X-Git-Url: http://gitweb.michael.orlitzky.com/?a=blobdiff_plain;f=mjo%2Fldlt.py;h=1c2663ea3d574360059b07d22f9af8163ebdc2e7;hb=fe85934cbfb26ff0dc7e5ef81411fe73eed739ff;hp=45615288085764c5bf59b3241343da030c948b6e;hpb=5aa2527f533c75cafe1daf9c030f4f465f0e291f;p=sage.d.git diff --git a/mjo/ldlt.py b/mjo/ldlt.py index 4561528..1c2663e 100644 --- a/mjo/ldlt.py +++ b/mjo/ldlt.py @@ -25,187 +25,49 @@ def is_positive_semidefinite_naive(A): return True # vacuously 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 +def is_positive_semidefinite(A): + r""" + A fast positive-semidefinite check based on the block-LDLT + factorization. SETUP:: - sage: from mjo.ldlt import ldlt_naive, is_positive_semidefinite_naive + sage: from mjo.ldlt import (is_positive_semidefinite, + ....: is_positive_semidefinite_naive) - EXAMPLES: + TESTS: - All three factors should be the identity when the original matrix is:: + Check that the naive and fast answers are the same, in general:: - sage: I = matrix.identity(QQ,4) - sage: P,L,D = ldlt_naive(I) - sage: P == I and L == I and D == I + sage: set_random_seed() + sage: F = NumberField(x^2 + 1, 'I') + sage: from sage.misc.prandom import choice + sage: ring = choice([ZZ,QQ,F]) + sage: A = matrix.random(ring, 10) + sage: is_positive_semidefinite(A) == is_positive_semidefinite_naive(A) True - TESTS: - - Ensure that a "random" positive-semidefinite matrix is factored correctly:: + Check that the naive and fast answers are the same for a Hermitian + matrix:: 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) + sage: F = NumberField(x^2 + 1, 'I') + sage: from sage.misc.prandom import choice + sage: ring = choice([ZZ,QQ,F]) + sage: A = matrix.random(ring, 10); A = A + A.conjugate_transpose() + sage: is_positive_semidefinite(A) == 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. """ - n = A.nrows() - ring = A.base_ring().fraction_field() - - A = A.change_ring(ring) - - # Don't try to store the results in the lower-left-hand corner of - # "A" itself; there lies madness. - L = copy(A.matrix_space().identity_matrix()) - D = copy(A.matrix_space().zero()) - - # Keep track of the permutations in a vector rather than in a - # matrix, for efficiency. - 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. - A.swap_columns(k,s) - A.swap_rows(k,s) - - # Have to do L, too, to keep track of the "P2.T" (which is 1 x - # P3.T which is 1 x P4 T)... in the recursive - # algorithm. There, we compute P2^T from the bottom up. Here, - # we apply the permutations one at a time, essentially - # building them top-down (but just applying them instead of - # building them. - L.swap_columns(k,s) - L.swap_rows(k,s) - - # Update the permutation "matrix" with the next swap. - 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.... - # Note: same as ``pivot``. - D[k,k] = alpha - - # 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+i,k+1+j] = A[k+1+i,k+1+j] - A[k,k+1+i]*A[k,k+1+j]/alpha - A[k+1+j,k+1+i] = A[k+1+i,k+1+j] # keep it symmetric! - - # Store the "new" (kth) column of L. - for i in range(n-k-1): - # Set the lower-left "half" from the upper-right "half"... - L[k+i+1,k] = A[k,k+1+i]/alpha - - I = A.matrix_space().identity_matrix() - P = matrix.column( I.row(p[j]) for j in range(n) ) - - return P,L,D + if not A.is_hermitian(): + return False + _,_,d = A._block_ldlt() + for d_i in d: + if d_i.nrows() == 1: + if d_i < 0: + return False + else: + # A 2x2 block indicates that it's indefinite + return False + return True