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 was not able to get to a contradiction. Maybe I better take another enumeration or something.

I am stuck and I don't know how to start, can anyone figure out the solution? Please I want help in this one, I really need it so much. Thanks in advance.

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.

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 $(v_n,v_1) \in E(T)$.

My attempt:

I already proved that $\forall i$ auch that $1 \le i \le n-1$ we have $(v_i,v_{i+1})\in E(T)$ by taking to contrary the enumeration $N'=v_1 \cdots v_{i-1}v_{i+1}v_iv_{i+2} \cdots v_n$, and proving that $i_{N'}(C_1)+ \cdots + i_{N'}(C_t)< i_N(C_1)+ \cdots + i_N(C_t)$.

But for proving $(v_n,v_1) \in E(T)$ I supposed that $(v_1,v_n) \in E(T)$ and took the enumeration $N''=v_nv_1\cdots v_{n-1}$ but I am not able to get to the contradiction. Maybe I better take another enumeration.

I am stuck and I don't know how to start, can anyone figure out the solution?

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 was not able to get to a contradiction. Maybe I better take another enumeration or something.

I am stuck and I don't know how to start, can anyone figure out the solution? Please I want help in this one, I really need it so much. Thanks in advance.