From 8f6fe9a17276099dc4cd82b752244036ce3aa523 Mon Sep 17 00:00:00 2001 From: Michael Orlitzky Date: Sun, 4 Oct 2020 12:11:25 -0400 Subject: [PATCH] mjo/ldlt.py: delete naive implementations; fix tests. --- mjo/ldlt.py | 335 ++-------------------------------------------------- 1 file changed, 8 insertions(+), 327 deletions(-) diff --git a/mjo/ldlt.py b/mjo/ldlt.py index 461dda3..6f3a628 100644 --- a/mjo/ldlt.py +++ b/mjo/ldlt.py @@ -26,328 +26,6 @@ def is_positive_semidefinite_naive(A): return A.is_hermitian() and all( v >= 0 for v in A.eigenvalues() ) -def ldlt_naive(A): - r""" - Perform a pivoted `LDL^{T}` factorization of the Hermitian - positive-semidefinite matrix `A`. - - This is a naive, recursive implementation that is inefficient due - to Python's lack of tail-call optimization. The pivot strategy is - to choose the largest diagonal entry of the matrix at each step, - and to permute it into the top-left position. Ultimately this - results in a factorization `A = PLDL^{T}P^{T}`, where `P` is a - permutation matrix, `L` is unit-lower-triangular, and `D` is - diagonal decreasing from top-left to bottom-right. - - ALGORITHM: - - The algorithm is based on the discussion in Golub and Van Loan, but with - some "typos" fixed. - - OUTPUT: - - A triple `(P,L,D)` such that `A = PLDL^{T}P^{T}` and where, - - * `P` is a permutaiton matrix - * `L` is unit lower-triangular - * `D` is a diagonal matrix whose entries are decreasing from top-left - to bottom-right - - SETUP:: - - sage: from mjo.ldlt import ldlt_naive, is_positive_semidefinite_naive - - EXAMPLES: - - All three factors should be the identity when the original matrix is:: - - sage: I = matrix.identity(QQ,4) - sage: P,L,D = ldlt_naive(I) - sage: P == I and L == I and D == I - True - - TESTS: - - Ensure that a "random" positive-semidefinite matrix is factored correctly:: - - sage: set_random_seed() - sage: n = ZZ.random_element(5) - sage: A = matrix.random(QQ, n) - sage: A = A*A.transpose() - sage: is_positive_semidefinite_naive(A) - True - sage: P,L,D = ldlt_naive(A) - sage: A == P*L*D*L.transpose()*P.transpose() - True - - """ - n = A.nrows() - - # Use the fraction field of the given matrix so that division will work - # when (for example) our matrix consists of integer entries. - ring = A.base_ring().fraction_field() - - if n == 0 or n == 1: - # We can get n == 0 if someone feeds us a trivial matrix. - P = matrix.identity(ring, n) - L = matrix.identity(ring, n) - D = A - return (P,L,D) - - A1 = A.change_ring(ring) - diags = A1.diagonal() - s = diags.index(max(diags)) - P1 = copy(A1.matrix_space().identity_matrix()) - P1.swap_rows(0,s) - A1 = P1.T * A1 * P1 - alpha1 = A1[0,0] - - # Golub and Van Loan mention in passing what to do here. This is - # only sensible if the matrix is positive-semidefinite, because we - # are assuming that we can set everything else to zero as soon as - # we hit the first on-diagonal zero. - if alpha1 == 0: - P = A1.matrix_space().identity_matrix() - L = P - D = A1.matrix_space().zero() - return (P,L,D) - - v1 = A1[1:n,0] - A2 = A1[1:,1:] - - P2, L2, D2 = ldlt_naive(A2 - (v1*v1.transpose())/alpha1) - - P1 = P1*block_matrix(2,2, [[ZZ(1), ZZ(0)], - [0*v1, P2]]) - L1 = block_matrix(2,2, [[ZZ(1), ZZ(0)], - [P2.transpose()*v1/alpha1, L2]]) - D1 = block_matrix(2,2, [[alpha1, ZZ(0)], - [0*v1, D2]]) - - return (P1,L1,D1) - - - -def ldlt_fast(A): - r""" - Perform a fast, pivoted `LDL^{T}` factorization of the Hermitian - positive-semidefinite matrix `A`. - - This function is much faster than ``ldlt_naive`` because the - tail-recursion has been unrolled into a loop. - """ - ring = A.base_ring().fraction_field() - A = A.change_ring(ring) - - # Keep track of the permutations in a vector rather than in a - # matrix, for efficiency. - n = A.nrows() - p = list(range(n)) - - for k in range(n): - # We need to loop once for every diagonal entry in the - # matrix. So, as many times as it has rows/columns. At each - # step, we obtain the permutation needed to put things in the - # right place, then the "next" entry (alpha) of D, and finally - # another column of L. - diags = A.diagonal()[k:n] - alpha = max(diags) - - # We're working *within* the matrix ``A``, so every index is - # offset by k. For example: after the second step, we should - # only be looking at the lower 3-by-3 block of a 5-by-5 matrix. - s = k + diags.index(alpha) - - # Move the largest diagonal element up into the top-left corner - # of the block we're working on (the one starting from index k,k). - # Presumably this is faster than hitting the thing with a - # permutation matrix. - # - # Since "L" is stored in the lower-left "half" of "A", it's a - # 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 - # are eventually applied to "v" in order. Here we're working - # from the top down, and rather than keep track of what - # permutations we need to perform, we just perform them as we - # go along. No recursion needed. - 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 largest diagonal is in the top-left corner of the - # block below and to the right of index k,k. When alpha is - # zero, we can just leave the rest of the D/L entries - # zero... which is exactly how they start out. - if alpha != 0: - # Update the "next" block of A that we'll work on during - # the following iteration. I think it's faster to get the - # entries of a row than a column here? - 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, being sure to set - # the lower-left "half" from the upper-right "half" - A[k+i+1,k] = A[k,k+1+i]/alpha - - MS = A.matrix_space() - P = MS.matrix(lambda i,j: p[j] == i) - D = MS.diagonal_matrix(A.diagonal()) - - for i in range(n): - A[i,i] = 1 - for j in range(i+1,n): - A[i,j] = 0 - - return P,A,D - - -def block_ldlt_naive(A, check_hermitian=False): - r""" - Perform a block-`LDL^{T}` factorization of the Hermitian - matrix `A`. - - This is a naive, recursive implementation akin to - ``ldlt_naive()``, where the pivots (and resulting diagonals) are - either `1 \times 1` or `2 \times 2` blocks. The pivots are chosen - using the Bunch-Kaufmann scheme that is both fast and numerically - stable. - - 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. - - """ - n = A.nrows() - - # Use the fraction field of the given matrix so that division will work - # when (for example) our matrix consists of integer entries. - ring = A.base_ring().fraction_field() - - 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 - return (P,L,D) - - alpha = (1 + ZZ(17).sqrt()) * ~ZZ(8) - A1 = A.change_ring(ring) - - # 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 = [ 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," - # which means that our matrix looks like, - # - # [ 1 0 ] - # [ 0 B ] - # - # 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_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): - # Perform a one-by-one pivot on "M," swapping row/columns "c". - # If "c" is None, no swap is performed. - if c is not None: - P1 = copy(M.matrix_space().identity_matrix()) - P1.swap_rows(0,c) - M = P1.T * M * P1 - - # The top-left entry is now our 1x1 pivot. - C = M[1:n,0] - B = M[1:,1:] - - P2, L2, D2 = block_ldlt_naive(B - (C*C.transpose())/M[0,0]) - - if c is None: - P1 = block_matrix(2,2, [[ZZ(1), ZZ(0)], - [0*C, P2]]) - else: - 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/M[0,0], L2]]) - D1 = block_matrix(2,2, [[M[0,0], ZZ(0)], - [0*C, D2]]) - - return (P1,L1,D1) - - - if A1[0,0].abs() > alpha*omega_1: - return pivot_one_by_one(A1) - - r = 1 + column_1_subdiag.index(omega_1) - - # If 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 A1[:r,r].list() ) - - 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: - # Higham step (3) - # Another 1x1 pivot, but this time swapping indices 0,r. - return pivot_one_by_one(A1,r) - - # Higham step (4) - # If we made it here, we have to do a 2x2 pivot. - 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,0:2] - 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 @@ -385,6 +63,10 @@ def block_ldlt(A): 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 @@ -409,7 +91,7 @@ def block_ldlt(A): [ 0|-4| 0] [--+--+--] [ 0| 0| 0] - sage: P.T*A*P == L*D*L.T + sage: P.transpose()*A*P == L*D*L.transpose() True This two-by-two matrix has no standard factorization, but it @@ -448,8 +130,7 @@ def block_ldlt(A): 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 = 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() @@ -463,7 +144,7 @@ def block_ldlt(A): 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() + sage: A == P*L*D*L.conjugate_transpose()*P.conjugate_transpose() True Ensure that a "random" complex positive-semidefinite matrix is @@ -476,7 +157,7 @@ def block_ldlt(A): 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() + sage: A == P*L*D*L.conjugate_transpose()*P.conjugate_transpose() True sage: diagonal_matrix(D.diagonal()) == D True -- 2.44.2