mjo/ldlt.py: various efficiency improvements to ldlt_fast().

diff --git a/mjo/ldlt.py b/mjo/ldlt.py
index 45615288085764c5bf59b3241343da030c948b6e..696d78d2053793164a49cd1ac7f25b452b1e2d2c 100644 (file)
--- a/mjo/ldlt.py
+++ b/mjo/ldlt.py
@@ -25,6 +25,7 @@ def is_positive_semidefinite_naive(A):
         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
@@ -135,18 +136,12 @@ def ldlt_fast(A):
     This function is much faster than ``ldlt_naive`` because the
     tail-recursion has been unrolled into a loop.
     """
-    n = A.nrows()
     ring = A.base_ring().fraction_field()
-
     A = A.change_ring(ring)
 
-    # Don't try to store the results in the lower-left-hand corner of
-    # "A" itself; there lies madness.
-    L = copy(A.matrix_space().identity_matrix())
-    D = copy(A.matrix_space().zero())
-
     # 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):
@@ -167,29 +162,27 @@ def ldlt_fast(A):
         # 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)
 
-        # Have to do L, too, to keep track of the "P2.T" (which is 1 x
-        # P3.T which is 1 x P4 T)... in the recursive
-        # algorithm. There, we compute P2^T from the bottom up. Here,
-        # we apply the permutations one at a time, essentially
-        # building them top-down (but just applying them instead of
-        # building them.
-        L.swap_columns(k,s)
-        L.swap_rows(k,s)
-
-        # Update the permutation "matrix" with the next swap.
+        # 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....
-        # Note: same as ``pivot``.
-        D[k,k] = alpha
-
-        # When alpha is zero, we can just leave the rest of the D/L entries
+        # 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
@@ -197,15 +190,21 @@ def ldlt_fast(A):
             # entries of a row than a column here?
             for i in range(n-k-1):
                 for j in range(i+1):
-                    A[k+1+i,k+1+j] = A[k+1+i,k+1+j] - A[k,k+1+i]*A[k,k+1+j]/alpha
-                    A[k+1+j,k+1+i] = A[k+1+i,k+1+j] # keep it symmetric!
+                    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!
 
-            # Store the "new" (kth) column of L.
             for i in range(n-k-1):
-                # Set the lower-left "half" from the upper-right "half"...
-                L[k+i+1,k] = A[k,k+1+i]/alpha
+                # 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())
 
-    I = A.matrix_space().identity_matrix()
-    P = matrix.column( I.row(p[j]) for j in range(n) )
+    for i in range(n):
+        A[i,i] = 1
+        for j in range(i+1,n):
+            A[i,j] = 0
 
-    return P,L,D
+    return P,A,D