# The top-left 2x2 submatrix is now our pivot.
E = A1[:2,:2]
- C = A1[2:n,0]
+ C = A1[2:n,0:2]
B = A1[2:,2:]
if B.nrows() == 0:
# 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()
p = list(range(n))
d = []
- def pivot1x1(M, k, s):
+ def swap_rows_columns(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.
+ 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.
- # 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)
+ # 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)
- # Update the permutation "matrix" with the swap we just did.
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]]]) )
# 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!
+ A[k+1+i,k+1+j] = ( A[k+1+i,k+1+j] -
+ A[k,k+1+i]*A[k,k+1+j]/A[k,k] )
+ A[k+1+j,k+1+i] = A[k+1+i,k+1+j] # keep it symmetric!
for i in range(n-k-1):
# Store the new (kth) column of "L" within the lower-
# 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() ]
+ 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:
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)
+ 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.
- k += 2
+ 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+i,k+2+j] = A[k+2+i,k+2+j] - schur_complement[i,j]
+ A[k+2+j,k+2+i] = A[k+2+j,k+2+i] - schur_complement[j,i]
+
+ # 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)
projects / sage.d.git / blobdiff