Timeline for Generalization of friendship theorem:n vertices, any m vertices have exactly one neighbor
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 8, 2013 at 10:09 | comment | added | Shahrooz | Oops, you are absolutely right. | |
| Feb 8, 2013 at 7:37 | comment | added | Aaron Meyerowitz | @Shahrooz In the definition of the friendship graph any two adjacent vertices have exactly one common neighbor but also any two non-adjacent vertices have exactly one common neighbor. In brief, every pair of vertices is connected by exactly one path of length 2. | |
| Feb 7, 2013 at 20:11 | answer | added | Aaron Meyerowitz | timeline score: 4 | |
| Feb 7, 2013 at 14:27 | comment | added | Shahrooz | Let $A_m$ be the set of all graphs with $m$ vertices and that they have not dominating vertex. So, why the cone over any subset of $A_m$ is not an answer? | |
| Feb 7, 2013 at 14:19 | comment | added | Shahrooz | But for your problem, the graph has a dominating vertex and also the graph is $P_4$ free. | |
| Feb 7, 2013 at 14:12 | comment | added | Shahrooz | I think if you want to generalize the friendship theorem, you have to see one of important parameter in its definition. In the definition of friendship graph, any two adjacent vertices have exactly one common neighbor. By your definition, you deleted this condition and it is not the generalization of friendship graph. | |
| Feb 7, 2013 at 12:00 | comment | added | Bigcrow | Hi joro! Yes I mean common neighbor. Sorry for the mistake and thanks! | |
| Feb 7, 2013 at 11:59 | history | edited | Bigcrow | CC BY-SA 3.0 |
added te word "common"
|
| Feb 7, 2013 at 11:27 | comment | added | joro | Do you mean "have exactly one common neighbor" ? | |
| Feb 7, 2013 at 9:54 | history | asked | Bigcrow | CC BY-SA 3.0 |