(see Proposition 3.20) isO(n):
Base case(n≤ 2):F(1) = 1 andF(2) = 2.
Induction step(n> 2): Assume claim true forn_n. Considern.F(n) =
F(n−2)+F(n−1). By induction,F(n−2) isO(n−2) andF(n−1) is
O(n−1). Then,F(n) isO((n−2)+(n−1)), by the identity presented in
Exercise R-3.11. Therefore,F(n) isO(n).
What is wrong with this “justification”?
Already registered? Login
Not Account? Sign up
Enter your email address to reset your password
Back to Login? Click here