On Wed, 2 Oct 2024 10:31:09 GMT, fabioromano1 <d...@openjdk.org> wrote:
> After changing `BigInteger.sqrt()` algorithm, this can be also used to speed > up `BigDecimal.sqrt()` implementation. Here is how I made it. > > The main steps of the algorithm are as follows: > first argument reduce the value to an integer using the following relations: > > x = y * 10 ^ exp > sqrt(x) = sqrt(y) * 10^(exp / 2) if exp is even > sqrt(x) = sqrt(y*10) * 10^((exp-1)/2) is exp is odd > > Then use BigInteger.sqrt() on the reduced value to compute the numerical > digits of the desired result. > > Finally, scale back to the desired exponent range and perform any adjustment > to get the preferred scale in the representation. Sure, but that does not explain the slow running times. Naïvely, this takes around 1s for the required precision of 1_000_000 decimal digits. `new BigDecimal(BigInteger.valueOf(121).multiply(BigInteger.TEN.pow(2_000_000)).sqrt(), 1_000_000).shortValue()` Of course, this is just a special case. Do I miss something? ------------- PR Comment: https://git.openjdk.org/jdk/pull/21301#issuecomment-2388759527
