Timeline for Is there an infinite number of combinatorial designs with $r=\lambda^{2}$
Current License: CC BY-SA 3.0
14 events
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| Dec 28, 2012 at 10:53 | vote | accept | Felix Goldberg | ||
| Dec 27, 2012 at 22:40 | comment | added | Yuichiro Fujiwara | @Felix ...and the condition given in Edit3 can be met only when $v = k(k-1)+1$, which implies that the $S(2,k,v)$ is symmetric, be the way. So if $v > k(k-1)+1$ (i.e., the design becomes asymmetric due to $v$ being too large if you will), an $S(2,k,v)$ has a pair of parallel blocks. So, for instance, if you pick $k=3$, using a large set of $S(2,3,v)$s will give you a simple $2$-design with a pair of parallel blocks while satisfying $r = \lambda^2$. If you don't need your $2$-design to be simple, simply copying the same $S(2,k,v)$ $\frac{v-1}{k-1}$ times does the job. | |
| Dec 27, 2012 at 21:57 | comment | added | Yuichiro Fujiwara | @Felix For large $v$, the construction works just fine. I'll write up the details as an edit to the answer. | |
| Dec 27, 2012 at 21:27 | comment | added | Felix Goldberg | @YuichiroFujiwara: Let me see if I get this right: the Fano plane is symmetric and so fails my condition (the correct one, sorry about that again), because every two blocks in it have an intersection. But for larger $v$ it works, with $S(2,k,v)$ ceasing to be symmetric? | |
| Dec 27, 2012 at 20:08 | history | edited | Felix Goldberg | CC BY-SA 3.0 |
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| Dec 27, 2012 at 17:59 | answer | added | Yuichiro Fujiwara | timeline score: 6 | |
| Dec 27, 2012 at 17:35 | comment | added | Yuichiro Fujiwara | Ah, I didn't F5 before posting the above comment. I'm sorry. | |
| Dec 27, 2012 at 17:22 | comment | added | Yuichiro Fujiwara | Didn't you forget some other restrictions on the kind of (block) design you want? If you're really ok with a $2$-design of any kind as long as $r = \lambda^2$, then infinitely many of them sure exist. For example, for any finite point set $V$, you take $B = V$ as its only block. Obviously $\lambda = 1$ because every pair of elements of $V$ appears exactly once in a block of the block set (which is a singleton consisting of a block of size $\vert V \vert$). You have $r = 1$ as well because every point appears exactly once for obvious reasons. | |
| Dec 27, 2012 at 17:13 | history | edited | Felix Goldberg | CC BY-SA 3.0 |
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| Dec 27, 2012 at 14:46 | comment | added | Ken W. Smith | @Felix: thanks! -- your more general question is probably VERY open! btw, even if a symmetric design does not exist (such as $(211,36,6)$), there is still a possibility that a design exists with the residual parameters. | |
| Dec 27, 2012 at 14:43 | answer | added | Ken W. Smith | timeline score: 1 | |
| Dec 27, 2012 at 14:35 | comment | added | Felix Goldberg | @Ken W. Smith: I am actually interested in designs where at least one pair of blocks has a nonempty intersection, so this rules out symmetric designs, but the residual idea is interesting. So thanks a million! | |
| Dec 27, 2012 at 14:12 | comment | added | Ken W. Smith | If we restrict the question just to symmetric designs (so that $v=b$) then there are examples $(7,4,2), (25,9,3), (61,16,4),(121,25,5)$ described in the CRC Handbook of Combinatorial Designs [link text][1] (The list does not go far enough to examine $\lambda=6$.) If one removes a block from such a design one obtains a residual design in which $r$ is still equal to $\lambda^2.$ So there are certainly a number of examples. It would be interesting to know if there are any infinite families. [1]: emba.uvm.edu/~jdinitz/hcd.html | |
| Dec 27, 2012 at 12:09 | history | asked | Felix Goldberg | CC BY-SA 3.0 |