Metadata-Version: 1.1
Name: munkres
Version: 1.0.10
Summary: munkres algorithm for the Assignment Problem
Home-page: http://software.clapper.org/munkres/
Author: Brian Clapper
Author-email: bmc@clapper.org
License: Apache Software License
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.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.
        
Platform: UNKNOWN
Classifier: Intended Audience :: Developers
Classifier: Intended Audience :: Science/Research
Classifier: License :: OSI Approved :: Apache Software License
Classifier: Operating System :: OS Independent
Classifier: Programming Language :: Python
Classifier: Topic :: Scientific/Engineering :: Mathematics
Classifier: Topic :: Software Development :: Libraries :: Python Modules
