Partition into strokes
N. J. A. Sloane
njas at research.att.com
Thu Aug 23 14:03:41 CEST 2007
----- Original Message -----
送信者：Max Alekseyev <maxale at gmail.com>
宛先：koh <zbi74583 at boat.zero.ad.jp>
回送先：seqfan at ext.jussieu.fr
件名：Re: RE : Partition into strokes
送信日時：2007年8月23日
>On 8/22/07, koh <zbi74583 at boat.zero.ad.jp> wrote:
>
>> > Or, "Partition of a graph G into strokes S_i" must satisfy the following conditions.
>> >
>> > o Union_{i} S_i = H
>> > o If not{i=j} -> S_i and S_j don't have the same edge
>> > o If not{i=j} -> S_i U S_j isn't a dipath
>> > o For all i S_i is a dipath
>> > Where H is a digraph on G
>
>[...]
>
>> "Partition of a graph G into strokes" means "Partition of a digraph H on graph G into strokes".
>
>But what is exactly "a digraph on graph" ?
>
Does the phrase "a digraph on graph" mean nothing?
If so then it is a problem of my English.
I want to say the following :
If all directed edges of a digraph H are replaced by ordinal edges then H becomes graph G.
I write this relationship between H and G as follows.
"a digraph H on graph G"
example : o->o is a digraph on o-o
>> See the four conditions.
>>
>> 2) n=3
>> o-o-o names of vertices 1-2-3
>>
>> Partitions into strokes :
>> 1->2->3
>> 3->2->1
>> 1->2, 3->2
>> 2->1, 2->3
>> So, a(3)=4
>
>I'm confused. All listed partitions represent partitions of
>*different* digraphs.
You are right.
I calculated the number of partitions into strokes of all different digraphs on the G_n.
>Say, the first partition is of the digraph ( 1 -> 2 -> 3 ) while the
>last partition is of the digraph ( 1 <- 2 -> 3 ). Clearly, these
>digraphs are different.
>So, what is H in this case and what exactly the equality "Union_{i}
>S_i = H" from your definition means?
>
I am sorry.
My description is ambiguous.
I should have written as H_i for each digraphs on G_n.
On this case, four digraphs H_i on G exist and each digraphs have one partition into strokes.
example : last partition
Union_{i|1<=i<=2} S_i = S_1 U S_2 = 2->1 U 2->3 = 1<-2->3 = H_4
>Regards,
>Max
>
Yasutoshi
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