Prove or disprove the following statements:

  1. $T\left( n \right) = 2T\left( {\frac{n}{2}} \right) + f\left( n \right),f\left( n \right) = \theta \left( {{n^2}} \right) $ then $ {\rm{ }}T\left( n \right) = \theta \left( {f\left( n \right)} \right) $ for all $ {\rm{ n = }}{{\rm{2}}^k}$

  2. $T\left( n \right) = 2T\left( {\frac{n}{2}} \right) + f\left( n \right),f\left( n \right) = \Omega \left( {{n^2}} \right) $ then $ {\rm{ }}T\left( n \right) = O\left( {f\left( n \right)} \right)$ for all $ {\rm{ n = }}{{\rm{2}}^k}$

I think I should use the third case of the master theorem to check these equations.

But I have not been able to check this constraint for these inequations:

$\qquad af\left( {\frac{n}{b}} \right) \le cf\left( n \right)$

How do I do that?

  • $\begingroup$ Just find a $c$ so that the inequality is fulfilled. This has been treated here and also here. $\endgroup$
    – Raphael
    Oct 8, 2012 at 10:35
  • $\begingroup$ And I know the first one is true,and second one is false. I just don't know why. $\endgroup$
    – sam
    Oct 9, 2012 at 6:23
  • $\begingroup$ Ok, the question is a bit more involved than if we have an $f$ given, and item 2. seems to be hard (are you sure it is wrong?). By the way, the assumption $n=2^k$ usually suggests that you should do something by hand, not with Master theorem. $\endgroup$
    – Raphael
    Oct 9, 2012 at 17:50
  • $\begingroup$ May I ask where you got this problem? It seems like possibly a good homework/exam question for a course I'm TAing and I'd like to know the source. $\endgroup$
    – SamM
    Oct 9, 2012 at 18:42
  • $\begingroup$ @Raphael here the $n=2^k$ may simply be because the author didn't want to deal with non-integer values when dividing by 2. (Note that there's no floor or ceiling in the recurrence). Of course that doesn't contradict what you said, but it's another possibility. $\endgroup$
    – SamM
    Oct 9, 2012 at 18:44

2 Answers 2


The following shows that the second statement is false.

Define $f$ as follows: $$ f(n) = \begin{cases} 4n^3 & n \text{ is a power of }4, \\ n^2 & \text{otherwise}. \end{cases} $$ Let $n = 2^{2k+1}$. Then $f(n) = n^2$ while $$ T(n) \geq 2T(n/2) \geq 2f(n/2) = n^3. $$

  • $\begingroup$ Checking a piecewise defined function only at few positions and then talking about asymptotics doesn't feel right. I guess it just needs more words: $f \in \Omega(n^2)$ is clear, and you show $T(n) \notin O(n^2)$ without ever saying what complexity $T$ has. Nice. $\endgroup$
    – Raphael
    Oct 9, 2012 at 22:14

For 1., trouble is that you don't know $f$; it could be any one of the functions in $\Theta(n^2)$. Therefore, applying the master theorem directly is not (immediately) possible.

So let's unfold $\Theta$! Let $f \in \Theta(n^2)$ arbitrary. We know by definition of $\Theta$ that there are $d_1, d_2, n_1, n_2 \in \mathbb{N}$ such that

$\qquad \displaystyle d_1n^2 \leq f(n) \leq d_2n^2$,

for all $n \geq n_0 := \max(n_1,n_2)$. We will ignore all $n < n_0$ in the sequel; consider the respective values of $T$ constants.

We can easily show with the master theorem (case 3) that for

$\qquad\begin{align} T_1(n) &= 2T_1(n/2) + d_1n^2 \text{ and} \\ T_2(n) &= 2T_2(n/2) + d_2n^2 \end{align}$

both $T_1 \in \Theta(n^2)$ and $T_2 \in \Theta(n^2)$. Since we have (asymptotically) that $T_1(n) \leq T(n) \leq T_2(n)$, we also get that $T \in \Theta(n^2) = \Theta(f)$ by squeeze theorem.

As you see, we needed both upper and lower bound to perform the proof. We don't have that in the second case where $f \in \Omega(n^2)$. That suggests the statement may be false (but does not prove it).

Once we suspect, we can choose a counter-example; see Yuval's answer for one.


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