Fixed Hamming distance property of binary deletion correcting codes - MathOverflow most recent 30 from http://mathoverflow.net2013-05-20T06:31:50Zhttp://mathoverflow.net/feeds/question/64476http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/64476/fixed-hamming-distance-property-of-binary-deletion-correcting-codesFixed Hamming distance property of binary deletion correcting codesFarzaneh2011-05-10T07:35:49Z2011-05-10T13:08:58Z
<p>Let $x=(x_1x_2...x_n)$ be a binary sequence of length $n$. The Varshamov-Tenengolts code $VT_0(n)$ consists of all binary vectors $(x_1, . . . , x_n)$ satisfying $\Sigma_{i=1}^n i*x_i \equiv0 \pmod {n+1} $.</p>
<p>Prove that $\forall$ $x,y \in VT_0(n)$ which has equal hamming weight the Hamming distance between $x$ and $y$ is exactly 4. For a binary vectore $x$ the hamming weight is $w$ if $\Sigma_{i=1}^n x_i= w $. </p>
http://mathoverflow.net/questions/64476/fixed-hamming-distance-property-of-binary-deletion-correcting-codes/64496#64496Answer by Gerry Myerson for Fixed Hamming distance property of binary deletion correcting codesGerry Myerson2011-05-10T13:08:58Z2011-05-10T13:08:58Z<p>Take $n=11$. 11000000100 and 00111000000 are both in the code and they both have Hamming weight 3 but the Hamming distance between them is 6. </p>