# scipy.optimize.linear_sum_assignment#

scipy.optimize.linear_sum_assignment()#

Solve the linear sum assignment problem.

Parameters:
cost_matrixarray

The cost matrix of the bipartite graph.

maximizebool (default: False)

Calculates a maximum weight matching if true.

Returns:
row_ind, col_indarray

An array of row indices and one of corresponding column indices giving the optimal assignment. The cost of the assignment can be computed as cost_matrix[row_ind, col_ind].sum(). The row indices will be sorted; in the case of a square cost matrix they will be equal to numpy.arange(cost_matrix.shape[0]).

Notes

The linear sum assignment problem [1] is also known as minimum weight matching in bipartite graphs. A problem instance is described by a matrix C, where each C[i,j] is the cost of matching vertex i of the first partite set (a ‘worker’) and vertex j of the second set (a ‘job’). The goal is to find a complete assignment of workers to jobs of minimal cost.

Formally, let X be a boolean matrix where $$X[i,j] = 1$$ iff row i is assigned to column j. Then the optimal assignment has cost

$\min \sum_i \sum_j C_{i,j} X_{i,j}$

where, in the case where the matrix X is square, each row is assigned to exactly one column, and each column to exactly one row.

This function can also solve a generalization of the classic assignment problem where the cost matrix is rectangular. If it has more rows than columns, then not every row needs to be assigned to a column, and vice versa.

This implementation is a modified Jonker-Volgenant algorithm with no initialization, described in ref. [2].

New in version 0.17.0.

References

[2]

DF Crouse. On implementing 2D rectangular assignment algorithms. IEEE Transactions on Aerospace and Electronic Systems, 52(4):1679-1696, August 2016, DOI:10.1109/TAES.2016.140952

Examples

>>> import numpy as np
>>> cost = np.array([[4, 1, 3], [2, 0, 5], [3, 2, 2]])
>>> from scipy.optimize import linear_sum_assignment
>>> row_ind, col_ind = linear_sum_assignment(cost)
>>> col_ind
array([1, 0, 2])
>>> cost[row_ind, col_ind].sum()
5