Timeline for Paths on Cartesian products of graphs satisfying linear constraints
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 20, 2014 at 18:05 | comment | added | ematsen | To expand on my previous note, we would actually like to be able to say things about the symmetric power of $G$ (the question concerning cospectrality in that paper is answered in this paper. It wouldn't be a big problem for us if we had to try every ordering of the points, but actually solving the symmetric product case would be ideal. | |
| Jun 20, 2014 at 17:12 | comment | added | ematsen | You are right on both counts. For the first statement, I thought it would be easier to find previous work if I went with a more general statement. For the second, you make a great point-- the vertices of $G^{\square r}$ are ordered, whereas our point configurations do not have to be. However, in both counts an answer to the problem as stated will give an answer to the more specific problem we care about.Do you think that it would be better to edit to make it more specific? Thanks, @BenBarber. | |
| Jun 20, 2014 at 15:56 | comment | added | Ben Barber | Here are some partial restatements to check I'm understanding correctly. Since $s$ is binary, the multiset of vertices occupied is actually a set, and the $i$th row of $C$ ensures that at most $m_i$ of the vertices in the set $C_i$ are occcupied. Am I right in thinking that there is no unique way of passing between a given configuration of this sort on $G$ and a vertex of $G^{\square r}$? | |
| Jun 20, 2014 at 15:56 | review | First posts | |||
| Jun 20, 2014 at 15:59 | |||||
| Jun 20, 2014 at 15:40 | history | asked | ematsen | CC BY-SA 3.0 |