I have been thinking of several properties which could distinguish binary codes from non binary codes. However I'm not sure I'm right, so please tell me your opinion:

  • The $(u|u+v)$ construction of Plotkin (1960): Given two codes $C_1$, $C_2$ with parameters $(n,M_i,d_i)$ we construct a new code $C_3$, containing all words that start with $u \in C_1$ and then comes the word $u+v$ where $v \in C_2$ and addition is done by coordinates (modulus the number $q$ of letters we use). If these are binary codes ($q=2$), we get a code with parameters $(2n,M_1M_2,\min\{2d_1,d_2\})$. Does it hold if $q \neq 2$? I tend to think it doesn't but cannot find a counter example.
  • Given a $q$-ary code $C$, we can list the differences between its different words - listing that there are $n_1$ pairs of words with $d=1$, $n_2$ words that differ on two characters and so on (Hamming distance). Given two codes with the same list, does it imply they are equivalent? It is not very difficult to find a counter example if we use ternary codes, but does it hold for binary codes?
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    $\begingroup$ For binary codes, properties of the Plotkin construction are discussed in Theorem 33, Chapter 2, of the MacWilliams and Sloane book Theory of Error-Correcting Codes, North-Holland 1978. Does the proof, which relies on the property $$\text{wt}(x+y) \geq \text{wt}(x) - \text{wt}(y)$$ of binary vectors, break down for nonbinary codes? $\endgroup$ – Dilip Sarwate Jul 15 '12 at 21:29
  • $\begingroup$ I know how to proof it holds for binary codes (using the fact that $d(x,y) = w(x+y)$ where $w(z)$ is the number of $1$'s in $z$). I don't know if you can modify the proof to include nonbinary codes. $\endgroup$ – Rob Jul 15 '12 at 21:42
  • $\begingroup$ Doesn't the statement that $d(x,y) = w(x-y)$ hold for both binary and nonbinary vectors since $x-y = x+y$ for binary vectors? But the proof of the property of the Plotkin construction needs a slightly different result (see my previous comment). $\endgroup$ – Dilip Sarwate Jul 15 '12 at 21:53
  • $\begingroup$ I'm sorry, I couldn't follow your train of thought. Could you please explain it once again? (What is $wt$?) $\endgroup$ – Rob Jul 15 '12 at 21:57
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    $\begingroup$ Good. Now write up your answer and post it. (Such actions are encouraged on this site). You can even accept your own answer as the best. $\endgroup$ – Dilip Sarwate Jul 15 '12 at 22:27

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