munkres module
Introduction
The Munkres module provides an implementation of the Munkres algorithm (also called the Hungarian algorithm or the KuhnMunkres algorithm), useful for solving the Assignment Problem.
Assignment Problem
Let C be an n by 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:
minval = sys.maxsize for indexes in results: cost = 0 for row, col in enumerate(indexes): 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://csclab.murraystate.edu/~bob.pilgrim/445/munkres.html
This version was written for Python by Brian Clapper from the 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(matrix, msg='Lowest cost through this 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.
Nonsquare 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.maxsize  col] cost_matrix += [cost_row] m = Munkres() indexes = m.compute(cost_matrix) print_matrix(matrix, msg='Highest profit through this 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 matrix:
import munkres cost_matrix = munkres.make_cost_matrix(matrix, lambda cost: sys.maxsize  cost)
So, the above profitcalculation 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.maxsize  cost) m = Munkres() indexes = m.compute(cost_matrix) print_matrix(matrix, msg='Lowest cost through this 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
Disallowed Assignments
You can also mark assignments in your cost or profit matrix as disallowed. Simply use the munkres.DISALLOWED constant.
from munkres import Munkres, print_matrix, make_cost_matrix, DISALLOWED matrix = [[5, 9, DISALLOWED], [10, DISALLOWED, 2], [8, 7, 4]] cost_matrix = make_cost_matrix(matrix, lambda cost: (sys.maxsize  cost) if (cost != DISALLOWED) else DISALLOWED) m = Munkres() indexes = m.compute(cost_matrix) print_matrix(matrix, msg='Lowest cost through this 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 this program produces:
Lowest cost through this matrix: [ 5, 9, D] [10, D, 2] [ 8, 7, 4] (0, 1) > 9 (1, 0) > 10 (2, 2) > 4 total profit=23
References

http://www.public.iastate.edu/~ddoty/HungarianAlgorithm.html

Harold W. Kuhn. The Hungarian Method for the assignment problem. Naval Research Logistics Quarterly, 2:8397, 1955.

Harold W. Kuhn. Variants of the Hungarian method for assignment problems. Naval Research Logistics Quarterly, 3: 253258, 1956.

Munkres, J. Algorithms for the Assignment and Transportation Problems. Journal of the Society of Industrial and Applied Mathematics, 5(1):3238, March, 1957.

http://en.wikipedia.org/wiki/Hungarian_algorithm
Copyright and License
Copyright 20082016 Brian M. Clapper
Licensed under the Apache License, Version 2.0 (the "License"); you may not use this file except in compliance with the License. You may obtain a copy of the License at
http://www.apache.org/licenses/LICENSE2.0
Unless required by applicable law or agreed to in writing, software distributed under the License is distributed on an "AS IS" BASIS, WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied. See the License for the specific language governing permissions and limitations under the License.
Module variables
var DISALLOWED
Functions
def make_cost_matrix(
profit_matrix, inversion_function)
Create a cost matrix from a profit matrix by calling 'inversion_function' to invert each value. The inversion function must take one numeric argument (of any type) and return another numeric argument which is presumed to be the cost inverse of the original profit.
This is a static method. Call it like this:
.. python:
cost_matrix = Munkres.make_cost_matrix(matrix, inversion_func)
For example:
.. python:
cost_matrix = Munkres.make_cost_matrix(matrix, lambda x : sys.maxsize  x)
:Parameters: profit_matrix : list of lists The matrix to convert from a profit to a cost matrix
inversion_function : function The function to use to invert each entry in the profit matrix
:rtype: list of lists :return: The converted matrix
Classes
class Munkres
Calculate the Munkres solution to the classical assignment problem. See the module documentation for usage.
Ancestors (in MRO)
Static methods
def make_cost_matrix(
profit_matrix, inversion_function)
DEPRECATED
Please use the module function make_cost_matrix()
.
Instance variables
var C
var Z0_c
var Z0_r
var col_covered
var marked
var n
var path
var row_covered
Methods
def __init__(
self)
Create a new instance
def compute(
self, cost_matrix)
Compute the indexes for the lowestcost pairings between rows and columns in the database. Returns a list of (row, column) tuples that can be used to traverse the matrix.
:Parameters:
cost_matrix : list of lists
The cost matrix. If this cost matrix is not square, it
will be padded with zeros, via a call to pad_matrix()
.
(This method does not modify the caller's matrix. It
operates on a copy of the matrix.)
**WARNING**: This code handles square and rectangular matrices. It does *not* handle irregular matrices.
:rtype: list
:return: A list of (row, column)
tuples that describe the lowest
cost path through the matrix
def pad_matrix(
self, matrix, pad_value=0)
Pad a possibly nonsquare matrix to make it square.
:Parameters: matrix : list of lists matrix to pad
pad_value : int value to use to pad the matrix
:rtype: list of lists :return: a new, possibly padded, matrix