Reccurrence T(n) = T(n^(1/2)) + 1
algorithm, analysis, big-o, math, recurrence
Solution
hint: assume n = 22m or m = log2log2n, and you know 22m-1 * 22m-1 = 22m so, if you define S(m)=T(n) your S will be:
S(m) = S(m-1)+1 → S(m) = Θ(m) → S(m)=T(n) = Θ(log2log2n)
extend it for the general case.
In recursion like T(n) = T(n/2) + 1, in each iteration, we reduce the height of the tree to half. This leads to Θ(logn). In this case, however, we divide the input number by a power of two (not by two) so it turns out to be Θ(log log n ).
Problem
I've been looking at this reccurrence and wanted to check if I was taking the right approach. ``` T(n) = T(n^(1/2)) + 1 = T(n^(1/4)) + 1 + 1 = T(n^(1/8)) + 1 + 1 + 1 ... = 1 + 1 + 1 + ... + 1 (a total of rad n times) = n^(1/2) ``` So the answer would come to theta bound of n^(1/2)