munkres algorithm for the Assignment Problem

## Project description

## Introduction

The Munkres module provides an implementation of the Munkres algorithm (also called the Hungarian algorithm or the Kuhn-Munkres algorithm), useful for solving the Assignment Problem.

## Assignment Problem

Let *C* be an *n*x*n* matrix representing the costs of each of *n* workers
to perform any of *n* jobs. The assignment problem is to assign jobs to
workers in a way that minimizes the total cost. Since each worker can perform
only one job and each job can be assigned to only one worker the assignments
represent an independent set of the matrix *C*.

One way to generate the optimal set is to create all permutations of the indexes necessary to traverse the matrix so that no row and column are used more than once. For instance, given this matrix (expressed in Python):

matrix = [[5, 9, 1], [10, 3, 2], [8, 7, 4]]

You could use this code to generate the traversal indexes:

def permute(a, results): if len(a) == 1: results.insert(len(results), a) else: for i in range(0, len(a)): element = a[i] a_copy = [a[j] for j in range(0, len(a)) if j != i] subresults = [] permute(a_copy, subresults) for subresult in subresults: result = [element] + subresult results.insert(len(results), result) results = [] permute(range(len(matrix)), results) # [0, 1, 2] for a 3x3 matrix

After the call to permute(), the results matrix would look like this:

[[0, 1, 2], [0, 2, 1], [1, 0, 2], [1, 2, 0], [2, 0, 1], [2, 1, 0]]

You could then use that index matrix to loop over the original cost matrix and calculate the smallest cost of the combinations:

n = len(matrix) minval = sys.maxint for row in range(n): cost = 0 for col in range(n): cost += matrix[row][col] minval = min(cost, minval) print minval

While this approach works fine for small matrices, it does not scale. It
executes in O(*n*!) time: Calculating the permutations for an *n*x*n*
matrix requires *n*! operations. For a 12x12 matrix, that’s 479,001,600
traversals. Even if you could manage to perform each traversal in just one
millisecond, it would still take more than 133 hours to perform the entire
traversal. A 20x20 matrix would take 2,432,902,008,176,640,000 operations. At
an optimistic millisecond per operation, that’s more than 77 million years.

The Munkres algorithm runs in O(*n*^3) time, rather than O(*n*!). This
package provides an implementation of that algorithm.

This version is based on http://www.public.iastate.edu/~ddoty/HungarianAlgorithm.html.

This version was written for Python by Brian Clapper from the (Ada) algorithm
at the above web site. (The `Algorithm::Munkres` Perl version, in CPAN, was
clearly adapted from the same web site.)

## Usage

Construct a Munkres object:

from munkres import Munkres m = Munkres()

Then use it to compute the lowest cost assignment from a cost matrix. Here’s a sample program:

from munkres import Munkres, print_matrix matrix = [[5, 9, 1], [10, 3, 2], [8, 7, 4]] m = Munkres() indexes = m.compute(matrix) print_matrix('Lowest cost through this matrix:', matrix) total = 0 for row, column in indexes: value = matrix[row][column] total += value print '(%d, %d) -> %d' % (row, column, value) print 'total cost: %d' % total

Running that program produces:

Lowest cost through this matrix: [5, 9, 1] [10, 3, 2] [8, 7, 4] (0, 0) -> 5 (1, 1) -> 3 (2, 2) -> 4 total cost=12

The instantiated Munkres object can be used multiple times on different matrices.

## Non-square Cost Matrices

The Munkres algorithm assumes that the cost matrix is square. However, it’s possible to use a rectangular matrix if you first pad it with 0 values to make it square. This module automatically pads rectangular cost matrices to make them square.

Notes:

- The module operates on a
*copy*of the caller’s matrix, so any padding will not be seen by the caller. - The cost matrix must be rectangular or square. An irregular matrix will
*not*work.

## Calculating Profit, Rather than Cost

The cost matrix is just that: A cost matrix. The Munkres algorithm finds the combination of elements (one from each row and column) that results in the smallest cost. It’s also possible to use the algorithm to maximize profit. To do that, however, you have to convert your profit matrix to a cost matrix. The simplest way to do that is to subtract all elements from a large value. For example:

from munkres import Munkres, print_matrix matrix = [[5, 9, 1], [10, 3, 2], [8, 7, 4]] cost_matrix = [] for row in matrix: cost_row = [] for col in row: cost_row += [sys.maxint - col] cost_matrix += [cost_row] m = Munkres() indexes = m.compute(cost_matrix) print_matrix('Lowest cost through this matrix:', matrix) total = 0 for row, column in indexes: value = matrix[row][column] total += value print '(%d, %d) -> %d' % (row, column, value) print 'total profit=%d' % total

Running that program produces:

Highest profit through this matrix: [5, 9, 1] [10, 3, 2] [8, 7, 4] (0, 1) -> 9 (1, 0) -> 10 (2, 2) -> 4 total profit=23

The `munkres` module provides a convenience method for creating a cost
matrix from a profit matrix. Since it doesn’t know whether the matrix contains
floating point numbers, decimals, or integers, you have to provide the
conversion function; but the convenience method takes care of the actual
creation of the cost matrix:

import munkres cost_matrix = munkres.make_cost_matrix(matrix, lambda cost: sys.maxint - cost)

So, the above profit-calculation program can be recast as:

from munkres import Munkres, print_matrix, make_cost_matrix matrix = [[5, 9, 1], [10, 3, 2], [8, 7, 4]] cost_matrix = make_cost_matrix(matrix, lambda cost: sys.maxint - cost) m = Munkres() indexes = m.compute(cost_matrix) print_matrix('Lowest cost through this matrix:', matrix) total = 0 for row, column in indexes: value = matrix[row][column] total += value print '(%d, %d) -> %d' % (row, column, value) print 'total profit=%d' % total

## References

- http://www.public.iastate.edu/~ddoty/HungarianAlgorithm.html
- Harold W. Kuhn. The Hungarian Method for the assignment problem.
*Naval Research Logistics Quarterly*, 2:83-97, 1955. - Harold W. Kuhn. Variants of the Hungarian method for assignment
problems.
*Naval Research Logistics Quarterly*, 3: 253-258, 1956. - Munkres, J. Algorithms for the Assignment and Transportation Problems.
*Journal of the Society of Industrial and Applied Mathematics*, 5(1):32-38, March, 1957. - http://en.wikipedia.org/wiki/Hungarian_algorithm

## Copyright and License

Copyright (c) 2008 Brian M. Clapper

This is free software, released under the following BSD-like license:

Redistribution and use in source and binary forms, with or without modification, are permitted provided that the following conditions are met:

Redistributions of source code must retain the above copyright notice, this list of conditions and the following disclaimer.

The end-user documentation included with the redistribution, if any, must include the following acknowlegement:

This product includes software developed by Brian M. Clapper (bmc@clapper.org, http://www.clapper.org/bmc/). That software is copyright (c) 2008 Brian M. Clapper.

Alternately, this acknowlegement may appear in the software itself, if and wherever such third-party acknowlegements normally appear.

THIS SOFTWARE IS PROVIDED **AS IS**, AND ANY EXPRESSED OR IMPLIED
WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE IMPLIED WARRANTIES OF
MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE ARE DISCLAIMED. IN NO
EVENT SHALL BRIAN M. CLAPPER BE LIABLE FOR ANY DIRECT, INDIRECT,
INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL DAMAGES (INCLUDING, BUT
NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS OR SERVICES; LOSS OF USE,
DATA, OR PROFITS; OR BUSINESS INTERRUPTION) HOWEVER CAUSED AND ON ANY
THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT LIABILITY, OR TORT
(INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY OUT OF THE USE OF
THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF SUCH DAMAGE.

`$Id: munkres.py 7768 2008-06-30 20:21:41Z bmc $`