1 votes 1 votes which of the following is correct? Algorithms logarithmic-function normal + – Anu asked May 18, 2015 • retagged Jun 26, 2022 by makhdoom ghaya Anu 1.2k views answer comment Share Follow See all 2 Comments See all 2 2 Comments reply Pranay Datta 1 commented Jun 7, 2015 reply Follow Share I think 1st one is correct and 2nd one is incorrect .... take a small example ....... (log2^8)^2 = 64 but loglog2^8 = 3 0 votes 0 votes Kabir5454 commented Jun 27, 2022 reply Follow Share It is definition specific . Different author use different notation . 0 votes 0 votes Please log in or register to add a comment.
3 votes 3 votes obviously 1st one is correct.basic arithmetic question . Refer any 10 std maths book sumit kumar answered May 18, 2015 sumit kumar comment Share Follow See 1 comment See all 1 1 comment reply Anu commented May 18, 2015 reply Follow Share Thank you :) since I have no 10 std maths book to refer,the 2 answer given here http://math.stackexchange.com/questions/159720/how-to-solve-this-recurrence-tn-2tn-2-n-log-n helped me 0 votes 0 votes Please log in or register to add a comment.
2 votes 2 votes 2nd one is correct.. Digvijay Pandey answered May 18, 2015 Digvijay Pandey comment Share Follow See all 10 Comments See all 10 10 Comments reply Show 7 previous comments spriti1991 commented May 19, 2015 reply Follow Share yeah second one is right ! 0 votes 0 votes Anu commented May 19, 2015 reply Follow Share @ spriti1991 I think i) is correct. sumit kumar said refer any 10 th std maths text.If you didn't have 10 th std text then I will help you :)please solve T(n)=2T(n/2)+nlogn using master theorem.We will get http://math.stackexchange.com/questions/159720/how-to-solve-this-recurrence-tn-2tn-2-n-log-n as shown in 2 nd answer.solve same recurrence relation using substitution we will get nT(1)+n((log(n)(1+log(n)))/2)=Θ(n*logn*logn) as shown in 1st answerso i) is correct 0 votes 0 votes spriti1991 commented May 19, 2015 reply Follow Share Yeah anu :) I will look after it Thanks :) 0 votes 0 votes Please log in or register to add a comment.
1 votes 1 votes definately first one is correct... log^k (n) = (log n)^k log .log n = log (logn) bgfbfg answered Jun 30, 2015 bgfbfg comment Share Follow See all 0 reply Please log in or register to add a comment.