Question

Question 6 !! Thanks

Order the following functions according to their order of growth (from the lowest to n!, n lg n, 8 lg (n + 10)^10, 2^3n, 3^2n, n^5 + 10 lg n Prove that a + lg(n^k + c) = Theta (lg n), for every fixed k > 0, a > 0 and c > 0. Determine the complexities of the following recursive functions, where c > 0 is the operations in the functions. (You may assume that n = 2^k). (a) Consider the recursive function, use the Big () notation to determine its

Answer 1

Ans 6 : g(n) = lg(n)

f(n)= a + lg(n^k + c)

a+lg(n^k *(1+c/n^k))

a+lg(n^k)+lg(1+c/n^k)

a+ klg(n)+lg(1+c/n^k)

as a,k,e all are >0 , hence we can neglect the value lg(1+c/n^k) as very small

hence the upper bound for the above function in O(lg n) (1)

similarly f(n) grows as fast as g(n), thtrefore , f(n)= Ω(g(n)) (2)

from above two points we can conclude that f(n)=Θ(g(n))