Timeline for When do the circuits of a matroid have a connected intersection graph?
Current License: CC BY-SA 4.0
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| Oct 31, 2020 at 3:21 | history | edited | Ethan Splaver | CC BY-SA 4.0 |
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| Oct 17, 2020 at 1:13 | history | became hot network question | |||
| Oct 16, 2020 at 19:54 | history | edited | Ethan Splaver | CC BY-SA 4.0 |
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| Oct 16, 2020 at 19:49 | history | edited | Ethan Splaver | CC BY-SA 4.0 |
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| Oct 16, 2020 at 19:31 | answer | added | Joao Costalonga | timeline score: 3 | |
| Oct 16, 2020 at 19:25 | history | edited | Ethan Splaver | CC BY-SA 4.0 |
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| Oct 16, 2020 at 18:37 | history | edited | Ethan Splaver | CC BY-SA 4.0 |
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| Oct 16, 2020 at 18:24 | history | edited | Ethan Splaver | CC BY-SA 4.0 |
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| Oct 16, 2020 at 18:09 | history | edited | Ethan Splaver | CC BY-SA 4.0 |
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| Oct 16, 2020 at 18:02 | comment | added | Ethan Splaver | @LorenzoNajt Sorry havn't slept in a while, you are correct. I also updated the condition on the indexing so it should be when is there a walk in the intersection graph hitting every vertex which is equivilent to connectedness | |
| Oct 16, 2020 at 17:59 | history | edited | Ethan Splaver | CC BY-SA 4.0 |
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| Oct 16, 2020 at 17:54 | history | edited | Ethan Splaver | CC BY-SA 4.0 |
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| Oct 16, 2020 at 17:54 | vote | accept | Ethan Splaver | ||
| Oct 16, 2020 at 17:51 | comment | added | Elle Najt | @Ethan Yeah, but that doesn't contradict what I'm saying. I mean that a unicycle gives an example of a graph with your property that is not 2-edge connected. (Maybe your 'iff' is meant to be an 'if'? Actually I'm not sure about that either, e.g. consider the wedge of 2 circuits at a single point. It's 2 edge connected, but the circuit graph consists of two isolated points.) | |
| Oct 16, 2020 at 17:46 | history | edited | Ethan Splaver | CC BY-SA 4.0 |
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| Oct 16, 2020 at 17:45 | comment | added | Ethan Splaver | @LorenzoNajt if its $2$-edge connected and a unicycle graph then it is a cycle graph, no? | |
| Oct 16, 2020 at 17:43 | comment | added | Elle Najt | @Ethan Is that condition about the graphic matroid right? For instance, a unicycle graph has a single circuit, so the intersection graph is connected, but it is not 2-edge connected. | |
| Oct 16, 2020 at 17:42 | comment | added | Tony Huynh | Your two conditions are still not equivalent. Your first condition is saying that the intersection graph has a Hamiltonian path, while the second condition is saying it is connected. I answered the connected version below. | |
| Oct 16, 2020 at 17:40 | history | edited | Ethan Splaver | CC BY-SA 4.0 |
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| Oct 16, 2020 at 17:39 | answer | added | Tony Huynh | timeline score: 5 | |
| Oct 16, 2020 at 17:28 | comment | added | Sam Hopkins | (Actually I guess there could be a connected component which has no circuits, i.e., a unique base.) | |
| Oct 16, 2020 at 17:26 | comment | added | Sam Hopkins | Here's a first stab: evidently the matroid itself must be connected. I don't know if that's sufficient. | |
| Oct 16, 2020 at 17:24 | comment | added | Ethan Splaver | @SamHopkins made a slight error, updated it | |
| Oct 16, 2020 at 17:24 | history | edited | Ethan Splaver | CC BY-SA 4.0 |
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| Oct 16, 2020 at 17:22 | comment | added | Sam Hopkins | I don't see why the two questions you posed are equivalent. The first seems to be saying that the intersection graph of circuits has no isolated vertices, which is different from it being connected. | |
| Oct 16, 2020 at 17:11 | history | asked | Ethan Splaver | CC BY-SA 4.0 |