Denote by $A(n,d)$ the maximal size of a binary code of length $n$ with distance $d$.

How to show that $A(3k, 2k)=4$? From Plotkin bound: $$2k > \dfrac{3k}{2} \Rightarrow A(3k, 2k) \leq 4$$ But I don't know why here equality. I also tried to use Gilbert–Varshamov lower bound. But it doesn't help to obtain $A(3k, 2k) \geq 4$.


1 Answer 1


Here is a (linear) code attaining the bound: $$ 000\\011\\101\\110 $$ Each bit represents a block of $k$ equal bits.


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