0 votes 0 votes (logn)1/2=O(loglogn) Algorithms asymptotic-notation + – piyushkr asked Dec 31, 2015 piyushkr 1.4k views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
Best answer 3 votes 3 votes (logn)1/2 = O(loglogn) => (log)1/2 ≤ loglogn Taking log both side 1/2 loglogn ≤ logloglogn ==> FALSE since loglogn > logloglogn . Hence loglogn ≤ (logn)1/2 => loglogn = O(√logn) Shashank Kumar answered Dec 31, 2015 • selected Dec 31, 2015 by अनुराग पाण्डेय Shashank Kumar comment Share Follow See all 0 reply Please log in or register to add a comment.
1 votes 1 votes Substitute logn = t . equation becomes t^(1/2) = O(logt). clearly a polynomial term cannot be upper bounded by a logarithmic term. FALSE Harsh181996 answered Aug 6, 2016 Harsh181996 comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes f(n) = √(logn) & g(n) = log(logn) let n = 22^10 f(n) = √(log22^10) = √210 = 25 g(n) = log(log22^10) = log(210) = 10 Thus, g(n) < f(n) so (loglogn) = O(logn)1/2 Thus, (logn)1/2=O(loglogn) is FALSE Morphine answered Dec 31, 2015 Morphine comment Share Follow See all 0 reply Please log in or register to add a comment.