RFR: 8077587: BigInteger Roots [v52]

[fabioromano1]

On Tue, 29 Jul 2025 12:06:37 GMT, fabioromano1 <duke at openjdk.org> wrote:

>> To the above, we can also add
>> 
>>                  * 
>>                  * Noting that x ≥ 2^(bl-1) and ex ≥ 0, similarly to the above we further get
>>                  *      x 2^(-sh') ≥ 2^(ex+P-1) ≥ 2^(P-1)
>>                  * which shows that ⌊x / 2^sh'⌋ has at least P bits of precision.
>
> @rgiulietti Yes, but why that complication, it is not more natural to prove it in the following way?
> 
> recalling x < 2^bl:
> x / 2^sh' = x / 2^(sh-ex) ≤ 2^bl / 2^(sh-ex) = 2^(bl - (sh-ex)) ≤ 2^ME < Double.MAX_VALUE
> ``` 
> relying on the fact that `bl - (sh-ex)` is the length of the shifted value?

> To the above, we can also add
> 
> ```
>                  * 
>                  * Noting that x ≥ 2^(bl-1) and ex ≥ 0, similarly to the above we further get
>                  *      x 2^(-sh') ≥ 2^(ex+P-1) ≥ 2^(P-1)
>                  * which shows that ⌊x / 2^sh'⌋ has at least P bits of precision.
> ```

And this should follow by the fact that `bl - (sh-ex) ≥ bl - sh = bl - (bl - P) = P`, since `ex ≥ 0`.

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PR Review Comment: https://git.openjdk.org/jdk/pull/24898#discussion_r2239646258