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"""
Solve the unique lowest-cost assignment problem using the
Hungarian algorithm (also known as Munkres algorithm).
"""
# Based on original code by Brain Clapper, adapted to NumPy by Gael Varoquaux.
# Heavily refactored by Lars Buitinck.
#
# TODO: a version of this algorithm has been incorporated in SciPy; use that
# when SciPy 0.17 is released.
# Copyright (c) 2008 Brian M. Clapper <bmc@clapper.org>, Gael Varoquaux
# Author: Brian M. Clapper, Gael Varoquaux
# LICENSE: BSD
import numpy as np
def linear_assignment(X):
"""Solve the linear assignment problem using the Hungarian algorithm.
The problem is also known as maximum weight matching in bipartite graphs.
The method is also known as the Munkres or Kuhn-Munkres algorithm.
Parameters
----------
X : array
The cost matrix of the bipartite graph
Returns
-------
indices : array
The pairs of (row, col) indices in the original array giving
the original ordering.
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. https://en.wikipedia.org/wiki/Hungarian_algorithm
"""
indices = _hungarian(X).tolist()
indices.sort()
# Re-force dtype to ints in case of empty list
indices = np.array(indices, dtype=int)
# Make sure the array is 2D with 2 columns.
# This is needed when dealing with an empty list
indices.shape = (-1, 2)
return indices
class _HungarianState(object):
"""State of one execution of the Hungarian algorithm.
Parameters
----------
cost_matrix : 2D matrix
The cost matrix. Does not have to be square.
"""
def __init__(self, cost_matrix):
cost_matrix = np.atleast_2d(cost_matrix)
# If there are more rows (n) than columns (m), then the algorithm
# will not be able to work correctly. Therefore, we
# transpose the cost function when needed. Just have to
# remember to swap the result columns back later.
transposed = (cost_matrix.shape[1] < cost_matrix.shape[0])
if transposed:
self.C = (cost_matrix.T).copy()
else:
self.C = cost_matrix.copy()
self.transposed = transposed
# At this point, m >= n.
n, m = self.C.shape
self.row_uncovered = np.ones(n, dtype=np.bool)
self.col_uncovered = np.ones(m, dtype=np.bool)
self.Z0_r = 0
self.Z0_c = 0
self.path = np.zeros((n + m, 2), dtype=int)
self.marked = np.zeros((n, m), dtype=int)
def _clear_covers(self):
"""Clear all covered matrix cells"""
self.row_uncovered[:] = True
self.col_uncovered[:] = True
def _hungarian(cost_matrix):
"""The Hungarian algorithm.
Calculate the Munkres solution to the classical assignment problem and
return the indices for the lowest-cost pairings.
Parameters
----------
cost_matrix : 2D matrix
The cost matrix. Does not have to be square.
Returns
-------
indices : 2D array of indices
The pairs of (row, col) indices in the original array giving
the original ordering.
"""
state = _HungarianState(cost_matrix)
# No need to bother with assignments if one of the dimensions
# of the cost matrix is zero-length.
step = None if 0 in cost_matrix.shape else _step1
while step is not None:
step = step(state)
# Look for the starred columns
results = np.array(np.where(state.marked == 1)).T
# We need to swap the columns because we originally
# did a transpose on the input cost matrix.
if state.transposed:
results = results[:, ::-1]
return results
# Individual steps of the algorithm follow, as a state machine: they return
# the next step to be taken (function to be called), if any.
def _step1(state):
"""Steps 1 and 2 in the Wikipedia page."""
# Step1: For each row of the matrix, find the smallest element and
# subtract it from every element in its row.
state.C -= state.C.min(axis=1)[:, np.newaxis]
# Step2: Find a zero (Z) in the resulting matrix. If there is no
# starred zero in its row or column, star Z. Repeat for each element
# in the matrix.
for i, j in zip(*np.where(state.C == 0)):
if state.col_uncovered[j] and state.row_uncovered[i]:
state.marked[i, j] = 1
state.col_uncovered[j] = False
state.row_uncovered[i] = False
state._clear_covers()
return _step3
def _step3(state):
"""
Cover each column containing a starred zero. If n columns are covered,
the starred zeros describe a complete set of unique assignments.
In this case, Go to DONE, otherwise, Go to Step 4.
"""
marked = (state.marked == 1)
state.col_uncovered[np.any(marked, axis=0)] = False
if marked.sum() < state.C.shape[0]:
return _step4
def _step4(state):
"""
Find a noncovered zero and prime it. If there is no starred zero
in the row containing this primed zero, Go to Step 5. Otherwise,
cover this row and uncover the column containing the starred
zero. Continue in this manner until there are no uncovered zeros
left. Save the smallest uncovered value and Go to Step 6.
"""
# We convert to int as numpy operations are faster on int
C = (state.C == 0).astype(np.int)
covered_C = C * state.row_uncovered[:, np.newaxis]
covered_C *= state.col_uncovered.astype(dtype=np.int, copy=False)
n = state.C.shape[0]
m = state.C.shape[1]
while True:
# Find an uncovered zero
row, col = np.unravel_index(np.argmax(covered_C), (n, m))
if covered_C[row, col] == 0:
return _step6
else:
state.marked[row, col] = 2
# Find the first starred element in the row
star_col = np.argmax(state.marked[row] == 1)
if not state.marked[row, star_col] == 1:
# Could not find one
state.Z0_r = row
state.Z0_c = col
return _step5
else:
col = star_col
state.row_uncovered[row] = False
state.col_uncovered[col] = True
covered_C[:, col] = C[:, col] * (
state.row_uncovered.astype(dtype=np.int, copy=False))
covered_C[row] = 0
def _step5(state):
"""
Construct a series of alternating primed and starred zeros as follows.
Let Z0 represent the uncovered primed zero found in Step 4.
Let Z1 denote the starred zero in the column of Z0 (if any).
Let Z2 denote the primed zero in the row of Z1 (there will always be one).
Continue until the series terminates at a primed zero that has no starred
zero in its column. Unstar each starred zero of the series, star each
primed zero of the series, erase all primes and uncover every line in the
matrix. Return to Step 3
"""
count = 0
path = state.path
path[count, 0] = state.Z0_r
path[count, 1] = state.Z0_c
while True:
# Find the first starred element in the col defined by
# the path.
row = np.argmax(state.marked[:, path[count, 1]] == 1)
if not state.marked[row, path[count, 1]] == 1:
# Could not find one
break
else:
count += 1
path[count, 0] = row
path[count, 1] = path[count - 1, 1]
# Find the first prime element in the row defined by the
# first path step
col = np.argmax(state.marked[path[count, 0]] == 2)
if state.marked[row, col] != 2:
col = -1
count += 1
path[count, 0] = path[count - 1, 0]
path[count, 1] = col
# Convert paths
for i in range(count + 1):
if state.marked[path[i, 0], path[i, 1]] == 1:
state.marked[path[i, 0], path[i, 1]] = 0
else:
state.marked[path[i, 0], path[i, 1]] = 1
state._clear_covers()
# Erase all prime markings
state.marked[state.marked == 2] = 0
return _step3
def _step6(state):
"""
Add the value found in Step 4 to every element of each covered row,
and subtract it from every element of each uncovered column.
Return to Step 4 without altering any stars, primes, or covered lines.
"""
# the smallest uncovered value in the matrix
if np.any(state.row_uncovered) and np.any(state.col_uncovered):
minval = np.min(state.C[state.row_uncovered], axis=0)
minval = np.min(minval[state.col_uncovered])
state.C[np.logical_not(state.row_uncovered)] += minval
state.C[:, state.col_uncovered] -= minval
return _step4