On Apr 16, 8:55 pm, Alec Mihailovs <alec.mihail...@gmail.com> wrote:
> Maybe, I still don't understand it right, but it seems as if you are
> saying, that the problem is to do a similar thing in cases when the
> solution is not unique, producing the list of all the solutions. That
> can be done similarly, just using the version of solve_left in
> 'decode1' producing the list of all solutions (plain solve_left gives
> only one solution even if there is more than 1), and then multiplying
> each of them by G, as in 'correct',
Here are the procedures for that -
def decode3(w,erasures=[]):
pos=[i for i in range(27) if not i in erasures]
G1=G.matrix_from_columns(pos)
return
G1.solve_left(vector(GF(3),w.list_from_positions(pos))),G1.kernel().basis_matrix()
def correct1(w,erasures=[],listed=False):
ans=decode3(w,erasures)
ans0,ans1=ans[0]*G,ans[1]*G
if not listed:
return ans0,ans1
else:
return [ans0+c*ans1 for c in VectorSpace(GF(3),ans1.nrows())]
For example,
V=VectorSpace(GF(3),10)
m=V.random_element(); m
(2, 1, 1, 0, 0, 0, 0, 1, 1, 0)
w=m*G; w
(1, 2, 1, 2, 2, 2, 0, 1, 1, 2, 1, 1, 2, 2, 1, 1, 2, 2, 0, 0,
1, 1, 1, 1, 2, 2, 2)
correct1(w,range(14),True)
[(1, 2, 1, 2, 2, 2, 0, 1, 1, 2, 1, 1, 2, 2, 1, 1, 2, 2, 0, 0, 1, 1, 1,
1, 2, 2, 2), (2, 0, 1, 2, 1, 1, 1, 2, 2, 0, 0, 0, 1, 1, 1, 1, 2, 2, 0,
0, 1, 1, 1, 1, 2, 2, 2), (0, 1, 1, 2, 0, 0, 2, 0, 0, 1, 2, 2, 0, 0, 1,
1, 2, 2, 0, 0, 1, 1, 1, 1, 2, 2, 2)]
correct1(w,range(14))
((1, 2, 1, 2, 2, 2, 0, 1, 1, 2, 1, 1, 2, 2, 1, 1, 2, 2, 0, 0, 1, 1, 1,
1, 2, 2, 2), [1 1 0 0 2 2 1 1 1 1 2 2 2 2 0 0 0 0 0 0 0 0 0 0 0 0 0])
correct1(w,[],True)
[(1, 2, 1, 2, 2, 2, 0, 1, 1, 2, 1, 1, 2, 2, 1, 1, 2, 2, 0, 0, 1, 1, 1,
1, 2, 2, 2)]
Alec
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