The simplicity index of tournaments

TL;DR

This paper investigates the maximum possible simplicity index of tournaments that are not doubly regular, especially focusing on cases where the number of vertices does not satisfy certain modular conditions.

Contribution

It characterizes the class of tournaments with the highest simplicity index for cases where doubly regular tournaments do not exist.

Findings

Maximal simplicity index achieved for specific tournament classes.

Established bounds for simplicity index when $n ot\equiv 3 \\pmod{4}$.

Provided structural insights into non-doubly regular tournaments.

Abstract

An $n$-tournament $T$ with vertex set $V$ is simple if there is no subset $M$ of $V$ such that $2\leq \left \vert M\right \vert \leq n-1$ and for every $x\in V\setminus M$, either $M\rightarrow x$ or $x \rightarrow M$. The simplicity index of an $n$-tournament $T$ is the minimum number $s(T)$ of arcs whose reversal yields a non-simple tournament. M\"{u}ller and Pelant (1974) proved that $s(T)\leq\frac{n-1}{2}$, and that equality holds if and only if $T$ is doubly regular. As doubly regular tournaments exist only if $n\equiv 3\pmod{4}$, $s(T)<\frac{n-1}{2}$ for $n\not\equiv3\pmod{4}$. In this paper, we study the class of $n$-tournaments with maximal simplicity index for $n\not\equiv3\pmod{4}$.

Equations

$$\sum _{z\in V}{\delta }_T^{+}(z)=\sum _{z\in V}{\delta }_T^{-}(z)=\frac{n\left(n-1\right)}{2}.$$

$$\left|N_T^{-}(x)\right|=\left|N_T^{-}(x)\cap N_T^{-}(y)\right|+\left|N_T^{-}(x)\cap N_T^{+}(y)\right|+\left|N_T^{-}(x)\cap \left\{y\right\}\right|$$

$$\left|N_T^{+}(y)\right|=\left|N_T^{+}(y)\cap N_T^{+}(x)\right|+\left|N_T^{+}(y)\cap N_T^{-}(x)\right|+\left|N_T^{+}(y)\cap \left\{x\right\}\right|.$$

$$\left|N_T^{-}(x)\cap N_T^{-}(y)\right|-\left|N_T^{+}(x)\cap N_T^{+}(y)\right|=\left|N_T^{-}(x)\right|-\left|N_T^{+}(y)\right|.\text{qed}$$

$${\delta }_T^{-}(z){\delta }_T^{+}(z)=\left|\{\{x,y\}\in (\genfrac{}{}{0ex}{}{V}{2}):z\in (N_T^{-}(x)\cap N_T^{+}(y))\cup (N_T^{+}(x)\cap N_T^{-}(y))\}\right|.$$

$$\sum _{z\in V}{\delta }_T^{+}(z){\delta }_T^{-}(z)=\sum _{\left\{x,y\right\}\in \left(\genfrac{}{}{0ex}{}{V}{2}\right)}{\sigma }_T(x,y).$$

$$\left|N_T^{+}(x)\cap N_T^{+}(y)\right|=\left|N_T^{-}(x)\cap N_T^{-}(y)\right|=\left|N_T^{+}(x)\cap N_T^{-}(y)\right|=k$$

$$\text{and}\left|N_T^{-}(x)\cap N_T^{+}(y)\right|=k+1.$$

$$s(T)\le \underset{x\in V}{\min }({\delta }_T^{+}(x),{\delta }_T^{-}(x))={\delta }_T.$$

$$B:=\left(\left\{x\right\}\times \left((N_T^{+}(x)\cap N_T^{-}(y)\right)\cup \left(N_T^{+}(y)\cap N_T^{-}(x)\right)\times \left\{x\right\}\right).$$

$$s(T)\le \left|B\right|=\left|N_T^{+}(x)\cap N_T^{-}(y)\right|+\left|N_T^{+}(y)\cap N_T^{-}(x)\right|={\sigma }_T(x,y).$$

$${\delta }_T\le \lfloor \frac{n-1}{2}\rfloor$$

$${\delta }_T\le \lfloor \frac{n-1}{2}\rfloor .$$

$${\sigma }_T\le \frac{1}{\left(\genfrac{}{}{0ex}{}{|V|}{2}\right)}\sum _{\left\{x,y\right\}\in \left(\genfrac{}{}{0ex}{}{V}{2}\right)}{\sigma }_T(x,y).$$

$${\sigma }_T$$

$$\le \frac{2}{n(n-1)}\sum _{z\in V}{\left(\frac{{\delta }_T^{+}(z)+{\delta }_T^{-}(z)}{2}\right)}^2$$

$$\le \frac{(n-1)}{2}.\text{qed}$$

$$s(T)\le {\delta }_T\le \lfloor \frac{n-1}{2}\rfloor =2k.$$

$$\sum _{\left\{x,y\right\}\in \left(\genfrac{}{}{0ex}{}{V}{2}\right)}{\sigma }_T(x,y)=\sum _{z\in V}{\delta }_T^{+}(z){\delta }_T^{-}(z)=8k^2(2k-1).$$

$${\sigma }_T$$

$$\le \frac{2}{4k(4k-1)}8k^2(2k-1)$$

$$\le (2k-1)+\frac{2k-1}{4k-1}$$

$$\le 2k-1.$$

$$\sum _{\left\{x,y\right\}\in \left(\genfrac{}{}{0ex}{}{V}{2}\right)}{\sigma }_T(x,y)\ge 2\left(\genfrac{}{}{0ex}{}{2k}{2}\right)(2k-1)+(2k{)}^2(2k)>8k^2(2k-1),$$

$$s(T)=\min \left\{s_C(T):2\le \left|C\right|\le n-1\right\}.$$

$$\mathrm{Inv}(T,\{(x,y)\in A:\{x,y\}\in E\})$$

$$\left|E\right|\ge (n-\left|C\right|)(\left|C\right|-(n-1-{\delta }_T)).$$

$${\delta }_G(x)$$

$$=\left|C\right|-\max (\left|N_T^{-}(x)\cap C\right|,\left|N_T^{+}(x)\cap C\right|).$$

$${\delta }_G(x)$$

$$\ge (\left|C\right|-(n-1-{\delta }_T)).$$

$$\left|E\right|=\sum _{x\in V\setminus C}{\delta }_G(x).$$

$$\left|E\right|$$

$$\ge (n-\left|C\right|)(\left|C\right|-(n-1-{\delta }_T)).$$

Full text

Keywords: Doubly regular tournament, arc reversal, module, simplicity index.

MSC Classification: 05C20, 05C12.

1 Introduction

A tournament $T$ consists of a finite set $V$ of vertices together with a set $A$ of ordered pairs of distinct vertices, called arcs, such that for all $x\not=y\in V$, $(x,y)\in A$ if and only if $(y,x)\notin A$. Such a tournament is denoted by $T=(V,A)$. Given $x\neq y\in V$, we say that $x$ dominates $y$ and we write $x\rightarrow y$ when $(x,y)\in A$. Similarly, given two disjoint subsets $X$ and $Y$ of $V$, we write $X\rightarrow Y$ if $x\rightarrow y$ holds for every $(x,y)\in X\times Y$. Throughout this paper, we mean by an $n$-tournament a tournament with $n$ vertices.

A tournament is regular if there is an integer $k\geq 1$ such that each vertex dominates exactly $k$ vertices. It is doubly regular if there is an integer $k\geq 1$ such that every unordered pair of vertices dominates exactly $k$ vertices.

A tournament is transitive, if for any vertices $x$, $y$ and $z$, $x\rightarrow y$ and $y\rightarrow z$ implies that $x\rightarrow z$. A tournament $T=(V,A)$ is reducible if $V$ admits a bipartition $\{X,Y\}$ such that $X\rightarrow Y$. The notion of simple tournament was introduced by Fried and Lakser [8], it was motivated by questions in algebra. It is closely related to modular decomposition [9] which involves the notion of module. Recall that a module of a tournament $T=(V,A)$ is a subset $M$ of $V$ such that for every $x\in V\setminus M$ either $M\rightarrow\left\{x\right\}$ or $\left\{x\right\}\rightarrow M$. For example, $\emptyset$, $\{x\}$, where $x\in V$, and $V$ are modules of $T$ called trivial modules. An $n$-tournament is simple [6, 15] (or prime [4] or primitive [5] or indecomposable [10, 17]) if $n\geq 3$ and all its modules are trivial. The simple tournaments with at most $5$ vertices are shown in Figure 1. A tournament is decomposable if it admits a non-trivial module.

Given an $n$-tournament $T$, the Slater index $i(T)$ of $T$ is the minimum number of arcs that must be reversed to make $T$ transitive [18]. It is not difficult to see that $i(T)\leq\frac{n(n-1)}{4}$. However, we do not know an exact determination of the upper bound of $i(T)$. Erdős and Moon [7] proved that this bound is asymptotically equal to $\frac{n^{2}}{4}$. Recently, Satake [16] proved that the Slater index of doubly regular $n$-tournaments is at least $\frac{n(n-1)}{4}-n^{\frac{3}{2}}\log_{2}(2n)$.

Kirkland [11] defined the reversal index $i_{R}(T)$ of a tournament $T$ as the minimum number of arcs whose reversal makes $T$ reducible. Clearly, $i_{R}(T)\leq i(T)$. Kirkland [11] proved that $i_{R}(T)\leq\left\lfloor\frac{n-1}{2}\right\rfloor$ and characterized all the tournaments for which equality holds.

The indices above can be interpreted in terms of distance between tournaments. The distance $d(T_{1},T_{2})$ between two tournaments $T_{1}$ and $T_{2}$ with the same vertex set is the number of pairs $\{x,y\}$ of vertices for which the arc between $x$ and $y$ has not the same direction in $T_{1}$ and $T_{2}$. Let $\mathcal{F}$ be a family of tournaments with vertex set $V$. The distance from a tournament $T$ to the family $\mathcal{F}$ is $d(T,\mathcal{F})=\min\{d(T,T^{\prime}):T^{\prime}\in\mathcal{F}\}$. If $\mathcal{F}$ is the family of transitive tournaments on $V$, then $i(T)=d(T,\mathcal{F})$. If $\mathcal{F}$ is the family of reducible tournaments on $V$, then $i_{R}(T)=d(T,\mathcal{F})$.

By considering the family of decomposable tournaments, we obtain the simplicity index introduced by Müller and Pelant [15]. Precisely, consider an $n$-tournament $T$, where $n\geq 3$. The * simplicity index* $s(T)$ of $T$ (also called the arrow-simplicity of $T$ in [15]) is the minimum number of arcs that must be reversed to make $T$ non-simple. For example, the tournaments shown in Figure 1 have simplicity index $1$. Obviously, $s(T)\leq i_{R}(T)$ and $s(T)\leq\frac{n-1}{2}$. Müller and Pelant proved that $s(T)=\frac{n-1}{2}$ if and only if $T$ is doubly regular.

A dual notion of the simplicity index is the decomposability index [2], which is obtained by considering the family of simple tournaments.

In this paper, we provide an upper bound for $s(T)$, where $T$ is an $n$-tournament for $n\not\equiv 3\pmod{4}$. More precisely, we obtain the following result.

Theorem 1.1**.**

Given an $n$-tournament $T$, the following statements hold

if $n=4k+2$, then $s(T)\leq 2k$; 2. 2.

if $n=4k+1$, then $s(T)\leq 2k-1$; 3. 3.

if $n=4k$, then $s(T)\leq 2k-2$.

To prove that the bounds in this theorem are the best possible, we use the double regularity as follows.

Theorem 1.2**.**

Let $l\in\{1,2,3\}$. Consider a doubly regular tournament $T$ of order $4k+3$, where $k\geq l$. The simplicity index of a tournament obtained from $T$ by removing $l$ vertices is $(2k+1)-l$.

As shown by the next result, the opposite direction in Theorem 1.2 holds when $l=1$.

Theorem 1.3**.**

Given a tournament $T$ with $4k+2$ vertices, where $k\geq 1$, if $s(T)=2k$, then $T$ is obtained from a doubly regular tournament by removing one vertex.

The existence of doubly regular tournaments is equivalent to the existence of skew-Hadamard matrices [3]. Wallis [20] conjectured that $n\times n$ skew-Hadamard matrices exist if and only if $n=2$ or $n$ is divisible by $4$. Infinite families of skew-Hadamard matrices can be found in [12].

The most known examples of a doubly regular tournament are obtained from Paley construction. For a prime power $q\equiv 3\pmod{4}$, the Paley tournament of order $q$ is the tournament whose vertex set is the finite field $\mathbb{F}_{q}$, such that $x$ dominates $y$ if and only if $x-y$ is a non-zero quadratic residue in $\mathbb{F}_{q}$.

2 Preliminaries

Let $T=(V,A)$ be an $n$-tournament and let $x\in V$. The * out-neighborhood* of $x$ is $N_{T}^{+}(x):=\left\{y\in V:x\rightarrow y\right\}$, and the in-neighborhood of $x$ is $N_{T}^{-}(x):=\left\{y\in V:y\rightarrow x\right\}$. The out-degree of $x$ ( resp. the in-degree of $x$) is $\delta_{T}^{+}(x):=\left|N_{T}^{+}(x)\right|$ (resp. $\delta_{T}^{-}(x):=\left|N_{T}^{-}(x)\right|$). The out-degree of $x$ is also called the * score* of $x$ in $T$. Recall that

[TABLE]

A tournament is near-regular if there exists an integer $k>0$ such that the out-degree of every vertex equals $k$ or $k-1$.

Remark 2.1**.**

Let $T$ be an $n$-tournament. It follows from (1) that

$T$* is regular if and only if $n$ is odd and every vertex has out-degree $\frac{(n-1)}{2}$;* 2. 2.

$T$* is near-regular if and only if $n$ is even and $T$ has $\frac{n}{2}$ vertices of out-degree $\frac{n}{2}$ and $\frac{n}{2}$ vertices of out-degree $\frac{(n-2)}{2}$.*

Notation 2.2**.**

Let $T=(V,A)$ be a near-regular tournament of order $4k+2$. We can partition $V$ into two $(2k+1)$-subsets, $V_{{\rm even}}:=\{z\in V,\delta_{T}^{+}(z)=2k\}$ and $V_{{\rm odd}}:=\{z\in V,\delta_{T}^{+}(z)=2k+1\}$.

Let $x,y$ be distinct vertices of an $n$-tournament $T=(V,A)$. The set $V\setminus\left\{x,y\right\}$ can be partitioned into four subsets:$\ N_{T}^{+}(x)\cap N_{T}^{+}(y)$, $N_{T}^{-}(x)\cap N_{T}^{-}(y)$, $N_{T}^{+}(x)\cap N_{T}^{-}(y)$ and $N_{T}^{-}(x)\cap N_{T}^{+}(y)$. The* out-degree* (resp. the in-degree) of $(x,y)$ is $\delta_{T}^{+}(x,y):=\left|N_{T}^{+}(x)\cap N_{T}^{+}(y)\right|$ (resp. $\delta_{T}^{-}(x,y):=\left|N_{T}^{-}(x)\cap N_{T}^{-}(y)\right|$). The elements of $(N_{T}^{+}(x)\cap N_{T}^{-}(y))\cup(N_{T}^{-}(x)\cap N_{T}^{+}(y))$ are called *separators *of $x,y$ and their number is denoted by $\sigma_{T}(x,y)$.

Lemma 2.3**.**

Let $T$ be an $n$-tournament with vertex set $V$. For any $x\neq y\in V$, we have

• •

$\sigma_{T}(x,y)+\delta_{T}^{-}(x,y)+\delta_{T}^{+}(x,y)=n-2$;

• •

$\delta_{T}^{-}(x,y)-\delta_{T}^{+}(x,y)=\delta_{T}^{-}(x)-\delta_{T}^{+}(y)$.

In particular, if $T$ is regular, then for any $x\neq y\in V$, $\delta_{T}^{-}(x,y)=\delta_{T}^{+}(x,y)$.

Proof.

The first assertion is obvious. For the second assertion, we have

[TABLE]

and

[TABLE]

Moreover, $y\in N_{T}^{-}(x)$ if and only if $x\in N_{T}^{+}(y)$. Then $\left|N_{T}^{-}(x)\cap\left\{y\right\}\right|=\left|N_{T}^{+}(y)\cap\left\{x\right\}\right|$ and hence

[TABLE]

Let $T=(V,A)$ be a tournament. For each vertex $z\in V$, we have

[TABLE]

By double-counting, we obtain

[TABLE]

In the next proposition, we give some basic properties of doubly regular tournaments. For the proof, see [15].

Proposition 2.4**.**

Let $T=(V,A)$ be a doubly regular $n$-tournament. There exists $k\geq 0$ such that $n=4k+3$, $T$ is regular, and for all $x,y\in V$ such that $x\rightarrow y$, we have

[TABLE]

[TABLE]

3 Proof of Theorem 1.1

Let $T=(V,A)$ be a tournament. Given a subset $B$ of $A$, we denote by ${\rm Inv}(T,B)$ the tournament obtained from $T$ by reversing all the arcs of $B$. We also use the following notation: $\delta_{T}^{+}$ $=\min\left\{\delta_{T}^{+}(x):x\in V\right\}$, $\delta_{T}^{-}$ $=\min\left\{\delta_{T}^{-}(x):x\in V\right\}$, $\delta_{T}=\min(\delta_{T}^{+},\delta_{T}^{-})$, and $\sigma_{T}=\min\{\sigma_{T}(x,y):x\neq y\in V\}$. The next proposition provides an upper bound of the simplicity index of a tournament.

Proposition 3.1**.**

For a tournament $T=(V,A)$ with at least $3$ vertices, we have $s(T)\leq\min(\delta_{T},\sigma_{T})$.

Proof.

Let $x\in V$. Clearly, the subset $V\setminus\left\{x\right\}$ is a non-trivial module of ${\rm Inv}(T,\left\{x\right\}\times N_{T}^{+}(x))$ and ${\rm Inv}(T,N_{T}^{-}(x)\times\left\{x\right\})$. It follows that

[TABLE]

Now, consider an unordered pair $\left\{x,y\right\}$ of vertices of $T$ and let

[TABLE]

Clearly, $\left\{x,y\right\}$ is a module of ${\rm Inv}(T,B)$. It follows that

[TABLE]

Hence, $s(T)\leq\sigma_{T}$. ∎

In addition to the previous proposition, the proof of Theorem 1.1 requires the following lemma.

Lemma 3.2**.**

Given an $n$-tournament $T=(V,A)$ with $n\geq$ $2$, we have

[TABLE]

Proof.

For every $x\in V$, we have $\min\left(\delta_{T}^{+}(x),\delta_{T}^{-}(x)\right)\leq\frac{n-1}{2}$. Thus,

[TABLE]

Now, to verify that $\sigma_{T}\leq\left\lfloor\dfrac{n-1}{2}\right\rfloor$, observe that

[TABLE]

It follows from (2) that

[TABLE]

Proof of Theorem 1.1.

For the first statement, suppose that $n=4k+2$. By Proposition 3.1 and Lemma 3.2, we have

[TABLE]

For the second statement, suppose that $n=4k+1$. By Proposition 3.1, $s(T)\leq\delta_{T}$. If $T$ is not regular, then $\delta_{T}<\frac{n-1}{2}$ and hence $s(T)\leq 2k-1$. Suppose that $T$ is regular and let $x\neq y\in V$. By Lemma 2.3, $\sigma_{T}(x,y)=n-2-$ $\delta_{T}^{-}(x,y)-\delta_{T}^{+}(x,y)$ and $\delta_{T}^{-}(x,y)=\delta_{T}^{+}(x,y)$. Therefore, $\sigma_{T}(x,y)$ is odd, and hence $\sigma_{T}$ is odd as well. By Lemma 3.2, $\sigma_{T}\leq\left\lfloor\frac{n-1}{2}\right\rfloor=2k$. Since $\sigma_{T}$ is odd, we obtain $\sigma_{T}\leq 2k-1$. It follows from Proposition 3.1 that $s(T)\leq 2k-1$.

For the third statement, suppose that $n=4k$. If $T$ is not near-regular, then $\delta_{T}<2k-1$, and hence $s(T)\leq 2k-2$ by Proposition 3.1. Suppose that $T$ is near-regular. By Remark 2.1, for every $z\in V$, $\delta_{T}^{+}(z)\in\{2k,2k-1\}$. It follows from (2) that

[TABLE]

Thus, we obtain

[TABLE]

Since $s(T)\leq\sigma_{T}$ by Proposition 3.1, we obtain $s(T)\leq\sigma_{T}\leq 2k-1$. Seeking a contradiction, suppose that $s(T)=2k-1$. We obtain $\sigma_{T}=2k-1$. Let $x\in V_{{\rm even}}$ and $y\in V_{{\rm odd}}$ (see Notation 2.2). It follows from Lemma 2.3 that $\sigma_{T}(x,y)$ is even and hence $\sigma_{T}(x,y)\geq 2k$. Thus, there are at least $(2k)^{2}$ unordered pairs $\{x,y\}$ satisfying $\sigma_{T}(x,y)\geq 2k$. For the other $2\binom{2k}{2}$ unordered pairs, we have $\sigma_{T}(x,y)\geq\sigma_{T}=2k-1$. It follows that

[TABLE]

which contradicts (3). Consequently, $s(T)\leq 2k-2$. ∎

4 Proof of Theorem 1.2

To begin, recall that a graph is defined by a vertex set $V$ and an edge set $E$. Two distinct vertices $x$ and $y$ of $G$ are * adjacent* if $\{x,y\}\in E$. For a vertex $x$ in $G$, the set $N_{G}(x):=\{y\in V:\{x,y\}\in E\}$ is called the neighborhood of $x$ in $G$. The degree of $x$ is $\delta_{G}(x):=\left|N_{G}(x)\right|$.

Let $T=(V,A)$ be a tournament. To each subset $C$ of $V$, we associate a graph in the following way. Denote by $s_{C}(T)$ the minimum number of arcs that must be reversed to make $C$ a module of $T$. Clearly,

[TABLE]

A graph $G=(V,E)$ is called a decomposability graph for $C$ if $\left|E\right|=s_{C}(T)$ and $C$ is a module of the tournament

[TABLE]

obtained from $T$ by reversing the arc between $x$ and $y$ for each edge $\{x,y\}$ of $G$. In the next lemma, we provide some of the properties of decomposability graphs.

Lemma 4.1**.**

Let $T=(V,A)$ be a $n$-tournament and let $C$ be a subset of $V$ such that $2\leq\left|C\right|\leq n-1$. Given a decomposability graph $G=(V,E)$ for $C$, the following assertions hold

• •

$G$* is bipartite with bipartition $\left\{C,V\setminus C\right\}$;*

• •

for each $x\in V\setminus C$, $N_{G}(x)=N_{T}^{+}(x)\cap C$ or $N_{G}(x)=N_{T}^{-}(x)\cap C$, and $\delta_{G}(x)=\min\left(\left|N_{T}^{-}(x)\cap C\right|,\left|N_{T}^{+}(x)\cap C\right|\right)$.

Proof.

The first assertion follows from the minimality of $\left|E\right|=s_{C}(T)$. For the second assertion, consider $x\in V\setminus C$. Since $C$ is a module of the tournament ${\rm Inv}(T,\{(x,y)\in A:\{x,y\}\in E\})$, we have $N_{G}(x)=N_{T}^{+}(x)\cap C$ or $N_{G}(x)=N_{T}^{-}(x)\cap C$. Furthermore, it follows from the minimality of $\left|E\right|=s_{C}(T)$ that $\delta_{G}(x)=\min\left(\left|N_{T}^{-}(x)\cap C\right|,\left|N_{T}^{+}(x)\cap C\right|\right)$. ∎

The next proposition is useful to prove Theorems 1.2 and 1.3.

Proposition 4.2**.**

Let $T=(V,A)$ be an $n$-tournament and let $C$ be a subset of $V$ such that $2\leq\left|C\right|\leq n-1$. Given a decomposability graph $G=(V,E)$ for $C$, the following statements hold

• •

if $n-\delta_{T}\leq\left|C\right|$, then $s_{C}(T)$ $\geq\delta_{T}$;

• •

if $\left|C\right|\leq\sigma_{T}$, then $s_{C}(T)\geq\sigma_{T}$.

Proof.

Before showing the first assertion, we establish

[TABLE]

Let $x\in V\smallsetminus C$. By the second assertion of Lemma 4.1

[TABLE]

Therefore, we obtain

[TABLE]

Since $G$ is bipartite with bipartition $\left\{C,V\setminus C\right\}$, we have

[TABLE]

It follows from (4) that

[TABLE]

Thus, (5) holds. Moreover, we have

[TABLE]

Now, to prove the first assertion, suppose that $n-\delta_{T}\leq\left|C\right|$. We obtain $(n-1-\left|C\right|)(\left|C\right|-(n-\delta_{T})\geq 0$, and hence $(n-\left|C\right|)(\left|C\right|-(n-1-\delta_{T}))\geq\delta_{T}$. It follows that $s_{C}(T)=\left|E\right|\geq\delta_{T}$.

Before showing the second assertion, we establish

[TABLE]

Consider two vertices $x\neq y\in C$. For convenience, denote by $\mathcal{S}_{T}(x,y)$ the set of separators of $\{x,y\}$. Clearly, we have $(\mathcal{S}_{T}(x,y)\setminus C)\subseteq N_{G}(x)\cup N_{G}(y)$. It follows that

[TABLE]

Consequently, we obtain

[TABLE]

Furthermore, observe that

[TABLE]

It follows from (8) that

[TABLE]

Therefore, we have

[TABLE]

Since $G$ is bipartite with bipartition $\left\{C,V\setminus C\right\}$, we have

[TABLE]

We obtain

[TABLE]

so (7) holds.

Finally, to prove the second assertion, suppose that $\left|C\right|\leq\sigma_{T}$. We obtain $\frac{\left|C\right|}{2}(2-\left|C\right|+\sigma_{T})\geq\sigma_{T}$. Since $s_{C}(T)=\left|E\right|$, it follows from (7) that $s_{C}(T)\geq\sigma_{T}$. ∎

Proof of Theorem 1.2.

Let $l\in\{1,2,3\}$. Consider a tournament $R$ from $T$ by removing $l$ vertices $v_{1},\ldots,v_{l}$. Set $V^{\prime}:=V\smallsetminus\{v_{1},\ldots,v_{l}\}$. It follows from Theorem 1.1 that $s(R)\leq(2k+1)-l$. It remains to show that $s(R)\geq(2k+1)-l$. By (4), it suffices to verify that $s_{C}(R)\geq(2k+1)-l$ for every subset $C$ of $V^{\prime}$ such that $2\leq\left|C\right|\leq(4k+2)-l$. Let $C\subseteq V^{\prime}$ such that $2\leq\left|C\right|\leq(4k+2)-l$. We distinguish the following three cases.

• •

Suppose that $2\leq\left|C\right|\leq(2k+1)-l$. Since $T$ is doubly regular, it follows from Proposition 2.4 that $\sigma_{T}=2k+1$. Therefore, $\sigma_{R}\geq(2k+1)-l$. Since $2\leq\left|C\right|\leq(2k+1)-l$, $\sigma_{R}\geq\left|C\right|$. It follows from Proposition 4.2 that $s_{C}(R)\geq\sigma_{R}$, and hence $s_{C}(R)\geq(2k+1)-l$.

• •

Suppose that $2k+2\leq\left|C\right|\leq(4k+2)-l$. Since $T$ is doubly regular, it follows from Proposition 2.4 that $T$ is regular. Thus, $\delta_{T}=2k+1$. It follows that $\delta_{R}\geq(2k+1)-l$. Since $2k+2\leq\left|C\right|\leq(4k+2)-l$, we obtain $\left|C\right|+\delta_{R}\geq(4k+3)-l$. It follows from Proposition 4.2 that $s_{C}(R)\geq\delta_{R}$, and hence $s_{C}(R)\geq(2k+1)-l$.

• •

Suppose that $(2k+2)-l\leq\left|C\right|\leq 2k+1$. Let $G=(E^{\prime},V^{\prime})$ be a decomposability graph for $C$. We verify that

[TABLE]

Otherwise, there exist $x\neq y\in V^{\prime}\smallsetminus C$ such that $\delta_{G}(x)=\delta_{G}(y)=0$. It follows from the second assertion of Lemma 4.1 applied to $R$ that $C$ is contained in one of the following intersections: $(N_{R}^{-}(x)\cap N_{R}^{+}(y))$, $(N_{R}^{-}(x)\cap N_{R}^{-}(y))$, $(N_{R}^{+}(x)\cap N_{R}^{+}(y))$, or $(N_{R}^{+}(x)\cap N_{R}^{-}(y))$. Thus, $C$ is contained in $(N_{T}^{-}(x)\cap N_{T}^{+}(y))$, $(N_{T}^{-}(x)\cap N_{T}^{-}(y))$, $(N_{T}^{+}(x)\cap N_{T}^{+}(y))$, or $(N_{T}^{+}(x)\cap N_{T}^{-}(y))$. It follows from Proposition 2.4 that $\left|C\right|\leq k+1$, which contradicts $\left|C\right|\geq(2k+2)-l$ because $k\geq l$. Consequently, (9) holds. Since $G$ is bipartite with bipartition $\left\{C,V^{\prime}\smallsetminus C\right\}$, we have $\left|E^{\prime}\right|=\sum_{x\in V^{\prime}\smallsetminus C}\delta_{G}(x)$. Since $\left|E^{\prime}\right|=s_{C}(R)$, we obtain

[TABLE]

5 Proof of Theorem 1.3

If a tournament $T$ is obtained from a doubly regular $(4k+3)$-tournament by deleting one vertex, then $T$ is near-regular and it follows from Proposition 2.4 that

• (C1)

if $x,y\in V_{{\rm even}}$ (see Notation 2.2) or $x,y\in V_{{\rm odd}}$, then $\sigma_{T}(x,y)=2k+1$.

• (C2)

if $x\in V_{{\rm even}}$ and $y\in V_{{\rm odd}}$, then $\sigma_{T}(x,y)=2k$.

Conversely, we have the following proposition.

Proposition 5.1**.**

Let $T=(V,A)$ be a near-regular tournament of order $4k+2$. If $T$ satisfies (C1) and (C2), then the tournament $U$ obtained from $T$ by adding a vertex $\omega$ which dominates $V_{{\rm odd}}$ and is dominated by $V_{{\rm even}}$ is doubly regular.

The proof of this proposition uses the following lemma.

Lemma 5.2**.**

Under the notation and conditions of Proposition 5.1, for every $x,y\in V$ such that $x\rightarrow y$, we have

• •

if $x,y\in V_{{\rm odd}}$, then

[TABLE]

• •

if $x,y\in V_{{\rm even}}$, then

[TABLE]

• •

if $x\in V_{{\rm odd}}$ and $y\in V_{{\rm even}}$, then

[TABLE]

• •

if $x\in V_{{\rm even}}$ and $y\in V_{{\rm odd}}$, then

[TABLE]

Proof.

We have

[TABLE]

By using Lemma 2.3, we obtain

[TABLE]

Using Assertions (C1) and (C2), we obtain the desired values of $|N_{T}^{-}(x)\cap N_{T}^{+}(y)|$. Then, $|N_{T}^{+}(x)\cap N_{T}^{-}(y)|$ follows immediately because $|N_{T}^{+}(x)\cap N_{T}^{-}(y)|=\sigma(x,y)-|N_{T}^{-}(x)\cap N_{T}^{+}(y)|$. ∎

Proof of Proposition 5.1.

Clearly, $U$ is regular. Furthermore, by Lemma 2.3,

[TABLE]

for distinct $x,y\in V\cup\{\omega\}$. Therefore, $U$ is doubly regular if and only if $\sigma_{U}(x,y)=2k+1$ for every $x,y\in V\cup\{\omega\}$. This equality follows directly from (C1) and (C2) when $x,y\in V$. Hence, it remains to prove that

[TABLE]

Consider $z\in V$. It is not difficult to see that

[TABLE]

Let $A_{{\rm odd}}:=(N_{T}^{+}(z)\cap V_{{\rm odd}})$, $A_{{\rm even}}:=(N_{T}^{+}(z)\cap V_{{\rm even}})$, $B_{{\rm odd}}:=(N_{T}^{-}(z)\cap V_{{\rm odd}})$, and $B_{{\rm even}}:=(N_{T}^{-}(z)\cap V_{{\rm even}})$. We determine $\left|A_{{\rm odd}}\right|$, $\left|A_{{\rm even}}\right|$, $\left|B_{{\rm odd}}\right|$, and $\left|B_{{\rm even}}\right|$ as follows.

To begin, suppose that $z\in V_{{\rm odd}}$. By counting the number of arcs from $N_{T}^{+}(z)$ to $N_{T}^{-}(z)$ in two ways, we get

[TABLE]

It follows from Lemma 5.2 that

[TABLE]

Since $z\in V_{{\rm odd}}$, we have $\left|A_{{\rm odd}}\right|+\left|A_{{\rm even}}\right|=2k+1$, $\left|B_{{\rm odd}}\right|+\left|B_{{\rm even}}\right|=2k$, $\left|A_{{\rm odd}}\right|+\left|B_{{\rm odd}}\right|=2k$, and $\left|A_{{\rm even}}\right|+\left|B_{{\rm even}}\right|=2k+1$. It follows that $\left|A_{{\rm odd}}\right|=\allowbreak k$, $\left|B_{{\rm odd}}\right|=k$, $\left|B_{{\rm even}}\right|=k$, and $\left|A_{{\rm even}}\right|=k+1$.

Similarly, if $z\in V_{{\rm even}}$, then $\left|A_{{\rm odd}}\right|=\allowbreak k$, $\left|B_{{\rm odd}}\right|=k+1$, $\left|B_{{\rm even}}\right|=k$, and $\left|A_{{\rm even}}\right|=k$.

Consequently, (12) holds whatever the parity of $\delta^{+}_{T}(z)$. ∎

Proof of Theorem 1.3.

Given $k\geq 1$, consider a tournament $T$, with $4k+2$ vertices, such that $s(T)=2k$. By Proposition 3.1, $\delta_{T}\geq 2k$. Thus, $T$ is near-regular. We conclude by applying Proposition 5.1. Therefore, it suffices to verify that (C1) and (C2) are satisfied.

By Proposition 3.1, $\sigma_{T}(x,y)\geq 2k$ for distinct $x,y\in V$. Moreover, it follows from Lemma 2.3 that if $x,y\in V_{{\rm even}}$ or $x,y\in V_{{\rm odd}}$ (see Notation 2.2), then $\sigma_{T}(x,y)$ is odd and hence $\sigma_{T}(x,y)\geq 2k+1$.

Lastly, seeking a contradiction, suppose that (C1) or (C2) are not satisfied. One of the following situations occurs

• •

there are distinct $x,y\in V_{{\rm even}}$ such that $\sigma_{T}(x,y)>2k+1$,

• •

there are distinct $x,y\in V_{{\rm odd}}$ such that $\sigma_{T}(x,y)>2k+1$,

• •

there are $x\in V_{{\rm even}}$ and $y\in V_{{\rm odd}}$ such that $\sigma_{T}(x,y)>2k$.

We obtain

[TABLE]

which contradicts (2). Consequently, (C1) and (C2) are satisfied. ∎

6 Concluding remarks

1. An $n$-tournament with $n=4k+1$ is called near-homogeneous [19] if every unordered pair of its vertices belongs to $k$ or $(k+1)$ 3-cycles. The existence of near-homogeneous tournaments is discussed in [19], [1], and [14]. For $n\equiv 1\pmod{4}$ or $n\equiv 0\pmod{4}$, the $n$-tournaments given in Theorem 1.2 are not the only ones with a maximal simplicity index. Indeed, let $T$ be a near-homogeneous tournament $T$ with $4k+1$ vertices. By adapting the proof of Theorem 1.2, we can verify that $s(T)=2k-1$. Moreover, by removing one vertex from $T$, we obtain a $(4k)$-tournament whose simplicity index is $2k-2$. Consequently, an analogue of Theorem 1.3 does not exist when $l=2$ or $3$.

2. The score vector of a $n$-tournament $T$ is the ordered sequence of the scores of $T$ listed in a non-decreasing order. Kirkland [11] proved that the reversal index of an $n$-tournament $T$ is equal to

[TABLE]

where $(s_{1},s_{2},\ldots,s_{n})$ is the score vector of $T$.

An equivalent form of this result was obtained earlier by Li and Huang [13]. As a consequence, two tournaments with the same score vector have the same reversal index. This fact is not true for the simplicity index. Indeed, for an odd number $n$, consider the $n$-tournament $R_{n}$ whose vertex set is the additive group $\mathbb{Z}_{n}=\{0,1,\ldots,n-1\}$ of integers modulo $n$, such that $i$ dominates $j$ if and only if $i-j\in\{1,\ldots,(n-1)/2\}$. It is not difficult to verify that the tournament $R_{n}$ is regular and simple. Moreover, by reversing the arc $\left(0,\frac{n-1}{2}\right)$, we obtain a non-simple tournament. Hence, the simplicity index of $R_{n}$ is $1$. If $n$ is prime and $n\equiv 3\pmod{4}$, the Paley tournament $P_{n}$ is also regular but its simplicity index is $\frac{n-1}{2}$.

Let $T$ be an $n$-tournament with vertex set $\{v_{1},\ldots,v_{n}\}$. The sequences $L_{1}=(\delta^{+}_{T}(v_{i}))_{1\leq i\leq n}$ and $L_{2}=(\delta^{+}_{T}(v_{i},v_{j}))_{1\leq i<j\leq n}$ are frequently used in our study of the simplicity index. It is natural to ask whether the simplicity index of $T$ can be expressed in terms of $L_{1}$ and $L_{2}$.