Math-1338 · mathematics · Combinatorics

Math-1338

Perfect 2-error-correcting codes over arbitrary finite alphabets.

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Statement

Definitions and mathematical statement

Let $Q$ be a finite alphabet of cardinality $q\ge2$ and let $n$ be a positive integer. A code is a subset $C\subseteq Q^n$; linearity, a field structure on $Q$, and prime-power alphabet size are not assumed. The Hamming distance $d(x,y)$ counts coordinates in which two words differ. Write $B_2(c)=\{x\in Q^n:d(x,c)\le2\}$. A perfect two-error-correcting code is a code for which these balls, as $c$ ranges over $C$, partition $Q^n$: every word has exactly one codeword at distance at most two.

The nontrivial classification problem asks whether every such code with $|C|>2$ is equivalent to the ternary Golay code of length $11$. Equivalence permits a permutation of coordinates and, independently in each coordinate, a permutation of the alphabet. The ternary Golay code is the linear $[11,6,5]$ code over $\mathbb F_3$, namely a six-dimensional subspace with minimum nonzero Hamming weight five; the standard code of these parameters is unique up to this equivalence.

In particular, the predicted parameters are $q=3$, $n=11$, and $|C|=3^6$. A necessary counting identity is \[ |C|\left(1+n(q-1)+\binom n2(q-1)^2\right)=q^n. \] Satisfying this identity alone does not produce a code: unique decoding requires the actual disjoint covering property. The question ranges over arbitrary finite alphabets and arbitrary, possibly nonlinear, codes.

Short English statement

Must every nontrivial perfect code correcting exactly two Hamming errors be a ternary Golay code?

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