#12179: Binomial of integer (mod n) returns integer
----------------------------------------------------+-----------------------
       Reporter:  scotts                            |         Owner:  
AlexGhitza                                 
           Type:  defect                            |        Status:  
needs_work                                 
       Priority:  major                             |     Milestone:  sage-5.7  
                                 
      Component:  basic arithmetic                  |    Resolution:            
                                 
       Keywords:  binomial coefficient modulo sd35  |   Work issues:            
                                 
Report Upstream:  N/A                               |     Reviewers:  Colton 
Pauderis, Johan Bosman, Marco Streng
        Authors:  Sam Scott                         |     Merged in:            
                                 
   Dependencies:  #11417                            |      Stopgaps:            
                                 
----------------------------------------------------+-----------------------
Description changed by mstreng:

Old description:

> {{{
> sage: R = Integers(6)
> sage: binomial(R(5), R(2))
> 10
> sage: binomial(R(5), R(2)).parent()
> Integer Ring
> }}}
>
> But {{{binomial(R(5), R(2))}}} is nonsense, both as an element of ZZ and
> as an element of R:
> {{{
> sage: binomial(5, 2)
> 10
> sage: binomial(11, 2)
> 55
> sage: binomial(5, 8)
> 0
> }}}
>
> On input {{{binomial(x, y)}}}, what Sage should do instead is the
> following:
>  * If the parent of y is Zmod(n) rather than ZZ, a `TypeError` should be
> raised.
>  * If factorial(y) is zero or a zero-divisor in the parent of x, a
> `ZeroDivisionError` should be raised. This is automatic if one computes
> binomial(x, y) simply as
>   {{{
>   x.parent(prod([x-k for k in range(y)]) / factorial(y))
>   }}}

New description:

 {{{
 sage: R = Integers(6)
 sage: binomial(R(5), R(2))
 10
 sage: binomial(R(5), R(2)).parent()
 Integer Ring
 }}}

 But {{{binomial(R(5), R(2))}}} is nonsense, both as an element of ZZ and
 as an element of R:
 {{{
 sage: binomial(5, 2)
 10
 sage: binomial(11, 2)
 55
 sage: binomial(5, 8)
 0
 }}}

 On input {{{binomial(x, y)}}}, what Sage should do instead is the
 following:
  * If the parent of y is Zmod(n) rather than ZZ, a `TypeError` should be
 raised.
  * If factorial(y) is zero or a zero-divisor in the parent of x, a
 `ZeroDivisionError` should be raised. This is automatic if one computes
 binomial(x, y) simply as
   {{{
   x.parent()(prod([x-k for k in range(y)]) / factorial(y))
   }}}

--

-- 
Ticket URL: <http://trac.sagemath.org/sage_trac/ticket/12179#comment:15>
Sage <http://www.sagemath.org>
Sage: Creating a Viable Open Source Alternative to Magma, Maple, Mathematica, 
and MATLAB

-- 
You received this message because you are subscribed to the Google Groups 
"sage-trac" group.
To unsubscribe from this group and stop receiving emails from it, send an email 
to [email protected].
To post to this group, send email to [email protected].
Visit this group at http://groups.google.com/group/sage-trac?hl=en.
For more options, visit https://groups.google.com/groups/opt_out.


Reply via email to