Timeline for Converting a digraph to an undirected graph in a reversible way
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 27, 2015 at 20:06 | answer | added | William | timeline score: 10 | |
| Jan 23, 2014 at 14:09 | vote | accept | Heterotic | ||
| Jan 16, 2014 at 18:39 | answer | added | Aaron | timeline score: -1 | |
| Jan 15, 2014 at 23:29 | answer | added | David Richerby | timeline score: 8 | |
| Jan 15, 2014 at 22:53 | answer | added | muede | timeline score: -1 | |
| Jan 15, 2014 at 19:45 | comment | added | Raphael | Please incorporate your updates into the question. At any point in time, SE posts should be readable top to bottom without wondering about history; that's archived separately. | |
| Jan 15, 2014 at 19:07 | history | edited | Heterotic | CC BY-SA 3.0 | added 511 characters in body |
| Jan 15, 2014 at 18:51 | comment | added | D.W.♦ | I think you should specify in the question what you mean by "the same graph". Do you mean that the vertices are labelled, or that the vertices are unlabelled? Do you mean that $(V,E)$ is the same for both, or that the two graphs are isomorphic? It sounds like you mean the latter. Are you sure that's a requirement in your application? If you're allowed to retain labels, the problem gets easier and AdrianN's answer works (because the edge $(3,4)$ is not the same as the edge $(1',2')$). | |
| Jan 15, 2014 at 17:39 | history | edited | Heterotic | CC BY-SA 3.0 | added 317 characters in body |
| Jan 15, 2014 at 15:49 | history | tweeted | twitter.com/#!/StackCompSci/status/423482127892889600 | ||
| Jan 15, 2014 at 13:11 | answer | added | adrianN | timeline score: 2 | |
| Jan 15, 2014 at 12:36 | comment | added | Heterotic | I can't really think of any constrains for now. I guess any way to encode the information of a directed graph into an undirected one would do, as long as it is reversible. I guess what I have in mind is the simplest type of undirected graphs, so I am looking for a solution that doesn't use colors either for the vertices or the edges. | |
| Jan 15, 2014 at 11:47 | comment | added | adrianN | I think this question is too broad. What are your constraints? | |
| Jan 15, 2014 at 10:30 | history | asked | Heterotic | CC BY-SA 3.0 |