def HammingCode(r,F):
r"""
Implements the Hamming codes.
The `r^{th}` Hamming code over `F=GF(q)` is an
`[n,k,d]` code with length `n=(q^r-1)/(q-1)`,
dimension `k=(q^r-1)/(q-1) - r` and minimum distance
`d=3`. The parity check matrix of a Hamming code has rows
consisting of all nonzero vectors of length r in its columns,
modulo a scalar factor so no parallel columns arise. A Hamming code
is a single error-correcting code.
INPUT:
- ``r`` - an integer 2
- ``F`` - a finite field.
OUTPUT: Returns the r-th q-ary Hamming code.
EXAMPLES::
sage: HammingCode(3,GF(2))
Linear code of length 7, dimension 4 over Finite Field of size 2
sage: C = HammingCode(3,GF(3)); C
Linear code of length 13, dimension 10 over Finite Field of size 3
sage: C.minimum_distance()
3
sage: C.minimum_distance(algorithm='gap') # long time, check d=3
3
sage: C = HammingCode(3,GF(4,'a')); C
Linear code of length 21, dimension 18 over Finite Field in a of size 2^2
"""
q = F.order()
n = (q**r-1)/(q-1)
k = n-r
MS = MatrixSpace(F,n,r)
X = ProjectiveSpace(r-1,F)
PFn = [list(p) for p in X.point_set(F).points(F)]
H = MS(PFn).transpose()
Cd = LinearCode(H)
# Hamming code always has distance 3, so we provide the distance.
return LinearCode(Cd.dual_code().gen_mat(), d=3)
def HammingCode(r, F):
r"""
Implements the Hamming codes.
The `r^{th}` Hamming code over `F=GF(q)` is an
`[n,k,d]` code with length `n=(q^r-1)/(q-1)`,
dimension `k=(q^r-1)/(q-1) - r` and minimum distance
`d=3`. The parity check matrix of a Hamming code has rows
consisting of all nonzero vectors of length r in its columns,
modulo a scalar factor so no parallel columns arise. A Hamming code
is a single error-correcting code.
INPUT:
- ``r`` - an integer 2
- ``F`` - a finite field.
OUTPUT: Returns the r-th q-ary Hamming code.
EXAMPLES::
sage: HammingCode(3,GF(2))
Linear code of length 7, dimension 4 over Finite Field of size 2
sage: C = HammingCode(3,GF(3)); C
Linear code of length 13, dimension 10 over Finite Field of size 3
sage: C.minimum_distance()
3
sage: C.minimum_distance(algorithm='gap') # long time, check d=3
3
sage: C = HammingCode(3,GF(4,'a')); C
Linear code of length 21, dimension 18 over Finite Field in a of size 2^2
"""
q = F.order()
n = (q**r - 1) / (q - 1)
k = n - r
MS = MatrixSpace(F, n, r)
X = ProjectiveSpace(r - 1, F)
PFn = [list(p) for p in X.point_set(F).points(F)]
H = MS(PFn).transpose()
Cd = LinearCode(H)
# Hamming code always has distance 3, so we provide the distance.
return LinearCode(Cd.dual_code().gen_mat(), d=3)
