For Prim's and Kruskal's Algorithm there are many implementations which will give different running times. However suppose our implementation of Prim's algorithm has runtime $O(|E| + |V|\cdot \log(|V|))$ and Kruskals's algorithm has runtime $O(|E|\cdot \log(|V|))$.

What is the base of the $\log$?

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    $\begingroup$ It doesn't matter. When they are inside O-notation like this, all logarithms with constant base are equivalent. $\endgroup$
    – David Eppstein
    Commented Oct 27, 2012 at 21:14
  • $\begingroup$ That said, if you want to know which base the logarithm is before O-ing it away, you'd have to look/give us the specific implementation. $\endgroup$
    – Raphael
    Commented Nov 1, 2012 at 22:49
  • $\begingroup$ See also the question Are log10(x) and log2(x) in the same big-O class of functions? $\endgroup$
    – Juho
    Commented Nov 1, 2012 at 23:38

1 Answer 1


To expand on what David Eppstein said, you can assume it's in any base.

If the log were initially in base $b$, it is asymptotically the same as if it were in base $k$:

$$O(\log_b n) = O\left(\frac{\log_k n}{\log_k b}\right) = O(\log_k n)$$

because $k$ and $b$ are both constants; $\log_k b = O(1)$.

The logs in any two bases are the same in big-O notation.


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