feasible and satisfy the complementary slackness conditions, i.e. it has to satisfy all the following conditions:
u_i' &\le 1 \quad \forall i\in A, \\
v_j' &\le 1 \quad \forall j\in B,\\
u_i'+v_j' &\le 0 \quad\forall ij\in J,\\
s_i(1-u_i') &= 0\quad\forall i\in A, \\
s_j(1-v_j') &= 0\quad\forall j\in B, \\
x_{ij}(u_i'+v_j') &= 0 \quad\forall ij\in J.
(The first three conditions are dual feasibility and the last three conditions are complementary slackness.) The last condition can be read: if $(i,j)$ are matched, then $u_i'+v_j'=0$.

If $i$ is exposed (i.e. $s_i=1$), then we must have $u_i'=1$ (for any $i\in V$).
Let $i_0$ be an exposed and consider any of its neighbors $j_0$ (in $J$). Clearly, $j$ must be matched (because otherwise we can add the edge $(i_0,j_0)$ to the current matching to obtain another matching, which is not possible since the current matching is already maximum). Since $u_{i_0}'+v_{j_0}' \le 0$ and $u_{i_0}'=1$, we have $v_{j_0}' \le -1$. Let $i_1$ be the mate of $j_0$, since $u_{i_1}'+v_{j_0}' = 0$ and $v_{j_0}' \le -1$ and $u_{i_1}' \le 1$, we must have $v_{j_0}' = -1$ and $u_{i_1}' = 1$. Thus, we have labeled the path $i_0,j_0,i_1$ with the values $1,-1,1$, where $j_0$ is any possible neighbor (in $J$) of $i_0$ and $i_1$ is the mate of $j_0$. Continue the same labeling strategy with $i_1$, and so on. By this, we have labeled all alternating paths from $i_0$. Note that these alternating paths are not augmenting because the given matching is maximum (c.f. topic on Maximum Matching Problem).

Two alternating paths generated from two different exposed nodes $i_0$ and $j_0$ (not necessarily in different partitions) do not have any node in common, otherwise we will have an augmenting path starting from $i_0$ and terminating at $j_0$.

For any exposed node $i_0\in V$: labeling the nodes on all maximal alternating paths (including the ones with a single node) starting from $i_0$ by alternating between $1$ and $-1$. Note: an alternating path is maximal if we cannot add any edge to it to create another alternating path.

Has the above labeling scheme already covered all the nodes? Not always. If some nodes remain, then obviously they are not connected to any of the already labeled nodes, and all of them are matched. For this subset of nodes, we label $1$ to all nodes on one side (e.g. $A$) and $-1$ to all nodes on the other side ($B$). It is easy to see that this labeling ensures the dual feasibility and the complementary slackness conditions. 


[1]     A. Schrijver, Combinatorial Optimization: Polyhedra and Efficiency, Springer 2012.

[2]     C. H. Papadimitriou and K. Steiglitz, Combinatorial Optimization: Algorithms and Complexity, Dover 1998.

[3]     M. X. Goemans and D. P. Williamson, The Primal-Dual Method for Approximation Algorithms and Its Application to Network Design Problems.

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Bipartite Minimum Cost (Non-perfect) Matching

The bipartite perfect matching problem is nice, but in practice (i.e. applications) we very often encounter non-perfect matching situations. I'll show how to reduce such problems to the perfect matching one.

Problem. Given a cost matrix $C = (c_{ij})_{1\le i \le m,1\le j \le n} \in \mathbb{R}^{m\times n}$, where $m$ and $n$ are given positive integers.
\begin{align} \mbox{minimize}\quad & \sum_{i=1}^m \sum_{j=1}^n c_{ij}x_{ij} \\ \mbox{subject to}\quad & \sum_{i=1}^m x_{ij} \le 1 \quad \forall j=\overline{1,n}, \\ & \sum_{j=1}^n x_{ij} \le 1 \quad \forall i=\overline{1,m},\\ & x_{ij} \in \set{0,1} \quad\forall i,j. \end{align}

First, if the two parts of the input bipartite graph are not of equal size, then we add dummy nodes to the smaller part to make them equal. 

Suppose that $m < n$, then define new cost values $c_{ij} = 0 \ \forall i=\overline{m+1,n}$. It is straightforward to show that the above problem is equivalent to the following problem: \begin{align} \mbox{minimize}\quad & \sum_{i=1}^n \sum_{j=1}^n c_{ij}x_{ij} \\ \mbox{subject to}\quad & \sum_{i=1}^n x_{ij} \le 1 \quad \forall j=\overline{1,n}, \\ & \sum_{j=1}^n x_{ij} \le 1 \quad \forall i=\overline{1,n},\\ & x_{ij} \in \set{0,1} \quad\forall i,j. \end{align}

Next, we define auxiliary variables to convert the inequality constraints to equality ones. Denote
\begin{align*} x_{(n+1),j} &= 1 - \sum_{i=1}^m x_{ij} \quad \forall j=\overline{1,n}, \\ x_{i,(n+1)} &= 1 - \sum_{j=1}^n x_{ij} \quad \forall i=\overline{1,n} \end{align*}
Define a new cost value $c_{ij}=0$ for each pair $(i,j) = (n+1,k)$ or $(i,j) = (k,n+1)$, where $1\le i,j \le n$, and a new cost value $c_{(n+1),(n+1)}$ that will be specified later. It is easy to see that the previous problem is equivalent to:
\begin{align} \mbox{minimize}\quad & \sum_{i=1}^{n+1} \sum_{j=1}^{n+1} c_{ij}x_{ij} \\ \mbox{subject to}\quad & \sum_{i=1}^{n+1} x_{ij} = 1 \quad \forall j=\overline{1,n+1}, \\ & \sum_{j=1}^n x_{ij} = 1 \quad \forall i=\overline{1,n+1},\\ & x_{(n+1),(n+1)} = 0 \\ & x_{ij} \in \set{0,1} \quad\forall i,j. \end{align}

Finally, we can remove the constraint $x_{(n+1),(n+1)} = 0$ by setting $c_{(n+1),(n+1)}$ to a very high value, then we obtain an instance of the bipartite perfect matching problem.

Therefore, any bipartite non-perfect matching can be converted to a perfect one.

Question: Is the converse true? I.e. can any bipartite minimum cost perfect matching problem be converted to a non-perfect one?

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