def f(n):
 i = 2
 while i < n:
  i = i * i

I evaluated an expression that becomes something like 2^n or 2^2^n and set that equal to when the loop breaks or (2^2^n ) (>=)n but can't figure out how to get a runtime. Can someone show how to do complete this analysis and get a runtime?


In the loop, $i$ is repeatedly squared so it will take the values $$ 2, 2^2=4, 4^2=16, 16^2=258, \dotsc $$ In other words, $i$ will be $$ 2^1, 2^2, 2^4, 2^8, \dots,2^{2^k} $$ so the loop will iterate $k$ times, until $2^{2^k}\ge n$. Taking the log of this inequality twice we'll have $k\ge \log \log n$ for our running time.


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