Timeline for Set system with different differences
Current License: CC BY-SA 3.0
10 events
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Dec 12, 2015 at 6:39 | history | edited | Daniel Soltész |
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Jul 24, 2014 at 16:03 | history | edited | Daniel Soltész |
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Jun 3, 2014 at 22:28 | answer | added | t3suji | timeline score: 6 | |
Jun 3, 2014 at 22:08 | answer | added | Kellen Myers | timeline score: 4 | |
Jun 3, 2014 at 20:33 | comment | added | Daniel Soltész | @eins6180 A little better but still simple upper bound would be that all the $2 \binom{|\mathcal{A}|}{2}$ differences are different so there can not be more than $2^n$ of them which means that $|\mathcal{A}|$ can not be asymptotically bigger than $2^{n/2}$. | |
Jun 3, 2014 at 20:26 | comment | added | eins6180 | Here is a simple and probably much too large upper bound: Any maximal antichain in the lattice of all subsets of $[n]$ together with two other subsets doesn't work. So the family $\mathcal{A}$ cannot be bigger than ${n \choose [n/2]} + 1$. | |
Jun 3, 2014 at 20:20 | comment | added | Daniel Soltész | @Seva Nice construction, you can also add [1,m] and [m+1,n] to obtain a system of $\lfloor n/2 \rfloor +2$ elements. | |
Jun 3, 2014 at 19:52 | comment | added | Seva | Constructing $m:=\lfloor n/2\rfloor$ sets is easy: assuming that the ground set is $[1,n]$, take $A_i:=\{i\}\cup[m+1,n]\setminus\{m+i\}\ (1\le i\le m)$. | |
Jun 3, 2014 at 19:37 | history | edited | Daniel Soltész | CC BY-SA 3.0 |
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Jun 3, 2014 at 18:04 | history | asked | Daniel Soltész | CC BY-SA 3.0 |