An $(n,M,d)$-code $C$ is an $(n,M,d-1)$-code

Question

Let $C$ be an $(n,M,d)$-code, where $n$ is its length, $M$ is its size and $d$ is its minimum distance.

*My question is: Is an $(n,M,d)$-code $C$ also an $(n,M,d-1)$-code? *

It seems that this doesn't hold. For example let $A:=\Bbb Z_2$ and consider the code $$C:=\{e_1:=(0,0,0),e_2:=(0,1,1),e_3:=(1,1,0)\}\subseteq A^3.$$ Then, $d(e_1,e_2)=2,d(e_1,e_3)=2,d(e_2,e_3)=2$. So, $d(C)=2$. Therefore, $C$ is a binary $(3,3,2)$-code.

So $C$ can't be an $(3,3,1)$-code, since there are not codewords in $C$ that differ in 1 digit.

Could you please give me help?

Answer 1

Why would you think this is true?

By definition, the minimal distance of your code is $d$. Thus the same code will never have minimal distance $d-1$.

Literally every code gives you a counterexample!