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