Question

The generator matrix of a \( (6,3) \) binary linear block code is given by 

The minimum Hamming distance \( d_{{min}} \) between codewords equals (answer in integer).

Hint:

To find the minimum Hamming distance for a linear block code, compute the pairwise Hamming distances between all codewords and identify the smallest distance.

Answer 1

Correct Answer: 3

Step 1: Form the codewords
The generator matrix \( G \) defines the codewords of the linear block code. The codewords are obtained by multiplying the message vector \( \mathbf{m} \) with the generator matrix \( G \). The message vector \( \mathbf{m} \) has 3 bits (since it is a \( (6,3) \) code), and can be represented as \( \mathbf{m} = [m_1 \, m_2 \, m_3] \). The corresponding codeword \( \mathbf{c} \) is given by: \[ \mathbf{c} = \mathbf{m} G. \] The possible message vectors \( \mathbf{m} \) are all 3-bit combinations, so we calculate the codewords for \( \mathbf{m} = [0 0 0] \), \( \mathbf{m} = [0 0 1] \), \( \mathbf{m} = [0 1 0] \), etc. 
Step 2: Compute the codewords 
For \( \mathbf{m} = [0 0 0] \): \[ \mathbf{c} = [0 0 0] G = [0 0 0 0 0 0] \] For \( \mathbf{m} = [0 0 1] \): \[ \mathbf{c} = [0 0 1] G = [0 0 1 1 1 0] \] For \( \mathbf{m} = [0 1 0] \): \[ \mathbf{c} = [0 1 0] G = [0 1 0 0 1 1] \] For \( \mathbf{m} = [0 1 1] \): \[ \mathbf{c} = [0 1 1] G = [0 1 1 1 0 1] \] For \( \mathbf{m} = [1 0 0] \): \[ \mathbf{c} = [1 0 0] G = [1 0 0 1 0 1] \] For \( \mathbf{m} = [1 0 1] \): \[ \mathbf{c} = [1 0 1] G = [1 0 1 1 1 0] \] For \( \mathbf{m} = [1 1 0] \): \[ \mathbf{c} = [1 1 0] G = [1 1 0 0 1 0] \] For \( \mathbf{m} = [1 1 1] \): \[ \mathbf{c} = [1 1 1] G = [1 1 1 1 0 1] \] Step 3: Compute the Hamming distance 
The Hamming distance between two codewords is the number of positions at which the corresponding symbols differ. To find \( d_{{min}} \), we need to calculate the pairwise Hamming distances between all codewords: The Hamming distance between codeword \( [0 0 0 0 0 0] \) and all other codewords is:
\( d([0 0 0 0 0 0], [0 0 1 1 1 0]) = 3 \)
\( d([0 0 0 0 0 0], [0 1 0 0 1 1]) = 3 \)
\( d([0 0 0 0 0 0], [0 1 1 1 0 1]) = 4 \)
\( d([0 0 0 0 0 0], [1 0 0 1 0 1]) = 3 \)
\( d([0 0 0 0 0 0], [1 0 1 1 1 0]) = 4 \)
\( d([0 0 0 0 0 0], [1 1 0 0 1 0]) = 4 \)
\( d([0 0 0 0 0 0], [1 1 1 1 0 1]) = 4 \)
Continuing this process for the remaining codewords, the minimum Hamming distance between any pair of codewords is found to be 3. Thus, the minimum Hamming distance \( d_{{min}} = 3 \).