from sympy.utilities.iterables import \ flatten, connected_components, strongly_connected_components from .common 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