Timeline for Isoperimetric inequality on the Hamming cube
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| May 7, 2015 at 5:11 | comment | added | Yoav Kallus | Indeed your lower bound is also an upper bound when m is the correct residue modulo 4, thanks to the result of Kleitman in the reference cited. | |
| May 6, 2015 at 21:23 | comment | added | rajatsen91 | I think we have to use Harper's theorem somehow, because the upper-bound we are looking for on $|X|$, is same as the cardinality of a sphere of radius $m/4$ that is less than $2^{0.81m}$ | |
| May 6, 2015 at 21:12 | comment | added | Christian Remling | That's right: $4^{1/4}(4/3)^{3/4}>2^{0.8}$, so the $m/4$ balls become too large by Stirling's formula. I actually did the same evaluation for myself a while ago, but then got the interpretation backwards apparently. | |
| May 6, 2015 at 21:12 | history | edited | Robert Israel | CC BY-SA 3.0 |
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| May 6, 2015 at 21:06 | history | edited | Robert Israel | CC BY-SA 3.0 |
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| May 6, 2015 at 21:00 | comment | added | Robert Israel | No, really: radius $m/4$, diameter $m/2$. | |
| May 6, 2015 at 20:58 | history | edited | Robert Israel | CC BY-SA 3.0 |
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| May 6, 2015 at 20:45 | comment | added | Robert Israel | Oops, yes. So the conjecture is false. | |
| May 6, 2015 at 20:39 | comment | added | Christian Remling | If you replace $0.8$ by any larger number, the conjecture becomes weaker. | |
| May 6, 2015 at 20:27 | history | edited | Robert Israel | CC BY-SA 3.0 |
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| May 6, 2015 at 20:03 | history | answered | Robert Israel | CC BY-SA 3.0 |