1
2
3
4
5
6 """
7 Introduction
8 ============
9
10 The Munkres module provides an implementation of the Munkres algorithm
11 (also called the Hungarian algorithm or the Kuhn-Munkres algorithm),
12 useful for solving the Assignment Problem.
13
14 Assignment Problem
15 ==================
16
17 Let *C* be an *n*\ x\ *n* matrix representing the costs of each of *n* workers
18 to perform any of *n* jobs. The assignment problem is to assign jobs to
19 workers in a way that minimizes the total cost. Since each worker can perform
20 only one job and each job can be assigned to only one worker the assignments
21 represent an independent set of the matrix *C*.
22
23 One way to generate the optimal set is to create all permutations of
24 the indexes necessary to traverse the matrix so that no row and column
25 are used more than once. For instance, given this matrix (expressed in
26 Python)::
27
28 matrix = [[5, 9, 1],
29 [10, 3, 2],
30 [8, 7, 4]]
31
32 You could use this code to generate the traversal indexes::
33
34 def permute(a, results):
35 if len(a) == 1:
36 results.insert(len(results), a)
37
38 else:
39 for i in range(0, len(a)):
40 element = a[i]
41 a_copy = [a[j] for j in range(0, len(a)) if j != i]
42 subresults = []
43 permute(a_copy, subresults)
44 for subresult in subresults:
45 result = [element] + subresult
46 results.insert(len(results), result)
47
48 results = []
49 permute(range(len(matrix)), results) # [0, 1, 2] for a 3x3 matrix
50
51 After the call to permute(), the results matrix would look like this::
52
53 [[0, 1, 2],
54 [0, 2, 1],
55 [1, 0, 2],
56 [1, 2, 0],
57 [2, 0, 1],
58 [2, 1, 0]]
59
60 You could then use that index matrix to loop over the original cost matrix
61 and calculate the smallest cost of the combinations::
62
63 n = len(matrix)
64 minval = sys.maxsize
65 for row in range(n):
66 cost = 0
67 for col in range(n):
68 cost += matrix[row][col]
69 minval = min(cost, minval)
70
71 print minval
72
73 While this approach works fine for small matrices, it does not scale. It
74 executes in O(*n*!) time: Calculating the permutations for an *n*\ x\ *n*
75 matrix requires *n*! operations. For a 12x12 matrix, that's 479,001,600
76 traversals. Even if you could manage to perform each traversal in just one
77 millisecond, it would still take more than 133 hours to perform the entire
78 traversal. A 20x20 matrix would take 2,432,902,008,176,640,000 operations. At
79 an optimistic millisecond per operation, that's more than 77 million years.
80
81 The Munkres algorithm runs in O(*n*\ ^3) time, rather than O(*n*!). This
82 package provides an implementation of that algorithm.
83
84 This version is based on
85 http://www.public.iastate.edu/~ddoty/HungarianAlgorithm.html.
86
87 This version was written for Python by Brian Clapper from the (Ada) algorithm
88 at the above web site. (The ``Algorithm::Munkres`` Perl version, in CPAN, was
89 clearly adapted from the same web site.)
90
91 Usage
92 =====
93
94 Construct a Munkres object::
95
96 from munkres import Munkres
97
98 m = Munkres()
99
100 Then use it to compute the lowest cost assignment from a cost matrix. Here's
101 a sample program::
102
103 from munkres import Munkres, print_matrix
104
105 matrix = [[5, 9, 1],
106 [10, 3, 2],
107 [8, 7, 4]]
108 m = Munkres()
109 indexes = m.compute(matrix)
110 print_matrix(matrix, msg='Lowest cost through this matrix:')
111 total = 0
112 for row, column in indexes:
113 value = matrix[row][column]
114 total += value
115 print '(%d, %d) -> %d' % (row, column, value)
116 print 'total cost: %d' % total
117
118 Running that program produces::
119
120 Lowest cost through this matrix:
121 [5, 9, 1]
122 [10, 3, 2]
123 [8, 7, 4]
124 (0, 0) -> 5
125 (1, 1) -> 3
126 (2, 2) -> 4
127 total cost=12
128
129 The instantiated Munkres object can be used multiple times on different
130 matrices.
131
132 Non-square Cost Matrices
133 ========================
134
135 The Munkres algorithm assumes that the cost matrix is square. However, it's
136 possible to use a rectangular matrix if you first pad it with 0 values to make
137 it square. This module automatically pads rectangular cost matrices to make
138 them square.
139
140 Notes:
141
142 - The module operates on a *copy* of the caller's matrix, so any padding will
143 not be seen by the caller.
144 - The cost matrix must be rectangular or square. An irregular matrix will
145 *not* work.
146
147 Calculating Profit, Rather than Cost
148 ====================================
149
150 The cost matrix is just that: A cost matrix. The Munkres algorithm finds
151 the combination of elements (one from each row and column) that results in
152 the smallest cost. It's also possible to use the algorithm to maximize
153 profit. To do that, however, you have to convert your profit matrix to a
154 cost matrix. The simplest way to do that is to subtract all elements from a
155 large value. For example::
156
157 from munkres import Munkres, print_matrix
158
159 matrix = [[5, 9, 1],
160 [10, 3, 2],
161 [8, 7, 4]]
162 cost_matrix = []
163 for row in matrix:
164 cost_row = []
165 for col in row:
166 cost_row += [sys.maxsize - col]
167 cost_matrix += [cost_row]
168
169 m = Munkres()
170 indexes = m.compute(cost_matrix)
171 print_matrix(matrix, msg='Highest profit through this matrix:')
172 total = 0
173 for row, column in indexes:
174 value = matrix[row][column]
175 total += value
176 print '(%d, %d) -> %d' % (row, column, value)
177
178 print 'total profit=%d' % total
179
180 Running that program produces::
181
182 Highest profit through this matrix:
183 [5, 9, 1]
184 [10, 3, 2]
185 [8, 7, 4]
186 (0, 1) -> 9
187 (1, 0) -> 10
188 (2, 2) -> 4
189 total profit=23
190
191 The ``munkres`` module provides a convenience method for creating a cost
192 matrix from a profit matrix. Since it doesn't know whether the matrix contains
193 floating point numbers, decimals, or integers, you have to provide the
194 conversion function; but the convenience method takes care of the actual
195 creation of the cost matrix::
196
197 import munkres
198
199 cost_matrix = munkres.make_cost_matrix(matrix,
200 lambda cost: sys.maxsize - cost)
201
202 So, the above profit-calculation program can be recast as::
203
204 from munkres import Munkres, print_matrix, make_cost_matrix
205
206 matrix = [[5, 9, 1],
207 [10, 3, 2],
208 [8, 7, 4]]
209 cost_matrix = make_cost_matrix(matrix, lambda cost: sys.maxsize - cost)
210 m = Munkres()
211 indexes = m.compute(cost_matrix)
212 print_matrix(matrix, msg='Lowest cost through this matrix:')
213 total = 0
214 for row, column in indexes:
215 value = matrix[row][column]
216 total += value
217 print '(%d, %d) -> %d' % (row, column, value)
218 print 'total profit=%d' % total
219
220 References
221 ==========
222
223 1. http://www.public.iastate.edu/~ddoty/HungarianAlgorithm.html
224
225 2. Harold W. Kuhn. The Hungarian Method for the assignment problem.
226 *Naval Research Logistics Quarterly*, 2:83-97, 1955.
227
228 3. Harold W. Kuhn. Variants of the Hungarian method for assignment
229 problems. *Naval Research Logistics Quarterly*, 3: 253-258, 1956.
230
231 4. Munkres, J. Algorithms for the Assignment and Transportation Problems.
232 *Journal of the Society of Industrial and Applied Mathematics*,
233 5(1):32-38, March, 1957.
234
235 5. http://en.wikipedia.org/wiki/Hungarian_algorithm
236
237 Copyright and License
238 =====================
239
240 This software is released under a BSD license, adapted from
241 <http://opensource.org/licenses/bsd-license.php>
242
243 Copyright (c) 2008 Brian M. Clapper
244 All rights reserved.
245
246 Redistribution and use in source and binary forms, with or without
247 modification, are permitted provided that the following conditions are met:
248
249 * Redistributions of source code must retain the above copyright notice,
250 this list of conditions and the following disclaimer.
251
252 * Redistributions in binary form must reproduce the above copyright notice,
253 this list of conditions and the following disclaimer in the documentation
254 and/or other materials provided with the distribution.
255
256 * Neither the name "clapper.org" nor the names of its contributors may be
257 used to endorse or promote products derived from this software without
258 specific prior written permission.
259
260 THIS SOFTWARE IS PROVIDED BY THE COPYRIGHT HOLDERS AND CONTRIBUTORS "AS IS"
261 AND ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE
262 IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE
263 ARE DISCLAIMED. IN NO EVENT SHALL THE COPYRIGHT HOLDER OR CONTRIBUTORS BE
264 LIABLE FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR
265 CONSEQUENTIAL DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF
266 SUBSTITUTE GOODS OR SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS
267 INTERRUPTION) HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN
268 CONTRACT, STRICT LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE)
269 ARISING IN ANY WAY OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE
270 POSSIBILITY OF SUCH DAMAGE.
271 """
272
273 __docformat__ = 'restructuredtext'
274
275
276
277
278
279 import sys
280 import copy
281
282
283
284
285
286 __all__ = ['Munkres', 'make_cost_matrix']
287
288
289
290
291
292
293 __version__ = "1.0.6"
294 __author__ = "Brian Clapper, bmc@clapper.org"
295 __url__ = "http://software.clapper.org/munkres/"
296 __copyright__ = "(c) 2008 Brian M. Clapper"
297 __license__ = "BSD-style license"
298
299
300
301
302
304 """
305 Calculate the Munkres solution to the classical assignment problem.
306 See the module documentation for usage.
307 """
308
310 """Create a new instance"""
311 self.C = None
312 self.row_covered = []
313 self.col_covered = []
314 self.n = 0
315 self.Z0_r = 0
316 self.Z0_c = 0
317 self.marked = None
318 self.path = None
319
321 """
322 **DEPRECATED**
323
324 Please use the module function ``make_cost_matrix()``.
325 """
326 import munkres
327 return munkres.make_cost_matrix(profit_matrix, inversion_function)
328
329 make_cost_matrix = staticmethod(make_cost_matrix)
330
332 """
333 Pad a possibly non-square matrix to make it square.
334
335 :Parameters:
336 matrix : list of lists
337 matrix to pad
338
339 pad_value : int
340 value to use to pad the matrix
341
342 :rtype: list of lists
343 :return: a new, possibly padded, matrix
344 """
345 max_columns = 0
346 total_rows = len(matrix)
347
348 for row in matrix:
349 max_columns = max(max_columns, len(row))
350
351 total_rows = max(max_columns, total_rows)
352
353 new_matrix = []
354 for row in matrix:
355 row_len = len(row)
356 new_row = row[:]
357 if total_rows > row_len:
358
359 new_row += [0] * (total_rows - row_len)
360 new_matrix += [new_row]
361
362 while len(new_matrix) < total_rows:
363 new_matrix += [[0] * total_rows]
364
365 return new_matrix
366
368 """
369 Compute the indexes for the lowest-cost pairings between rows and
370 columns in the database. Returns a list of (row, column) tuples
371 that can be used to traverse the matrix.
372
373 :Parameters:
374 cost_matrix : list of lists
375 The cost matrix. If this cost matrix is not square, it
376 will be padded with zeros, via a call to ``pad_matrix()``.
377 (This method does *not* modify the caller's matrix. It
378 operates on a copy of the matrix.)
379
380 **WARNING**: This code handles square and rectangular
381 matrices. It does *not* handle irregular matrices.
382
383 :rtype: list
384 :return: A list of ``(row, column)`` tuples that describe the lowest
385 cost path through the matrix
386
387 """
388 self.C = self.pad_matrix(cost_matrix)
389 self.n = len(self.C)
390 self.original_length = len(cost_matrix)
391 self.original_width = len(cost_matrix[0])
392 self.row_covered = [False for i in range(self.n)]
393 self.col_covered = [False for i in range(self.n)]
394 self.Z0_r = 0
395 self.Z0_c = 0
396 self.path = self.__make_matrix(self.n * 2, 0)
397 self.marked = self.__make_matrix(self.n, 0)
398
399 done = False
400 step = 1
401
402 steps = { 1 : self.__step1,
403 2 : self.__step2,
404 3 : self.__step3,
405 4 : self.__step4,
406 5 : self.__step5,
407 6 : self.__step6 }
408
409 while not done:
410 try:
411 func = steps[step]
412 step = func()
413 except KeyError:
414 done = True
415
416
417 results = []
418 for i in range(self.original_length):
419 for j in range(self.original_width):
420 if self.marked[i][j] == 1:
421 results += [(i, j)]
422
423 return results
424
426 """Return an exact copy of the supplied matrix"""
427 return copy.deepcopy(matrix)
428
430 """Create an *n*x*n* matrix, populating it with the specific value."""
431 matrix = []
432 for i in range(n):
433 matrix += [[val for j in range(n)]]
434 return matrix
435
437 """
438 For each row of the matrix, find the smallest element and
439 subtract it from every element in its row. Go to Step 2.
440 """
441 C = self.C
442 n = self.n
443 for i in range(n):
444 minval = min(self.C[i])
445
446
447 for j in range(n):
448 self.C[i][j] -= minval
449
450 return 2
451
453 """
454 Find a zero (Z) in the resulting matrix. If there is no starred
455 zero in its row or column, star Z. Repeat for each element in the
456 matrix. Go to Step 3.
457 """
458 n = self.n
459 for i in range(n):
460 for j in range(n):
461 if (self.C[i][j] == 0) and \
462 (not self.col_covered[j]) and \
463 (not self.row_covered[i]):
464 self.marked[i][j] = 1
465 self.col_covered[j] = True
466 self.row_covered[i] = True
467
468 self.__clear_covers()
469 return 3
470
472 """
473 Cover each column containing a starred zero. If K columns are
474 covered, the starred zeros describe a complete set of unique
475 assignments. In this case, Go to DONE, otherwise, Go to Step 4.
476 """
477 n = self.n
478 count = 0
479 for i in range(n):
480 for j in range(n):
481 if self.marked[i][j] == 1:
482 self.col_covered[j] = True
483 count += 1
484
485 if count >= n:
486 step = 7
487 else:
488 step = 4
489
490 return step
491
493 """
494 Find a noncovered zero and prime it. If there is no starred zero
495 in the row containing this primed zero, Go to Step 5. Otherwise,
496 cover this row and uncover the column containing the starred
497 zero. Continue in this manner until there are no uncovered zeros
498 left. Save the smallest uncovered value and Go to Step 6.
499 """
500 step = 0
501 done = False
502 row = -1
503 col = -1
504 star_col = -1
505 while not done:
506 (row, col) = self.__find_a_zero()
507 if row < 0:
508 done = True
509 step = 6
510 else:
511 self.marked[row][col] = 2
512 star_col = self.__find_star_in_row(row)
513 if star_col >= 0:
514 col = star_col
515 self.row_covered[row] = True
516 self.col_covered[col] = False
517 else:
518 done = True
519 self.Z0_r = row
520 self.Z0_c = col
521 step = 5
522
523 return step
524
526 """
527 Construct a series of alternating primed and starred zeros as
528 follows. Let Z0 represent the uncovered primed zero found in Step 4.
529 Let Z1 denote the starred zero in the column of Z0 (if any).
530 Let Z2 denote the primed zero in the row of Z1 (there will always
531 be one). Continue until the series terminates at a primed zero
532 that has no starred zero in its column. Unstar each starred zero
533 of the series, star each primed zero of the series, erase all
534 primes and uncover every line in the matrix. Return to Step 3
535 """
536 count = 0
537 path = self.path
538 path[count][0] = self.Z0_r
539 path[count][1] = self.Z0_c
540 done = False
541 while not done:
542 row = self.__find_star_in_col(path[count][1])
543 if row >= 0:
544 count += 1
545 path[count][0] = row
546 path[count][1] = path[count-1][1]
547 else:
548 done = True
549
550 if not done:
551 col = self.__find_prime_in_row(path[count][0])
552 count += 1
553 path[count][0] = path[count-1][0]
554 path[count][1] = col
555
556 self.__convert_path(path, count)
557 self.__clear_covers()
558 self.__erase_primes()
559 return 3
560
562 """
563 Add the value found in Step 4 to every element of each covered
564 row, and subtract it from every element of each uncovered column.
565 Return to Step 4 without altering any stars, primes, or covered
566 lines.
567 """
568 minval = self.__find_smallest()
569 for i in range(self.n):
570 for j in range(self.n):
571 if self.row_covered[i]:
572 self.C[i][j] += minval
573 if not self.col_covered[j]:
574 self.C[i][j] -= minval
575 return 4
576
578 """Find the smallest uncovered value in the matrix."""
579 minval = sys.maxsize
580 for i in range(self.n):
581 for j in range(self.n):
582 if (not self.row_covered[i]) and (not self.col_covered[j]):
583 if minval > self.C[i][j]:
584 minval = self.C[i][j]
585 return minval
586
588 """Find the first uncovered element with value 0"""
589 row = -1
590 col = -1
591 i = 0
592 n = self.n
593 done = False
594
595 while not done:
596 j = 0
597 while True:
598 if (self.C[i][j] == 0) and \
599 (not self.row_covered[i]) and \
600 (not self.col_covered[j]):
601 row = i
602 col = j
603 done = True
604 j += 1
605 if j >= n:
606 break
607 i += 1
608 if i >= n:
609 done = True
610
611 return (row, col)
612
614 """
615 Find the first starred element in the specified row. Returns
616 the column index, or -1 if no starred element was found.
617 """
618 col = -1
619 for j in range(self.n):
620 if self.marked[row][j] == 1:
621 col = j
622 break
623
624 return col
625
627 """
628 Find the first starred element in the specified row. Returns
629 the row index, or -1 if no starred element was found.
630 """
631 row = -1
632 for i in range(self.n):
633 if self.marked[i][col] == 1:
634 row = i
635 break
636
637 return row
638
640 """
641 Find the first prime element in the specified row. Returns
642 the column index, or -1 if no starred element was found.
643 """
644 col = -1
645 for j in range(self.n):
646 if self.marked[row][j] == 2:
647 col = j
648 break
649
650 return col
651
653 for i in range(count+1):
654 if self.marked[path[i][0]][path[i][1]] == 1:
655 self.marked[path[i][0]][path[i][1]] = 0
656 else:
657 self.marked[path[i][0]][path[i][1]] = 1
658
660 """Clear all covered matrix cells"""
661 for i in range(self.n):
662 self.row_covered[i] = False
663 self.col_covered[i] = False
664
666 """Erase all prime markings"""
667 for i in range(self.n):
668 for j in range(self.n):
669 if self.marked[i][j] == 2:
670 self.marked[i][j] = 0
671
672
673
674
675
677 """
678 Create a cost matrix from a profit matrix by calling
679 'inversion_function' to invert each value. The inversion
680 function must take one numeric argument (of any type) and return
681 another numeric argument which is presumed to be the cost inverse
682 of the original profit.
683
684 This is a static method. Call it like this:
685
686 .. python::
687
688 cost_matrix = Munkres.make_cost_matrix(matrix, inversion_func)
689
690 For example:
691
692 .. python::
693
694 cost_matrix = Munkres.make_cost_matrix(matrix, lambda x : sys.maxsize - x)
695
696 :Parameters:
697 profit_matrix : list of lists
698 The matrix to convert from a profit to a cost matrix
699
700 inversion_function : function
701 The function to use to invert each entry in the profit matrix
702
703 :rtype: list of lists
704 :return: The converted matrix
705 """
706 cost_matrix = []
707 for row in profit_matrix:
708 cost_matrix.append([inversion_function(value) for value in row])
709 return cost_matrix
710
712 """
713 Convenience function: Displays the contents of a matrix of integers.
714
715 :Parameters:
716 matrix : list of lists
717 Matrix to print
718
719 msg : str
720 Optional message to print before displaying the matrix
721 """
722 import math
723
724 if msg is not None:
725 print(msg)
726
727
728 width = 0
729 for row in matrix:
730 for val in row:
731 width = max(width, int(math.log10(val)) + 1)
732
733
734 format = '%%%dd' % width
735
736
737 for row in matrix:
738 sep = '['
739 for val in row:
740 sys.stdout.write(sep + format % val)
741 sep = ', '
742 sys.stdout.write(']\n')
743
744
745
746
747
748 if __name__ == '__main__':
749
750 matrices = [
751
752 ([[400, 150, 400],
753 [400, 450, 600],
754 [300, 225, 300]],
755 850),
756
757
758 ([[400, 150, 400, 1],
759 [400, 450, 600, 2],
760 [300, 225, 300, 3]],
761 452),
762
763
764
765 ([[10, 10, 8],
766 [9, 8, 1],
767 [9, 7, 4]],
768 18),
769
770
771 ([[10, 10, 8, 11],
772 [9, 8, 1, 1],
773 [9, 7, 4, 10]],
774 15)]
775
776 m = Munkres()
777 for cost_matrix, expected_total in matrices:
778 print_matrix(cost_matrix, msg='cost matrix')
779 indexes = m.compute(cost_matrix)
780 total_cost = 0
781 for r, c in indexes:
782 x = cost_matrix[r][c]
783 total_cost += x
784 print('(%d, %d) -> %d' % (r, c, x))
785 print('lowest cost=%d' % total_cost)
786 assert expected_total == total_cost
787