Timeline for A criterion for rooted trees to be isomorphic based on walks
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Jan 12, 2016 at 21:23 | history | edited | Chris Godsil | CC BY-SA 3.0 |
edited body
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| Jan 12, 2016 at 19:30 | comment | added | Brendan McKay | @Ilya Bogdanov : Chris and I found this example in our first paper together (1975), see Fig 7 at users.cecs.anu.edu.au/~bdm/papers/GodsilMcKayComputational.pdf . I'm sure it is the first example where the two rooted trees are isomorphic if the root is ignored, but plausibly there are smaller examples without that property. | |
| Jan 12, 2016 at 16:52 | vote | accept | batconjurer | ||
| Jan 12, 2016 at 15:04 | comment | added | Chris Godsil | @Ilya Bogdanov: thanks, I've made the appropriate edits. | |
| Jan 12, 2016 at 15:03 | history | edited | Chris Godsil | CC BY-SA 3.0 |
fixed mistake
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| Jan 12, 2016 at 14:39 | comment | added | batconjurer | I'm a bit confused by your answer. First of all, $T$ was formed by deleting vertex 8, so I don't know what it means for $T$ to have no automorphism taking 5 to 8. If $T$ is the graph with 8 still in, then indeed $T\setminus 5\cong T\setminus 8$, but no roots are specified for me to check the condition. Clarification would be greatly appreciated. (Also, I should apologize as I added a clarification as a comment earlier that I should have put in the question. Walks are not assumed to be simple. I don't know if that changes your example) | |
| Jan 12, 2016 at 13:39 | history | answered | Chris Godsil | CC BY-SA 3.0 |