#12179: Binomial of integer (mod n) returns integer
-------------------------------------+-------------------------------------
Reporter: scotts | Owner: AlexGhitza
Type: defect | Status: needs_work
Priority: major | Milestone: sage-6.4
Component: basic arithmetic | Resolution:
Keywords: binomial | Merged in:
coefficient modulo sd35 | Reviewers: Colton Pauderis,
Authors: Sam Scott, Marco | Johan Bosman, Marco Streng
Streng | Work issues:
Report Upstream: N/A | Commit:
Branch: | Stopgaps:
Dependencies: |
-------------------------------------+-------------------------------------
Changes (by mstreng):
* status: needs_review => needs_work
* milestone: sage-duplicate/invalid/wontfix => sage-6.4
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))
> }}}
>
> Apply:
>
> * [attachment:12179_new.patch]
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.
* (This seems to be fixed by #17852) 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))
}}}
Apply:
* [attachment:12179_new.patch]
--
Comment:
Replying to [comment:23 vdelecroix]:
> sage: R = Integers(21)
> sage: binomial(R(5), R(2))
> 10
This should be {{{TypeError}}}, because {{{binomial(x,y)}}} makes no sense
when y is an element of {{{R}}}. It only makes sense when y is an integer.
For example, binomial(5, 2) = 10, but binomial(5, 2+21) = 0.
So of the two points in the ticket description, the work in #17852 fixes
the second one, but the first one is still open.
--
Ticket URL: <http://trac.sagemath.org/ticket/12179#comment:24>
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