Now consider the following recurrence T(n): T(n) = { 210 HD+ v ifnas if n no Since the input size n will always be power of 8, the base case will be always be the input size is of 1. Base Case:T(1) = 1 2 an =a as 1 as long as this condition is met, base case holds. Inductive Hypothesis: Let input n>8 and n is power of 8. T(K) = a*k, 15 k sn Inductive Step: T(n) = 211%) + n 2 2a1) + n = an + na an 4 an = n3 > an 3 = -a *n 4 4 Not sure what to do from here ... | Now consider the following recurrence T(n): T(n) = { 210 HD+ v ifnas if n no Since the input size n will always be power of 8, the base case will be always be the input size is of 1. Base Case:T(1) = 1 2 an =a as 1 as long as this condition is met, base case holds. Inductive Hypothesis: Let input n>8 and n is power of 8. T(K) = a*k, 15 k sn Inductive Step: T(n) = 211%) + n 2 2a1) + n = an + na an 4 an = n3 > an 3 = -a *n 4 4 Not sure what to do from here ... |