Re: A divide by zero bug in lib/math/rational.c (with triggering input)

Andy Shevchenko
2021-05-21T09:56:58Z

+Cc: Daniel (here is a real case for test cases!)

On Fri, May 21, 2021 at 12:20 PM Trent Piepho <tpiepho@gmail.com> wrote:
> On Fri, May 21, 2021 at 12:55 AM Yiyuan guo <yguoaz@gmail.com> wrote:
> >
> > Thanks for your timely response.
> >
> > I am not familiar with the theorem. But any input satisfying the
> > condition below will
> > trigger a divide by zero at the first loop iteration:
> >
> > (given_numerator / given_denominator > max_numerator) || (1 +
> > given_numerator / given_denominator > max_denominator)
>
> I think the error can only occur when the loop exits on the 1st
> iteration, when d1 is still zero. In this case the prior convergent,
> n1/d1 = 1/0, does not really exist as this is the 1st iteration. The
> actual series of convergents generated will never have zero terms,
> because we stop at zero, so there will never be zero from the prior
> iteration as we would have stopped there.

This is my conclusion as well, but you beat me to it.
And below is exactly my understanding of what's going on.

> I think the prior version of the code, which did not consider
> semi-convergents, would have determined the 1st convergent, 314/1,
> exceeded the bounds and would return the prior one, 1/0, without
> generating an exception but also not a correct answer, since 1/0 isn't
> really part of the series, it's just an initial value to make the math
> that generates the series work (d2 = a * d1 + d0).
>
> With semi-convergents, this can actually get the correct answer. The
> best semi-convergent term is correctly found, (max_numerator - n0) /
> n1 = 255. Using this would return 255/1, which is in this case the
> best answer.
>
> But the "is semi-convergent better than prior convergent" test does
> not consider what I think is a special case of there being no prior
> convergent. In this case it should always select the semi-convergent.
>
> I think this handles it:
>
> if ((n2 > max_numerator) || (d2 > max_denominator)) {
> unsigned long t = (max_numerator - n0) / n1;
> if (!d1 || (t = min(t, max_denominator - d0) / d1)) ||
> 2u * t > a || (2u * t == a && d0 * dp > d1 * d)) {
> n1 = n0 + t * n1;
> d1 = d0 + t * d1;
> }
> break;
> }
>
> Above !d1 is the special case. I don't like that, but I'm not seeing
> a way to think about the problect that doesn't involve one.

Let me think about it.

> > I think such a condition is rather complex and may not be enforced by
> > all callers of this function.
> >
> > On Fri, May 21, 2021 at 3:42 PM Andy Shevchenko
> > <andy.shevchenko@gmail.com> wrote:
> > > On Friday, May 21, 2021, Andy Shevchenko <andy.shevchenko@gmail.com> wrote:
> > >> On Friday, May 21, 2021, Yiyuan guo <yguoaz@gmail.com> wrote:
> > >>>
> > >>> In the file lib/math/rational.c, the function
> > >>> rational_best_approximation has the following
> > >>> code:
> > >>>
> > >>> void rational_best_approximation(
> > >>> unsigned long given_numerator, unsigned long given_denominator,
> > >>> unsigned long max_numerator, unsigned long max_denominator,
> > >>> unsigned long *best_numerator, unsigned long *best_denominator) {
> > >>> ...
> > >>> if ((n2 > max_numerator) || (d2 > max_denominator)) {
> > >>> unsigned long t = min((max_numerator - n0) / n1,
> > >>> (max_denominator - d0) / d1);
> > >>> ...
> > >>> }
> > >>>
> > >>> d1 may be equal to zero when performing the division, leading to a
> > >>> divide by zero problem.
> > >>>
> > >>> One input to trigger the divide by zero bug is:
> > >>> rational_best_approximation(31415, 100, (1 << 8) - 1, (1 << 5) - 1, &n, &d)
> > >>
> > >> Have you read a theorem about this? TL;DR; as far as I can see the input data is not suitable for this function.
> > >
> > > I think we may add the proper check and saturate the output which in your case should be (255,1).

--
With Best Regards,
Andy Shevchenko