4 votes 4 votes Algorithms recurrence-relation algorithms + – Deep99 asked Jun 18, 2016 • retagged Jul 6, 2022 by Lakshman Bhaiya Deep99 34.8k views answer comment Share Follow See 1 comment See all 1 1 comment reply garam_masala_ commented Jul 29, 2016 reply Follow Share plz give me ans by back substitution..... 0 votes 0 votes Please log in or register to add a comment.
1 votes 1 votes T(n) = 7T(n/2) + n^2 comparing with the equation (MASTER THEOREM) T(n) = aT(n/b) + ⊖(n^k $\log_{n}p$) we get ,a=7,b=2,k=2,p=0 now it satisfies a>b^k, so case first of master theorem T(n) = ⊖(n^$\log_{b}a$) T(n)=⊖(n^3) sourav. answered Jun 18, 2016 sourav. comment Share Follow See all 3 Comments See all 3 3 Comments reply acak1994 commented Jun 21, 2016 reply Follow Share T(n) = n^2.801 0 votes 0 votes cse23 commented Jun 21, 2016 reply Follow Share we know by master theoram T(n) = aT(n/b) + n^klogp(base n) where a>=1 , b>1 , k>=0 and p is a real number In above question a =7, b=2 , k=2 and p=0 if a>b^k then Tc = O(n^loga(base b)) = O(n^log7(base 2)) =O(n^3) 1 votes 1 votes air1ankit commented Jun 1, 2018 reply Follow Share so what will be the answer, according to the extended master theorem, T(n) = Θ (n2.80735) that is approximate value of θ(n3) 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes Compare with master theorem ,a=7,b=2 and f(n)=n2 T(n)=nlog27 =n2.81 i.e f(n) is polynomially smaller than t(n), therefore it is first case therefore, ans is ⊖(nlog7) or ⊖(n2.81) Pravin Paikrao answered Jul 10, 2016 Pravin Paikrao comment Share Follow See all 0 reply Please log in or register to add a comment.