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devel / comp.lang.apl / Nars floor of a big float

SubjectAuthor
* Nars floor of a big floatRosario19
+* Re: Nars floor of a big floatJames Cloos
|`- Re: Nars floor of a big floatRosario19
`- Re: Nars floor of a big floatBob Smith

1
Nars floor of a big float

<tl8etih5ldjfoufu2oig49ne05r2g60vak@4ax.com>

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From: Ros@invalid.invalid (Rosario19)
Newsgroups: comp.lang.apl
Subject: Nars floor of a big float
Date: Thu, 22 Feb 2024 11:34:06 +0100
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 by: Rosario19 - Thu, 22 Feb 2024 10:34 UTC

In Nars is right that the floor of log10 of 9999999999999.9v with
fpc=128 result 13?

30??10 ? 9999999999999.9v
13.000000000000000000000000000000

but here result 12

(4) -> floor( log10 ( 9999999999999.9))
(4) 12.0

Re: Nars floor of a big float

<m3bk88y04t.fsf@nitrogen.jhcloos.org>

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From: cloos@jhcloos.com (James Cloos)
Newsgroups: comp.lang.apl
Subject: Re: Nars floor of a big float
Date: Thu, 22 Feb 2024 10:44:18 -0500
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 by: James Cloos - Thu, 22 Feb 2024 15:44 UTC

>>>>> "R" == Rosario19 <Ros@invalid.invalid> writes:

R> In Nars is right that the floor of log10 of 9999999999999.9v with
R> fpc=128 result 13?

R> 30??10 ? 9999999999999.9v
R> 13.000000000000000000000000000000

I don't know nars, you are probably being hit by floating point
precision issues.

with arbitrary precision, the log10 of 9999999999999.9 is:

+12.999999999999995657055180967460008764615648097801793308999084...

Using ieee 64bit floats, you are likely to get 12.999999999999995,
but with 32bit floats you'll get 13.0.

Even with 64bit floats, rounding issues along the way could lead to a
result of 13 rather than 12.99999999...

Hense the difference in floor()'s output.

-JimC
--
James Cloos <cloos@jhcloos.com>
OpenPGP: https://jhcloos.com/0x997A9F17ED7DAEA6.asc

Re: Nars floor of a big float

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From: Ros@invalid.invalid (Rosario19)
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Subject: Re: Nars floor of a big float
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 by: Rosario19 - Thu, 22 Feb 2024 21:41 UTC

On Thu, 22 Feb 2024 10:44:18 -0500, James Cloos wrote:

>>>>>> "R" == Rosario19 <Ros@invalid.invalid> writes:
>
>R> In Nars is right that the floor of log10 of 9999999999999.9v with
>R> fpc=128 result 13?
>
>R> 30??10 ? 9999999999999.9v
>R> 13.000000000000000000000000000000
>
>I don't know nars, you are probably being hit by floating point
>precision issues.
>
>with arbitrary precision, the log10 of 9999999999999.9 is:
>
> +12.999999999999995657055180967460008764615648097801793308999084...
>
>Using ieee 64bit floats, you are likely to get 12.999999999999995,
>but with 32bit floats you'll get 13.0.
>
>Even with 64bit floats, rounding issues along the way could lead to a
>result of 13 rather than 12.99999999...
>
>Hense the difference in floor()'s output.
>
>-JimC

ok pheraps you are right for 64 and 32 bit numbers, but
9999999999999.9v with fpc of 128 it should mean it is one float of 128
bits

Re: Nars floor of a big float

<65d7e9bf$0$708$14726298@news.sunsite.dk>

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From: bsmith@sudleydeplacespam.com (Bob Smith)
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 by: Bob Smith - Fri, 23 Feb 2024 00:41 UTC

On 2/22/2024 5:34 AM, Rosario19 wrote:
> In Nars is right that the floor of log10 of 9999999999999.9v with
> fpc=128 result 13?
>
> 30??10 ? 9999999999999.9v
> 13.000000000000000000000000000000
>
> but here result 12
>
> (4) -> floor( log10 ( 9999999999999.9))
> (4) 12.0
>

The answer depends upon your setting for ⎕CT. If you are using the
default value of 3E¯15, then yes the floor of that multi-precision
number is 13. That's because the algorithm first looks for an integer
within Comparison Tolerance of the number. If it finds one, then that's
the answer, as in

a←10 ⍟ 9999999999999.9v
a+a×⎕CT
13.0000000000000347

--
_________________________________________
Bob Smith -- bsmith@sudleydeplacespam.com
http://www.sudleyplace.com - http://www.nars2000.org

To reply to me directly, delete "despam".

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