chore: 添加虚拟环境到仓库
- 添加 backend_service/venv 虚拟环境 - 包含所有Python依赖包 - 注意：虚拟环境约393MB，包含12655个文件
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from sympy.utilities.iterables import \
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flatten, connected_components, strongly_connected_components
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from .exceptions import NonSquareMatrixError
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def _connected_components(M):
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"""Returns the list of connected vertices of the graph when
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a square matrix is viewed as a weighted graph.
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Examples
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========
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>>> from sympy import Matrix
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>>> A = Matrix([
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... [66, 0, 0, 68, 0, 0, 0, 0, 67],
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... [0, 55, 0, 0, 0, 0, 54, 53, 0],
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... [0, 0, 0, 0, 1, 2, 0, 0, 0],
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... [86, 0, 0, 88, 0, 0, 0, 0, 87],
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... [0, 0, 10, 0, 11, 12, 0, 0, 0],
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... [0, 0, 20, 0, 21, 22, 0, 0, 0],
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... [0, 45, 0, 0, 0, 0, 44, 43, 0],
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... [0, 35, 0, 0, 0, 0, 34, 33, 0],
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... [76, 0, 0, 78, 0, 0, 0, 0, 77]])
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>>> A.connected_components()
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[[0, 3, 8], [1, 6, 7], [2, 4, 5]]
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Notes
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=====
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Even if any symbolic elements of the matrix can be indeterminate
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to be zero mathematically, this only takes the account of the
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structural aspect of the matrix, so they will considered to be
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nonzero.
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"""
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if not M.is_square:
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raise NonSquareMatrixError
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V = range(M.rows)
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E = sorted(M.todok().keys())
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return connected_components((V, E))
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def _strongly_connected_components(M):
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"""Returns the list of strongly connected vertices of the graph when
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a square matrix is viewed as a weighted graph.
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Examples
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========
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>>> from sympy import Matrix
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>>> A = Matrix([
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... [44, 0, 0, 0, 43, 0, 45, 0, 0],
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... [0, 66, 62, 61, 0, 68, 0, 60, 67],
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... [0, 0, 22, 21, 0, 0, 0, 20, 0],
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... [0, 0, 12, 11, 0, 0, 0, 10, 0],
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... [34, 0, 0, 0, 33, 0, 35, 0, 0],
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... [0, 86, 82, 81, 0, 88, 0, 80, 87],
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... [54, 0, 0, 0, 53, 0, 55, 0, 0],
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... [0, 0, 2, 1, 0, 0, 0, 0, 0],
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... [0, 76, 72, 71, 0, 78, 0, 70, 77]])
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>>> A.strongly_connected_components()
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[[0, 4, 6], [2, 3, 7], [1, 5, 8]]
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"""
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if not M.is_square:
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raise NonSquareMatrixError
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# RepMatrix uses the more efficient DomainMatrix.scc() method
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rep = getattr(M, '_rep', None)
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if rep is not None:
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return rep.scc()
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V = range(M.rows)
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E = sorted(M.todok().keys())
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return strongly_connected_components((V, E))
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def _connected_components_decomposition(M):
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"""Decomposes a square matrix into block diagonal form only
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using the permutations.
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Explanation
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===========
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The decomposition is in a form of $A = P^{-1} B P$ where $P$ is a
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permutation matrix and $B$ is a block diagonal matrix.
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Returns
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=======
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P, B : PermutationMatrix, BlockDiagMatrix
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*P* is a permutation matrix for the similarity transform
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as in the explanation. And *B* is the block diagonal matrix of
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the result of the permutation.
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If you would like to get the diagonal blocks from the
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BlockDiagMatrix, see
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:meth:`~sympy.matrices.expressions.blockmatrix.BlockDiagMatrix.get_diag_blocks`.
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Examples
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========
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>>> from sympy import Matrix, pprint
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>>> A = Matrix([
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... [66, 0, 0, 68, 0, 0, 0, 0, 67],
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... [0, 55, 0, 0, 0, 0, 54, 53, 0],
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... [0, 0, 0, 0, 1, 2, 0, 0, 0],
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... [86, 0, 0, 88, 0, 0, 0, 0, 87],
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... [0, 0, 10, 0, 11, 12, 0, 0, 0],
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... [0, 0, 20, 0, 21, 22, 0, 0, 0],
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... [0, 45, 0, 0, 0, 0, 44, 43, 0],
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... [0, 35, 0, 0, 0, 0, 34, 33, 0],
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... [76, 0, 0, 78, 0, 0, 0, 0, 77]])
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>>> P, B = A.connected_components_decomposition()
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>>> pprint(P)
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PermutationMatrix((1 3)(2 8 5 7 4 6))
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>>> pprint(B)
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[[66 68 67] ]
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[[ ] ]
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[[86 88 87] 0 0 ]
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[[ ] ]
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[[76 78 77] ]
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[ ]
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[ [55 54 53] ]
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[ [ ] ]
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[ 0 [45 44 43] 0 ]
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[ [ ] ]
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[ [35 34 33] ]
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[ ]
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[ [0 1 2 ]]
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[ [ ]]
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[ 0 0 [10 11 12]]
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[ [ ]]
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[ [20 21 22]]
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>>> P = P.as_explicit()
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>>> B = B.as_explicit()
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>>> P.T*B*P == A
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True
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Notes
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=====
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This problem corresponds to the finding of the connected components
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of a graph, when a matrix is viewed as a weighted graph.
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"""
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from sympy.combinatorics.permutations import Permutation
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from sympy.matrices.expressions.blockmatrix import BlockDiagMatrix
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from sympy.matrices.expressions.permutation import PermutationMatrix
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iblocks = M.connected_components()
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p = Permutation(flatten(iblocks))
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P = PermutationMatrix(p)
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blocks = []
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for b in iblocks:
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blocks.append(M[b, b])
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B = BlockDiagMatrix(*blocks)
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return P, B
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def _strongly_connected_components_decomposition(M, lower=True):
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"""Decomposes a square matrix into block triangular form only
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using the permutations.
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Explanation
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===========
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The decomposition is in a form of $A = P^{-1} B P$ where $P$ is a
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permutation matrix and $B$ is a block diagonal matrix.
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Parameters
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==========
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lower : bool
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Makes $B$ lower block triangular when ``True``.
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Otherwise, makes $B$ upper block triangular.
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Returns
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=======
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P, B : PermutationMatrix, BlockMatrix
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*P* is a permutation matrix for the similarity transform
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as in the explanation. And *B* is the block triangular matrix of
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the result of the permutation.
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Examples
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========
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>>> from sympy import Matrix, pprint
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>>> A = Matrix([
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... [44, 0, 0, 0, 43, 0, 45, 0, 0],
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... [0, 66, 62, 61, 0, 68, 0, 60, 67],
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... [0, 0, 22, 21, 0, 0, 0, 20, 0],
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... [0, 0, 12, 11, 0, 0, 0, 10, 0],
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... [34, 0, 0, 0, 33, 0, 35, 0, 0],
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... [0, 86, 82, 81, 0, 88, 0, 80, 87],
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... [54, 0, 0, 0, 53, 0, 55, 0, 0],
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... [0, 0, 2, 1, 0, 0, 0, 0, 0],
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... [0, 76, 72, 71, 0, 78, 0, 70, 77]])
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A lower block triangular decomposition:
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>>> P, B = A.strongly_connected_components_decomposition()
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>>> pprint(P)
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PermutationMatrix((8)(1 4 3 2 6)(5 7))
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>>> pprint(B)
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[[44 43 45] [0 0 0] [0 0 0] ]
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[[ ] [ ] [ ] ]
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[[34 33 35] [0 0 0] [0 0 0] ]
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[[ ] [ ] [ ] ]
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[[54 53 55] [0 0 0] [0 0 0] ]
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[ ]
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[ [0 0 0] [22 21 20] [0 0 0] ]
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[ [ ] [ ] [ ] ]
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[ [0 0 0] [12 11 10] [0 0 0] ]
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[ [ ] [ ] [ ] ]
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[ [0 0 0] [2 1 0 ] [0 0 0] ]
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[ ]
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[ [0 0 0] [62 61 60] [66 68 67]]
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[ [ ] [ ] [ ]]
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[ [0 0 0] [82 81 80] [86 88 87]]
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[ [ ] [ ] [ ]]
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[ [0 0 0] [72 71 70] [76 78 77]]
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>>> P = P.as_explicit()
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>>> B = B.as_explicit()
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>>> P.T * B * P == A
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True
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An upper block triangular decomposition:
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>>> P, B = A.strongly_connected_components_decomposition(lower=False)
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>>> pprint(P)
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PermutationMatrix((0 1 5 7 4 3 2 8 6))
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>>> pprint(B)
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[[66 68 67] [62 61 60] [0 0 0] ]
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[[ ] [ ] [ ] ]
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[[86 88 87] [82 81 80] [0 0 0] ]
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[[ ] [ ] [ ] ]
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[[76 78 77] [72 71 70] [0 0 0] ]
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[ ]
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[ [0 0 0] [22 21 20] [0 0 0] ]
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[ [ ] [ ] [ ] ]
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[ [0 0 0] [12 11 10] [0 0 0] ]
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[ [ ] [ ] [ ] ]
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[ [0 0 0] [2 1 0 ] [0 0 0] ]
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[ ]
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[ [0 0 0] [0 0 0] [44 43 45]]
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[ [ ] [ ] [ ]]
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[ [0 0 0] [0 0 0] [34 33 35]]
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[ [ ] [ ] [ ]]
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[ [0 0 0] [0 0 0] [54 53 55]]
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>>> P = P.as_explicit()
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>>> B = B.as_explicit()
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>>> P.T * B * P == A
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True
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"""
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from sympy.combinatorics.permutations import Permutation
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from sympy.matrices.expressions.blockmatrix import BlockMatrix
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from sympy.matrices.expressions.permutation import PermutationMatrix
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iblocks = M.strongly_connected_components()
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if not lower:
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iblocks = list(reversed(iblocks))
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p = Permutation(flatten(iblocks))
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P = PermutationMatrix(p)
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rows = []
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for a in iblocks:
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cols = []
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for b in iblocks:
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cols.append(M[a, b])
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rows.append(cols)
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B = BlockMatrix(rows)
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return P, B
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