Talk:Crown graph: Difference between revisions
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I think that the crown graph on 2''n'' vertices can also be regarded as the [[Tensor product of graphs|tensor product]] of [[Cycle graph|C<sub>''n''</sub>]] and [[Complete graph|K<sub>2</sub>]]. Unless someone points out that I am mistaken, I shall add this to the first paragraph. [[User:Maproom|Maproom]] ([[User talk:Maproom|talk]]) 21:34, 16 February 2011 (UTC) |
I think that the crown graph on 2''n'' vertices can also be regarded as the [[Tensor product of graphs|tensor product]] of [[Cycle graph|C<sub>''n''</sub>]] and [[Complete graph|K<sub>2</sub>]]. Unless someone points out that I am mistaken, I shall add this to the first paragraph. [[User:Maproom|Maproom]] ([[User talk:Maproom|talk]]) 21:34, 16 February 2011 (UTC) |
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:I don't think so. Tensor product with K2 (also known as the [[bipartite double cover]]) preserves the degree of a vertex. Cn x K2 is always either C<sub>2n</sub> (if n is odd) or two disjoint copies of Cn (if n is even). Maybe you mean some other kind of product? —[[User:David Eppstein|David Eppstein]] ([[User talk:David Eppstein|talk]]) 21:59, 16 February 2011 (UTC) |
:I don't think so. Tensor product with K2 (also known as the [[bipartite double cover]]) preserves the degree of a vertex. Cn x K2 is always either C<sub>2n</sub> (if n is odd) or two disjoint copies of Cn (if n is even). Maybe you mean some other kind of product? —[[User:David Eppstein|David Eppstein]] ([[User talk:David Eppstein|talk]]) 21:59, 16 February 2011 (UTC) |
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::You are right. I mean [[Cartesian product of graphs|Cartesian product]], still of [[Cycle graph|C<sub>''n''</sub>]] and [[Complete graph|K<sub>2</sub>]]. [[User:Maproom|Maproom]] ([[User talk:Maproom|talk]]) 22:23, 16 February 2011 (UTC) |
Revision as of 22:23, 16 February 2011
Mathematics B‑class Low‑priority | ||||||||||
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Greedy worst case lower bound
I attributed the greedy worst case lower bound to Johnson, which is what Kubale (2004) does. That’s the oldest reference I’ve been able to find for this construction. If the Füredi et al. (2008) paper was only used for that reference, I think it can be removed. (But I’m slightly confused about that paper, so maybe I’m misunderstanding something.) Thore Husfeldt (talk) 09:41, 14 January 2009 (UTC)
Polyhedral resemblance
The crown graph with an octagonal boundary looks a bit like a square cupola... Professor M. Fiendish, Esq. 13:01, 9 September 2009 (UTC)
Tensor product
I think that the crown graph on 2n vertices can also be regarded as the tensor product of Cn and K2. Unless someone points out that I am mistaken, I shall add this to the first paragraph. Maproom (talk) 21:34, 16 February 2011 (UTC)
- I don't think so. Tensor product with K2 (also known as the bipartite double cover) preserves the degree of a vertex. Cn x K2 is always either C2n (if n is odd) or two disjoint copies of Cn (if n is even). Maybe you mean some other kind of product? —David Eppstein (talk) 21:59, 16 February 2011 (UTC)
- You are right. I mean Cartesian product, still of Cn and K2. Maproom (talk) 22:23, 16 February 2011 (UTC)