Binomial of integer (mod n) returns integer

[sagetrac-scotts]

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 Cleanup in rings.arith and rings.integer #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

Component: basic arithmetic

Keywords: binomial coefficient modulo sd35

Stopgaps: todo

Author: Sam Scott, Marco Streng

Reviewer: Colton Pauderis, Johan Bosman, Marco Streng

Issue created by migration from https://trac.sagemath.org/ticket/12179

[sagetrac-scotts]

Attachment: 12179.patch.gz

[sagetrac-scotts]

comment:1

Attached patch should add this functionality with no impact on speed of standard integer/float/etc binomial calculation.

[sagetrac-cpauderis]

comment:2

Looks good to me: positive review.

[sagetrac-scotts]

Author: Sam Scott

[sagetrac-scotts]

comment:4

Attachment: 12179.2.patch.gz

Patch added to deal with the case where the input was a primitive python int.

Thanks to Colton (cpauderis) for catching that one.

[sagetrac-johanbosman]

comment:5

The line

 Test of integers modulo n:
}}{
should end in two colons (for docbuilding).  Furthermore, 
{{{
 return x.parent()(binomial(ZZ(x), m, **kwds)) 
}}}
is inefficient when computing modulo a small number n: the binomial coefficient over ZZ may be huge, whereas its reduction modulo n will be small.  It is therefore better to to the entire calculation modulo n.