On Wed, 2 Oct 2024 13:49:29 GMT, Raffaello Giulietti <rgiulie...@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. > > Looking at the code, the divisors are quite limited, so my hypothesis above > and the suggestions are out of place. @rgiulietti The fundamental problem is that `BigDecimal` numbers store the magnitude in binary, using `BigInteger`s objects. A solution could have been to store the magnitude in radix 10 rather than in binary, but obviously this would require to redesign the entire class `BigDecimal`, which is undesirable. ------------- PR Comment: https://git.openjdk.org/jdk/pull/21301#issuecomment-2388722321
