lap: Linear Assignment Problem solver

lap is a linear assignment problem solver using Jonker-Volgenant algorithm for dense (LAPJV [1]) or sparse (LAPMOD [2]) matrices.

Both algorithms are implemented from scratch based solely on the papers [1,2] and the public domain Pascal implementation provided by A. Volgenant [3].

In my tests the LAPMOD implementation seems to be faster than the LAPJV implementation for matrices with a side of more than ~5000 and with less than 50% finite coefficients.

[1] R. Jonker and A. Volgenant, "A Shortest Augmenting Path Algorithm for Dense and Sparse Linear Assignment Problems", Computing 38, 325-340 (1987)
[2] A. Volgenant, "Linear and Semi-Assignment Problems: A Core Oriented Approach", Computer Ops Res. 23, 917-932 (1996)
[3] http://www.assignmentproblems.com/LAPJV.htm

Usage

cost, x, y = lap.lapjv(C)

The function lapjv(C) returns the assignment cost (cost) and two arrays, x, y. If cost matrix C has shape N x M, then x is a size-N array specifying to which column is row is assigned, and y is a size-M array specifying to which row each column is assigned. For example, an output of x = [1, 0] indicates that row 0 is assigned to column 1 and row 1 is assigned to column 0. Similarly, an output of x = [2, 1, 0] indicates that row 0 is assigned to column 2, row 1 is assigned to column 1, and row 2 is assigned to column 0.

Note that this function does not return the assignment matrix (as done by scipy's linear_sum_assignment and lapsolver's solve dense). The assignment matrix can be constructed from x as follows:

A = np.zeros((N, M))
for i in range(N):
    A[i, x[i]] = 1

Equivalently, we could construct the assignment matrix from y:

A = np.zeros((N, M))
for j in range(M):
    A[y[j], j] = 1

Finally, note that the outputs are redundant: we can construct x from y, and vise versa:

x = [np.where(y == i)[0][0] for i in range(N)]
y = [np.where(x == j)[0][0] for j in range(M)]