Timeline for a conjecture about Hamiltonian graph
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 23, 2013 at 12:33 | comment | added | user40096 | The order in $\{a_1,...,a_n,b_1,...,b_n\}$ has nothing to do with the cycle order. | |
| Oct 22, 2013 at 13:04 | comment | added | jon | but u should have to know a well-order(@ adjacentcy of the vertices of $C$),then u can find a Hamiltonian cycle. In above ans : without loss of generality I had define the edges of the graph $C$ and then we show that $G-S$ is a Hamiltonian graph.So in general u can also claim that. | |
| Oct 22, 2013 at 12:28 | comment | added | user40096 | No,I just want to say the set of vertices of $C$ is $\{a_1,...,a_n,b_1,...b_n\}$. | |
| Oct 22, 2013 at 11:49 | comment | added | jon | So u want to say in the cycle $C$ $a_1$ adjacent to $b_2$ and $b_n$? | |
| Oct 22, 2013 at 11:49 | comment | added | user40096 | fedja,Thank you for your explaining! | |
| Oct 22, 2013 at 11:43 | comment | added | fedja | You do not know the edges except they form a cycle (the enumeration $a_j,b_j$ has nothing to do with the cycle order). | |
| Oct 22, 2013 at 11:19 | review | First posts | |||
| Oct 22, 2013 at 11:22 | |||||
| Oct 22, 2013 at 11:03 | history | answered | jon | CC BY-SA 3.0 |