Comment by pavpanchekha | Hacker News Reader

Comment by pavpanchekha | Hacker News Reader

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What do you mean, take care of the limited precision? The formula you gave is only correct for

    n = [1, 3, 5, 7, 9, 10, 11, 12, 13, 14, 15, 16, 17]

The reason the explicit floating point formula doesn't get used is because you need as many bits of floating point as you expect to get bits of answer. Which means that you get the wrong answer pretty quickly. If the goal was to calculate an approximation to the fibonacci numbers, sure, but then you don't need the (ceil) call.

Oh yeah you're right

Although, wouldn't variable overflow be the case for any algorithm, since already the 100th or so Fibonacci number takes more space than 2^64

And if you were to use the full float precision of sqr5 (and not just 4 digits like I did), you would run into 64bit float overflow way before that algorithm stops being accurate

Everything else is talking about bigints.

If you were dealing with fixed-length integers, just use a lookup table.

Also, you're wrong regarding floating-point. On my machine, fib(101) is off from your approximation by ~2 million (!). Well before hitting the limits of a 64-bit float. It starts erroring at n=71, and only gets worse from there. Worse, the relative delta grows as n increases.

Hmm ok

Thanks for actually testing it

Python is wonderful for that sort of thing.

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