Timeline for A certain matrix associated to graphs
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 31, 2015 at 22:40 | comment | added | Dima Pasechnik | yes, indeed, please see my 2nd answer. | |
| Mar 31, 2015 at 14:10 | comment | added | Robert Wilms | I mean, I have an undirected graph and I choose cycles and for every cycle an orientation. The orientation of the cycles induces orientations for the edges, but these orientations depends on the cycle, thus two different cycles can induce two different orientations on the same edge, let's call this case a bad choice of orientations for the cycles. Now there are spanning trees, such that there are only bad choices of orientations for the fundamental cycles. Are there always spanning trees, such that there are good choices of orientations? | |
| Mar 31, 2015 at 13:07 | comment | added | Dima Pasechnik | I don't see how you can choose orientation of cycles independently. You can choose orientations for edges, but that's all, no? | |
| Mar 31, 2015 at 12:25 | comment | added | Robert Wilms | Yes, it looks different. But I think, this works, if one starts with the $b\times |E|$ matrix $$a_{j,e}=\begin{cases}0&e\notin \tau_j\\ \sqrt{T_e}&e\in\tau_j\end{cases}$$ and if the cycles are chosen to have the same directions on their intersections. Can I always choose a spanning tree, such that the fundamental base of cycles satisfies this? | |
| Mar 31, 2015 at 11:41 | comment | added | Dima Pasechnik | Please double-check. The diagonal entries in particular look different to me, and perhaps either $A$ or $A^\top$ should be prefilled with $\pm T_{e}$ rather than $\pm 1$. | |
| Mar 31, 2015 at 11:22 | vote | accept | Robert Wilms | ||
| Mar 31, 2015 at 11:22 | comment | added | Robert Wilms | Thank you. That is exactly the case, I am interested in. | |
| Mar 31, 2015 at 9:59 | history | answered | Dima Pasechnik | CC BY-SA 3.0 |