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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.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 *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.

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.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

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

==========

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

2. Harold W. Kuhn. The Hungarian Method for the assignment problem.

*Naval Research Logistics Quarterly*, 2:83-97, 1955.

3. Harold W. Kuhn. Variants of the Hungarian method for assignment

problems. *Naval Research Logistics Quarterly*, 3: 253-258, 1956.

4. Munkres, J. Algorithms for the Assignment and Transportation Problems.

*Journal of the Society of Industrial and Applied Mathematics*,

5(1):32-38, March, 1957.

5. http://en.wikipedia.org/wiki/Hungarian_algorithm

Copyright and License

=====================

Copyright 2008-2016 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/LICENSE-2.0

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.

============

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.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 *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.

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.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

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

==========

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

2. Harold W. Kuhn. The Hungarian Method for the assignment problem.

*Naval Research Logistics Quarterly*, 2:83-97, 1955.

3. Harold W. Kuhn. Variants of the Hungarian method for assignment

problems. *Naval Research Logistics Quarterly*, 3: 253-258, 1956.

4. Munkres, J. Algorithms for the Assignment and Transportation Problems.

*Journal of the Society of Industrial and Applied Mathematics*,

5(1):32-38, March, 1957.

5. http://en.wikipedia.org/wiki/Hungarian_algorithm

Copyright and License

=====================

Copyright 2008-2016 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/LICENSE-2.0

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.

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