Timeline for Maximum number of tuples from $n$ numbers such that no pair is repeated [duplicate]
Current License: CC BY-SA 4.0
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Aug 19, 2020 at 17:25 | history | closed |
Emil Jeřábek CommunityBot |
Duplicate of maximum size of intersecting set families | |
Aug 19, 2020 at 16:18 | comment | added | მამუკა ჯიბლაძე | @BrendanMcKay ...which further links to mathoverflow.net/q/161159/41291 which in turn links to mathoverflow.net/q/160787/41291 | |
Aug 19, 2020 at 14:46 | comment | added | Brendan McKay | This question is a special case of mathoverflow.net/questions/175969/… . | |
Aug 19, 2020 at 14:36 | comment | added | vidyarthi | @MaxAlekseyev no, that was ok. I asked about my answer posted below. If not, could you give me an example for, say $n=8$ | |
Aug 19, 2020 at 14:34 | comment | added | Max Alekseyev | @vidyarthi: Yes. Each number appear in at most $\lfloor (n-1)/(k-1)\rfloor$ blocks. Then the number of blocks is at most $$\left\lfloor \frac{n}{k}\left\lfloor \frac{n-1}{k-1}\right\rfloor\right\rfloor.$$ You did achieve this bound in your example for $n=6$. | |
Aug 19, 2020 at 14:27 | comment | added | vidyarthi | @MaxAlekseyev so then, the maximum given by my answer is right? | |
Aug 19, 2020 at 14:25 | comment | added | Max Alekseyev | @vidyarthi: It does not because not every pair appears in your example. E.g., there is no pair $1,6$. | |
Aug 19, 2020 at 14:23 | comment | added | vidyarthi | @MaxAlekseyev but, $r=2$ does not satisfy $\lambda(n-1)=r(k-1)$ right? | |
Aug 19, 2020 at 14:12 | comment | added | Max Alekseyev | @vidyarthi: Block design represents a uniform construction. Like in your example for $n=6$, you have $r=2$. | |
Aug 19, 2020 at 13:12 | comment | added | vidyarthi | @MaxAlekseyev thanks. what is $r$ of the block design here? I dont put any condition on number of blocs containing any number. And, by the way, isnt my answer right? | |
Aug 19, 2020 at 12:54 | comment | added | Max Alekseyev | See en.wikipedia.org/wiki/Block_design | |
Aug 19, 2020 at 11:54 | history | edited | vidyarthi | CC BY-SA 4.0 |
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Aug 19, 2020 at 9:30 | history | asked | vidyarthi | CC BY-SA 4.0 |