Timeline for Jordan-like cycles in graphs
Current License: CC BY-SA 3.0
8 events
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Jan 21, 2013 at 18:17 | comment | added | Will Sawin | No. Indeed, it's clear that, for polyhedral graphs, every cycle is either a Jordan curve in each embedding or a curve whose complement has one component, depending on whether it bounds a single face or not. | |
Jan 21, 2013 at 18:03 | comment | added | Hans-Peter Stricker | @Will: Things might look different, if we allow one of the two components to be empty. Does it still hold then, that every planar graphs contains a cycle which does not separate the graph into two components? | |
Jan 21, 2013 at 17:02 | comment | added | Will Sawin | One might also ask if every graph such that each Jordan curve in one planar embedding is a Jordan curve in another planar embedding has this property. | |
Jan 21, 2013 at 16:35 | comment | added | Will Sawin | No such graphs exist, because every planar graph contains a cycle which does not separate the graph into two components, let alone do so in every embedding. I think the right thing to consider is graphs tha thave only one embedding inthe sphere, up to orientation-preserving homotopy. I think it's clear that these are polyhedral graphs, with the edges subdivided, or otherwise something really trivial like a path. | |
Jan 21, 2013 at 8:59 | comment | added | Hans-Peter Stricker | @Will: I'd like to understand better the connection between 3-connected planar graphs and planar graphs in which every cycle is a Jordan cycle. Does it hold - eventually - that in a 3-connected planar graph every cycle is a Jordan cycle? Or can you give a counter-example? | |
Jan 14, 2013 at 19:16 | comment | added | Ben Barber | "Falling prey to unchecked details may be regarded as a misfortune; contradicting the given example looks like carelessness." | |
Jan 14, 2013 at 19:05 | history | edited | Will Sawin | CC BY-SA 3.0 |
added 418 characters in body
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Jan 14, 2013 at 18:46 | history | answered | Will Sawin | CC BY-SA 3.0 |