[seqfan] Re: partition of a circle
wl at particle.uni-karlsruhe.de
wl at particle.uni-karlsruhe.de
Fri May 11 17:35:57 CEST 2012
Neil, but then what about n=4 and the graph with b=2, i=4 internal
nodes Q1 to Q4 with e=7 internal edges F1 to F7. See below. This is
not among the 15 paper drawings. If Q1 and Q3 would be connected to
some two new circle nodes should it be admitted for n=6?
Cycles for inner nodes are allowed (see n=4 paper examples).
A sketch without the circle line:
P1 | Q2
/ \ Q3---Q1 \ / Q4
| P2
Besides the Euler rule you gave earlier (n = e-i+1, here 4 = 7 - 4 + 1)
one can also use the degree check formula: 2*|edges| = sum over the
degrees of all nodes, that is
2(b+e) =sum(dP_j,j=1..b) + sum(dQ_j,j=1..i).
Here: 2*(2 + 7) = 18 = 2*3 + 4*3 = 6 + 12.
(In your grid example (actually a 12-gon with 16 part, where the four
square nodes of degree 2 have been deleted) it was 2*(12 + 24) = 72 =
12*3 + 9*4 ).
Wolfdieter
Quoting Neil Sloane <njasloane at gmail.com>:
> Wolfdieter, I don't think that rule is needed.
> I think the following is a legitimate dissection (into 16 regions):
> draw a square grid with 5 horizontal lines and 5 vertical lines.
> The boundary is a square, but topologically that is the same as
> a circle. There are 12 P's and 9 Q's.
> I asked Tudor Z. if he knew the next term, but he said no.
> Neil
>
>
> On Fri, May 11, 2012 at 3:10 AM, <wl at particle.uni-karlsruhe.de> wrote:
>
>> Maybe one should also include the rule:
>> There has to be at least one line segment (edge) connecting each internal
>> node Q to a boundary node P. This would exclude Andrews n=3 and n=5
>> graphs, and also Andrew's n=5 graph with all of the four 'outer' Qs
>> connected to some Ps (now n=8), but the 'central' node Q is not connected
>> to any P. Such a central Q is never connectable to any P (planarity).
>>
>> I have written yesterday to one of the authors (T. Zamfirescu at Dortmund
>> TH) my zeroth order guess of the rules, and asked him for a definition of
>> 'Jordan domain' on a euclidean plane E_2 (maybe some fans can here help
>> also). I am waiting for his answer.
>>
>> Wolfdieter Lang
>>
>>
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>>
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>
>
>
> --
> Dear Friends, I will soon be retiring from AT&T. New coordinates:
>
> Neil J. A. Sloane, President, OEIS Foundation
> 11 South Adelaide Avenue, Highland Park, NJ 08904, USA
> Phone: 732 828 6098; home page: http://NeilSloane.com
> Email: njasloane at gmail.com
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