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Analysis

T(n) = T(n-1) + $\Theta(n)$ in the worst case

= cn + c(n-1) + c(n-2) + ... + c

= $c \sum_{i=1}^n i$ = $c \frac{n(n+1)}{2}$ = $\Theta(n^2)$

T(n) = T(n/2) + $\Theta(n)$ in the best case

a=1, b=2, f(n) = $\Omega(n^{log_b a + \epsilon})$ , $\epsilon$ = 1

Case 3: T(n) = $\Theta(n)$

Show: 1*f(n/2) $\leq$ cf(n), c < 1

f(n/2) $\leq$ c*f(n)

c >= 1/2

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