On Mon, 25 Nov 2024 19:36:04 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.
>
> fabioromano1 has updated the pull request incrementally with one additional
> commit since the last revision:
>
> Added tests for exact results path
Thanks.
What about the coverage of the other paths in the `sqrt()` code?
test/jdk/java/math/BigDecimal/SquareRootTests.java line 200:
> 198: // mc.roundingMode != RoundingMode.UNNECESSARY &&
> mc.precision == 0
> 199: try {
> 200: result = BigDecimal.TEN.sqrt(arbitrary);
Suggestion:
result = input.sqrt(arbitrary);
-------------
PR Review: https://git.openjdk.org/jdk/pull/21301#pullrequestreview-2461649513
PR Review Comment: https://git.openjdk.org/jdk/pull/21301#discussion_r1858609104
