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Oct
16
comment Computational complexity proof
Ah ok, you advice to try base case with n=4, and to prove by induction.
Oct
16
comment Worst case analysis of a basic algorithm
Thanks guys, I just don't understand how do you find out this result for the second innermost loop ?
Oct
16
comment Worst case analysis of a basic algorithm
No, condition might not be part of the input, it's just to show that there's a test performed I think.
Oct
8
comment inequality $ \frac{x}{2}< 2^{\lfloor\log_2 (x) \rfloor} \leq x$
Yes, I started with the base case, x=1, but I can't get it for x+1
Oct
8
comment inequality $ \frac{x}{2}< 2^{\lfloor\log_2 (x) \rfloor} \leq x$
Yes, I mean log base 2 :) I don't really see how to prove that...