Timeline for Strong tournaments
Current License: CC BY-SA 4.0
26 events
when toggle format | what | by | license | comment | |
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Oct 28, 2019 at 20:12 | vote | accept | Fareed Abi Farraj | ||
Oct 28, 2019 at 14:27 | comment | added | Gabe Conant | Finally, since there is at least one circuit $C$ such that $(v_1,v_n)\in E(C)$, the total sum with respect to $N'$ is strictly less than the total sum with respect to $N$. A contradiction. | |
Oct 28, 2019 at 14:27 | comment | added | Gabe Conant | Case 3: $(v_1,v_n)\in E(C)$. Then there is $1<i<n$ such that $(v_i,v_1)\in E(C)$. When computing $i_N(C)$, $(v_1,v_n)$ contributes $+0$ and $(v_i,v_1)$ contributes $+1$. When computing $i_{N'}(C)$, $(v_1,v_n)$ and $(v_i,v_1)$ both contribute $+0$. The rest of the edges in $C$ contribute the same counts to $i_{N}(C)$ and $i_{N'}(C)$ since they do not involve $v_1$. So $i_{N'}(C)=i_{N}(C)-1$. | |
Oct 28, 2019 at 14:27 | comment | added | Gabe Conant | Case 2: $C$ involves $v_1$, but $(v_1,v_n)\not\in E(C)$. Then there are $1<i,j<n$ such that $(v_i,v_1),(v_1,v_j)\in E(C)$. When computing $i_N(C)$, $(v_i,v_1)$ contributes $+1$ and $(v_1,v_j)$ contributes $+0$. When computing $i_{N'}(C)$, $(v_i,v_1)$ contributes $+0$ and $(v_1,v_j)$ contributes $+1$. The rest of the edges in $C$ contribute the same counts to $i_{N}(C)$ and $i_{N'}(C)$ since they do not involve $v_1$. So $i_N(C)=i_{N'}(C)$. | |
Oct 28, 2019 at 14:27 | comment | added | Gabe Conant | Case 1: $C$ does not involve $v_1$. Then clearly $i_N(C)=i_{N'}(C)$ since the enumeration is the same on all other vertices. | |
Oct 28, 2019 at 14:27 | comment | added | Gabe Conant | But in any case, the enumeration $N'$ in the answer below does provide a contradiction to the assumption $(v_n,v_1)\not\in E(T)$. Indeed, given a circuit $C$, there are three cases: | |
Oct 28, 2019 at 14:26 | comment | added | Gabe Conant | Then I don't understand your concern below (in the comments to the proposed answer) about the possibility that $(v_1,v_2),(v_1,v_3),(v_1,v_4)\in E(C)$. If this were to happen then the circuit $C$ would pass through $v_1$ several times. | |
Oct 28, 2019 at 12:30 | comment | added | Fareed Abi Farraj | No the circuits pass through a single vertex only 1 time. | |
Oct 27, 2019 at 14:45 | comment | added | Gabe Conant | Do you allow your circuits to visit the same vertex numerous times? If so, what restrictions do you put on the circuits in order to justify the finite enumeration $C_1,\ldots,C_t$? | |
Oct 9, 2019 at 4:29 | history | edited | Fareed Abi Farraj | CC BY-SA 4.0 |
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Oct 7, 2019 at 10:52 | answer | added | Erlang Wiratama Surya | timeline score: 4 | |
Oct 6, 2019 at 19:00 | history | edited | YCor | CC BY-SA 4.0 |
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Oct 6, 2019 at 18:41 | history | edited | Fareed Abi Farraj | CC BY-SA 4.0 |
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Sep 25, 2019 at 12:53 | history | edited | Fareed Abi Farraj | CC BY-SA 4.0 |
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Aug 20, 2019 at 19:39 | history | edited | Fareed Abi Farraj | CC BY-SA 4.0 |
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Aug 4, 2019 at 16:51 | history | edited | Fareed Abi Farraj | CC BY-SA 4.0 |
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Aug 4, 2019 at 16:49 | comment | added | Fareed Abi Farraj | Yes you're right I'll change that | |
Aug 4, 2019 at 16:48 | comment | added | Gabe Conant | You seem to be using $E$ for an enumeration of the vertices, and also for the set of directed edges. | |
Aug 4, 2019 at 16:28 | history | edited | Fareed Abi Farraj | CC BY-SA 4.0 |
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Jul 30, 2019 at 18:08 | history | edited | Fareed Abi Farraj | CC BY-SA 4.0 |
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Jul 29, 2019 at 17:27 | comment | added | Fareed Abi Farraj | @András, Actually it is not from a book. Do you have an idea on how to solve it? | |
Jul 29, 2019 at 15:48 | history | edited | Fareed Abi Farraj | CC BY-SA 4.0 |
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Jul 29, 2019 at 11:49 | history | edited | Fareed Abi Farraj | CC BY-SA 4.0 |
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Jul 29, 2019 at 10:58 | comment | added | András Bátkai | Is this an excercise from a book? You should give a reference wher you get your question from. | |
Jul 29, 2019 at 10:25 | review | First posts | |||
Jul 29, 2019 at 10:58 | |||||
Jul 29, 2019 at 10:23 | history | asked | Fareed Abi Farraj | CC BY-SA 4.0 |