\(\rm Fun\left( n \right)\)

{

if \(\rm \left( {n \le 1} \right)\left\{ {return;} \right\}\)

else

{

Statement;

Fun \(\rm \left( {\sqrt n } \right)\)

return;

}

Assume “Statement” takes \(\rm \theta \left( 1 \right)\) time, Further assume that \(\rm n = {2^m}\) , then the time complexity of \(\rm Fun( )\) is:

1
\(\rm O\left( {\log \log n} \right)\)
2
\(\rm O\left( {\sqrt n } \right)\)
3
\(\rm O\left( {\sqrt {\log n} } \right)\)
4
\(\rm O\left( {\log \sqrt {\log n} } \right)\)

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