return A.is_hermitian() and all( v >= 0 for v in A.eigenvalues() )
-def ldlt_naive(A):
+def is_positive_semidefinite(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
+ A fast positive-semidefinite check based on the block-LDLT
+ factorization.
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
+ sage: from mjo.ldlt import (is_positive_semidefinite,
+ ....: is_positive_semidefinite_naive)
TESTS:
- Ensure that a "random" positive-semidefinite matrix is factored correctly::
+ Check that the naive and fast answers are the same, in general::
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()
+ 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
- """
- 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]])
+ Check that the naive and fast answers are the same for a Hermitian
+ matrix::
- 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.
+ 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); A = A + A.conjugate_transpose()
+ sage: is_positive_semidefinite(A) == is_positive_semidefinite_naive(A)
+ 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.
- # 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]])
+ 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:
- 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]
- 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".
- 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 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.
-
- Note that ``A`` is passed in by reference here, so it doesn't
- matter if we shadow the name ``A`` with itself.
- """
- if s > k:
- # s == k would swap row/column k with itself, and we don't
- # actually want to perform the identity permutation.
- # We don't have to permute "L" separately so long as "L"
- # is stored within "A".
- 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 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,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(A1,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.
- 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)
+ # A 2x2 block indicates that it's indefinite
+ return False
+ return True
projects / sage.d.git / blobdiff