Re: [9front] seq: fix infinite loop

[Eckard Brauer <eckard.brauer@gmx.de>]

Am Mon, 16 Aug 2021 16:45:04 -0500
schrieb Amavect <amavect@gmail.com>:

> On Sun, 15 Aug 2021 15:02:38 -0700
> Sean Hinchee <henesy.dev@gmail.com> wrote:
>  [...]  
> At this range, 2e16 and 20000000000000002 are represented by the same
> double floating point value.
> 
>  [...]  
> We can cover more cases.
> For example, your patch does not fix the following:
> seq 9007199254740992 9007199254740996
> The issue is with floating point precision.
> 
> A double has a 52 bit mantissa, meaning that it allows 2^52 different
> values before the exponent is incremented (thus, floating point.)
> A double has a 11 bit exponent, and 1 sign bit.
> Exponents range from −1022 to +1023, with exponents −1023 and
> +1024 reserved for special numbers.
> 
> At an exponent e, the mantissa represents a value between
> [2^e,2^(e+1)). The precision is range/buckets, so we can calculate
> the precision. At an exponent e and mantissa m, the precision is
> 2^(e-m) (2^(e+1)-2^e)/2^m = 2^e/2^m = 2^(e-m)
> 
> Between [1, 2), the exponent is 0.
> The precision is 2^(0-52) ≈ 0.0000000000000002220446
> Between [2^52, 2^53), the exponent is 52.
> The precision is 2^(52-52) = 1.
> 
> 2e16 is between [2^54, 2^55), so the exponent is 54.
> The precision is 2^(54-52) = 4.
> Halfway ties round to the even bucket.
> 
> This is demonstrated by the following.
> 
> # infinite loop
> seq 2e16 19999999999999998
> 
> # 2^53-1, halts
> seq 9007199254740991 9007199254740991
> 
> # 2^53, infinite loop due to halfway tie rounding down to even bucket.
> seq 9007199254740992 9007199254740992
> # halts, halfway tie rounds up to even bucket, 9007199254740996
> seq 9007199254740994 9007199254740994
> 
> # different due to rounding error in loss of precision
> seq 5000000000000000 1.1 5000000000000002
> seq 0 1.1 2
> 
> 
> I don't believe that there is an ideal resolution to rounding error in
> that last example (libmp would be too complex), but we can cover more
> cases if we check that the previous val value is different from the
> current val.
> The correct thing to do is to throw an error warning about precision.
> 
> Thanks,
> Amavect

IMHO the args could easily be double, the internal counter can never.
seq could be kind of that:

if (argc == 3)
	delta = argv[2];
else
	delta = 1;
if (!(argv[1] + delta > argv[1]))
	error(...);
for (ulong i=0; i<MAXULONG && val <= argv[argc], i++)
	fprintf("%whatever\n", val = (argv[1]+i*delta));

or do i have too much beer tonight...?

-- 
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