Next: Up: Previous:

Worst Case Revisited

Assume we do not know what the worst partition is.

T(n) = $\max_{1 \leq q \leq n-1}$ (T(q) + T(n-q)) + $\Theta(n)$ .

By the substitution method (since we know the answer), try T(n) $\leq$ cn².

$\begin{displaymath}T(n) \;\leq\; \max_{1 \leq q \leq n-1} (cq^2 \;+\; c(n-q)^2) \;+\; \Theta(n)\end{displaymath}$

$\begin{displaymath}=\; c * \max_{1 \leq q \leq n-1} (q^2 \;+\; (n-q)^2) \;+\; \Theta(n)\end{displaymath}$

$\frac{d}{dq} (q^2 \;+\; (n-q)^2)$ = 2q - 2(n-q) = 2q - 2n + 2q = 4q - 2n

$\frac{d}{dq} (4q - 2n) = 4$ .

For q=1: $1^2 \;+\; (n-1)^2 \;=\; n^2 \;-\; 2n \;+\; 2$ .

For q=n-1: $(n-1)^2 \;+\; 1^2 \;=\; n^2 \;-\; 2n \;+\; 2$ .

$\psfig{figure=figures/f3-7.ps,width=3.5in,height=1.5in}$

T(n) $\leq$ cn² - 2c(n - 1) + $\Theta(n)$ $\leq$ cn². Picking a large enough c,

T(n) = $\Theta(n^2)$

Next: Up: Previous: