91Ó°ÊÓ

The problem requires finding the minimum Hamming distance between codewords in the given set \( C \). Each codeword is a 4-bit binary vector.

The Hamming distance between two codewords is the number of positions at which the corresponding bits are different. For example, the Hamming distance between \( \begin{bmatrix} 0 \ 1 \ 0 \ 1 \end{bmatrix} \) and \( \begin{bmatrix} 1 \ 1 \ 0 \ 0 \end{bmatrix} \) is 3 because they differ in the first, third, and fourth positions.

We will compute the Hamming distance between every possible pair of codewords in the set \( C \). There are 4 codewords, so there are \( \binom{4}{2} = 6 \) pairs to compare. The pairs are: \( (c_1, c_2) \), \( (c_1, c_3) \), \( (c_1, c_4) \), \( (c_2, c_3) \), \( (c_2, c_4) \), \( (c_3, c_4) \).

1. \( c_1 = \begin{bmatrix} 0 \ 0 \ 1 \ 1 \end{bmatrix} \) and \( c_2 = \begin{bmatrix} 1 \ 1 \ 0 \ 0 \end{bmatrix} \): Distance = 42. \( c_1 = \begin{bmatrix} 0 \ 0 \ 1 \ 1 \end{bmatrix} \) and \( c_3 = \begin{bmatrix} 1 \ 0 \ 1 \ 0 \end{bmatrix} \): Distance = 23. \( c_1 = \begin{bmatrix} 0 \ 0 \ 1 \ 1 \end{bmatrix} \) and \( c_4 = \begin{bmatrix} 0 \ 1 \ 0 \ 1 \end{bmatrix} \): Distance = 24. \( c_2 = \begin{bmatrix} 1 \ 1 \ 0 \ 0 \end{bmatrix} \) and \( c_3 = \begin{bmatrix} 1 \ 0 \ 1 \ 0 \end{bmatrix} \): Distance = 35. \( c_2 = \begin{bmatrix} 1 \ 1 \ 0 \ 0 \end{bmatrix} \) and \( c_4 = \begin{bmatrix} 0 \ 1 \ 0 \ 1 \end{bmatrix} \): Distance = 26. \( c_3 = \begin{bmatrix} 1 \ 0 \ 1 \ 0 \end{bmatrix} \) and \( c_4 = \begin{bmatrix} 0 \ 1 \ 0 \ 1 \end{bmatrix} \): Distance = 4

From the calculated distances, the values are 4, 2, 2, 3, 2, and 4. The minimum Hamming distance among these values is 2.