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The master theorem didn't work here. I tried to do the substitution method but I ended up with an additional term: $2Σ(i \cdot 3^i)$. Also I should find the solution $g(n)$ such as $f=\Theta(g)$.

dkaeae
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Ely Toy
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2 Answers2

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The given recurrence is: $T(n) = 3T(n/4) + \log n$

Using the substitution method, we get:

\begin{align} T(n) &= 3^2T \left(n/4^2\right) + \log n + 3\log (n/4) \\ &= 3^3T \left(n/4^3\right) + \log n + 3\log(n/4) + 9 \log(n/4^2)\\ \end{align} and so on.

Can you see a pattern now?

Gokul
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  • I already found that T(n) = $log(n)Σ3^i - 2Σi*3^i$, the first sum is geometrical but I don't know how to calculate the second one? – Ely Toy Nov 29 '18 at 17:48
  • The second term is an AGP - arithmetico geometric series. You can Google that and you'll find methods to calculate the summation. The trick is to multiply the summation with a term (usually the common difference) and then do some algebraic manipulations. – Gokul Nov 29 '18 at 17:50
  • @ElyToy Check this link – Gokul Nov 29 '18 at 17:53
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I use a simpler master theorem for scholar recurrences. Consider the general form $T(n) = aT(n/b) + f(n)$. Call $c=log_ba$. Now you have 3 cases:

  1. $\theta(n^c) = f(n)$ then $T(n) = \theta(f(n)log^k(n)) \text{ where }k\geq1$
  2. $\theta(n^c) \gt f(n)$ then $T(n) = \theta(n^c)$
  3. $\theta(n^c) \lt f(n)$ then $T(n) = \theta(f(n))$

In your exercise you have the 2nd case, so $T(n) = \theta(n^{log_43})$.
However this method works only with simple recurrences.

Emanuele
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