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May 7, 2015 at 4:46 vote accept rajatsen91
May 7, 2015 at 3:36 answer added Yoav Kallus timeline score: 6
May 6, 2015 at 21:11 comment added rajatsen91 The $1/2$ is not a typo I think. But it is trivial to prove this for $\delta = 1/10$ . The cited reference is a 1966 paper by Kleitman, but I am having trouble using the results there to prove this.
May 6, 2015 at 21:03 comment added Benoît Kloeckner @rajatsen91: did you have a look at the cited reference (Kleitman 1966)? Moreover, reading below the lemma it is written that other factors than $1/2$ in the distance lower bound would do for that paper, and the choice cited is $1/10$. Given Robert Israel's answer, I would bet that the $1/2$ in the paper is a typo.
May 6, 2015 at 20:36 comment added Christian Remling @YoavKallus: I think this gets you to about (not quite) $D=m/3$, but it doesn't seem good enough for the full claim.
May 6, 2015 at 20:05 comment added rajatsen91 @BenoîtKloeckner : I found this stated without a proof in a paper. math.washington.edu/~rothvoss/publications/…
May 6, 2015 at 20:03 answer added Robert Israel timeline score: 5
May 6, 2015 at 19:25 comment added Yoav Kallus Did you try the following: Let $D$ be the diameter of $X$, i.e., $\max_{x,y\in X} ||x-y||$, then $X$ is in the intersection of two Hamming balls of radius $D$ with centers a distance $D$ apart. Does this give you a tight enough bound?
May 6, 2015 at 19:04 comment added Benoît Kloeckner Why "isoperimetric inequality"? This looks like something that should follow from the concentration of measure phenomenon. Why are you interested in this?
May 6, 2015 at 17:58 history edited GH from MO CC BY-SA 3.0
fixed spelling
May 6, 2015 at 15:55 history asked rajatsen91 CC BY-SA 3.0