Timeline for Counting triples family with double shared elements
Current License: CC BY-SA 4.0
18 events
| when toggle format | what | by | license | comment | |
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| Aug 29, 2018 at 21:59 | comment | added | John Machacek | Removing the condition on intersection seems to make the question about 3-regular 3-uniform hypergraphs. I am not sure what is known or not known about enumerating such hypergraphs. | |
| Aug 28, 2018 at 19:34 | comment | added | Gerhard Paseman | The case where the intersection is at most two differs significantly from the case that the intersection is exactly two or exactly zero. Further, if three is changed to four, the nature of the solution also changes. I would recommend a new question. Gerhard "Rather Like The Old Question" Paseman, 2018.08.28. | |
| Aug 28, 2018 at 19:24 | vote | accept | Mohammad Al-Turkistany | ||
| Aug 28, 2018 at 19:24 | history | edited | Mohammad Al-Turkistany | CC BY-SA 4.0 |
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| Aug 28, 2018 at 19:23 | comment | added | Mohammad Al-Turkistany | @JohnMachacek Should I post a new question about the case where two triples intersect at most in two elements? | |
| Aug 28, 2018 at 19:21 | comment | added | Mohammad Al-Turkistany | @JohnMachacek Oops, my bad. | |
| Aug 28, 2018 at 19:16 | comment | added | John Machacek | In the proposed counter example it looks like there are two sets whose intersection has one element, namely [9,8,6] and [1,4,8]. | |
| Aug 28, 2018 at 19:13 | history | edited | Mohammad Al-Turkistany | CC BY-SA 4.0 |
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| Aug 28, 2018 at 18:55 | history | edited | Mohammad Al-Turkistany | CC BY-SA 4.0 |
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| Jun 30, 2018 at 9:29 | vote | accept | Mohammad Al-Turkistany | ||
| Aug 28, 2018 at 18:45 | |||||
| Jun 29, 2018 at 8:21 | history | edited | Mohammad Al-Turkistany | CC BY-SA 4.0 |
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| Jun 29, 2018 at 3:15 | answer | added | James | timeline score: 3 | |
| Jun 29, 2018 at 0:00 | answer | added | Gerhard Paseman | timeline score: 4 | |
| Jun 28, 2018 at 20:27 | comment | added | Gerhard Paseman | I am now thinking that the answer is that up to isomorphism, there is exactly one such structure for N=4k, and no structures otherwise. You should be able to prove this for yourself by combinatorial reasoning. Gerhard "Not Many Ways To Group" Paseman, 2018.06.28. | |
| Jun 28, 2018 at 20:15 | comment | added | Gerhard Paseman | There is up to isomorphism a unique such structure on N=4 elements. I recommend asking the question for N not a multiple of 3. Gerhard "You May Get More Answers" Paseman, 2018.06.28. | |
| Jun 28, 2018 at 18:55 | history | edited | Mohammad Al-Turkistany | CC BY-SA 4.0 |
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| Jun 28, 2018 at 18:43 | history | edited | Mohammad Al-Turkistany | CC BY-SA 4.0 |
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| Jun 28, 2018 at 18:28 | history | asked | Mohammad Al-Turkistany | CC BY-SA 4.0 |