58 views

Which of the following two is correct?

• If f(n) = Ο(g(n)) then h(f(n)) = Ο(h(g(n)))
• If f(n) ≠ Ο(g(n)) then g(n) = Ο(f(n))
second only.

Both of them are incorrect.

For first one, let's assume $f(n) = n$ and $g(n) = n^2$. Clearly, $f(n) = O(g(n))$. Now consider $h(n) = 1/n$.

$h(f(n)) = 1/f(n) = 1/n$ and $h(g(n)) = 1/g(n) = 1/n^2$. Now for no constants $c$ and $n_0$, $0 \leq h(f(n)) \leq ch(g(n))$ for all $n > n_0$. (You can confirm this by plotting both $1/n$ and $1/n^2$).

So first statement is incorrect.

For second statement.

CLRS says "Not all functions are asymptotically comparable." That is, for two functions $f(n)$ and $g(n)$, it may be the case
that neither $f(n) = O(g(n))$ nor $f(n) = \Omega(g(n))$ holds. An example would be $f(n) = n$ and $g(n) = n^{1+\sin{n}}$

Hence, both of the statements are incorrect.
edited

please explain Not all functions are asymptotically comparable. bit more.

That statement means that it is possible for two functions $f(n)$ and $g(n)$ to exist such that, you can't apply the definitions of any of $O$, or, $\Omega$, or, $\Theta$, or, $o$, or, $\omega$ between them.

Considering the same $f(n) = n$ and $g(n) = n ^ {1 + \sin{n}}$, Since $1 + \sin{n}$ oscillates between 0 and 2, $g(n)$ oscillates between $n^0$ and $n^2$, hence it's not possible to find two constants, $c$ and $n_0$ such that $0 \leq f(n) \leq cg(n)$ for all $n > n_0$ or similarly for other definitions.

You can get more insights from here

1
+1 vote
2