Time complexity of an algorithm: Is it important to state the base of the logarithm?

Question

Since there is only a constant between bases of logarithms, isn't it just alright to write $f(n) = \Omega(\log{n})$, as opposed to $\Omega(\log_2{n})$, or whatever the base might be?

Answer 1

It depends where the logarithm is. If it is just a factor, then it doesn't make a difference, because big-O or $\theta$ allows you to multiply by any constant.

If you take $O(2^{\log n})$ then the base is important. In base 2 you would have just $O(n)$, in base 10 it's about $O(n^{0.3010})$.

Answer 2

Because asymptotic notation is oblivious of constant factors, and any two logarithms differ by a constant factor, the base makes no difference: $\log_a n = \Theta(\log_b n)$ for all $a,b > 1$. So there is no need to specify the base of a logarithm when using asymptotic notation.

Answer 3

As $\log_xy = \frac{1}{\log_y{x}}$ and $\log_x{y} = \frac{\log_z{y}}{\log_z{x}}$, so $\frac{\log_a{n}}{\log_b{n}} = \frac{\log_n{b}}{\log_n{a}} = \log_a{b}$. As $\log_a{b}$ is positive constant (for all $a,b > 1$), so $\log_a{n} = \Theta(\log_b{n})$.

Answer 4

In most cases, it's safe to drop the base of the logarithm because, as other answers have pointed out, the change-of-basis formula for logarithms means that all logarithms are constant multiples of one another.

There are some cases where this isn't safe to do. For example, @gnasher729 has pointed out that if you have a logarithm in an exponent, then the logarithmic base is indeed significant.

I wanted to point out another case where the base of the logarithm is significant, and that's cases where the base of the logarithm depends directly on a parameter specified as input to the problem. For example, the radix sort algorithm works by writing out numbers in some base $b$, decomposing the input numbers into their base-$b$ digits, then using counting sort to sort those numbers one digit at a time. The work done per round is then $\Theta(n + b)$ and there are roughly $\log_b U$ rounds (where $U$ is the maximum input integer), so the total runtime is $O((n + b) \log_b U)$. For any fixed integer $b$ this simplifies to $O(n \log U)$. However, what happens if $b$ isn't a constant? A clever technique is to pick $b = n$, in which case the runtime simplifies to $O(n + \log_n U)$. Since $\log_n U$ = $\frac{\log U}{\log n}$, the overall expression simplifies to $O(\frac{n \log U}{\log n})$. Notice that, in this case, the base of the logarithm is indeed significant because it isn't a constant with respect to the input size. There are other algorithms that have similar runtimes (an old analysis of disjoint-set forests ended up with a term of $\log_{m/2 + 2}$ somewhere, for example), in which case dropping the log base would interfere with the runtime analysis.

Another case in which the log base matters is one in which there's some externally-tunable parameter to the algorithm that control the logarithmic base. A great example of this is the B-tree, which requires some external parameter $b$. The height of a B-tree of order $b$ is $\Theta(\log_b n)$, where the base of the logarithm is significant in that $b$ is not a constant.

To summarize, in the case where you have a logarithm with a constant base, you can usually (subject to exceptions like what @gnasher729 has pointed out) drop the base of the logarithm. But when the base of the logarithm depends on some parameter to the algorithm, it's usually not safe to do so.

Answer 5

Let $f_k(n)=\log_kn$,$k>2$, and let $t(n)=\log_2n$.

You can rewrite $f_k(n)$ as follow $$f_k(n)=\frac{\log_2n}{\log_2k}$$ Now $k$ have two states:

1. $k$ is constant:

$$\lim_{n\to \infty}\frac{f_k(n)}{t(n)}=\lim_{n\to \infty}\frac{\frac{\log_2n}{\log_2k}}{\log_2n}=c$$ that $c$ is a constant$>0$, hence $$t(n)=\Theta(f_k(n)).$$ So there is no difference between $f_k(n)$ , and $t(n)$ in terms of asymptotic notation.

2. $k$ isn't constant:

$$\lim_{n\to \infty}\frac{f_k(n)}{t(n)}=\lim_{n\to \infty}\frac{\frac{\log_2n}{\log_2k}}{\log_2n}=0$$

So $t(n)$ have faster growth rate and $$t(n)=\omega(f_k(n)).$$