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Solve the recurrence relation: $T(n)=T(n/4)+T(3n/4)+n$. Specify the best asymptotic running time.

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closed as off-topic by Did, Behaviour, Mathmo123, rogerl, Ivo Terek Oct 13 at 0:29

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Hint: try $T(n) = c n \log n$ for appropriate constant $c$. –  Robert Israel Oct 18 '12 at 0:49
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@Robert Israel: Your trying can find the particular solution part of the recurrence relation. –  doraemonpaul Aug 12 '13 at 5:03

1 Answer 1

This calls for the use of the Akra-Bazzi Method. Given that $T(n) = T(n/4) + T(3n/4) +n$, we have that
\begin{align} g(n)= n,\\ a_1 = 1, a_2 = 1 ~\mbox{and}\\ b_1= 1/4, b_2 = 3/4 \end{align}

We first need to solve for $p$ subject to $(1/4)^p + (3/4)^p = 1$ , giving $p= 1$.The method now gives $T(x) \in \Theta(f(x))$, where

\begin{align} f(x) = x^p (1+ \int _{[1,x]} \frac{g(u)}{u^{1+p}} du) \\ = x(1+ \log(x)) \end{align}

Thus $T(n) = \Theta(n \log(n))$.

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That's really nice. I hadn't heard of that result. Thanks. –  marty cohen Oct 18 '12 at 1:42

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