From: Michael Orlitzky Date: Mon, 5 Oct 2020 17:55:19 +0000 (-0400) Subject: mjo/ldlt.py: move implementation to SageMath proper. X-Git-Url: https://gitweb.michael.orlitzky.com/?a=commitdiff_plain;h=b1cae89cc2e8049c43638d7a8bed2256a72f1650;p=sage.d.git mjo/ldlt.py: move implementation to SageMath proper. The actual block-LDLT code is now on the u/mjo/ticket/10332 branch. --- diff --git a/mjo/ldlt.py b/mjo/ldlt.py index 682eecf..a86dbbe 100644 --- a/mjo/ldlt.py +++ b/mjo/ldlt.py @@ -24,405 +24,3 @@ def is_positive_semidefinite_naive(A): if A.nrows() == 0: return True # vacuously return A.is_hermitian() and all( v >= 0 for v in A.eigenvalues() ) - - -def _block_ldlt(A): - r""" - Perform a user-unfriendly block-`LDL^{T}` factorization of the - Hermitian matrix `A` - - This function is used internally to compute the factorization for - the user-friendly ``block_ldlt`` function. Whereas that function - returns three nice matrices, this one returns - - * A list ``p`` of the first ``n`` natural numbers, permuted. - * A matrix whose lower-triangular portion is ``L``, but whose - * (strict) upper-triangular portion is junk. - * A list of the block-diagonal entries of ``D`` - - This is mainly useful to avoid havinf to "undo" the construction - of the matrix ``D`` when we don't need it. For example, it's much - easier to compute the inertia of a matrix from the list of blocks - than it is from the block-diagonal matrix itself, because given a - block-diagonal matrix, you first have to figure out where the - blocks are! - """ - ring = A.base_ring().fraction_field() - A = A.change_ring(ring) - MS = A.matrix_space() - - # The magic constant used by Bunch-Kaufman - alpha = (1 + ZZ(17).sqrt()) * ~ZZ(8) - - # Keep track of the permutations and diagonal blocks in a vector - # rather than in a matrix, for efficiency. - n = A.nrows() - p = list(range(n)) - d = [] - - def swap_rows_columns(M, k, s): - r""" - Swap rows/columns ``k`` and ``s`` of the matrix ``M``, and update - the list ``p`` accordingly. - """ - if s > k: - # s == k would swap row/column k with itself, and we don't - # actually want to perform the identity permutation. If - # you work out the recursive factorization by hand, you'll - # notice that the rows/columns of "L" need to be permuted - # as well. A nice side effect of storing "L" within "A" - # itself is that we can skip that step. The first column - # of "L" is hit by all of the transpositions in - # succession, and the second column is hit by all but the - # first transposition, and so on. - M.swap_columns(k,s) - M.swap_rows(k,s) - - p_k = p[k] - p[k] = p[s] - p[s] = p_k - - # No return value, we're only interested in the "side effects" - # of modifing the matrix M (by reference) and the permutation - # list p (which is in scope when this function is defined). - return - - - def pivot1x1(M, k, s): - r""" - Perform a 1x1 pivot swapping rows/columns `k` and `s >= k`. - Relies on the fact that matrices are passed by reference, - since for performance reasons this routine should overwrite - its argument. Updates the local variables ``p`` and ``d`` as - well. - """ - swap_rows_columns(M,k,s) - - # Now the pivot is in the (k,k)th position. - d.append( matrix(ring, 1, [[A[k,k]]]) ) - - # Compute the Schur complement that we'll work on during - # the following iteration, and store it back in the lower- - # right-hand corner of "A". - for i in range(n-k-1): - for j in range(i+1): - A[k+1+i,k+1+j] = ( A[k+1+i,k+1+j] - - A[k+1+i,k]*A[k,k+1+j]/A[k,k] ) - A[k+1+j,k+1+i] = A[k+1+i,k+1+j].conjugate() # stay hermitian! - - for i in range(n-k-1): - # Store the new (kth) column of "L" within the lower- - # left-hand corner of "A". - A[k+i+1,k] /= A[k,k] - - # No return value, only the desired side effects of updating - # p, d, and A. - return - - k = 0 - while k < n: - # At each step, we're considering the k-by-k submatrix - # contained in the lower-right half of "A", because that's - # where we're storing the next iterate. So our indices are - # always "k" greater than those of Higham or B&K. Note that - # ``n == 0`` is handled by skipping this loop entirely. - - if k == (n-1): - # Handle this trivial case manually, since otherwise the - # algorithm's references to the e.g. "subdiagonal" are - # meaningless. The corresponding entry of "L" will be - # fixed later (since it's an on-diagonal element, it gets - # set to one eventually). - d.append( matrix(ring, 1, [[A[k,k]]]) ) - k += 1 - continue - - # Find the largest subdiagonal entry (in magnitude) in the - # kth column. This occurs prior to Step (1) in Higham, - # but is part of Step (1) in Bunch and Kaufman. We adopt - # Higham's "omega" notation instead of B&K's "lambda" - # because "lambda" can lead to some confusion. - column_1_subdiag = [ a_ki.abs() for a_ki in A[k+1:,k].list() ] - omega_1 = max([ a_ki for a_ki in column_1_subdiag ]) - - if omega_1 == 0: - # In this case, our matrix looks like - # - # [ a 0 ] - # [ 0 B ] - # - # and we can simply skip to the next step after recording - # the 1x1 pivot "a" in the top-left position. The entry "a" - # will be adjusted to "1" later on to ensure that "L" is - # (block) unit-lower-triangular. - d.append( matrix(ring, 1, [[A[k,k]]]) ) - k += 1 - continue - - if A[k,k].abs() > alpha*omega_1: - # This is the first case in Higham's Step (1), and B&K's - # Step (2). Note that we have skipped the part of B&K's - # Step (1) where we determine "r", since "r" is not yet - # needed and we may waste some time computing it - # otherwise. We are performing a 1x1 pivot, but the - # rows/columns are already where we want them, so nothing - # needs to be permuted. - pivot1x1(A,k,k) - k += 1 - continue - - # Now back to Step (1) of Higham, where we find the index "r" - # that corresponds to omega_1. This is the "else" branch of - # Higham's Step (1). - r = k + 1 + column_1_subdiag.index(omega_1) - - # Continuing the "else" branch of Higham's Step (1), and onto - # B&K's Step (3) where we find the largest off-diagonal entry - # (in magniture) in column "r". Since the matrix is Hermitian, - # we need only look at the above-diagonal entries to find the - # off-diagonal of maximal magnitude. - omega_r = max( a_rj.abs() for a_rj in A[r,k:r].list() ) - - if A[k,k].abs()*omega_r >= alpha*(omega_1**2): - # Step (2) in Higham or Step (4) in B&K. - pivot1x1(A,k,k) - k += 1 - continue - - if A[r,r].abs() > alpha*omega_r: - # This is Step (3) in Higham or Step (5) in B&K. Still a 1x1 - # pivot, but this time we need to swap rows/columns k and r. - pivot1x1(A,k,r) - k += 1 - continue - - # If we've made it this far, we're at Step (4) in Higham or - # Step (6) in B&K, where we perform a 2x2 pivot. - swap_rows_columns(A,k+1,r) - - # The top-left 2x2 submatrix (starting at position k,k) is now - # our pivot. - E = A[k:k+2,k:k+2] - d.append(E) - - C = A[k+2:n,k:k+2] - B = A[k+2:,k+2:] - - # We don't actually need the inverse of E, what we really need - # is C*E.inverse(), and that can be found by setting - # - # X = C*E.inverse() <====> XE = C. - # - # Then "X" can be found easily by solving a system. Note: I - # do not actually know that sage solves the system more - # intelligently, but this is still The Right Thing To Do. - CE_inverse = E.solve_left(C) - - schur_complement = B - (CE_inverse*C.conjugate_transpose()) - - # Compute the Schur complement that we'll work on during - # the following iteration, and store it back in the lower- - # right-hand corner of "A". - for i in range(n-k-2): - for j in range(i+1): - A[k+2+i,k+2+j] = schur_complement[i,j] - A[k+2+j,k+2+i] = schur_complement[j,i] - - # The on- and above-diagonal entries of "L" will be fixed - # later, so we only need to worry about the lower-left entry - # of the 2x2 identity matrix that belongs at the top of the - # new column of "L". - A[k+1,k] = 0 - for i in range(n-k-2): - for j in range(2): - # Store the new (k and (k+1)st) columns of "L" within - # the lower-left-hand corner of "A". - A[k+i+2,k+j] = CE_inverse[i,j] - - - k += 2 - - for i in range(n): - # We skipped this during the main loop, but it's necessary for - # correctness. - A[i,i] = 1 - - return (p,A,d) - -def block_ldlt(A): - r""" - Perform a block-`LDL^{T}` factorization of the Hermitian - matrix `A`. - - The standard `LDL^{T}` factorization of a positive-definite matrix - `A` factors it as `A = LDL^{T}` where `L` is unit-lower-triangular - and `D` is diagonal. If one allows row/column swaps via a - permutation matrix `P`, then this factorization can be extended to - some positive-semidefinite matrices `A` via the factorization - `P^{T}AP = LDL^{T}` that places the zeros at the bottom of `D` to - avoid division by zero. These factorizations extend easily to - complex Hermitian matrices when one replaces the transpose by the - conjugate-transpose. - - However, we can go one step further. If, in addition, we allow `D` - to potentially contain `2 \times 2` blocks on its diagonal, then - every real or complex Hermitian matrix `A` can be factored as `A = - PLDL^{*}P^{T}`. When the row/column swaps are made intelligently, - this process is numerically stable over inexact rings like ``RDF``. - Bunch and Kaufman describe such a "pivot" scheme that is suitable - for the solution of Hermitian systems, and that is how we choose - our row and column swaps. - - OUTPUT: - - If the input matrix is Hermitian, we return a triple `(P,L,D)` - such that `A = PLDL^{*}P^{T}` and - - * `P` is a permutation matrix, - * `L` is unit lower-triangular, - * `D` is a block-diagonal matrix whose blocks are of size - one or two. - - If the input matrix is not Hermitian, the output from this function - is undefined. - - SETUP:: - - sage: from mjo.ldlt import block_ldlt - - EXAMPLES: - - This three-by-three real symmetric matrix has one positive, one - negative, and one zero eigenvalue -- so it is not any flavor of - (semi)definite, yet we can still factor it:: - - sage: A = matrix(QQ, [[0, 1, 0], - ....: [1, 1, 2], - ....: [0, 2, 0]]) - sage: P,L,D = block_ldlt(A) - sage: P - [0 0 1] - [1 0 0] - [0 1 0] - sage: L - [ 1 0 0] - [ 2 1 0] - [ 1 1/2 1] - sage: D - [ 1| 0| 0] - [--+--+--] - [ 0|-4| 0] - [--+--+--] - [ 0| 0| 0] - sage: P.transpose()*A*P == L*D*L.transpose() - True - - This two-by-two matrix has no standard factorization, but it - constitutes its own block-factorization:: - - sage: A = matrix(QQ, [ [0,1], - ....: [1,0] ]) - sage: block_ldlt(A) - ( - [1 0] [1 0] [0 1] - [0 1], [0 1], [1 0] - ) - - The same is true of the following complex Hermitian matrix:: - - sage: A = matrix(QQbar, [ [ 0,I], - ....: [-I,0] ]) - sage: block_ldlt(A) - ( - [1 0] [1 0] [ 0 I] - [0 1], [0 1], [-I 0] - ) - - TESTS: - - All three factors should be the identity when the original matrix is:: - - sage: set_random_seed() - sage: n = ZZ.random_element(6) - sage: I = matrix.identity(QQ,n) - sage: P,L,D = block_ldlt(I) - sage: P == I and L == I and D == I - True - - Ensure that a "random" real symmetric matrix is factored correctly:: - - sage: set_random_seed() - sage: n = ZZ.random_element(6) - sage: A = matrix.random(QQ, n) - sage: A = A + A.transpose() - sage: P,L,D = block_ldlt(A) - sage: A == P*L*D*L.transpose()*P.transpose() - True - - Ensure that a "random" complex Hermitian matrix is factored correctly:: - - sage: set_random_seed() - sage: n = ZZ.random_element(6) - sage: F = NumberField(x^2 +1, 'I') - sage: A = matrix.random(F, n) - sage: A = A + A.conjugate_transpose() - sage: P,L,D = block_ldlt(A) - sage: A == P*L*D*L.conjugate_transpose()*P.conjugate_transpose() - True - - Ensure that a "random" complex positive-semidefinite matrix is - factored correctly and that the resulting block-diagonal matrix is - in fact diagonal:: - - sage: set_random_seed() - sage: n = ZZ.random_element(6) - sage: F = NumberField(x^2 +1, 'I') - sage: A = matrix.random(F, n) - sage: A = A*A.conjugate_transpose() - sage: P,L,D = block_ldlt(A) - sage: A == P*L*D*L.conjugate_transpose()*P.conjugate_transpose() - True - sage: diagonal_matrix(D.diagonal()) == D - True - - The factorization should be a no-op on diagonal matrices:: - - sage: set_random_seed() - sage: n = ZZ.random_element(6) - sage: A = matrix.diagonal(random_vector(QQ, n)) - sage: I = matrix.identity(QQ,n) - sage: P,L,D = block_ldlt(A) - sage: P == I and L == I and A == D - True - - """ - - # We have to make at least one copy of the input matrix so that we - # can change the base ring to its fraction field. Both "L" and the - # intermediate Schur complements will potentially have entries in - # the fraction field. However, we don't need to make *two* copies. - # We can't store the entries of "D" and "L" in the same matrix if - # "D" will contain any 2x2 blocks; but we can still store the - # entries of "L" in the copy of "A" that we're going to make. - # Contrast this with the non-block LDL^T factorization where the - # entries of both "L" and "D" overwrite the lower-left half of "A". - # - # This grants us an additional speedup, since we don't have to - # permute the rows/columns of "L" *and* "A" at each iteration. - p,L,d = _block_ldlt(A) - MS = L.matrix_space() - P = MS.matrix(lambda i,j: p[j] == i) - - # Warning: when n == 0, this works, but returns a matrix - # whose (nonexistent) entries are in ZZ rather than in - # the base ring of P and L. - D = block_diagonal_matrix(d) - - # Overwrite the (strict) upper-triangular part of "L", since a - # priori it contains the same entries as "A" did after _block_ldlt(). - n = L.nrows() - for i in range(n): - for j in range(i+1,n): - L[i,j] = 0 - - return (P,L,D)