Re: [Maxima-discuss] Mod function

[Stavros M. (Σ. Μ. <mac...@al...>]

Though float(mod(rationalize(x),rationalize(y))) is the easiest exact
calculation, mod(a*2^m,b*2^n) can easily be calculated without expanding
the powers, since it is equal to mod(a*2^(m-min(m,n)), b)*2^min(m,n), where
the modular exponentiation can be performed using cexpt (in rat3a).

The Maxima equivalent is:

modfloat(a,m,b,n) :=
   block([q:min(m,n), modulus: b, t],
 t: ratdisrep(a*rat(2)^(m-q)),
 2^q * (if t>=0 then t else t:t+b))


        -s

On Wed, Mar 30, 2016 at 11:13 AM, Raymond Toy <toy...@gm...> wrote:

> >>>>> "Robert" == Robert Dodier <rob...@gm...> writes:
>
>     Robert> On 2016-03-21, Stavros Macrakis <mac...@al...> wrote:
>     >> So we have a choice between making float mod exact (i.e., the
> correctly
>     >> rounded result of the infinite-precision result) or making it
> consistent
>     >> with floating division. And if we want to make it consistent with
>     >> floating-point division, it is outside the nominal range of mod. Is
> it
>     >> correct to cheatXXX round up in this case and define
> mod(6.3,2.1)=0.0?
>
>     Robert> I have spent some time thinking about this, and I've come to
> the
>     Robert> conclusion that we should make mod exact. This might be a
> little more
>     Robert> work, but it's much easier to understand, and therefore it's
> much
>     Robert> easier to figure out what to do in any given case.
>
>     Robert> As part of this, I'd like to implement a two-argument floor
> function
>     Robert> which will be consistent with mod, that is, such that x -
> floor(x,y)*y =
>     Robert> mod(x,y) (since it is possible to construct examples such that
>     Robert> x - floor(x/y)*y != mod(x,y) when mod(x,y) is exact).
>
> It seems to me the only way to make it exact is to convert all floats
> to the corresponding rational number.  How will that work with bfloats
> where the exponent could be so large that no rational will fit in the
> machine?
>
> I personally am happy with float behaving as they do today.  Well,
> except for x/y being computed as x*(1/y), which you've fixed.
>
> --
> Ray
>
>
>
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