Question

This function is O(log(n)). Why? Isn’t it looping up to n?

function fxn($n) {
    for ($i = 1; $i <= $n; $i *= 2)
        echo $i;
}

I’m pretty new to O(n) analysis by the way. This function sure looks O(n) though since it loops up to n.

Answer 1

This loop would be O(n):

function fxn($n) {
    for ($i = 0; $i <= $n; $i++)
        echo $i;
}

Because$i takes on the values:

1, 2, 3, 4, 5, 6, 7, ..., n

But this loop is only O(log(n)):

function fxn($n) {
    for ($i = 1; $i <= $n; $i *= 2)
        echo $i;
}

Because $i takes on the values:

1, 2, 4, 8, 16, 32, 64, ..., n

And a sequence that grows in that manner is called “logarithmic”.