283 lines
9.6 KiB
Python
283 lines
9.6 KiB
Python
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# Hungarian algorithm (Kuhn-Munkres) for solving the linear sum assignment
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# problem. Taken from scikit-learn. Based on original code by Brian Clapper,
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# adapted to NumPy by Gael Varoquaux.
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# Further improvements by Ben Root, Vlad Niculae and Lars Buitinck.
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#
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# Copyright (c) 2008 Brian M. Clapper <bmc@clapper.org>, Gael Varoquaux
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# Author: Brian M. Clapper, Gael Varoquaux
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# License: 3-clause BSD
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import numpy as np
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def linear_sum_assignment(cost_matrix):
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"""Solve the linear sum assignment problem.
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The linear sum assignment problem is also known as minimum weight matching
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in bipartite graphs. A problem instance is described by a matrix C, where
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each C[i,j] is the cost of matching vertex i of the first partite set
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(a "worker") and vertex j of the second set (a "job"). The goal is to find
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a complete assignment of workers to jobs of minimal cost.
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Formally, let X be a boolean matrix where :math:`X[i,j] = 1` iff row i is
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assigned to column j. Then the optimal assignment has cost
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.. math::
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\\min \\sum_i \\sum_j C_{i,j} X_{i,j}
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s.t. each row is assignment to at most one column, and each column to at
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most one row.
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This function can also solve a generalization of the classic assignment
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problem where the cost matrix is rectangular. If it has more rows than
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columns, then not every row needs to be assigned to a column, and vice
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versa.
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The method used is the Hungarian algorithm, also known as the Munkres or
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Kuhn-Munkres algorithm.
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Parameters
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----------
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cost_matrix : array
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The cost matrix of the bipartite graph.
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Returns
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-------
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row_ind, col_ind : array
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An array of row indices and one of corresponding column indices giving
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the optimal assignment. The cost of the assignment can be computed
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as ``cost_matrix[row_ind, col_ind].sum()``. The row indices will be
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sorted; in the case of a square cost matrix they will be equal to
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``numpy.arange(cost_matrix.shape[0])``.
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Notes
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-----
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.. versionadded:: 0.17.0
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Examples
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--------
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>>> cost = np.array([[4, 1, 3], [2, 0, 5], [3, 2, 2]])
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>>> from scipy.optimize import linear_sum_assignment
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>>> row_ind, col_ind = linear_sum_assignment(cost)
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>>> col_ind
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array([1, 0, 2])
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>>> cost[row_ind, col_ind].sum()
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5
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References
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----------
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1. http://csclab.murraystate.edu/bob.pilgrim/445/munkres.html
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2. Harold W. Kuhn. The Hungarian Method for the assignment problem.
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*Naval Research Logistics Quarterly*, 2:83-97, 1955.
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3. Harold W. Kuhn. Variants of the Hungarian method for assignment
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problems. *Naval Research Logistics Quarterly*, 3: 253-258, 1956.
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4. Munkres, J. Algorithms for the Assignment and Transportation Problems.
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*J. SIAM*, 5(1):32-38, March, 1957.
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5. https://en.wikipedia.org/wiki/Hungarian_algorithm
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"""
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cost_matrix = np.asarray(cost_matrix)
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if len(cost_matrix.shape) != 2:
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raise ValueError("expected a matrix (2-d array), got a %r array"
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% (cost_matrix.shape,))
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if not (np.issubdtype(cost_matrix.dtype, np.number) or
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cost_matrix.dtype == np.dtype(np.bool)):
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raise ValueError("expected a matrix containing numerical entries, got %s"
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% (cost_matrix.dtype,))
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if np.any(np.isinf(cost_matrix) | np.isnan(cost_matrix)):
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raise ValueError("matrix contains invalid numeric entries")
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if cost_matrix.dtype == np.dtype(np.bool):
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cost_matrix = cost_matrix.astype(np.int)
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# The algorithm expects more columns than rows in the cost matrix.
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if cost_matrix.shape[1] < cost_matrix.shape[0]:
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cost_matrix = cost_matrix.T
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transposed = True
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else:
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transposed = False
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state = _Hungary(cost_matrix)
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# No need to bother with assignments if one of the dimensions
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# of the cost matrix is zero-length.
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step = None if 0 in cost_matrix.shape else _step1
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while step is not None:
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step = step(state)
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if transposed:
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marked = state.marked.T
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else:
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marked = state.marked
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return np.where(marked == 1)
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class _Hungary(object):
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"""State of the Hungarian algorithm.
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Parameters
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----------
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cost_matrix : 2D matrix
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The cost matrix. Must have shape[1] >= shape[0].
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"""
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def __init__(self, cost_matrix):
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self.C = cost_matrix.copy()
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n, m = self.C.shape
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self.row_uncovered = np.ones(n, dtype=bool)
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self.col_uncovered = np.ones(m, dtype=bool)
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self.Z0_r = 0
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self.Z0_c = 0
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self.path = np.zeros((n + m, 2), dtype=int)
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self.marked = np.zeros((n, m), dtype=int)
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def _clear_covers(self):
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"""Clear all covered matrix cells"""
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self.row_uncovered[:] = True
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self.col_uncovered[:] = True
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# Individual steps of the algorithm follow, as a state machine: they return
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# the next step to be taken (function to be called), if any.
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def _step1(state):
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"""Steps 1 and 2 in the Wikipedia page."""
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# Step 1: For each row of the matrix, find the smallest element and
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# subtract it from every element in its row.
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state.C -= state.C.min(axis=1)[:, np.newaxis]
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# Step 2: Find a zero (Z) in the resulting matrix. If there is no
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# starred zero in its row or column, star Z. Repeat for each element
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# in the matrix.
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for i, j in zip(*np.where(state.C == 0)):
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if state.col_uncovered[j] and state.row_uncovered[i]:
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state.marked[i, j] = 1
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state.col_uncovered[j] = False
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state.row_uncovered[i] = False
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state._clear_covers()
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return _step3
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def _step3(state):
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"""
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Cover each column containing a starred zero. If n columns are covered,
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the starred zeros describe a complete set of unique assignments.
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In this case, Go to DONE, otherwise, Go to Step 4.
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"""
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marked = (state.marked == 1)
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state.col_uncovered[np.any(marked, axis=0)] = False
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if marked.sum() < state.C.shape[0]:
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return _step4
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def _step4(state):
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"""
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Find a noncovered zero and prime it. If there is no starred zero
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in the row containing this primed zero, Go to Step 5. Otherwise,
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cover this row and uncover the column containing the starred
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zero. Continue in this manner until there are no uncovered zeros
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left. Save the smallest uncovered value and Go to Step 6.
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"""
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# We convert to int as numpy operations are faster on int
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C = (state.C == 0).astype(int)
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covered_C = C * state.row_uncovered[:, np.newaxis]
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covered_C *= np.asarray(state.col_uncovered, dtype=int)
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n = state.C.shape[0]
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m = state.C.shape[1]
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while True:
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# Find an uncovered zero
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row, col = np.unravel_index(np.argmax(covered_C), (n, m))
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if covered_C[row, col] == 0:
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return _step6
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else:
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state.marked[row, col] = 2
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# Find the first starred element in the row
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star_col = np.argmax(state.marked[row] == 1)
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if state.marked[row, star_col] != 1:
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# Could not find one
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state.Z0_r = row
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state.Z0_c = col
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return _step5
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else:
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col = star_col
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state.row_uncovered[row] = False
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state.col_uncovered[col] = True
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covered_C[:, col] = C[:, col] * (
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np.asarray(state.row_uncovered, dtype=int))
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covered_C[row] = 0
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def _step5(state):
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"""
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Construct a series of alternating primed and starred zeros as follows.
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Let Z0 represent the uncovered primed zero found in Step 4.
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Let Z1 denote the starred zero in the column of Z0 (if any).
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Let Z2 denote the primed zero in the row of Z1 (there will always be one).
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Continue until the series terminates at a primed zero that has no starred
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zero in its column. Unstar each starred zero of the series, star each
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primed zero of the series, erase all primes and uncover every line in the
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matrix. Return to Step 3
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"""
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count = 0
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path = state.path
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path[count, 0] = state.Z0_r
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path[count, 1] = state.Z0_c
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while True:
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# Find the first starred element in the col defined by
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# the path.
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row = np.argmax(state.marked[:, path[count, 1]] == 1)
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if state.marked[row, path[count, 1]] != 1:
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# Could not find one
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break
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else:
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count += 1
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path[count, 0] = row
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path[count, 1] = path[count - 1, 1]
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# Find the first prime element in the row defined by the
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# first path step
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col = np.argmax(state.marked[path[count, 0]] == 2)
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if state.marked[row, col] != 2:
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col = -1
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count += 1
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path[count, 0] = path[count - 1, 0]
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path[count, 1] = col
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# Convert paths
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for i in range(count + 1):
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if state.marked[path[i, 0], path[i, 1]] == 1:
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state.marked[path[i, 0], path[i, 1]] = 0
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else:
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state.marked[path[i, 0], path[i, 1]] = 1
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state._clear_covers()
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# Erase all prime markings
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state.marked[state.marked == 2] = 0
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return _step3
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def _step6(state):
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"""
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Add the value found in Step 4 to every element of each covered row,
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and subtract it from every element of each uncovered column.
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Return to Step 4 without altering any stars, primes, or covered lines.
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"""
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# the smallest uncovered value in the matrix
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if np.any(state.row_uncovered) and np.any(state.col_uncovered):
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minval = np.min(state.C[state.row_uncovered], axis=0)
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minval = np.min(minval[state.col_uncovered])
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state.C[~state.row_uncovered] += minval
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state.C[:, state.col_uncovered] -= minval
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return _step4
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