X-Git-Url: http://gitweb.michael.orlitzky.com/?a=blobdiff_plain;f=mjo%2Fldlt.py;h=696d78d2053793164a49cd1ac7f25b452b1e2d2c;hb=f471b4e3c164cbc6ed79594a30540f4ef07911bc;hp=8f39198d19a2007a91667baa34f89fc72b7c73db;hpb=48e1ebc120b5b3c068b7a7f86950a4b106ac89eb;p=sage.d.git diff --git a/mjo/ldlt.py b/mjo/ldlt.py index 8f39198..696d78d 100644 --- a/mjo/ldlt.py +++ b/mjo/ldlt.py @@ -25,6 +25,7 @@ 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 @@ -124,3 +125,86 @@ def ldlt_naive(A): [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 permuts "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