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Let $A$ and $B$ be two binary $(n,M,d)$ codes. We define $a_i = \#\{(w_1,w_2) \in A^2:\:d(w_1,w_2) = i\}$, and same for $b_i$. If $a_i = b_i$ for all $i$, can one deduct that $A$ and $B$ are equivalent, i.e. equal up to permutation of positions and permutation of letters in a fixed position?

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Isn't this the same as mathoverflow.net/questions/102339/…, which has been answered? –  Gerry Myerson Jul 17 '12 at 0:12
    
@Gerry: looks like it is, but I don't seem to find a real example in any of the papers? –  Coder Jul 17 '12 at 18:22
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2 Answers

See Example 3.5.4 and the "other examples ... given in Section IV-4.9."

EDIT: Here are two inequivalent binary codes with the same distance enumerator. Code A consists of the codewords $a=00000000000$, $b=11110000000$, $c=11111111100$, $d=11000110011$; code R consists of the codewords $r=0000000000$, $s=1111000000$, $t=0111111111$, $u=1000111100$. Writing $xy$ for the Hamming distance between words $x$ and $y$, we have $ab=rs=4$, $bc=ru=5$, $cd=tu=6$, $ac=rt=9$, and $ad=bd=st=su=7$, so the distance enumerators are identical. But code A has the 7-7-4 triangle $adb$, while code R has the 7-7-6 triangle $tsu$, so there is no distance-preserving bijection between the two codes.

I have also posted this as an answer to the MO question. I'm sure there are shorter examples.

MORE EDIT: a much shorter example:

Code A consists of the codewords $a=000$, $b=100$, $c=011$, $d=111$; code R consists of the codewords $r=0000$, $s=1000$, $t=0100$, $u=0011$. Both codes have two pairs of words at each of the distances 1, 2, and 3; R has a 1-1-2 triangle, A doesn't.

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As an other example I proffer families of codes like Delsarte-Goethals codes. These are Grey images of codes that are linear over $\mathbf{Z}_4$, so they are not linear binary codes (even though they are binary).

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