Timeline for Intersecting 4-sets
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Nov 6, 2012 at 13:55 | comment | added | Patricia Hersh | @Noam: aha! Now that you mention it, what I was doing also can be extended to this -- I was thinking of numbers in a $4\times 2$ grid with $2i-1,2i$ comprising row $i$. I was picking 2 of 4 rows in their entirety to get 6 sets, then getting a further set by taking one element from each row. But I guess you can get 8 further sets by choosing 1 from each row in any way meeting your parity constraint. I suppose this is equivalent to your cleaner description. | |
| Nov 6, 2012 at 6:05 | comment | added | Noam D. Elkies | Actually $14$ sets for $n=8$, from the extended Hamming code: label the 8 coordinates by $({\bf Z}/2{\bf Z})^3$ and consider all subsets consisting of four elements that sum to zero. | |
| Nov 6, 2012 at 3:46 | comment | added | Gerhard Paseman | Looks like I did not refresh soon enough. Hope the example works for you. Gerhard "Slow On The Uptake Tonight" Paseman, 2012.11.05 | |
| Nov 6, 2012 at 3:45 | comment | added | Gerhard Paseman | I believe there is an example of 7 sets on 8 points. The first three sets share points 1 and 2, the next two share 1 and 3, and the last share 2 and r. It should be a routine problem in combinatorial design theory. Gerhard "Ask Me About Binary Matrices" Paseman, 2012.11.05 | |
| Nov 6, 2012 at 3:29 | comment | added | Patricia Hersh | @Brendan: good point! | |
| Nov 6, 2012 at 3:25 | comment | added | Brendan McKay | @Patricia: We can get 7 sets in this case. Take the complements of the lines of a Fano geometry. | |
| Nov 6, 2012 at 3:25 | history | edited | Patricia Hersh | CC BY-SA 3.0 |
added 277 characters in body
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| Nov 6, 2012 at 3:17 | comment | added | Patricia Hersh | Here's a bigger example for $n=7$, which makes me wonder if for $n$ odd, maybe you just need to change the floor into a ceiling? Take $S_1=1345, S_2=2356, S_3=5617, S_4=4527, S_5=2713, S_6=1264$. | |
| Nov 6, 2012 at 2:43 | comment | added | Brendan McKay | @Patricia: Can you do it with $n$ even? That's the case I actually need and I added in the odd case as an afterthought. | |
| Nov 6, 2012 at 2:39 | comment | added | Patricia Hersh | Maybe this just means you want to tweak your hypotheses? | |
| Nov 6, 2012 at 2:35 | history | answered | Patricia Hersh | CC BY-SA 3.0 |