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I understand that Time Complexity is the measure of how an algorithm scales with respect to the input size.
Let us say an algorithm is having a runtime of O($3^n$) + O($n^{20}$)

In order to identify the Non-Dominant term, I followed my intuition for a sample of values:
$3^n$ = {1, 3, 9, 27, ...} for n={0, 1, 2, ...}
$n^{20}$ = {0, 1, 1048576, 3486784401, ...} for n={0, 1, 2, ...}

I see that $n^{20}$ scales up more quickly than $3^n$ for all values of n>=2
Can I say in this regard that the Non-Dominant term is $3^n$?

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2 Answers 2

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No, when we talk about dominant terms in big-O notation we talk about asymptotic domination, i.e. how terms scale as $n \to \infty$. To see this you can think about the limit: $$\lim_{n \to \infty}\frac{f(n)}{g(n)} = C \ .$$

If this value is $C \in \mathbb{R}$, then we have that $f(n)=O(g(n))$ and so the sum $O(f(n))+O(g(n))=O(g(n))$.

If instead, $C=+\infty$, then it's the opposite, i.e. $g(n)=O(f(n))$ and so the sum will be $O(f(n))+O(g(n))=O(f(n))$.

In your case, even if for small values of $n$ it seems that $n^{20}$ grows faster than $3^n$, this is not the case.

Indeed, $\lim_{n\to \infty} \frac{3^n}{n^{20}} = + \infty $.

You can see it in this way: $3^n=(3^{\log_3 n})^{n/\log_3 n}=n^{n/\log_3 n}$ which grows obviously bigger than $n^{20}$, in particular it surpasses $n^{20}$ as soon as the exponent $\frac{n}{\log_3 n}$ surpasses $20$.

It it worth to notice that asymptotic behavior isn't everything. In practice, polynomials with large exponents or large multiplicative constants can be bigger than exponentials for values of $n$ from real world applications.

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  • $\begingroup$ Could you elaborate or share a reference to understand about 'If this value is C∈R , then we have that f(n)=O(g(n)) and so the sum O(f(n))+O(g(n))=O(g(n))' ? $\endgroup$ Commented Aug 12, 2023 at 10:38
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    $\begingroup$ @PolamreddyVivekReddy the fact that if the limit is a real value than $f(n)=O(g(n))$ comes directly from the definition of big-O notation (en.wikipedia.org/wiki/Big_O_notation). The fact that the sum $O(f(n))+O(g(n))$ is also a consequence of the definition (assume $g$ is the dominant term, than the quotient between $(f+g)/g$ is limited). It is better to write $O(f(n)+g(n))$, which is the correct notation. Regarding the last question, re-writing a function does not change its behavior, it is just another way of writing the same thing where the functions we use are well-defined $\endgroup$
    – SilvioM
    Commented Aug 12, 2023 at 10:45
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Numerical examples can be misleading.

$$100^{20}=10000000000000000000000000000000000000000, \\3^{100}=515377520732011331036461129765621272702107522001.$$


When you consider the ratios $\dfrac{T(n+1)}{T(n)}$, you get

$$\frac{(n+1)^{20}}{n^{20}}=\left(1+\frac1n\right)^{20}\sim1+\frac{20}n\to1$$ and $$\frac{3^{n+1}}{3^n}=3.$$

So in fact the second function grows much faster !


Another argument is by taking the $20^{\text{th}}$ roots of both functions, to get

$$n$$

vs.

$$\sqrt[20]{3^n}=(\sqrt[20]3)^n.$$

The first function is linear and the second exponential.

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  • $\begingroup$ If we re-write original functions as mentioned in the answer, Aren't we changing their natural behavior because their graphical plots seem different to me? $\endgroup$ Commented Aug 12, 2023 at 10:43
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    $\begingroup$ @PolamreddyVivekReddy: that does not matter. $\endgroup$
    – user16034
    Commented Aug 13, 2023 at 14:18

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