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 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 ]
+ #
+ B = A1[1:,1:]
+ P2, L2, D2 = 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)
+
+ if A1[0,0].abs() > alpha*omega_1:
+ # Higham step (1)
+ # The top-left entry is our 1x1 pivot.
+ C = A1[1:n,0]
+ B = A1[1:,1:]
+
+ P2, L2, D2 = block_ldlt_naive(B - (C*C.transpose())/A1[0,0])
+
+ P1 = block_matrix(2,2, [[ZZ(1), ZZ(0)],
+ [0*C, P2]])
+ L1 = block_matrix(2,2, [[ZZ(1), ZZ(0)],
+ [P2.transpose()*C/A1[0,0], L2]])
+ D1 = block_matrix(2,2, [[A1[0,0], ZZ(0)],
+ [0*C, D2]])
+
+ return (P1,L1,D1)
+
+
+ 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):
+ # Higham step (2)
+ # The top-left entry is our 1x1 pivot.
+ C = A1[1:n,0]
+ B = A1[1:,1:]
+
+ P2, L2, D2 = block_ldlt_naive(B - (C*C.transpose())/A1[0,0])
+
+ P1 = block_matrix(2,2, [[ZZ(1), ZZ(0)],
+ [0*C, P2]])
+ L1 = block_matrix(2,2, [[ZZ(1), ZZ(0)],
+ [P2.transpose()*C/A1[0,0], L2]])
+ D1 = block_matrix(2,2, [[A1[0,0], ZZ(0)],
+ [0*C, D2]])
+
+ return (P1,L1,D1)
+
+
+ if A1[r,r].abs() > alpha*omega_r:
+ # Higham step (3)
+ # Another 1x1 pivot, but this time swapping indices 0,r.
+ P1 = copy(A1.matrix_space().identity_matrix())
+ P1.swap_rows(0,s)
+ A1 = P1.T * A1 * P1
+
+ # The top-left entry is now our 1x1 pivot.
+ C = A1[1:n,0]
+ B = A1[1:,1:]
+
+ P2, L2, D2 = block_ldlt_naive(B - (C*C.transpose())/A1[0,0])
+
+ 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/A1[0,0], L2]])
+ D1 = block_matrix(2,2, [[A1[0,0], ZZ(0)],
+ [0*C, D2]])
+
+ return (P1,L1,D1)
+
+ # Higham step (4)
+ # If we made it here, we have to do a 2x2 pivot.
+ return None
projects / sage.d.git / blobdiff