As we know, the time complexity of $\gcd(x,y)$ is $O(\log \min(x,y))$ by using Euclidean algorithm. Now we fix a constant $n$ and consider the average time complexity of $\gcd(x,n)$.

Formally, let $f(x)$ be the number of divisions when calculating $\gcd(x,n)$. How to give a bound of $\frac 1 n \sum_{x=1}^n f(x)$?

I don't konw whether it is strictly $o(\log n)$ or $\Theta(\log n)$. How to prove the bound?

I list some of $n$ and the coresponding value here.

$n$ ; $\frac 1 n \sum_{x=1}^n f(x)$

  • $\begingroup$ Are you interested in the cost of the Euclidian algorithm, or in the complexity of the problem? $\endgroup$ – Raphael Oct 14 '18 at 18:52
  • $\begingroup$ Note that the averaging sum you give is not what we commonly call "average case". The latter averages about all inputs of the same size (typically $n$) while your sum averages over all numbers up to value (!) $n$, i.e. including all smaller inputs as well. $\endgroup$ – Raphael Oct 14 '18 at 18:54
  • $\begingroup$ What have you tried and where did you get stuck? $\endgroup$ – Raphael Oct 14 '18 at 18:54
  • $\begingroup$ Yes, I just want to fix a variable and sum over the other. I have written a program to check the value, as listed above. $\endgroup$ – zbh2047 Oct 15 '18 at 2:48
  • $\begingroup$ do you see the pattern $10^x$ gives $\approx 2 x$ $\endgroup$ – kelalaka Oct 15 '18 at 5:59

Since $n\geq x$ in your sum, $O(\log\min(x,n))$ simplifies to $O(\log x)$. Being very careless with constants here, but $$\frac{1}{n}\sum\limits_{x=1}^n f(x)=\frac{1}{n}\left(n\cdot O(\log n)\right)=O(\log n)$$

I'm not sure what this is calculating though; average runtime of all input sizes less than $n$?

  • $\begingroup$ Yes, $O(\log n)$ is obvious, but I am curious about the lower bound. $\endgroup$ – zbh2047 Oct 15 '18 at 13:28
  • $\begingroup$ Looking at your calculations, it seems that the values do correspond to about $2\log n$. So without a thorough analysis (just looking at the calculations), I would say it is $\Theta(\log n)$. $\endgroup$ – ydh28 Oct 15 '18 at 23:07

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