For an integer $q\ge2$, let $A_q(n,\{2,4\})$ be the largest cardinality of an unrestricted $q$-ary code of length $n$ in which any two distinct words have Hamming distance $2$ or $4$. We prove that \[ A_q(n,\{2,4\})=1+\binom n2 \qquad(q\in\{3,4\},\ n\ge6), \] thereby settling the two remaining alphabet sizes in the conjecture of Boyvalenkov, Delchev, Zinoviev, and Zinoviev; the binary case was settled previously. The quaternary result implies the ternary result by alphabet inclusion. For $n\ge16$, the quaternary upper bound is proved combinatorially. After normalizing one word to zero, the weight-four supports form a $2$-intersecting family; exact support intersections force symbol synchronization and restrict support multiplicities. We combine the nontrivial intersection theorem with matching arguments and a three-coordinate projection lemma. The finite lengths $6\le n\le13$ are handled by a complete Hamming-isometry reduction to exact clique decisions, and $n=14,15$ by a smaller exact weighted-clique relaxation. All finite decisions are reproduced by two standalone integer-only C++17 verifiers provided as supplementary source files.

翻译：暂无翻译