munkres algorithm for the Assignment Problem
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.
Let C be an nxn 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.maxsize 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 nxn 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.)
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.
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.
- 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 cost matrix:
import munkres cost_matrix = munkres.make_cost_matrix(matrix, lambda cost: sys.maxsize - 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.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
- 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.
Copyright and License
Copyright 2008-2016 Brian M. Clapper
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