Let $T$ be a strong tournament, and let $N=v\_1v\_2 \cdots v\_n$ be an enumeration of $V(T)$. Let $C$ be a circuit in $T$. We define $i\_N(C)=|\{(v\_i,v\_j) \in E(C); i>j\}|$. Suppose that $N$ is chosen in such a way that $i\_N(C\_1)+ \cdots + i\_N(C\_t)$ is minimum, where $C\_1, \cdots, C\_t$ are all the circuits of $T$.


Prove that $\forall i$ such that $1 \le i \le n-1$, $(v\_i,v\_{i+1}) \in E(T)$ and that $(v\_n,v\_1) \in E(T)$.


**My attempt:**


I already proved that $\forall i$ such that $1 \le i \le n-1$ we have $(v\_i,v\_{i+1})\in E(T)$.


I first assumed that $(v\_i,v\_{i+1}) \not \in E(T)$, and this gives that $(v\_{i+1},v\_{i})\in E(T)$, so I took the enumeration $N'=v\_1 \cdots v\_{i-1}v\_{i+1}v\_iv\_{i+2} \cdots v\_n$, and proved that $i\_{N'}(C\_1)+ \cdots + i\_{N'}(C\_t)< i\_N(C\_1)+ \cdots + i\_N(C\_t)$, which is a contradiction.


But for proving $(v\_n,v\_1) \in E(T)$ I supposed that $(v\_1,v\_n) \in E(T)$ and tried to take the enumeration $N''=v\_nv\_1\cdots v\_{n-1}$ but I wasn't able to get to a contradiction, since to get to a contradiction from this enumeration I must be sure that the number of forward edges going to $v\_n$ was less than that of the backward edges from $v\_n$ in the first enumeration, can I prove this?Or do I take 2 cases if the number of forward edges was less or more than that of the backward edges? Or is there another enumeration that can finish it?


Please help, and thanks in advance.



Strong tournaments
As you suspected, taking a better enumeration suffices. If $(v\_n,v\_1)\not\in E$ then consider the enumeration $N'=(v\_2,v\_3,\cdots,v\_{n-1},v\_1,v\_n)$. It is easy to see that for any circuit $C$ such that $(v\_1,v\_n)\not\in E(C)$ we have $i\_N(C)=i\_{N'}(C)$. For any circuit $C$ such that $(v\_1,v\_n)\in E(C)$ (and there is one since $T$ is strong) we get $i\_{N'}(C)=i\_N(C)-1$, contradicting our assumption on $N$.


