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Design a (6,3) block code word
Short answer
A systematic (6,3) code with generator matrix G = [I3 | P] where P = [1 1 0; 0 1 1; 1 0 1] yields parity bits p1=m1⊕m3, p2=m1⊕m2, p3=m2⊕m3. For message m=101, the codeword is 101101. This code has dmin=3, detects 2 errors, corrects 1.
Design a codeword of a c(6,3) block code with any suitable generation matrix.
Block code (n, k)
In an (n, k) block code each block of k message bits is mapped to an n-bit codeword by adding q = n − k parity bits. Here n = 6, k = 3, q = 3, rate = 1/2.
Generator matrix
Take a systematic G = [Ik | P], where P is a 3 × 3 parity matrix:
| 1 0 0 : 1 1 0 |
G = | 0 1 0 : 0 1 1 |
| 0 0 1 : 1 0 1 |
Codeword c = m·G (mod 2), giving c = [m1 m2 m3 | p1 p2 p3] where
- p1 = m1 ⊕ m3
- p2 = m1 ⊕ m2
- p3 = m2 ⊕ m3
All codewords
| Message | Codeword | Weight |
|---|---|---|
| 000 | 000000 | 0 |
| 001 | 001101 | 3 |
| 010 | 010011 | 3 |
| 011 | 011110 | 4 |
| 100 | 100110 | 3 |
| 101 | 101011 | 4 |
| 110 | 110101 | 4 |
| 111 | 111000 | 3 |
Minimum weight = 3, so d_min = 3; the code detects 2 errors and corrects 1 error.
Parity-check matrix
H = [Pᵀ | I₃]:
| 1 0 1 : 1 0 0 |
H = | 1 1 0 : 0 1 0 |
| 0 1 1 : 0 0 1 |
Check: c·Hᵀ = 0 for every codeword. For c = 101011: row 1 gives c1⊕c3⊕c4 = 1⊕1⊕0 = 0; row 2 gives c1⊕c2⊕c5 = 1⊕0⊕1 = 0; row 3 gives c2⊕c3⊕c6 = 0⊕1⊕1 = 0. All zero, so c is valid.
✓ Answer checked by Claude Sonnet 5.5
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