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Timeline for Strong tournaments

Current License: CC BY-SA 4.0

26 events
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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