Asymptotically, the question with more coconuts is more interesting. I haven't verified this for sure (there are some assumptions), but: With the "naive" asymp

Asymptotically, the question with more coconuts is more interesting. I haven't verified this for sure (there are some assumptions), but: With the "naive" asymptotically correct solution (go up a constant number k of floors, then linearly search the remainder), two coconuts gives us a worst case cost of: n/k + k minimize for k, we get a cost of 2n^(1/2) With 3 coconuts, we have n/k + 2k^(1/2) minimize for k, we get 3n^(1/3) What happens when we get log(n) coconuts? log(n)n^(1/log(n)) = O(log(n)) Sweet! So at least we can see that it converges to binary search as we add coconuts.