\[ T(n) = aT(n/b)+f(n) \]
Then $ T(n) $ has the following asympotic bounds:
- If $f(n) = O (n^{\log_b a - \epsilon})$ for some constant $\epsilon>0$, Then $T(n) = \Theta(n^{log_b a})$
- If $f(n) = \Theta (n^{log_b a})$, Then $T(n) = \Theta(n^{log_b a}\lg n)$
- If $f(n) = \Omega(n^{log_b a + \epsilon})$ for some constant $\epsilon>0$, and if $af(n/b) \leq cf(n)$ for some constant $c \leq 1$ and all sufficiently large $n$, then $T(n)=\Theta(f(n))$
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