Question

 Factor x5 - 1 over equation. Give the generator polynomials of all cyclic codes of length five over equation. 

09 Jan 2026
Answer :
Word Count : 230
For (\mathcal C_1) the generator matrix is [ G_1=\begin{bmatrix} 1&0&0&1\ 0&1&0&0\ 0&0&1&1 \end{bmatrix}. ] All codewords are obtained as (\mathbf{u}=a(1,0,0,1)+b(0,1,0,0)+c(0,0,1,1)) with (a,b,c\in{0,1}). Thus [ \mathbf{u}=(a,b,c,a+c). ] Listing the nonzero choices: [ \begin{aligned} (1,0,0)&\to(1,0,0,1),; w=2,\ (0,1,0)&\to(0,1,0,0),; w=1,\ (0,0,1)&\to(0,0,1,1),; w=2,\ (1,1,0)&\to(1,1,0,1),; w=3,\ (1,0,1)&\to(1,0,1,0),; w=2,\ (0,1,1)&\to(0,1,1,1),; w=3,\ (1,1,1)&\to(1,1,1,0),; w=3. \end{aligned} ] The smallest nonzero weight is (1). Hence [ d(\mathcal C_1)=1. ] For (\mathcal C_2) the generator matrix is [ G_2=\begin{bmatrix} 1&0&0&1\ 0&1&1&0 \end{bmatrix}. ] A general codeword is [ \mathbf{v}=a(1,0,0,1)+b(0,1,1,0)=(a,b,b,a).
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