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Significance of dmin; (7,4) cyclic code for 1101 with g(x) = 1 + x + x²

Short answer

Minimum Hamming distance d_min determines error detection (up to d_min-1) and correction (up to floor((d_min-1)/2)) capability. For (7,4) cyclic code with message 1101 and generator 1+x+x^3 (corrected), non-systematic vector is 1010001, systematic is 0001101.

What is the significance of (dmin) minimum hamming distance? Determine the systematic and non-systematic code vector for a (7,4) cyclic code for message vector (m₀, m₁, m₂, m₃) = (1101) for a given generator polynomial g(x) = 1 + x + x².

Significance of dmind_{min}

The minimum Hamming distance dmind_{min} is the smallest number of positions in which any two codewords differ (for a linear code, the smallest non-zero codeword weight). It fixes the power of the code:

  • Detects up to ss errors if dmin≥s+1d_{min}\ge s+1.
  • Corrects up to tt errors if dmin≥2t+1d_{min}\ge 2t+1, i.e. t=⌊(dmin−1)/2⌋t=\lfloor(d_{min}-1)/2\rfloor.

For a (7,4) Hamming code dmin=3d_{min}=3: detects 2, corrects 1 error.

Note on the generator polynomial

A generator for a (7,4) cyclic code must have degree n−k=3n-k=3 and divide x7+1=(1+x)(1+x+x3)(1+x2+x3)x^7+1=(1+x)(1+x+x^3)(1+x^2+x^3). The printed g(x)=1+x+x2g(x)=1+x+x^2 has degree 2 and does not divide x7+1x^7+1, so it is taken as a misprint of the standard g(x)=1+x+x3g(x)=1+x+x^3.

Message (m0m1m2m3)=1101(m_0m_1m_2m_3)=1101, ascending powers:

m(x)=1+x+x3m(x)=1+x+x^3

Non-systematic code vector

c(x)=m(x)g(x)=(1+x+x3)(1+x+x3)=1+x2+x6(cross terms cancel in mod 2: (a+b+c)2=a2+b2+c2)\begin{aligned} c(x)&=m(x)g(x)=(1+x+x^3)(1+x+x^3)\\ &=1+x^2+x^6\quad(\text{cross terms cancel in mod 2: }(a+b+c)^2=a^2+b^2+c^2) \end{aligned}

(c0c1…c6)(c_0c_1\dots c_6) = 1010001

Systematic code vector

x3m(x)=x3+x4+x6x^3m(x)=x^3+x^4+x^6

Divide by g(x)=1+x+x3g(x)=1+x+x^3:

              x^3
          -----------------------
x^3+x+1 ) x^6 + x^4 + x^3
          x^6 + x^4 + x^3
          ---------------
                          0

Remainder b(x)=0b(x)=0, parity bits (b0b1b2)=000(b_0b_1b_2)=000.

c(x)=b(x)+x3m(x)=x3+x4+x6c(x)=b(x)+x^3m(x)=x^3+x^4+x^6

(b0b1b2 m0m1m2m3)(b_0b_1b_2\,m_0m_1m_2m_3) = 000 1101

This happens because m(x)m(x) itself equals g(x)g(x), so x3m(x)x^3m(x) is already a multiple of g(x)g(x).

FormCode vectorWeight
Non-systematic10100013
Systematic00011013

Both have weight 3, equal to dmind_{min} of the code.

✓ Answer checked by Claude Opus 5.5 (high effort)

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