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    How can I find the cyclic codes of length 6 and dimension 3
    I've been told I need to find the irreducible factorisation of x^6 - 1 in F_2[x]
    So I've got (x+1)(x-1)(x^2 + x +1)(x^2 - x + 1)
    How do I then form the generator matrix from this?
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    Okay so I've just realised that in F2[x]
    (x+1)(x-1)(x^2 + x +1)(x^2 - x + 1)
    becomes
    (x-1)^2 (x^2 + x + 1)^2
    So where would I go from here?
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    (Original post by Bruce Harrisface)
    Okay so I've just realised that in F2[x]
    (x+1)(x-1)(x^2 + x +1)(x^2 - x + 1)
    becomes
    (x-1)^2 (x^2 + x + 1)^2
    So where would I go from here?
    So, to get a code of dimension 3, you need to chose from these factors so that you get a polynomial of degree 6-3 = 3. (i.e. if you had a 7th degree poly then for dimension 3 you'd need to factors to get a poly of degree 7-3 = 4).

    So work out that poly in the form g_0 + g_1 x + g_2 x^2 + g_3 x^3

    Your generator matrix will then have the form

    \begin{pmatrix}g_0 & g_1 & g_2 & g_3 & 0 & 0 \\ 0 & g_0 & g_1 & g_2 & g_3 & 0 \\ 0 & 0 & g_0 & g_1 & g_2 & g_3\end{pmatrix}

    Disclaimer: I should perhaps say that I didn't really know anything about this topic, I've just googled the terms. But I think this is relatively straightforward.
 
 
 
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