The codes in details

The code C₁

It is a (28,24,5) code over the field with 256 elements.

If x is a 24 byte word, the code word of C₁ that corresponds to it is the 28 bytes word equal to (x,x R₁^t), where the matrix R₁ of dimension (4,24)   is defined by

R1 =

[ [ [ [

a6 a192 a142 a159 a99 a88 a104 a144 a55 a180 a174 a101 a111 a118 a169 a107 a132 a25 a167 a239 a168 a188 a111 a9 a45 a108 a248 a131 a64 a221 a100 a235 a147 a45 a198 a21 a228 a186 a231 a56 a68 a81 a46 a32 a60 a225 a134 a226 a50 a52 a59 a132 a186 a81 a128 a126 a133 a32 a213 a195 a43 a198 a194 a13 a167 a167 a252 a61 a3 a12 a66 a144 a42 a136 a153 a93 a82 a98 a138 a49 a174 a168 a95 a105 a112 a163 a101 a126 a19 a161 a233 a162 a182 a105 a3 a226

] ] ] ]

The code C₂

It is a (32,28,5) code over the field with 256 elements.

If x is a 28 byte word, the code word of C₂ that corresponds to it is the 32 byte word equal to (x,x R₂^T), where the (4,28) matrix R₂ is : 

Correction of one error in C₂

The code C₂ has a minimal distance of 5. In theory, it can correct up to 2 errors, or detect 4 errors. It can also correct 1 error and detect 3, which will be used here.

Let set H₂=(R₂|I), where I is the 4× 4 identity matrix , and for every vector v of length 32, S₂(v) = H₂ v^T.
One easily checks that every code word y of C₂ satisfies S₂(y)=0 (remembering that F₂₅₆ is a field of characteristic 2). If y has been modified at one position, for example z=y+e, where e is a vector of length 32 with only one non zero coordinate. Then S₂(z)=S₂(y)+S₂(e)=S₂(e). Thus if S₂(z) is proportional to the i-th column of H₂, then the error is located at the i-th position and its value is the proportion coefficient with S₂(z).

For example, if one gets one can notice that S₂(z) is equal to a times the first column of R₂ (and also of H₂ too). One can deduce that y=z+(a,0,...,0) is in C₂. Adding a to the first coordinate of z, we just fixed one error.

Correction of four erasures in C₁

The code C₁ has a minimal distance of 5. If can correct 4 erasures. One erasur is an error which position is known.

H₁=(R₁|I), where I is the 4× 4 identity matrix , and for every vector v of length 28, S₁(v) = H₁ v^tr. We suppose that we know a vector z of length 28 with coefficients 3, 8, 16 et 25 (and no other) are possibly erroneous. We consider the 4× 4 matrix formed by the corresponding columns of H₁ Let e=(e₀,e₁,...,e₂₇) be the vector of length 28 such that e_i=0 if iÏ{3,8,16,25} and We have S₁(e)=S₁(z) and therefore y=z+e is an element of C₁. We just corrected 4 erasures at the positions 3, 8, 16 et 25.