Introduction to Algorithms, 3rd edition (p.95) has an example of how to solve the recurrence

$$\displaystyle T(n)= 3T\left(\frac{n}{4}\right) + n\cdot \log(n)$$

by applying the Master Theorem.

I am very confused by how it is done. So, $a=3, b=4, f(n) = n\cdot \log(n)$
First step is to compare $n^{\log_b a} = n^{\log_4 3}= O(n^{0.793})$ with $f(n)$.

I have no clue on how they compared this. The book explains:

$f(n) = \Omega (n^{\log_4 3+\epsilon })$, where $\epsilon \approx 0.2$, case 3 applies if we can show that the regularity condition holds for $f(n).$

Followed by:

For sufficiently large n, we have that: $af\left(\frac{n}{b}\right) = 3\left(\frac{n}{4}\right)\log\left(\frac{n}{5}\right) \le\left(\frac{3}{4}\right)n \log n = cf(n)~ for~ c=\frac{3}{4}.$

Where did $3\left(\frac{n}{4}\right)$ come from?


2 Answers 2


If our recurrence takes the form $T(n) = aT(n/b) + f(n)$, then to use the "third case" of the Master method we must have the following hold:

$f(n) = \Omega(n^{\log_b a + \epsilon})$ for some $\epsilon > 0$ and if $a f(n/b) \leq c f(n)$ for some constant $c < 1$ and all sufficiently large $n$, then $T(n) = \Theta(f(n))$

Our recurrence is defined as $$T(n) = 3T(n/4) + n \log n.$$

By definition we have, $a = 3, b = 4, f(n) = n \log n$.

We now need to show that $f(n)$ is polynomially larger than $n^{\log_b a}$. That is the "$f(n) = \Omega(n^{\log_b a + \epsilon})$" part from above. Defining $\epsilon \approx 0.2$ achieves this. The reason being that $\log_4 3 \approx 0.793$ and $f(n) = \Omega(n^{0.793 + \epsilon})$.

We are left to show that $f(n)$ satisfies the regularity condition. That is the "$a f(n/b) \leq c f(n)$ for some constant $c < 1$ and all sufficiently large $n$" part.

We simply plug in our values of $a,b$ to get: $$a f(n/b) = 3(n/4) \log (n/4).$$ All we did was take the "$n$" in $f(n)$ and plug in "$n/4$".

To make this easier to see, let $k = n/4$ and observe that $a f(k) = a k \log k$. Swapping $k$ with $n/4$ we have $$3(n/4) \log (n/4) \leq (3/4)n \log n = c f(n)$$ where $c = 3/4$.

This finishes our necessary requirements and we have that $T(n) = \Theta(n \log n)$.

  • 1
    $\begingroup$ Thank you for detailed response. So, $n^{0.793}$ is polynomially $\leq$ than $ n^{1}\log n$ ? I probably misunderstood the meaning of "polynomially" $\endgroup$
    – newprint
    Sep 11, 2012 at 19:54
  • 1
    $\begingroup$ The polynomial in this case is $n^{1-0.793} = n^{0.217}$. $\endgroup$
    – JeffE
    Sep 11, 2012 at 22:00
  • $\begingroup$ @newprint Polynomially larger just means that the difference between the two is at least a (possibly fractional) power of $n$, not some smaller function, such as $(\log n)$. Specifically, it is exactly the condition for the "third case", that $f(n) = \Omega(n^{\log_b n} \cdot n^\epsilon)$. It is that extra $n^\epsilon$ factor that makes it polynomially larger. And as JeffE pointed out, in this case, $\epsilon = .217$. $\endgroup$
    – Joe
    Sep 11, 2012 at 22:29
  • $\begingroup$ Could we have chosen $\epsilon = 0.00000001$? As far as I understand $f(n)$ is already $\Omega(n^{0.793})$ so why do we need to add $0.2$? $\endgroup$
    – Yos
    Apr 14, 2017 at 16:54
  • $\begingroup$ Isn't big Omega a lower bound on running time so that should say is "asymptotocially equal to or greater than" instead of " is polynomially larger than.." ? $\endgroup$ May 17, 2020 at 18:58

enter image description here

This is the solution. we have to compare with the general equation of master theorem


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