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Let $F=\{0,1\}$; $F$ is a field and let $x$, $y$, $z$ be words in $F^n$ that form an "equilateral triangle" that is: $d(x,y)=d(x,z)=d(y,z)=2t$. Show that there is exactly one word $v$ that belongs to $F^n$ such that $d(x,v)=d(y,v)=d(z,v)=t$.

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  • $\begingroup$ This is a part of Problem 20 in Chapter 1 of MacWilliams and Sloane, Theory of Error-Correcting Codes, North-Holland Elsevier 1978. $\endgroup$ – Dilip Sarwate Dec 16 '13 at 0:14
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look at the $k^{th}$ bit for each of $x,y,z$.

choose the $k^{th}$ bit for the equidistant point $v$ as follows:

if all three vertices have the same bit value, then take this value. if they differ, take the majority value.

do this for each of the $n$ bits

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    $\begingroup$ This does work (+1). Getting the uniqueness takes a bit more I think? $\endgroup$ – Jyrki Lahtonen Dec 15 '13 at 12:54
  • $\begingroup$ good point. however with this hint, perhaps OP will have a go at the uniqueness? $\endgroup$ – David Holden Dec 15 '13 at 12:57
  • $\begingroup$ Let's hope so! ${}$ $\endgroup$ – Jyrki Lahtonen Dec 15 '13 at 12:58

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