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
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]
[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
+ matrix `A`.
+
+ 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.
+ """
+
+ # We have to make at least one copy of the input matrix so that we
+ # can change the base ring to its fraction field. Both "L" and the
+ # intermediate Schur complements will potentially have entries in
+ # the fraction field. However, we don't need to make *two* copies.
+ # We can't store the entries of "D" and "L" in the same matrix if
+ # "D" will contain any 2x2 blocks; but we can still store the
+ # 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()
+
+ # The magic constant used by Bunch-Kaufman
+ alpha = (1 + ZZ(17).sqrt()) * ~ZZ(8)
+
+ # Keep track of the permutations and diagonal blocks in a vector
+ # rather than in a matrix, for efficiency.
+ n = A.nrows()
+ p = list(range(n))
+ d = []
+
+ def swap_rows_columns(M, k, s):
+ r"""
+ 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. 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)
+
+ p_k = p[k]
+ p[k] = p[s]
+ p[s] = p_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
+
+
+ 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+j,k+1+i] = ( A[k+1+j,k+1+i] -
+ A[k,k+1+j]*A[k,k+1+i]/A[k,k] )
+ 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]/A[k,k]
+
+ # No return value, only the desired side effects of updating
+ # p, d, and A.
+ return
+
+ k = 0
+ while k < n:
+ # At each step, we're considering the k-by-k submatrix
+ # contained in the lower-right half of "A", because that's
+ # where we're storing the next iterate. So our indices are
+ # always "k" greater than those of Higham or B&K. Note that
+ # ``n == 0`` is handled by skipping this loop entirely.
+
+ if k == (n-1):
+ # Handle this trivial case manually, since otherwise the
+ # algorithm's references to the e.g. "subdiagonal" are
+ # meaningless.
+ d.append( matrix(ring, 1, [[A[k,k]]]) )
+ k += 1
+ continue
+
+ # Find the largest subdiagonal entry (in magnitude) in the
+ # 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,k+1:].list() ]
+ omega_1 = max([ a_ki for a_ki in column_1_subdiag ])
+
+ if omega_1 == 0:
+ # In this case, our matrix looks like
+ #
+ # [ a 0 ]
+ # [ 0 B ]
+ #
+ # and we can simply skip to the next step after recording
+ # the 1x1 pivot "1" in the top-left position.
+ d.append( matrix(ring, 1, [[A[k,k]]]) )
+ k += 1
+ continue
+
+ if A[k,k].abs() > alpha*omega_1:
+ # This is the first case in Higham's Step (1), and B&K's
+ # Step (2). Note that we have skipped the part of B&K's
+ # Step (1) where we determine "r", since "r" is not yet
+ # needed and we may waste some time computing it
+ # otherwise. We are performing a 1x1 pivot, but the
+ # rows/columns are already where we want them, so nothing
+ # needs to be permuted.
+ pivot1x1(A,k,k)
+ k += 1
+ continue
+
+ # Now back to Step (1) of Higham, where we find the index "r"
+ # that corresponds to omega_1. This is the "else" branch of
+ # Higham's Step (1).
+ r = k + 1 + column_1_subdiag.index(omega_1)
+
+ # Continuing the "else" branch of Higham's Step (1), and onto
+ # 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() )
+
+ if A[k,k].abs()*omega_r >= alpha*(omega_1**2):
+ # Step (2) in Higham or Step (4) in B&K.
+ pivot1x1(A,k,k)
+ k += 1
+ continue
+
+ 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(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.
+ 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
+ #
+ # C*E.inverse() == X <====> XE == C.
+ #
+ # The latter can be found much more 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.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+j,k+2+i] = A[k+2+j,k+2+i] - schur_complement[j,i]
+ A[k+2+i,k+2+j] = A[k+2+j,k+2+i] # keep it symmetric!
+
+ # 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", being sure to set
+ # the lower-left entries from the upper-right ones to
+ # avoid collisions.
+ 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)
+
+ # Warning: when n == 0, this works, but returns a matrix
+ # whose (nonexistent) entries are in ZZ rather than in
+ # the base ring of P and L.
+ D = block_diagonal_matrix(d)
+
+ # Overwrite the diagonal and upper-right half of "A",
+ # since we're about to return it as the unit-lower-
+ # triangular "L".
+ for i in range(n):
+ A[i,i] = 1
+ for j in range(i+1,n):
+ A[i,j] = 0
+
+ return (P,A,D)
projects / sage.d.git / blobdiff