Timeline for BEST theorem for Eulerian paths with open ends
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| Jul 22, 2011 at 23:54 | comment | added | Gerhard Paseman | I am not sure what you are counting. The edge removal does matter if you are counting a traversal of an Eulerian tour with a given start/endpoint. If you consider two traversals equivalent if the sequence of edges traveled differs by a cyclic permutation, then I think the (set of) equivalence classes of traversals is equinumerous with the paths. If you make clear what is being counted (I thought it was equiv. classes), then I may adjust my answer as needed. Gerhard "Ask Me About System Design" Paseman, 2011.07.22 | |
| Jul 22, 2011 at 13:34 | comment | added | tomate | Wait, now I see why this is wrong. It's not true that the correspondence is 1-1. Eulerian paths on G with open ends are 1-1 with those eulerian cycles on G+(v,u) whose final edge is (v,u)! But internal cycles in an eulerian cycle can be walked in any desired order! I'm more and more convinced that my guess is correct. | |
| Jul 6, 2011 at 8:30 | comment | added | Gerhard Paseman | Sleep on it for a night. If it still doesn't work out tomorrow, try a new question to ask for help with the sticky bits. Gerhard "Email Me About System Design" Paseman, 2011.07.06 | |
| Jul 6, 2011 at 8:21 | comment | added | tomate | It seems to work. Then my formula is wrong, since deletion-contraction formulas for $t_v(G')$ imply that I'm forgetting a piece; I'll have to make up my mind where Stanley's proof fails for unbalanced digraphs. | |
| Jul 6, 2011 at 8:16 | vote | accept | tomate | ||
| Jul 5, 2011 at 22:02 | history | answered | Gerhard Paseman | CC BY-SA 3.0 |