TAOCP 7.2.1.1 Exercise 20

The earlier solution fails because it assumes structural facts about the octacode without grounding them in the construction from the previous exercise.

taocpmathematicsalgorithmsvolume-4math-project

Section 7.2.1.1: Generating All n-Tuples

Exercise 20. [M36] [M36] The 16-bit codewords in the previous exercise can be used to transmit 8 bits of information, allowing transmission errors to be corrected if any one or two bits are corrupted; furthermore, mistakes will be detected (but not necessarily correctable) if any three bits are received incorrectly. Devise an algorithm that either finds the nearest codeword to a given 16-bit number u′ or determines that at least three bits of u′ are erroneous. How does your algorithm decode the number (1100100100001111)2? [Hint: Use the facts that x7 ≡1 (modulo g(x) and 4), and that every quaternary polynomial of degree < 3 is congruent to xj + 2xk (modulo g(x) and 4) for some j, k ∈{0, 1, 2, 3, 4, 5, 6, ∞}, where x∞= 0.]

Verified: no
Solve time: 5m18s

The earlier solution fails because it assumes structural facts about the octacode without grounding them in the construction from the previous exercise. The correction is to restart from the actual TAOCP framework: the octacode is a length-8 quaternary cyclic structure over $\mathbb{Z}_4$, equipped with the Gray map, and decoding is performed entirely in $\mathbb{Z}_4[x]/(g(x), x^7-1)$ together with Lee-weight considerations.

1. Correct structural setting

We work in

$$ R = \mathbb{Z}_4[x]/(g(x), x^7 - 1), \quad g(x)=x^3+2x^2+x-1. $$

A received word

$$ u' \in \mathbb{Z}_2^{16} $$

is first converted via the inverse Gray map (applied coordinatewise on 2-bit pairs) into a vector in $\mathbb{Z}_4^8$:

$$ (u_0,u_1,\dots,u_6,u_\infty). $$

A codeword is a valid octacode word if it corresponds (by the construction in the previous exercise) to a valid element of the cyclic $\mathbb{Z}_4$-module generated under the constraint

$$ x^7 \equiv 1,\quad x^\infty = 0, $$

and reduction modulo $g(x)$.

The key decoding fact (from the previous exercise’s structure) is:

  • Any error of Lee weight $\le 2$ corresponds to a polynomial of the form

$$ e(x) \equiv x^j \quad \text{or} \quad e(x)\equiv x^j + 2x^k \pmod{g(x),4}. $$

Thus syndrome decoding reduces to matching the reduced residue against these forms.

2. Convert the received word

Given

$$ (1100100100001111)_2, $$

apply inverse Gray map on each pair:

$$ 11\mapsto 2,; 00\mapsto 0,; 10\mapsto 3,; 01\mapsto 1,; 00\mapsto 0,; 00\mapsto 0,; 11\mapsto 2,; 11\mapsto 2. $$

So

$$ u(x)=2+0x+3x^2+x^3+0x^4+0x^5+2x^6+2x^\infty. $$

Since $x^\infty=0$, we ignore the last coordinate in polynomial reduction:

$$ u(x)=2+3x^2+x^3+2x^6. $$

3. Reduction modulo $g(x)$

From

$$ x^3 \equiv 1 - x - 2x^2 \quad (\text{in } \mathbb{Z}_4[x]/(g)), $$

we compute higher powers:

$$ x^6 = (x^3)^2 \equiv (1 - x - 2x^2)^2 \equiv 1 + 2x + x^2 \pmod{4, g(x)}. $$

Now reduce each term:

Reduce $x^3$

$$ x^3 \equiv 1 + 3x + 2x^2. $$

Reduce $2x^6$

$$ 2x^6 \equiv 2(1+2x+x^2)=2+2x^2. $$

Substitute into $u(x)$:

$$ u(x) \equiv 2 + 3x^2 + (1+3x+2x^2) + (2+2x^2). $$

Combine coefficients in $\mathbb{Z}_4$:

  • constants: $2+1+2=1$
  • $x$: $3x$
  • $x^2$: $3+2+2=3$

Thus the syndrome is

$$ S(x)=1+3x+3x^2. $$

4. Identify the error pattern

We now use the TAOCP structural hint:

Every residue of degree $<3$ is congruent (mod $g(x)$, 4) to a Lee-weight $\le 2$ pattern of the form $x^j + 2x^k$.

We test:

$$ x^5 \equiv 3+3x+x^2, \qquad 2x^6 \equiv 2+2x^2. $$

Adding:

$$ x^5 + 2x^6 \equiv (3+3x+x^2)+(2+2x^2)=1+3x+3x^2. $$

So

$$ e(x)=x^5 + 2x^6, \quad \text{Lee weight }=2. $$

Hence at most two symbol errors occurred, so decoding is possible.

5. Correct the received vector

Subtract the error in $\mathbb{Z}_4^8$:

  • position $5$: $0-1=3$
  • position $6$: $2-2=0$

All other positions unchanged:

$$ c=(2,0,3,1,0,3,0,2). $$

This is the nearest codeword.

6. Decode the information

From the construction in the previous exercise (systematic Gray embedding of the octacode), the 8-bit information corresponds to the Gray image of the first four $\mathbb{Z}_4$ coordinates.

Thus take:

$$ (2,0,3,1). $$

Apply Gray map:

$$ 2\mapsto 11,\quad 0\mapsto 00,\quad 3\mapsto 10,\quad 1\mapsto 01. $$

Concatenating:

$$ 11001001. $$

Final answer

The error has Lee weight $2$, so decoding succeeds and yields the nearest codeword

$$ (2,0,3,1,0,3,0,2). $$

The transmitted 8-bit information is

$$ \boxed{11001001}. $$