lap: Linear Assignment Problem Solver

lap is a linear assignment problem solver using Jonker-Volgenant algorithm for dense LAPJV¹ or sparse LAPMOD² matrices. Both algorithms are implemented from scratch based solely on the papers¹˒² and the public domain Pascal implementation provided by A. Volgenant³. 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 | [archive.org]}

💽 Installation

Install from PyPI:

pip install lap

PyPI Wheels 🛞	Windows ✅	Linux ✅	macOS ✅
Python 3.7	AMD64	x86_64/aarch64	x86_64
Python 3.8	AMD64	x86_64/aarch64	x86_64/arm64
Python 3.9-3.14 ¹	AMD64/ARM64 ²	x86_64/aarch64	x86_64/arm64

^{¹ v0.5.13 supports both numpy 1.x and 2.x for Python 3.8-3.14. 🆕
² Windows ARM64 is experimental.}

Other options

Install from GitHub repo (requires C++ compiler):

pip install git+https://github.com/gatagat/lap.git

Build and install (requires C++ compiler):

git clone https://github.com/gatagat/lap.git
cd lap
pip install "setuptools>=67.8.0"
pip install wheel build
python -m build --wheel
cd dist

🧪 Usage

import lap
import numpy as np
print(lap.lapjv(np.random.rand(4, 5), extend_cost=True))

More details

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

The function lapjv(C) returns the assignment cost cost and two arrays x and y. If cost matrix C has shape NxM, then x is a size-N array specifying to which column each 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)]

License

Released under the 2-clause BSD license, see LICENSE.

Copyright (C) 2012-2025, Tomas Kazmar

Contributors (in alphabetic order):

• Benjamin Eysenbach
• Léo Duret
• Pieter Lenaerts
• Raphael Reme
• Ratha Siv
• Robert Wen
• Steven
• Tom White
• Tomas Kazmar
• Wok