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Jun 14, 2017 at 12:30 vote accept Fedor Petrov
Jun 14, 2017 at 11:44 answer added Peter Mueller timeline score: 9
Jun 12, 2017 at 13:06 comment added Dima Pasechnik @FedorPetrov - oops, right. I was thinking about upper bounds too much...
Jun 12, 2017 at 10:39 comment added Shahrooz Gilbert bound is true if the code is not linear, but Varshamov bound is true for linear codes. also, binary cyclic codes are better than the Gilbert- Varshamov bound for lengths up to 1023. For the case $q=6$, we must analyse $Z_6$ codes at the first step.
Jun 12, 2017 at 10:37 comment added Fedor Petrov @DimaPasechnik It is lower bound: it claims that somehow large code exists. Moreover, even a linear code of such size exists. Of course, not linear code also exists. Your question would be reasonable for upper bounds on the size of a code.
Jun 12, 2017 at 10:28 comment added Dima Pasechnik @FedorPetrov Varshamov bound is a bound on the code size in terms of $d$, $n$ and $q$, which holds for linear codes. Does it hold for non-linear codes?
Jun 12, 2017 at 9:05 comment added Fedor Petrov @Dima what do you mean? The claim "linear code exists" implies a claim "code exists".
Jun 12, 2017 at 8:39 comment added Dima Pasechnik In my answer below I remark that perhaps a more natural question is whether Varshamov's bound holds for non-linear codes.
Jun 12, 2017 at 8:36 answer added Dima Pasechnik timeline score: 2
Jun 12, 2017 at 6:36 history edited Fedor Petrov CC BY-SA 3.0
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Jun 12, 2017 at 1:35 answer added kodlu timeline score: 2
Jun 11, 2017 at 20:28 comment added Fedor Petrov @WillSawin Any set of cardinality $q$. We need no other structure for defining Hamming metric on $\Sigma^n$.
Jun 11, 2017 at 19:09 comment added Will Sawin What is $\Sigma$? A cyclic group? Another abelian group?
Jun 11, 2017 at 13:23 history edited Fedor Petrov CC BY-SA 3.0
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Jun 11, 2017 at 12:32 history asked Fedor Petrov CC BY-SA 3.0