Timeline for A criterion for rooted trees to be isomorphic based on walks
Current License: CC BY-SA 3.0
9 events
when toggle format | what | by | license | comment | |
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Jan 12, 2016 at 16:52 | vote | accept | batconjurer | ||
Jan 12, 2016 at 14:40 | history | edited | batconjurer | CC BY-SA 3.0 |
added 208 characters in body
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Jan 12, 2016 at 13:39 | answer | added | Chris Godsil | timeline score: 8 | |
Jan 12, 2016 at 13:35 | comment | added | batconjurer | Sure, I can give that a shot. | |
Jan 12, 2016 at 13:13 | comment | added | Ilya Bogdanov | As for me, possible counterexanples probably can be found among pairs of isospectral trees. Do you have any software to check whether your condition is satisfied? If so, I would suggest to test some pairs from, e.g., pubs.acs.org/doi/pdf/10.1021/ci970242r (parhaps, those on Fig. 5?). | |
Jan 12, 2016 at 12:53 | comment | added | batconjurer | Just a sequence of edges $(v_1,v_2),(v_2,v_3),\dots,(v_{k-2},v_{k-1})$. And I'm also requiring that $v_1$ be the root. The paths are allowed to end wherever, and any amount of backtracking is allowed; these walks do not have to be simple. They can repeat edges and vertices. It's quite possible I missed some easy counterexample. | |
Jan 12, 2016 at 12:51 | review | Close votes | |||
Jan 12, 2016 at 13:42 | |||||
Jan 12, 2016 at 12:35 | comment | added | Vladimir Dotsenko | Can you be a bit more specific about what you mean by a walk of length $k$? some obvious interpretations admit obvious counterexamples, so some clarity would help. | |
Jan 12, 2016 at 11:59 | history | asked | batconjurer | CC BY-SA 3.0 |