0 votes 0 votes If we arrange the following according to increasing asymptotic complexity, which quantity comes in the second position; which one comes in the fourth position? $(\sqrt{2}^{\log n}, n^2, 2^{\sqrt{2 \log n} }, e^n, (\log n)!, n!$ Algorithms algorithms asymptotic-notation iisc-interview + – go_editor asked Jun 8, 2016 retagged Jan 10 by Hira Thakur go_editor 425 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
1 votes 1 votes TAKE LOG ON EACH TERMS AND EVALUATE... asu answered Jun 8, 2016 asu comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes i think n^2 comes in second position and e^n comes in fourth [position because we know the growth of functions are like that- 2^2^n > n! >4^n >2^n >n^2 >nlogn >logn! >n >2^logn >logn^2 >underoot logn> loglogn>1 anshul namdeo answered Jun 8, 2016 anshul namdeo comment Share Follow See all 0 reply Please log in or register to add a comment.