RFR: 8341402: BigDecimal's square root optimization [v16]

[Raffaello Giulietti]

On Tue, 22 Oct 2024 14:07:30 GMT, fabioromano1 <duke at 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:
> 
>   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