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# Hungarian algorithm (Kuhn-Munkres) for solving the linear sum assignment # problem. Taken from scikit-learn. Based on original code by Brian Clapper, # adapted to NumPy by Gael Varoquaux. # Further improvements by Ben Root, Vlad Niculae and Lars Buitinck. # # Copyright (c) 2008 Brian M. Clapper <bmc@clapper.org>, Gael Varoquaux # Author: Brian M. Clapper, Gael Varoquaux # License: 3-clause BSD
"""Solve the linear sum assignment problem.
The linear sum assignment problem is also known as minimum weight matching in bipartite graphs. A problem instance is described by a matrix C, where each C[i,j] is the cost of matching vertex i of the first partite set (a "worker") and vertex j of the second set (a "job"). The goal is to find a complete assignment of workers to jobs of minimal cost.
Formally, let X be a boolean matrix where :math:`X[i,j] = 1` iff row i is assigned to column j. Then the optimal assignment has cost
.. math:: \\min \\sum_i \\sum_j C_{i,j} X_{i,j}
s.t. each row is assignment to at most one column, and each column to at most one row.
This function can also solve a generalization of the classic assignment problem where the cost matrix is rectangular. If it has more rows than columns, then not every row needs to be assigned to a column, and vice versa.
The method used is the Hungarian algorithm, also known as the Munkres or Kuhn-Munkres algorithm.
Parameters ---------- cost_matrix : array The cost matrix of the bipartite graph.
Returns ------- row_ind, col_ind : array An array of row indices and one of corresponding column indices giving the optimal assignment. The cost of the assignment can be computed as ``cost_matrix[row_ind, col_ind].sum()``. The row indices will be sorted; in the case of a square cost matrix they will be equal to ``numpy.arange(cost_matrix.shape[0])``.
Notes ----- .. versionadded:: 0.17.0
Examples -------- >>> cost = np.array([[4, 1, 3], [2, 0, 5], [3, 2, 2]]) >>> from scipy.optimize import linear_sum_assignment >>> row_ind, col_ind = linear_sum_assignment(cost) >>> col_ind array([1, 0, 2]) >>> cost[row_ind, col_ind].sum() 5
References ---------- 1. http://csclab.murraystate.edu/bob.pilgrim/445/munkres.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. *J. SIAM*, 5(1):32-38, March, 1957.
5. https://en.wikipedia.org/wiki/Hungarian_algorithm """ cost_matrix = np.asarray(cost_matrix) if len(cost_matrix.shape) != 2: raise ValueError("expected a matrix (2-d array), got a %r array" % (cost_matrix.shape,))
if not (np.issubdtype(cost_matrix.dtype, np.number) or cost_matrix.dtype == np.dtype(np.bool)): raise ValueError("expected a matrix containing numerical entries, got %s" % (cost_matrix.dtype,))
if np.any(np.isinf(cost_matrix) | np.isnan(cost_matrix)): raise ValueError("matrix contains invalid numeric entries")
if cost_matrix.dtype == np.dtype(np.bool): cost_matrix = cost_matrix.astype(np.int)
# The algorithm expects more columns than rows in the cost matrix. if cost_matrix.shape[1] < cost_matrix.shape[0]: cost_matrix = cost_matrix.T transposed = True else: transposed = False
state = _Hungary(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)
if transposed: marked = state.marked.T else: marked = state.marked return np.where(marked == 1)
"""State of the Hungarian algorithm.
Parameters ---------- cost_matrix : 2D matrix The cost matrix. Must have shape[1] >= shape[0]. """
self.C = cost_matrix.copy()
n, m = self.C.shape self.row_uncovered = np.ones(n, dtype=bool) self.col_uncovered = np.ones(m, dtype=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)
"""Clear all covered matrix cells""" self.row_uncovered[:] = True self.col_uncovered[:] = True
# Individual steps of the algorithm follow, as a state machine: they return # the next step to be taken (function to be called), if any.
"""Steps 1 and 2 in the Wikipedia page."""
# Step 1: 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] # Step 2: 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
""" 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
""" 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(int) covered_C = C * state.row_uncovered[:, np.newaxis] covered_C *= np.asarray(state.col_uncovered, dtype=int) 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 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] * ( np.asarray(state.row_uncovered, dtype=int)) covered_C[row] = 0
""" 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 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
""" 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[~state.row_uncovered] += minval state.C[:, state.col_uncovered] -= minval return _step4 |