OK. This makes sense given that A is relatively small.
On Thu, Oct 21, 2010 at 8:40 AM, Alexander Hans <[email protected]> wrote: > > One point that I don't quite see is how you accumulate partial sums of > > inv(X' * X). It seems that the inverse makes this difficult. > > This is orthogonal to the map-reduce architecture. How are you doing > > this? > > Simple answer: The inv() is done last (by the reducer), the partial sums > are enclosed by the brackets, i.e., inv(X' * X) = inv(A) = inv(A_1 + A_2 + > ... A_N).
