Timeline for Is the domination number of a combinatorial design determined by the design parameters?
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Mar 18, 2014 at 23:33 | vote | accept | Felix Goldberg | ||
| Mar 10, 2014 at 10:59 | comment | added | Dima Pasechnik | for 3-dimensional projective spaces $PG(3,q)$ and their hyperplane designs, probably the best one can do is $\gamma=2(q+1)$: take the points and the hyperplanes on a line. | |
| Mar 10, 2014 at 10:32 | comment | added | Dima Pasechnik | OK, I see: take a line $\ell_0$ and a point $p_0\in\ell_0$. Then set $P$ to be the points of $\ell_0$ except $p_0$ and $L$ to be the lines on $p_0$ except $\ell_0$. Nice :-) | |
| Mar 10, 2014 at 10:23 | comment | added | Dima Pasechnik | A dominating set in this case is a subset $(P,L)$ of points and lines such that each line on the plane not in $L$ intersects $P$ and each point of the plane not in $P$ lies on some $\ell\in L$, right? I see how $\gamma=2q$ works for $q=2$, but for bigger values, I don't... | |
| Mar 9, 2014 at 21:24 | comment | added | Felix Goldberg | @DimaPasechnik Oops, sorry, I meant $2q$. The $q$th order plane is a $(q^2+q+1,q+1,1)$-design and I meant "twice of (one less the degree of the design)"... | |
| Mar 9, 2014 at 21:01 | comment | added | Dima Pasechnik | Hmm, for $q=2$ this gives $\gamma=2$. This cannot be right, as the size of the set of neighbours of such a set is at most 6, but the graph has 14 vertices. | |
| Mar 9, 2014 at 14:22 | comment | added | Dima Pasechnik | What is the value of $\gamma$ for a projective plane of order $q$? | |
| Mar 9, 2014 at 13:54 | answer | added | Gordon Royle | timeline score: 7 | |
| Mar 9, 2014 at 9:54 | answer | added | Peter Dukes | timeline score: 2 | |
| Mar 9, 2014 at 9:47 | comment | added | Felix Goldberg | @GordonRoyle On the other hand, if the dominating set is of the form $P \cup B$, then the points in $P$ do not have to be dominated by the blocks in $B$. So resolvability may have much of an effect on $\gamma$. But I'm going to think about this angle for sure. | |
| Mar 7, 2014 at 3:01 | comment | added | Gordon Royle | My guess is that the domination number will depend on the structure. If the design is resolvable, then you will need exactly $v/k$ blocks to dominate the points, while if it is not resolvable, you will need more. I can't see how this would alter the number of points needed to dominate the blocks (but perhaps it does?). | |
| Mar 7, 2014 at 0:25 | history | asked | Felix Goldberg | CC BY-SA 3.0 |