Timeline for Intersecting 4-sets
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
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| Nov 6, 2012 at 17:38 | comment | added | Gerhard Paseman | Actually, maybe no more typos. The style leads me down the opposite direction. Gerhard "In Eye Of The Beholder" Paseman, 2012.11.06 | |
| Nov 6, 2012 at 17:35 | comment | added | Gerhard Paseman | You're welcome. Gerhard "At Least Two Are Left" Paseman, 2012.11.06 | |
| Nov 6, 2012 at 17:28 | comment | added | Tony Huynh | @Gerhard. Thanks! Typos corrected. | |
| Nov 6, 2012 at 17:27 | history | edited | Tony Huynh | CC BY-SA 3.0 |
edited body; edited body
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| Nov 6, 2012 at 17:06 | comment | added | Gerhard Paseman | It looks like a few typos are present, Tony. =1 should be =2, in should be not in, and 345 should be 561. Otherwise it looks good. Gerhard "Of Course I've Missed Some" Paseman, 2012.11.06 | |
| Nov 6, 2012 at 16:47 | history | edited | Tony Huynh | CC BY-SA 3.0 |
included proof of Brendan McKay; added 14 characters in body
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| Nov 6, 2012 at 16:36 | comment | added | Gerhard Paseman | Brendan, sorry for the confusion. I am taking Tony's assumption of more than N sets which leads to at least k 4-sets sharing d, which is one of 2k points, and then removing d and focussing on the system of (1-intersecting) 3-sets and their union. They either all share a point c, making their union too big, or there are at most 7 of them, with their union being size 7. From there I should have said k at most 7, but I was thinking 2k at most 7+1. This fragment along with Tony's analysis and one for the 14 set system appeals to me. Gerhard "Noam's Post Seems OK Too" Paseman, 2012.11.06 | |
| Nov 6, 2012 at 9:23 | comment | added | Brendan McKay | Continuing: Actually Gordon and Noam's example of 14 sets in 8 points also beats $N$ for $n=10$, i.e. $k=5$. So only $k=6,7$ remain. | |
| Nov 6, 2012 at 9:17 | comment | added | Brendan McKay | Consider 3-sets that pairwise intersect in 1 point. If three of the 3-sets form a triangle, there are no more than 7 3-sets and the only way to do that is a Fano plane (which has 7 points). We can add isolated points as well. This beats $k-1$ 3-sets in $2k-1$ points for $k\le 7$ and equals it for $k=8$. So I think this argument shows that $N$ is optimal for $k\ge 8$. I suspect that looking in more detail at $k=5,6,7$ will show they can't occur, but I'm not sure. @Gerhard: I don't get your argument. | |
| Nov 6, 2012 at 8:51 | comment | added | Gordon Royle | Nice answer, Tony. If you build a set of 3-subsets with each pair meeting in a unique point, then you're basically building a projective plane. You either get the Fano plane, with 7 points/lines or you get a degenerate projective plane with all lines through a common point. | |
| Nov 6, 2012 at 5:37 | comment | added | Gerhard Paseman | Although mildly tedious, there is an argument which shows k cannot be larger than 4. Let A and B be two 3-sets which intersect in point c. Now there are at most 4 other 3-sets which intersect both A and B (and each other) in one point and miss c. Given there is at least one which misses c, use symmetry to get k is at most 4. Otherwise all the 3-sets contain c. Gerhard "And Robert's Your Mother's Brother" Paseman, 2012.11.05 | |
| Nov 6, 2012 at 4:51 | history | undeleted | Tony Huynh | ||
| Nov 6, 2012 at 4:51 | history | edited | Tony Huynh | CC BY-SA 3.0 |
added 238 characters in body
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| Nov 6, 2012 at 3:49 | history | deleted | Tony Huynh | ||
| Nov 6, 2012 at 3:47 | history | answered | Tony Huynh | CC BY-SA 3.0 |