X-Git-Url: http://gitweb.michael.orlitzky.com/?a=blobdiff_plain;f=mjo%2Fldlt.py;h=461dda3062f1de5e1f0d2acac07d9589d28f8ca5;hb=3abfe3dd31271b6863d52b4fe23993b92359bed9;hp=6b9908341add2dfaadb99c17a76dc7c4c19913dc;hpb=a2334e928c769d00d59ae43e2cad8abd61389066;p=sage.d.git diff --git a/mjo/ldlt.py b/mjo/ldlt.py index 6b99083..461dda3 100644 --- a/mjo/ldlt.py +++ b/mjo/ldlt.py @@ -325,7 +325,7 @@ def block_ldlt_naive(A, check_hermitian=False): # The top-left 2x2 submatrix is now our pivot. E = A1[:2,:2] - C = A1[2:n,0] + C = A1[2:n,0:2] B = A1[2:,2:] if B.nrows() == 0: @@ -353,14 +353,144 @@ def block_ldlt(A): 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: - A triple `(P,L,D)` such that `A = PLDL^{T}P^{T}` and where, + 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 + * `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. + + 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.T*A*P == L*D*L.T + 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: F = NumberField(x^2 +1, 'I') + sage: A = matrix.random(F, 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.transpose()*P.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.transpose()*P.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 @@ -372,6 +502,9 @@ def block_ldlt(A): # 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. ring = A.base_ring().fraction_field() A = A.change_ring(ring) MS = A.matrix_space() @@ -385,52 +518,64 @@ def block_ldlt(A): p = list(range(n)) d = [] - def pivot1x1(M, k, s): + def swap_rows_columns(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. + 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. - # 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) + # 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) - # 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]]]) ) + # 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 - # 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 + 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! - # No return value, only the desired side effects of updating - # p, d, and A. - return + 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: @@ -443,7 +588,9 @@ def block_ldlt(A): if k == (n-1): # Handle this trivial case manually, since otherwise the # algorithm's references to the e.g. "subdiagonal" are - # meaningless. + # 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 @@ -452,10 +599,8 @@ def block_ldlt(A): # 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() ] + # 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: @@ -465,7 +610,9 @@ def block_ldlt(A): # [ 0 B ] # # and we can simply skip to the next step after recording - # the 1x1 pivot "1" in the top-left position. + # 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 @@ -491,9 +638,8 @@ def block_ldlt(A): # 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() ) + # 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. @@ -504,14 +650,55 @@ def block_ldlt(A): 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) + 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. - k += 2 + 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 MS = A.matrix_space() P = MS.matrix(lambda i,j: p[j] == i)