30 votes 30 votes Let $T(n)$ be a function defined by the recurrence $T(n) = 2T(n/2) + \sqrt n$ for $n \geq 2$ and $T(1) = 1$ Which of the following statements is TRUE? $T(n) = \Theta(\log n)$ $T(n) = \Theta(\sqrt n)$ $T(n) = \Theta(n)$ $T(n) = \Theta(n \log n)$ Algorithms gateit-2005 algorithms recurrence-relation easy + – Ishrat Jahan asked Nov 3, 2014 • edited Nov 8, 2017 by kenzou Ishrat Jahan 9.4k views answer comment Share Follow See all 8 Comments See all 8 8 Comments reply Show 5 previous comments Pradip13 commented Dec 3, 2020 reply Follow Share @piyushwm Your recursion tree is wrong. Check once again. 0 votes 0 votes Overflow04 commented Oct 16, 2022 reply Follow Share @gatecse @Abhrajyoti00 @Kabir5454Somebody explain How( see comment of @piyushwm)$\sqrt{2}^{logn} = \sqrt{n}$andIn the gp , how you came to know that number of terms = logn not logn +1 0 votes 0 votes Kabir5454 commented Oct 17, 2022 reply Follow Share (considering base of $logn$ is $2$) . (1 --logn)-→ total logn terms. (0—logn)→ total logn+1 terms 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes T(n)=O(n) jay__sadhu answered Oct 17, 2022 jay__sadhu comment Share Follow See all 0 reply Please log in or register to add a comment.