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Challenge: find the first value where two functions differ


On Fri, Aug 4, 2017 at 11:59 AM, Ian Kelly <ian.g.kelly at gmail.com> wrote:
> On Fri, Aug 4, 2017 at 11:50 AM, Chris Angelico <rosuav at gmail.com> wrote:
>> My logic was that floating point rounding is easiest to notice when
>> you're working with a number that's very close to something, and since
>> we're working with square roots, "something" should be a perfect
>> square. The integer square root of n**2 is n, the ISR of n**2+1 is
>> also n, and the ISR of n**2-1 should be n-1. I actually wanted to
>> start at 2**53, but being an odd power, that doesn't have an integer
>> square root, so I started at 2**52, which has an ISR of 2**26.
>
> A slight irony here is that it actually would have taken your script a
> *very* long time to get to 2**53 having started at 2**52, even only
> iterating over the perfect squares.

Never mind, I just tried it and it's a few seconds. I guess there are
not as many perfect squares in that range as I thought.