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The following recurrence relation,

$$T(n)=16T(\frac{n}{4}) + n^2$$

has been given to me to be solved via the Master Theorem. I'm pretty sure this is a case 2 situation, since

$$\log_4{16} = 2$$ and

$$\log_n{n^2} = 2$$

Where $b = 4$, $a = 16$, and $c = 2$, such that $f(n) = n^c$

What I'm not sure about is the end runtime complexity.

What I've come up with so far is

$$T(n) = \theta(n^2log_2{n})$$

...since the Master Theorem states that, for case 2 situations, $f(n)$ must be within the set of $O(n^clog^kn)$, where $k$ is some constant in which $k \ge 0$.

My reasoning was to substitute 0 for $k$, which led to $\log_2^1{n}$ in the result, as the result requires that $T(n) = \theta(n^clog_2^{k+1}{n})$.

In a nutshell, while I know the first portion of the analysis is correct, and that this is a case 2 situation, am I correct in the reasoning behind the logarithmic portion in the ending runtime analysis?

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    $\begingroup$ Either your proof works, or there is some gap. In both cases, you should be able to find out yourself. We have a policy of not grading homework on this site. $\endgroup$ – Yuval Filmus Apr 24 '14 at 0:28
  • $\begingroup$ Does our reference question help? @YuvalFilmus, I think this question contains a (just barely) specific enough question about the attempt to be allowable. $\endgroup$ – Raphael Apr 24 '14 at 6:14
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    $\begingroup$ "What I'm not sure about is the end runtime complexity." A recurrence is just a way of defining a function: it has nothing to do with runtime complexity per se, since that function could be used to measure anything at all. It's a bit like saying "I'm trying to work out what the total distance is in 753+3858" -- the numbers in the sum could stand for anything. $\endgroup$ – David Richerby Apr 24 '14 at 8:03
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The Master theorem (roughly) states that in case two,

$\qquad\displaystyle f \in \Theta\left( n^{\log_b a} \log^{k} n \right) \implies T \in \Theta\left( n^{\log_b a} \log^{k+1} n \right)$.

That is, once you have identified the recurrence as case two by finding $k \geq 0$ so that $f$ is in this $\Theta$-class, the theorem automagically spits out the final result -- no further reasoning on your part necessary.

To be specific, in your case $a=16$, $b=4$ and $f : n \mapsto n^2$; therefore, $f \in \Theta\bigl( n^{\log_b a} \log^{k} n \bigr) = \Theta\bigl( n^2 \log^{k} n \bigr)$ with $k=0$ and you get

$\qquad\displaystyle T \in \Theta(n^2 \log n)$.

Two comments:

  • You use base $2$ for the logarithm in the final result. It should be $b=4$ (check the proof of the Master theorem) but it luckily does not matter; all logarithms share the same $\Theta$-class.
  • You talk about "runtime analysis"; this is not specific to runtime functions, and the recurrence at hand is unlikely to accurately describe the runtime of any algorithm. See also here.
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