Re: [OF miscellaneous] Hilbert curve: recursion faster than loop?

[Juan Pablo Carbajal]

Wow, I tried

    z  = complex (zeros (nz(end),order+1));
    for k = 1:order
      idx              = 1:nz(k);
      w                = i * conj (z(idx,k));
      z(1:nz(k+1),k+1) = [w-a; z(idx,k)-b; z(idx,k)+a; -w+b] / 2;
    endfor

that is I use the columns of the matrix to store the intermediate
results and it is even more expensive!

24 ms [3 ms]

Any clues who's the culprit? Is it the assignation?