1 votes 1 votes H(n) = H(2/3(n-1)) + 1 H(1) = 0 solve this recurrence relation A_i_$_h asked Aug 18, 2017 A_i_$_h 284 views answer comment Share Follow See all 5 Comments See all 5 5 Comments reply Show 2 previous comments A_i_$_h commented Aug 22, 2017 reply Follow Share Can u please show me the way u solved it 0 votes 0 votes Shubhanshu commented Aug 22, 2017 reply Follow Share I do it as: Put m= n-1 n=m-1 Eqn H(n) = H(2/3(n-1)) + 1 Become H(m-1) = H(2m/3) + 1 H(m) - H(1) = H(2m/3) + 1 H(m) = H(2m/3) + 1 as H(1) = 0 Then do substitution you get O(log3/2n). 2 votes 2 votes A_i_$_h commented Aug 22, 2017 reply Follow Share thank u :) 0 votes 0 votes Please log in or register to add a comment.