mjo/ldlt.py: get fast block_ldlt() working.

diff --git a/mjo/ldlt.py b/mjo/ldlt.py
index 4955d9fa9248f2bde240f70a526062f927aae9b5..6db747fee7095900d9c2c45ea2000f4a323de5f5 100644 (file)
--- a/mjo/ldlt.py
+++ b/mjo/ldlt.py
@@ -372,6 +372,9 @@ def block_ldlt(A):
     # 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()
@@ -385,30 +388,44 @@ def block_ldlt(A):
     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]]]) )
 
@@ -455,7 +472,7 @@ def block_ldlt(A):
         # 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:
@@ -504,14 +521,51 @@ def block_ldlt(A):
         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:]
+
+        # TODO: don't invert, there are better ways to get the C*E^(-1)
+        # that we need.
+        E_inverse = E.inverse()
+
+        schur_complement = B - (C*E_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+j,k+2+i] = A[k+2+j,k+2+i] - schur_complement[j,i]
+                A[k+2+i,k+2+j] = A[k+2+j,k+2+i] # keep it symmetric!
+
+        # 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] = (C*E_inverse)[i,j]
+
+
+        k += 2
 
     MS = A.matrix_space()
     P = MS.matrix(lambda i,j: p[j] == i)