Timeline for What is the independence number of this graph which is a generalization of a Kneser graph?
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
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| Feb 19, 2016 at 1:53 | answer | added | Ilya Bogdanov | timeline score: 0 | |
| Feb 17, 2016 at 16:19 | comment | added | Tony Huynh | Based on this chat, I edited the post to clarify the definition of the graph. Feel free to rollback if I am mistaking what you mean. | |
| Feb 17, 2016 at 16:18 | history | edited | Tony Huynh | CC BY-SA 3.0 |
Clarified the definition of the graph.
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| Feb 17, 2016 at 15:51 | comment | added | Gerhard Paseman | What of the other nine edges you build from 4 through 8? Aren't they adjacent to B as well? Gerhard "Is This Listed On Wikipedia?" Paseman, 2016.02.17. | |
| Feb 17, 2016 at 14:53 | comment | added | Maryam | In fact all adjacent vertices of $B$ are $(\{1,2,3\},\{67\}),(\{1,2,3\},\{68\}),(\{1,2,3\},\{7,8\}),(\{6,7,8\},\{4,5\}),$ | |
| Feb 17, 2016 at 14:43 | comment | added | Maryam | @ Brendan, No they are not adjacent because of 3 in $B$ and 6 in $C$. I think if we consider this graph as a generalization of usual Kneser graph, the definition is much clear. | |
| Feb 17, 2016 at 14:02 | comment | added | Brendan McKay | Maryam, are $B=(\lbrace1,2,3\rbrace,\lbrace4,5\rbrace)$ and $C=(\lbrace1,2,6\rbrace,\lbrace4,5\rbrace)$ adjacent? In the new version of the question they are adjacent ($B$ and $C$ are sets of size 2 whose intersection is $\lbrace4,5\rbrace$). Or do you mean us to consider the intersection of $\lbrace1,2,3,4,5\rbrace$ and $\lbrace1,2,6,4,5\rbrace$? | |
| Feb 17, 2016 at 5:18 | history | edited | Maryam | CC BY-SA 3.0 |
deleted 3 characters in body
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| Feb 17, 2016 at 1:04 | comment | added | Brendan McKay | You need to clarify the question. See Aaron's answer and the comments after it. | |
| Feb 16, 2016 at 19:35 | history | edited | user9072 | CC BY-SA 3.0 |
added 25 characters in body; edited tags; edited title
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| Feb 16, 2016 at 19:16 | history | edited | Maryam | CC BY-SA 3.0 |
added 18 characters in body
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| Feb 16, 2016 at 19:09 | answer | added | Aaron Meyerowitz | timeline score: 1 | |
| Feb 16, 2016 at 16:39 | comment | added | Gerhard Paseman | Note that B and C have an edge iff their union has 7 or 8 elements. So an independent set of vertices can't have two sets with a union of more than 7 elements. You can now choose either a set of four vertices all sharing 4 elements, or all 6 vertices contained in a set of six elements. Gerhard "I Would Go With Six" Paseman, 2016.02.16. | |
| Feb 16, 2016 at 15:59 | history | edited | Maryam | CC BY-SA 3.0 |
added 97 characters in body; edited title
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| Feb 16, 2016 at 15:51 | history | asked | Maryam | CC BY-SA 3.0 |