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Given: $T(n) = T(\sqrt{n}) + 1$ (base case $T(x) = 1$ for $x<=2$)

How do you solve such a recurrence?

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  • $\begingroup$ Have a look here. $\endgroup$ – Juho Dec 16 '12 at 19:50
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    $\begingroup$ Hint: Let $n = 2^{2^k}$, so $\sqrt{n} = 2^{2^{k-1}}$, and take it from there. $\endgroup$ – Yuval Filmus Dec 16 '12 at 20:04
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For the recurrence, $$ T(n) = T(\sqrt{n}) + 1 $$ Let $n = 2^{2^{k}}$, therefore we can write the recurrence as:

$$ T(2^{2^{k}}) = T(2^{2^{k-1}}) + 1 \\ T(2^{2^{k-1}}) = T(2^{2^{k-2}}) + 1 \\\ldots\\\\\ldots\\ T(2^{2^{k-k}}) = 1 $$

, i.e. $ k * O(1) $ work or linear in $k$. We can express $k$ in terms of $n$: $$ \log{n} = 2^{k} \\ \log{\log{n}} = k $$

Hence, the recurrence solves $T(n) = O(\log{\log{n}}) $

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