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I recently came across the iterated algorithm function denoted as $\lg^* n$.

But I am having a hard time understanding this statement:

$\lg^* n = \min \{i \ge 0: \lg^{(i)} n ≤ 1\}$

I could not understand what $\min$ set is used to signify. Also, could someone please provide an example of how this works. I also came across examples like:

$\lg^* 4= 2$

But could not understand the computation, specifically how did they take up the values of i?

Thank you.

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$\lg^* n$ is just the minimum number of times you need to apply the $\lg$ function to $n$ in order to obtain a number that is smaller than or equal to $1$.

For example, assuming that you are working with base-2 logarithms and that $n=65536$ you have the following:

  • $\lg^{(0)} 65536 = 65536$,
  • $\lg^{(1)} 65536 = \lg 65536 = 16$,
  • $\lg^{(2)} 65536 = \lg 16 = 4$,
  • $\lg^{(3)} 65536 = \lg 4 = 2$,
  • $\lg^{(4)} 65536 = \lg 2 = 1$.

As you can see $4$ is the smallest integer value of $i \ge 0$ for which $\lg^{(i)} 65536 \le 1$. Therefore $\lg^* 65536 = 4$.

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  • $\begingroup$ Okay. noted. But how could we get to define a recurrence based on it ? BTW, Thank you for the clear and concise explaination. $\endgroup$ – Sachin Bahukhandi Jun 14 '20 at 10:35
  • $\begingroup$ What do you mean by "define a recurrence based on it"? Do you want a recurrence relation that has $\Theta( \log^* n )$ as a solution? $\endgroup$ – Steven Jun 14 '20 at 10:36
  • $\begingroup$ Reurrence like stated here: en.wikipedia.org/wiki/Iterated_logarithm $\endgroup$ – Sachin Bahukhandi Jun 14 '20 at 10:37
  • $\begingroup$ So you want a recursive definition of $\log^* n$? There is one right in the page you linked. $\endgroup$ – Steven Jun 14 '20 at 10:38
  • $\begingroup$ Yeah. But how could we get to it. $\endgroup$ – Sachin Bahukhandi Jun 14 '20 at 10:41

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