# Exercise 4.4.5

Use a reccursion tree to determine a good asymptotic upper bound on the recurrence $T(n) = T(n-1) + T(n/2) + n$. Use the substitution method to verify your answer.

This is a curious one. The tree makes it look like it is exponential in the worst case. The tree is not full (not a complete binary tree of height $n$), but it is not polynomial either. It's easy to show $\Omega(n^2)$ and $O(2^n)$:

We guess $T(n) \le c2^n - 4n$:

\begin{aligned} T(n) & \le c2^{n-1} - 4(n-1) + c2^{n/2} - 4n/2 + n \\ & \le c(2^{n-1} + 2^{n/2}) - 5n + 1 & (n > 1/4) \\ & \le c(2^{n-1} + 2^{n/2}) - 4n & (n > 2)\\ & \le c(2^{n-1} + 2^{n-1}) - 4n \\ & \le c2^n - 4n \\ & = O(2^n) \end{aligned}

We guess $T(n) \ge cn^2$:

\begin{aligned} T(n) & \ge c(n - 1)^2 + c(n/2)^2 + n \\ & \ge cn^2 - 2cn + 1 + cn^2/4 + n \\ & \ge (5/4)cn^2 + (1 - 2c)n + 1 \\ & \ge cn^2 + (1 - 2c)n + 1 & (c < 1/2)\\ & \ge cn^2 \\ & = O(n^2) \end{aligned}

As to more details - I am lost.