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

[sage.d.git] / mjo / ldlt.py

diff --git a/mjo/ldlt.py b/mjo/ldlt.py
index 8f39198d19a2007a91667baa34f89fc72b7c73db..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
@@ -124,3 +125,86 @@ def ldlt_naive(A):
                             [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