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Hadamard code conditions and 8×8 construction

Short answer

Hadamard code conditions: square N×N matrix, first row all zeros, every other row has N/2 zeros and N/2 ones, and any two rows differ in N/2 positions. Sylvester construction H2n=[Hn Hn; Hn Hn'] gives H8 with binary rows 00000000, 01010101, 00110011, 01100110, 00001111, 01011010, 00111100, 01101001.

Write the steps to satisfy condition for Hadamard code. Construct 8 by 8 Hadamard code that satisfies above conditions.

Conditions (steps) for a Hadamard code

A Hadamard (Walsh) code is a set of N codewords of length N (N = 2ᵏ) taken from the rows of a Hadamard matrix. The conditions it must satisfy are:

  1. Square matrix: N rows and N columns, entries only 0/1 (or +1/−1).
  2. First row (and first column) all zeros (all +1), i.e. the matrix is normalized.
  3. Equal weight: every row except the first has exactly N/2 zeros and N/2 ones.
  4. Orthogonality: any two different rows agree in exactly N/2 places and differ in N/2 places. In ±1 form, the inner product of two different rows is 0, so H·Hᵀ = N·I.

Steps to construct (Sylvester method) – these steps guarantee the above conditions:

  1. Start with H₁ = [0].
  2. Form H₂ₙ from Hₙ by placing Hₙ in the top-left, top-right and bottom-left blocks and its complement Hₙ′ in the bottom-right block:
 H2n = | Hn  Hn  |
       | Hn  Hn' |
  1. Repeat until the required size N is reached (1 → 2 → 4 → 8).

Construction of 8 × 8 Hadamard code

 H1 = [0]

 H2 = | H1  H1 |  = | 0 0 |
      | H1  H1'|    | 0 1 |

 H4 = | H2  H2 |  = | 0 0 0 0 |
      | H2  H2'|    | 0 1 0 1 |
                    | 0 0 1 1 |
                    | 0 1 1 0 |

 H8 = | H4  H4 |      (H' = complement of H)
      | H4  H4'|

Writing H8 in full:

         c1 c2 c3 c4 c5 c6 c7 c8
 W0:      0  0  0  0  0  0  0  0
 W1:      0  1  0  1  0  1  0  1
 W2:      0  0  1  1  0  0  1  1
 W3:      0  1  1  0  0  1  1  0
 W4:      0  0  0  0  1  1  1  1
 W5:      0  1  0  1  1  0  1  0
 W6:      0  0  1  1  1  1  0  0
 W7:      0  1  1  0  1  0  0  1

Verification:

  • Row W0 is all zeros; every other row has four 0s and four 1s.
  • Example: W1 = 01010101 and W2 = 00110011 agree in positions 1, 4, 5, 8 and differ in 2, 3, 6, 7, i.e. 4 agreements and 4 disagreements, so they are orthogonal. The same holds for every pair (checked for all 28 pairs).
  • Minimum distance between codewords = N/2 = 4.

These 8 orthogonal codewords are used as Walsh spreading codes to separate channels in CDMA (IS-95 uses the 64 × 64 version).

✓ Answer checked by Claude Opus 5.5 (high effort)

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