How to calculate the square root of a "long"

Question

so I've got some fairly large numbers to work with, and I need a way to calculate the square root of a given "long".

I cannot use Math.sqrt(x) as it is only for doubles

Hmm. So you are worried about the loss of precision? How significant would that be? — Thilo, Mar 15 '16 at 04:13
Related (about BigInteger): http://stackoverflow.com/questions/4407839/how-can-i-find-the-square-root-of-a-java-biginteger — Thilo, Mar 15 '16 at 04:15
I suppose you could use the [Babylonian method](https://en.wikipedia.org/wiki/Methods_of_computing_square_roots#Babylonian_method) and calculate it yourself. — Elliott Frisch, Mar 15 '16 at 04:15
Over here, they say that the loss of precision is insignificant: http://stackoverflow.com/a/18501209/14955 — Thilo, Mar 15 '16 at 04:16
you can use mine [sqrt by binary search without multiplication in C++](http://stackoverflow.com/a/34657972/2521214) just change the variable names, iterations count and LUT to your `bitcount/2` — Spektre, Mar 15 '16 at 07:41
@AvanishKumar the cast is unnecessary. And the problem here is `long` has far more precision than `double` (64 vs 53). Some adjustment is needed to get the correct result as in Thilo's link — phuclv, Apr 07 '16 at 05:53

score 0 · Answer 1 · answered Apr 07 '16 at 05:40

You can extract one bit at a time, as demonstrated withthis C program: http://www.realitypixels.com/turk/opensource/index.html#FractSqrt This assumes that you have a fixed-point number with 30 fractional bits. If you have 0 fractional bits, then a 2N bit number would have an N bit square root, achieved by setting count to 15.

score -4 · Answer 2 · edited Mar 15 '16 at 10:51

-4

go cast it:

Math.sqrt((double)x)

edited Mar 15 '16 at 10:51

Floern

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answered Mar 15 '16 at 04:17

Jacob

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You don't need a cast. Java will cast `long` to `double` without being told to. – ajb Mar 15 '16 at 04:39

How to calculate the square root of a "long"

2 Answers2