Timeline for Does every finite bridgeless cubic planar simple undirected graph admit a 2-factorization with at most two components each of which has even order?
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 26, 2017 at 20:56 | answer | added | Peter Heinig | timeline score: 1 | |
| S Aug 25, 2017 at 18:15 | history | bounty ended | Jimmy Dillies | ||
| S Aug 25, 2017 at 18:15 | history | notice removed | Jimmy Dillies | ||
| Aug 25, 2017 at 15:14 | vote | accept | Jimmy Dillies | ||
| Aug 25, 2017 at 4:22 | answer | added | Ian Agol | timeline score: 4 | |
| S Aug 24, 2017 at 15:53 | history | bounty started | Jimmy Dillies | ||
| S Aug 24, 2017 at 15:53 | history | notice added | Jimmy Dillies | Draw attention | |
| Aug 24, 2017 at 13:10 | review | Suggested edits | |||
| Aug 24, 2017 at 13:36 | |||||
| S Aug 23, 2017 at 9:01 | history | suggested | Peter Heinig | CC BY-SA 3.0 |
The title clearly was too general. I made the *question* in the OP the new title. To speak of the 'length of a component' just *happened* to be sensical here, yet sounds incorrect.
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| Aug 23, 2017 at 8:51 | answer | added | Peter Heinig | timeline score: 1 | |
| Aug 23, 2017 at 7:45 | comment | added | Peter Heinig | If one takes the question strictly literally, the answer is no: the infinite hexagonal grid is cubic bridgeless planar, yet does not admit any 2-factorization with an even number of components, for the boring cardinality reason that each 'factor' is by definition finite, so there must be $\aleph_0$-many components in any factorization, and $\aleph_0$ is not usually considered 'even'. Of course, this is not what the OP intends. I will not make this an answer, rather edit the OP. Yet it shows that any proof must make use of finiteness here. | |
| Aug 23, 2017 at 7:33 | comment | added | Peter Heinig | Dear @JimmyDillies: I made several edits to the OP. If you have reasons to object, please correct or roll back. Most seriously, the paragraph starting with "Stated otherwise" was confusing (to me), at least one the level of English-composition: it left it unclear whether you claimed that what comes next is equivalent to what comes before (which it not really is). | |
| Aug 23, 2017 at 7:27 | review | Suggested edits | |||
| S Aug 23, 2017 at 9:01 | |||||
| Aug 23, 2017 at 1:44 | comment | added | Jimmy Dillies | A 2-factor is a regular subgraph of degree 2 containing all vertices of the original graph. The complement is indeed a 1 factor. | |
| Aug 22, 2017 at 23:55 | comment | added | fidbc | What is your definition of 2-factorization? My understanding is that a 2-factorization is a partition of the edge set into 2-factors. This does not seem possible in a cubic graph, since once you find a 2-factor and remove the edges you are left with a 1-factor. | |
| Aug 22, 2017 at 14:51 | history | asked | Jimmy Dillies | CC BY-SA 3.0 |