On Tue, 22 Oct 2024 14:07:30 GMT, fabioromano1 <[email protected]> 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:
>
> Code simplification
src/java.base/share/classes/java/math/BigDecimal.java line 2190:
> 2188: resultScale = strippedScale >> 1;
> 2189: } else {
> 2190: working = working.multiply(10L);
While this is correct, there's useless work that is being performed.
After the multiplication by 10, `working` cannot be an exact square, so this
will fail later at L.2199.
I wonder if this can be simplified to avoid the `multiply()` and the following
`sqrtAndRemainder()` when `strippedScale` is odd.
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PR Review Comment: https://git.openjdk.org/jdk/pull/21301#discussion_r1843763055
