大师法：分而治之

Question

According to my evaluation the overall asymptotic running time of the below algorithm is O(n) ,since x (number of recursions) is 1, and y is ( the number of splits) is 2 , and finally z ( the power of of amount of work done outside of the recursion call) is 1, hence x<y^{d}, but my answer is wrong . Why?

    FastPower(a,b) :
    if b = 1
      return a
    else
      c := a*a
      ans := FastPower(c,[b/2])
    if b is odd
      return a*ans
    else return ans
  end

Answer 1

So, first of all, it's not immediately clear what you mean by n , but I'm guessing |b| would be the most natural (as a doesn't affect run-time). Your analysis was mostly correct, but the mistake was saying z=1. There is constant work being done outside of the recursion, but that doesn't mean z, "the power of the amount of work done outside of the recursion call", is 1. To get z, you express that work as a polynomial function of n: f(n)=n^z=1, so z=0.

So that means x = y^d (1 = 2^0) rather than x < y^d, changing the case of the Master Theorem you apply. Instead of T(n) = O(n), you get T(n) = n^0*log(n) = O(log(n)), which should be what you expect as the problem is being divided in half at every recursive call, with only constant working being done in each call.