mjo/ldlt.py

   1 from sage.all import *
   2
   3 def is_positive_semidefinite_naive(A):
   4     r"""
   5     A naive positive-semidefinite check that tests the eigenvalues for
   6     nonnegativity.  We follow the sage convention that positive
   7     (semi)definite matrices must be symmetric or Hermitian.
   8
   9     SETUP::
  10
  11         sage: from mjo.ldlt import is_positive_semidefinite_naive
  12
  13     TESTS:
  14
  15     The trivial matrix is vaciously positive-semidefinite::
  16
  17         sage: A = matrix(QQ, 0)
  18         sage: A
  19         []
  20         sage: is_positive_semidefinite_naive(A)
  21         True
  22
  23     """
  24     if A.nrows() == 0:
  25         return True # vacuously
  26     return A.is_hermitian() and all( v >= 0 for v in A.eigenvalues() )
  27
  28 def ldlt_naive(A):
  29     r"""
  30     Perform a pivoted `LDL^{T}` factorization of the Hermitian
  31     positive-semidefinite matrix `A`.
  32
  33     This is a naive, recursive implementation that is inefficient due
  34     to Python's lack of tail-call optimization. The pivot strategy is
  35     to choose the largest diagonal entry of the matrix at each step,
  36     and to permute it into the top-left position. Ultimately this
  37     results in a factorization `A = PLDL^{T}P^{T}`, where `P` is a
  38     permutation matrix, `L` is unit-lower-triangular, and `D` is
  39     diagonal decreasing from top-left to bottom-right.
  40
  41     ALGORITHM:
  42
  43     The algorithm is based on the discussion in Golub and Van Loan, but with
  44     some "typos" fixed.
  45
  46     OUTPUT:
  47
  48     A triple `(P,L,D)` such that `A = PLDL^{T}P^{T}` and where,
  49
  50       * `P` is a permutaiton matrix
  51       * `L` is unit lower-triangular
  52       * `D` is a diagonal matrix whose entries are decreasing from top-left
  53         to bottom-right
  54
  55     SETUP::
  56
  57         sage: from mjo.ldlt import ldlt_naive, is_positive_semidefinite_naive
  58
  59     EXAMPLES:
  60
  61     All three factors should be the identity when the original matrix is::
  62
  63         sage: I = matrix.identity(QQ,4)
  64         sage: P,L,D = ldlt_naive(I)
  65         sage: P == I and L == I and D == I
  66         True
  67
  68     TESTS:
  69
  70     Ensure that a "random" positive-semidefinite matrix is factored correctly::
  71
  72         sage: set_random_seed()
  73         sage: n = ZZ.random_element(5)
  74         sage: A = matrix.random(QQ, n)
  75         sage: A = A*A.transpose()
  76         sage: is_positive_semidefinite_naive(A)
  77         True
  78         sage: P,L,D = ldlt_naive(A)
  79         sage: A == P*L*D*L.transpose()*P.transpose()
  80         True
  81
  82     """
  83     n = A.nrows()
  84
  85     # Use the fraction field of the given matrix so that division will work
  86     # when (for example) our matrix consists of integer entries.
  87     ring = A.base_ring().fraction_field()
  88
  89     if n == 0 or n == 1:
  90         # We can get n == 0 if someone feeds us a trivial matrix.
  91         P = matrix.identity(ring, n)
  92         L = matrix.identity(ring, n)
  93         D = A
  94         return (P,L,D)
  95
  96     A1 = A.change_ring(ring)
  97     diags = A1.diagonal()
  98     s = diags.index(max(diags))
  99     P1 = copy(A1.matrix_space().identity_matrix())
 100     P1.swap_rows(0,s)
 101     A1 = P1.T * A1 * P1
 102     alpha1 = A1[0,0]
 103
 104     # Golub and Van Loan mention in passing what to do here. This is
 105     # only sensible if the matrix is positive-semidefinite, because we
 106     # are assuming that we can set everything else to zero as soon as
 107     # we hit the first on-diagonal zero.
 108     if alpha1 == 0:
 109         P = A1.matrix_space().identity_matrix()
 110         L = P
 111         D = A1.matrix_space().zero()
 112         return (P,L,D)
 113
 114     v1 = A1[1:n,0]
 115     A2 = A1[1:,1:]
 116
 117     P2, L2, D2 = ldlt_naive(A2 - (v1*v1.transpose())/alpha1)
 118
 119     P1 = P1*block_matrix(2,2, [[ZZ(1), ZZ(0)],
 120                                [0*v1,  P2]])
 121     L1 = block_matrix(2,2, [[ZZ(1),                    ZZ(0)],
 122                             [P2.transpose()*v1/alpha1, L2]])
 123     D1 = block_matrix(2,2, [[alpha1, ZZ(0)],
 124                             [0*v1,   D2]])
 125
 126     return (P1,L1,D1)
 127
 128
 129
 130 def ldlt_fast(A):
 131     r"""
 132     Perform a fast, pivoted `LDL^{T}` factorization of the Hermitian
 133     positive-semidefinite matrix `A`.
 134
 135     This function is much faster than ``ldlt_naive`` because the
 136     tail-recursion has been unrolled into a loop.
 137     """
 138     n = A.nrows()
 139     ring = A.base_ring().fraction_field()
 140
 141     A = A.change_ring(ring)
 142
 143     # Don't try to store the results in the lower-left-hand corner of
 144     # "A" itself; there lies madness.
 145     L = copy(A.matrix_space().identity_matrix())
 146     D = copy(A.matrix_space().zero())
 147
 148     # Keep track of the permutations in a vector rather than in a
 149     # matrix, for efficiency.
 150     p = list(range(n))
 151
 152     for k in range(n):
 153         # We need to loop once for every diagonal entry in the
 154         # matrix. So, as many times as it has rows/columns. At each
 155         # step, we obtain the permutation needed to put things in the
 156         # right place, then the "next" entry (alpha) of D, and finally
 157         # another column of L.
 158         diags = A.diagonal()[k:n]
 159         alpha = max(diags)
 160
 161         # We're working *within* the matrix ``A``, so every index is
 162         # offset by k. For example: after the second step, we should
 163         # only be looking at the lower 3-by-3 block of a 5-by-5 matrix.
 164         s = k + diags.index(alpha)
 165
 166         # Move the largest diagonal element up into the top-left corner
 167         # of the block we're working on (the one starting from index k,k).
 168         # Presumably this is faster than hitting the thing with a
 169         # permutation matrix.
 170         A.swap_columns(k,s)
 171         A.swap_rows(k,s)
 172
 173         # Have to do L, too, to keep track of the "P2.T" (which is 1 x
 174         # P3.T which is 1 x P4 T)... in the recursive
 175         # algorithm. There, we compute P2^T from the bottom up. Here,
 176         # we apply the permutations one at a time, essentially
 177         # building them top-down (but just applying them instead of
 178         # building them.
 179         L.swap_columns(k,s)
 180         L.swap_rows(k,s)
 181
 182         # Update the permutation "matrix" with the next swap.
 183         p_k = p[k]
 184         p[k] = p[s]
 185         p[s] = p_k
 186
 187         # Now the largest diagonal is in the top-left corner of
 188         # the block below and to the right of index k,k....
 189         # Note: same as ``pivot``.
 190         D[k,k] = alpha
 191
 192         # When alpha is zero, we can just leave the rest of the D/L entries
 193         # zero... which is exactly how they start out.
 194         if alpha != 0:
 195             # Update the "next" block of A that we'll work on during
 196             # the following iteration. I think it's faster to get the
 197             # entries of a row than a column here?
 198             v = vector(ring, A[k,k+1:n].list())
 199             b = v.column()*v.row()/alpha
 200             for i in range(n-k-1):
 201                 for j in range(i+1):
 202                     # Something goes wrong if I try to access the kth row/column
 203                     # of A to save the intermediate "b" here...
 204                     A[k+1+i,k+1+j] = A[k+1+i,k+1+j] - b[i,j]
 205                     A[k+1+j,k+1+i] = A[k+1+i,k+1+j] # keep it symmetric!
 206
 207             # Store the "new" (kth) column of L.
 208             for i in range(n-k-1):
 209                 L[k+i+1,k] = v[i]/alpha
 210
 211     I = A.matrix_space().identity_matrix()
 212     P = matrix.column( I.row(p[j]) for j in range(n) )
 213
 214     return P,L,D