Timeline for Strong tournaments
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 28, 2019 at 20:12 | vote | accept | Fareed Abi Farraj | ||
| Oct 10, 2019 at 16:12 | comment | added | Erlang Wiratama Surya | Yes that case is possible but it doesn't change my argument. My answer asses 1 circuit at a time. | |
| Oct 10, 2019 at 14:35 | comment | added | Fareed Abi Farraj | Not quite well actually, because for example $2$ circuits might have the same forward edge to $v_1$ but 2 different backward edges from $v_1$ (considering your enumeration). Isn't this case possible? If it is, then the number of backward edges in your enumeration is more than that in the original enumeration, right? | |
| Oct 9, 2019 at 17:17 | comment | added | Erlang Wiratama Surya | @FareedAF What I was saying is that for every forward edge incident to $v_1$ in any circuit $C$ you will get one backward edge right next to it in the same circuit $C$, so intuitively they will cancel out. Can you see now why the third sentence in my answer is true? | |
| Oct 9, 2019 at 15:15 | comment | added | Plain_Dude_Sleeping_Alone | Are you Erlang of the Bronze medalist IMO? | |
| Oct 8, 2019 at 15:48 | comment | added | Fareed Abi Farraj | You mean I was wrong by saying "...all..."? Or is it something else that you mean it? But even if they weren't all forward, how can we know that the number of forward edges was more than the number of backward edges in the first enumeration?(Because if this is right then it is done by the enumeration you've given) Maybe I didn't get your point exactly, please explain it more to me. | |
| Oct 8, 2019 at 14:14 | comment | added | Erlang Wiratama Surya | Note that any circuit $C$ that contains $v_1$ but not $(v_1,v_n)$ will contain the edges $(v_i,v_1), (v_1,v_j)$ for some $i,j>1$. | |
| Oct 8, 2019 at 14:01 | comment | added | Fareed Abi Farraj | How about if $v_1$ was connected to all the other vertices by a forward edge? I mean $(v_1,v_2), (v_1,v_3), (v_1,v_4) ... \in E(C)$ ? Then they will all be back edges in your enumeration | |
| Oct 8, 2019 at 13:33 | comment | added | Erlang Wiratama Surya | There can be more than 1. What I meant to say was there is at least one. | |
| Oct 7, 2019 at 14:52 | comment | added | Fareed Abi Farraj | Why there is only 1? Is the only circuit you're talking about the one passing through all the vertices? How can we prove that? | |
| Oct 7, 2019 at 11:09 | review | Late answers | |||
| Oct 7, 2019 at 11:17 | |||||
| Oct 7, 2019 at 10:55 | review | First posts | |||
| Oct 7, 2019 at 11:42 | |||||
| Oct 7, 2019 at 10:52 | history | answered | Erlang Wiratama Surya | CC BY-SA 4.0 |