from sage.all import * def is_positive_semidefinite_naive(A): r""" A naive positive-semidefinite check that tests the eigenvalues for nonnegativity. We follow the sage convention that positive (semi)definite matrices must be symmetric or Hermitian. SETUP:: sage: from mjo.ldlt import is_positive_semidefinite_naive TESTS: The trivial matrix is vaciously positive-semidefinite:: sage: A = matrix(QQ, 0) sage: A [] sage: is_positive_semidefinite_naive(A) True """ if A.nrows() == 0: 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 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 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 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 entries are decreasing from top-left to bottom-right and 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. 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 = A1[1:,0].list() omega_1 = max([ a_i1.abs() 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:] 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]]) 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. return None