Timeline for Have this generalization of Indifference graphs been studied before?
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 14, 2017 at 3:57 | vote | accept | j.s. | ||
| Jan 14, 2017 at 3:57 | comment | added | j.s. | @Ben Barber: Thank you so much for this wonderful answer. | |
| Jan 13, 2017 at 23:06 | comment | added | Ben Barber | Given a permutation, we can draw its graph (in the sense that functions have graphs) as a subset of $\mathbb R^2$. Then inversions correspond precisely to pairs of points joined by a line of negative slope. Note that the actual $x$ and $y$ coordinates aren't important: all that matters is their relative order. This is only obvious if you've seen it before! | |
| Jan 13, 2017 at 22:31 | comment | added | j.s. | @Ben Barber: A Genius Answer. thanks. But It's not perfectly clear to me why the class of permutation graphs are equivalent to graphs constructed by assigning a point in plane to each vertex and connecting two vertices by an edge when the line joining them has negative slope. Is this obvious? | |
| Jan 13, 2017 at 13:47 | comment | added | T. Amdeberhan | Quite interesting. | |
| Jan 13, 2017 at 13:34 | history | answered | Ben Barber | CC BY-SA 3.0 |