RFR: 8077587: BigInteger Roots [v23]

[fabioromano1]

On Wed, 23 Jul 2025 19:49:51 GMT, Raffaello Giulietti <rgiulietti at openjdk.org> wrote:

>> @rgiulietti I've changed the formula that computes the initial estimate, now it does not use `Math.pow()` anymore, but `Math.exp()` and `Math.log()` instead, which are guaranteed to have always an error less than 1 ulp  by the fdlibm.
>> 
>> Anyway, the proof that the Newton's recurrence should output an overestimate if the input is an underestimate should follow by two facts:
>> - $f(x) = x^n - C$ with domain $x \in [0, +\infty)$ is an increasing function and its derivative is increasing;
>> - Newton's recurrence computes the next approximation $x_{i+1}$ by finding the zero of the line with gradient $f'(x_i)$ that passes through the point $(x_i, f(x_i))$.
>> 
>> So, Newton's method approximates the curve of $f(x)$ with the tangent in the point $(x_i, f(x_i))$. Since the curve is increasing and its derivative is increasing, if $x_i$ is less than the zero of $f(x)$, then the zero of the line must be greater than the zero of $f(x)$ (because the line increases more slowly than the curve of $f(x)$ ).
>
> I noticed the usage of exp() and log(), thanks for the change.
> 
> My concerns about using transcendental are rooted in [this paper](https://members.loria.fr/PZimmermann/papers/accuracy.pdf).
> The Javadoc claims an error of 1 ulp for pow(), but it turns out to be plainly wrong: the worst _known_ error is 636 ulps! (In that paper, see the column for OpenLibm, a derivative work of fdlibm.)
> 
> On a more positive side, that paper also shows the worst _known_ error for exp() and log() to be around 0.95 ulps. But again, it could be much worse, who knows?
> 
> The Brent & Zimmermann paper assumes an initial estimate $u \ge \lfloor x^{1/n}\rfloor$, probably for a (unstated) reason.
> The proof of the algorithm makes use of Newton's formula $x_{i+1} = f(x_i)$, where $f$ is the real-valued counterpart of the integer recurrence formula in the algorithm.
> It is straightforward to see that $x_{i+1} > x^{1/n}$ when $0 < x_i < x^{1/n}$.
> But it's less clear to me that the same applies to the _integer_ recurrence formula of the algorithm.
> 
> Given all of the above, we must ensure that the initial estimate meets the requirements of the BZ paper, or we need a proof that an underestimate in the 1st iteration is harmless because it will become an overestimate in the 2nd, i.e., that the reasoning which holds for the real-valued $f$ also holds with the integer-valued analogous formula.

@rgiulietti
> The Brent & Zimmermann paper assumes an initial estimate u ≥ ⌊ x 1 / n ⌋ , probably for a (unstated) reason.

The reason is the condition to stop the loop, since it terminates when the estimate does not decrease anymore.

> 
> Given all of the above, we must ensure that the initial estimate meets the requirements of the BZ paper, or we need a proof that an underestimate in the 1st iteration is harmless because it will become an overestimate in the 2nd, i.e., that the reasoning which holds for the real-valued f also holds with the integer-valued analogous formula.

In Brent & Zimmermann proof, we have the following definition: $f(t) := \lfloor (n-1) t + m/t^{n-1}  \rfloor / n$, with $t \in (0; +\infty)$. Since $f(t)' < 0$ if $t < \sqrt[n]{m}$ and $f(t)' > 0$ if $t > \sqrt[n]{m}$, then $\sqrt[n]{m}$ is a point of minimum, so $f(t) \ge f(\sqrt[n]{m}) = \sqrt[n]{m}$, hence $\lfloor f(t) \rfloor \ge \lfloor \sqrt[n]{m} \rfloor$ for any $t$ in the domain. This proves that, when $\lfloor f(t) \rfloor \le \lfloor \sqrt[n]{m} \rfloor$ becomes true (it does because the sequence of estimates is strictly decreasing), we get $\lfloor f(t) \rfloor = \lfloor \sqrt[n]{m} \rfloor$ and the loop stops at the next iteration.

> The proof of the algorithm makes use of Newton's formula x i + 1 = f ( x i ) , where f is the real-valued counterpart of the integer recurrence formula in the algorithm. It is straightforward to see that x i + 1 > x 1 / n when 0 < x i < x 1 / n . But it's less clear to me that the same applies to the _integer_ recurrence formula of the algorithm.

It should be clear if we note that the integer-valued analogous formula simply discard the fraction part of the real-valued counterpart:
$\lfloor f(t) \rfloor = \lfloor  \lfloor (n-1) t + m/t^{n-1}  \rfloor / n \rfloor$ = $\lfloor ((n-1) t + m/t^{n-1}) / n \rfloor$

So, if $x_i < \lfloor \sqrt[n]{m} \rfloor$, then $((n-1) x_i + m/x_i^{n-1}) / n > \sqrt[n]{m}$, hence $\lfloor f(x_i) \rfloor \ge \lfloor \sqrt[n]{m} \rfloor$.

>From what has just been shown, it follows that it is sufficient to perform an initial iteration before starting the loop to get an overestimate.

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PR Review Comment: https://git.openjdk.org/jdk/pull/24898#discussion_r2226698680