mjo/ldlt.py: fix a dimension in block_ldlt_naive().

[sage.d.git] / mjo / ldlt.py

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 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".
    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)