Example: Deleting from a B-Tree (not to be confused with binary tree) has Big-O complexity of $ O(\log_t n) $ (where $t \in \mathbb{N}$ is the order of the tree).

There was one true/false question on exam which asks if the Big-O complexity of the operation mentioned in the example above is $ O(\log_2 n) $.

I am beginner in this topic but I understand that both $ O(\log_2 n) $ and $ O(\log_t n) $ belong to the same Big-O complexity category of $ O(\log n) $. The only thing which confuses me is whether it matters if the base of a logarithm is given as a constant or as a variable.

Additionally: Would the answer change if we swap the bases from the example and the question?

Edit (if relevant): The mentioned complexity is related to the number of disk accesses and it is not the time complexity.

  • $\begingroup$ Do you know that $\log_b(n) = \ln(n) / \ln(b)$? $\endgroup$ Oct 29, 2020 at 12:05
  • $\begingroup$ Yes, I do know that. $\endgroup$
    – Bashyar
    Oct 29, 2020 at 12:21
  • $\begingroup$ (I don't want to think about $O(\log_n n)$.) $\endgroup$
    – greybeard
    Oct 29, 2020 at 13:18
  • $\begingroup$ Hmm. You do know, I trust, that $log_nn=1$ wherever it's defined. $\endgroup$ Oct 29, 2020 at 14:37

1 Answer 1


Let's look at the root of question: we have $\log_2 n=\frac{\ln n}{\ln 2}$ and $\log_t n=\frac{\ln n}{\ln t}$, so $\log_2 n = \frac{\ln t}{\ln 2} \log_t n$ and when $t$ is constant, then logarithms are in same complexity class.

Bit if $t=t(n)$, then we can ruin this relation and create any class $O(f(n))$ of complexity taking $t(n)=e^{\frac{\ln n}{f(n)}}$ we have $\log_t n=\frac{\ln n}{\ln t}=f(n)$.

  • $\begingroup$ I am not sure that I get your point. Does this mean they are unequal in both cases? $\endgroup$
    – Bashyar
    Oct 29, 2020 at 14:05
  • $\begingroup$ @Bashyar The classes coincide for constant $t$. They do not coincide, in general, if $t$ is not a constant. $\endgroup$
    – Steven
    Oct 29, 2020 at 14:28
  • $\begingroup$ I come later and confirm Steven's comment. $\endgroup$
    – zkutch
    Oct 29, 2020 at 16:20
  • $\begingroup$ @Steven Thanks for the comment. This makes things much clearer to me. In the meantime I have also asked my professor and from his answer I could conclude the same. The source of my confusion was that I thought that $t$ is a variable, when in fact it is a constant. This opens a new question for me: How to tell difference between a constant and a variable? $\endgroup$
    – Bashyar
    Oct 29, 2020 at 16:41
  • $\begingroup$ @Bashyar in the context of data structures, if a parameter doesn't change through the life-time of the data structure, then it is a constant. For example, changing $t$ requires you to rebuild the B-Tree from scratch. On the other hand, insert/delete operations causes some parameters (like number of nodes and depth of the tree) to vary — these are variables. $\endgroup$ Nov 10, 2020 at 6:25

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