IntegerModRing representation
How can I generate the elements of IntegerModring(n) in symmetric representation? For example Integers(6) = {0, 1,-1,2, -2, 3}
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4 Answers
Well, if you want to change the representation of the objects, so when you type Integers(6)(5) it returns -1, I think that it is not implemented in sage.
However, if computing the symmetric representant of an element in Integers(n) can be done with the following code:
I use this function in some multimodular algorithms.
def lift_pm(p):
"""
Compute a representative of the element p mod n in [-n/2 , n/2]
INPUT:
- ``p`` an integer mod n
OUTPUT:
- An integer r such that -n/2 < r <= n/2
EXAMPLES::
sage: p = Mod(2,4)
sage: lift_pm(p)
2
sage: p = Mod(3,4)
sage: lift_pm(p)
-1
"""
r = p.lift()
if r*2 > p.modulus():
r -= p.modulus()
return r
Comments
Nowadays, the IntegerMod_abstract class has a .lift_centered() method to compute the "symmetric" or "centered" representative for a given residue class:
sage: [x.lift_centered() for x in Zmod(10)]
[0, 1, 2, 3, 4, 5, -4, -3, -2, -1]
Surprisingly in IntegerModrings Sage uses positive representations, but in polynomialrings over these it uses symmetric representation. So we can make the computations with the residues in the polynomialring.
Comments
See the file sage/rings/polynomial/multi_polynomial_libsingular.pyx,
in particular the comment at the beginning that Singular is used for
multipolynomial ring over GF(p) when p is prime,
and the _repr_ method at
You could use Python's list comprehension to make a list of the numbers you want:
sage: N = 5
sage: [(-1)^(n+1)*floor((n+1)/2) for n in range(N)]
[0, 1, -1, 2, -2]
sage: N = 6
sage: [(-1)^(n+1)*floor((n+1)/2) for n in range(N)]
[0, 1, -1, 2, -2, 3]
Is this the kind of thing you were looking for, or something else?
Comments
But it is not teh solition. I think, My question was'nt clear. The positive representation for mod 6 is {1,2,3,4,5,6}. The symmetric representation is {1,2,3,-2,-1}. The problem is'nt that, how can generate this remainder set, but the Integers(6) ring how can use it permanently.
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Asked: 2010-10-10 10:35:27 +0100
Seen: 3,359 times
Last updated: Jan 07
