With a finite range, you can bisect directly by splitting in the middle of the range.
With infinite ranges, you can't do that; so the usual way is to start with a small number and increase exponentially until you find a number that is too large; which is what is done here. When you got that number, it becomes the upper bound of a finite interval.
So that's a two step process, which we can see here. The first 32 is in the exponential growth step (so is 64), and the second one is in the bisect step.
This will always happen exactly once (unless the expected result is 0) and for only the first pivot in the bisect, so it's not that bad; but indeed, they could get rid of it by bisecting on [1/n; n] instead of [0; n], as they already know that 1/n (and numbers lower than 1/n) isn't a valid candidate from the first step.
> they could get rid of it by bisecting on [1/n; n] instead of [0; n], as they already know that 1/n (and numbers lower than 1/n) isn't a valid candidate from the first step.
With infinite ranges, you can't do that; so the usual way is to start with a small number and increase exponentially until you find a number that is too large; which is what is done here. When you got that number, it becomes the upper bound of a finite interval.
So that's a two step process, which we can see here. The first 32 is in the exponential growth step (so is 64), and the second one is in the bisect step.
This will always happen exactly once (unless the expected result is 0) and for only the first pivot in the bisect, so it's not that bad; but indeed, they could get rid of it by bisecting on [1/n; n] instead of [0; n], as they already know that 1/n (and numbers lower than 1/n) isn't a valid candidate from the first step.