From: Michael Orlitzky Date: Fri, 2 Oct 2020 22:16:06 +0000 (-0400) Subject: mjo/ldlt.py: begin fast block-LDLT implementation. X-Git-Url: https://gitweb.michael.orlitzky.com/?a=commitdiff_plain;h=a2334e928c769d00d59ae43e2cad8abd61389066;p=sage.d.git mjo/ldlt.py: begin fast block-LDLT implementation. --- diff --git a/mjo/ldlt.py b/mjo/ldlt.py index c6c4be3..6b99083 100644 --- a/mjo/ldlt.py +++ b/mjo/ldlt.py @@ -164,7 +164,7 @@ def ldlt_fast(A): # permutation matrix. # # Since "L" is stored in the lower-left "half" of "A", it's a - # good thing that we need to permuts "L," too. This is due to + # good thing that we need to permute "L," too. This is due to # how P2.T appears in the recursive algorithm applied to the # "current" column of L There, P2.T is computed recusively, as # 1 x P3.T, and P3.T = 1 x P4.T, etc, from the bottom up. All @@ -227,9 +227,9 @@ def block_ldlt_naive(A, check_hermitian=False): * `P` is a permutation matrix * `L` is unit lower-triangular - * `D` is a block-diagonal matrix whose entries are decreasing - from top-left to bottom-right and whose blocks are of size + * `D` is a block-diagonal matrix whose blocks are of size one or two. + """ n = A.nrows() @@ -239,6 +239,7 @@ def block_ldlt_naive(A, check_hermitian=False): if n == 0 or n == 1: # We can get n == 0 if someone feeds us a trivial matrix. + # For block-LDLT, n=2 is a base case. P = matrix.identity(ring, n) L = matrix.identity(ring, n) D = A @@ -250,8 +251,8 @@ def block_ldlt_naive(A, check_hermitian=False): # Bunch-Kaufmann step 1, Higham step "zero." We use Higham's # "omega" notation instead of Bunch-Kaufman's "lamda" because # lambda means other things in the same context. - column_1_subdiag = A1[1:,0].list() - omega_1 = max([ a_i1.abs() for a_i1 in column_1_subdiag ]) + column_1_subdiag = [ a_i1.abs() for a_i1 in A1[1:,0].list() ] + omega_1 = max([ a_i1 for a_i1 in column_1_subdiag ]) if omega_1 == 0: # "There's nothing to do at this step of the algorithm," @@ -263,13 +264,11 @@ def block_ldlt_naive(A, check_hermitian=False): # We could still do a pivot_one_by_one() here, but it would # pointlessly subract a bunch of zeros and multiply by one. B = A1[1:,1:] + one = matrix(ring, 1, 1, [1]) P2, L2, D2 = block_ldlt_naive(B) - P1 = block_matrix(2,2, [[ZZ(1), ZZ(0)], - [ZZ(0), P2]]) - L1 = block_matrix(2,2, [[ZZ(1), ZZ(0)], - [ZZ(0), L2]]) - D1 = block_matrix(2,2, [[ZZ(1), ZZ(0)], - [ZZ(0), D2]]) + P1 = block_diagonal_matrix(one, P2) + L1 = block_diagonal_matrix(one, L2) + D1 = block_diagonal_matrix(one, D2) return (P1,L1,D1) def pivot_one_by_one(M, c=None): @@ -310,7 +309,7 @@ def block_ldlt_naive(A, check_hermitian=False): # diagonal entries to find the off-diagonal of maximal magnitude. omega_r = max( a_rj.abs() for a_rj in A1[:r,r].list() ) - if A1[0,0].abs()*omega_r >= alpha*(omega_1^2): + if A1[0,0].abs()*omega_r >= alpha*(omega_1**2): return pivot_one_by_one(A1) if A1[r,r].abs() > alpha*omega_r: @@ -320,4 +319,214 @@ def block_ldlt_naive(A, check_hermitian=False): # Higham step (4) # If we made it here, we have to do a 2x2 pivot. - return None + P1 = copy(A1.matrix_space().identity_matrix()) + P1.swap_rows(1,r) + A1 = P1.T * A1 * P1 + + # The top-left 2x2 submatrix is now our pivot. + E = A1[:2,:2] + C = A1[2:n,0] + B = A1[2:,2:] + + if B.nrows() == 0: + # We have a two-by-two matrix that we can do nothing + # useful with. + P = matrix.identity(ring, n) + L = matrix.identity(ring, n) + D = A1 + return (P,L,D) + + P2, L2, D2 = block_ldlt_naive(B - (C*E.inverse()*C.transpose())) + + P1 = P1*block_matrix(2,2, [[ZZ(1), ZZ(0)], + [0*C, P2]]) + + L1 = block_matrix(2,2, [[ZZ(1), ZZ(0)], + [P2.transpose()*C*E.inverse(), L2]]) + D1 = block_diagonal_matrix(E,D2) + + return (P1,L1,D1) + + +def block_ldlt(A): + r""" + Perform a block-`LDL^{T}` factorization of the Hermitian + matrix `A`. + + OUTPUT: + + A triple `(P,L,D)` such that `A = PLDL^{T}P^{T}` and where, + + * `P` is a permutation matrix + * `L` is unit lower-triangular + * `D` is a block-diagonal matrix whose blocks are of size + one or two. + """ + + # 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". + 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 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. + + Note that ``A`` is passed in by reference here, so it doesn't + matter if we shadow the name ``A`` with itself. + """ + if s > k: + # s == k would swap row/column k with itself, and we don't + # actually want to perform the identity permutation. + # We don't have to permute "L" separately so long as "L" + # is stored within "A". + A.swap_columns(k,s) + A.swap_rows(k,s) + + # Update the permutation "matrix" with the swap we just did. + p_k = p[k] + p[k] = p[s] + p[s] = p_k + + # 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+j,k+1+i] = ( A[k+1+j,k+1+i] - + A[k,k+1+j]*A[k,k+1+i]/alpha ) + A[k+1+i,k+1+j] = A[k+1+j,k+1+i] # keep it symmetric! + + for i in range(n-k-1): + # Store the new (kth) column of "L" within the lower- + # left-hand corner of "A", being sure to set the lower- + # left entries from the upper-right ones to avoid + #collisions. + A[k+i+1,k] = A[k,k+1+i]/alpha + + # 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. + 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. Beware: + # the subdiagonals of our matrix are being overwritten! + # So we actually use the corresponding row entries instead. + column_1_subdiag = [ a_ki.abs() for a_ki in A[k,1:].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 "1" in the top-left position. + 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. (Beware: the subdiagonal + # entries are being overwritten.) + omega_r = max( a_rj.abs() for a_rj in A[:r,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(A1,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. + k += 2 + + + MS = A.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 diagonal and upper-right half of "A", + # since we're about to return it as the unit-lower- + # triangular "L". + for i in range(n): + A[i,i] = 1 + for j in range(i+1,n): + A[i,j] = 0 + + return (P,A,D)