# The $\\{-3\\}$-reconstruction and the $\\{-3\\}$-self duality of tournaments

###### Abstract

Let $T=(V,A)$ be a (finite) tournament and $k$ be a non negative integer. For
every subset $X$ of $V$ is associated the subtournament $T[X]=(X,A\cap(X\times
X))$ of $T$, induced by $X$. The dual tournament of $T$, denoted by
$T^{\ast}$, is the tournament obtained from $T$ by reversing all its arcs. The
tournament $T$ is self dual if it is isomorphic to its dual. $T$ is
$\\{-k\\}$-self dual if for each set $X$ of $k$ vertices, $T[V\setminus X]$ is
self dual. $T$ is strongly self dual if each of its induced subtournaments is
self dual. A subset $I$ of $V$ is an interval of $T$ if for $a,b\in I$ and for
$x\in V\setminus I$, $(a,x)\in A$ if and only if $(b,x)\in A$. For instance,
$\emptyset$, $V$ and $\\{x\\}$, where $x\in V$, are intervals of $T$ called
trivial intervals. $T$ is indecomposable if all its intervals are trivial;
otherwise, it is decomposable. A tournament $T^{\prime}$, on the set $V$, is
$\\{-k\\}$-hypomorphic to $T$ if for each set $X$ on $k$ vertices,
$T[V\setminus X]$ and $T^{\prime}[V\setminus X]$ are isomorphic. The
tournament $T$ is $\\{-k\\}$-reconstructible if each tournament
$\\{-k\\}$-hypomorphic to $T$ is isomorphic to it.
Suppose that $T$ is decomposable and $\mid V\mid\geq 9$. In this paper, we
begin by proving the equivalence between the $\\{-3\\}$-self duality and the
strong self duality of $T$. Then we characterize each tournament
$\\{-3\\}$-hypomorphic to $T$. As a consequence of this characterization, we
prove that if there is no interval $X$ of $T$ such that $T[X]$ is
indecomposable and $\mid V\setminus X\mid\leq 2$, then $T$ is
$\\{-3\\}$-reconstructible. Finally, we conclude by reducing the
$\\{-3\\}$-reconstruction problem to the indecomposable case (between a
tournament and its dual). In particular, we find and improve, in a less
complicated way, the results of [6] found by Y. Boudabbous and A. Boussaïri.

Mouna Achour(a), Youssef Boudabbous(a), Abderrahim Boussaïri(b)

(a) Département de Mathématiques, Faculté des Sciences de Sfax, Université de
Sfax, BP $1171$, $3038$ Sfax, Tunisie. Fax: (00216) 74.27.44.37
E-mail: mouna<EMAIL_ADDRESS>youssef<EMAIL_ADDRESS>(b) Faculté des
Sciences Aïn Chock, Département de Mathématiques et Informatique, Km 8 route
d’El Jadida, BP 5366 Maarif, Casablanca, Maroc. E-mail<EMAIL_ADDRESS>
##  1 Introduction

###  1.1 Preliminaries on tournaments

A (finite) _tournament_ $T$ consists of a finite set $V$ of _vertices_ with a
prescribed collection $A$ of ordered pairs of distinct vertices, called the
set of _arcs_ of $T$, which satisfies: for $x,y\in V$ with $x\neq y$,
$(x,y)\in A$ if and only if $(y,x)\not\in A$. Such a tournament is denoted by
$(V,A)$. If $(x,y)$ is an arc of $T$, then we say that $x$ _dominates_ $y$
(symbolically $x\rightarrow y$). The _dual_ of the tournament $T$ is the
tournament $T^{\ast}=(V,A^{\ast})$ defined by: for all $x$, $y\in V$,
$(y,x)\in A^{\ast}$ if and only if $(x,y)\in A$. The tournament $T$ is
_transitive_ or a _linear order_ provided that for any $x,y,z\in V$, if
$(x,y)\in A$ and $(y,z)\in A$, then $(x,z)\in A$. For example, a total order
on a finite set $E$ can be identified to a transitive tournament with a vertex
set $E$ in the following way: for $x,y\in E$ with $x\neq y$, $x\rightarrow y$
if and only if $x<y$. The tournament corresponding to the usual order on
$\\{1,\ldots,n\\}$ (where $n\in{\mathbb{N}}^{\ast}$) is denoted by $O_{n}$. An
_almost transitive tournament_ is a tournament obtained from a transitive
tournament with at least three vertices by reversing the arc formed by its two
extremal vertices.

For every finite sets $E$ and $F$, we denote $E\subset F$ when $E$ is a subset
of $F$ and $\mid E\mid$ the cardinality of $E$.

Given a tournament $T=(V,A)$, for each subset $X$ of $V$ we associate the
_subtournament_ of $T$ induced by $X$, that is the tournament
$T[X]=(X,A\cap\\\ (X\times X))$. For convenience, the subtournament
$T[V\backslash X]$ is denoted by $T-X$, and by $T-x$ whenever $X=\\{x\\}$.

Let $T=(V,A)$ be a tournament, a subset $I$ of $V$ is an _interval_ of $T$ if
for every $x\in V\setminus I$, $x$ dominates or is dominated by all elements
of $I$. For instance, $\emptyset$, $V$ and $\\{x\\}$, where $x\in V$, are
intervals of $T$ called _trivial_ intervals. A tournament is _indecomposable_
if all its intervals are trivial; otherwise, it is _decomposable_. For
example, the tournament $C_{3}=(\\{1,2,3\\},\\{(1,2),(2,3),(3,1)\\})$ is
indecomposable, whereas, the tournaments
$C_{4}=(\\{1,2,3,4\\},\\{(1,2),(2,3),(3,4),(4,1),(3,1),(2,4)\\})$,
$\delta^{+}=(\\{1,2,3,4\\},\\\ \\{(1,2),(2,3),(3,1),(1,4),(2,4),(3,4)\\})$ and
$\delta^{-}=(\delta^{+})^{\ast}$ are decomposable.

Given two tournaments $T=(V,A)$ and $T^{\prime}=(V^{\prime},A^{\prime})$, an
_isomorphism_ from $T$ onto $T^{\prime}$ is a bijection $f$ from $V$ onto
$V^{\prime}$ satisfying: for any $x,y\in V$, $(x,y)\in A$ if and only if
$(f(x),f(y))\in A^{\prime}$. The tournaments $T$ and $T^{\prime}$ are
_isomorphic_ if there exists an isomorphism from one onto the other. This is
denoted by $T\sim T^{\prime}$. A tournament $T^{\prime}$ _embeds_ into a
tournament $T$ (or $T$ _embeds_ $T^{\prime}$ ), if $T^{\prime}$ is isomorphic
to a subtournament of $T$. A _$3$ -cycle_ (resp. _$4$ -cycle_) is a tournament
which is isomorphic to $C_{3}$ (resp. $C_{4}$). Moreover, a _positive diamond_
(resp. _negative diamond_) is a tournament that is isomorphic to $\delta^{+}$
(resp. $\delta^{-}$). A _diamond_ is a positive or a negative diamond. For
convenience, a set $X$ of vertices of a tournament $T$ is called diamond of
$T$ if $T[X]$ is a diamond.

###  1.2 Self duality and reconstruction

A tournament $T$ on a set $V$ is _self dual_ if $T$ and $T^{\ast}$ are
isomorphic, it’s _strongly self dual_ if for every subset $X$ of $V$, $T[X]$
and $T^{\ast}[X]$ are isomorphic. For each non negative integer $k$, the
tournament $T$ is _$(\leq k)$ -self dual_ whenever for every set $X$ of at
most $k$ vertices, the subtournament $T[X]$ is self dual. It is easy to see
that a transitive tournament or an almost transitive tournament is strongly
self dual. Conversely, Reid and Thomassen [23] was proved that a strongly self
dual tournament with at least $8$ vertices is transitive or almost transitive.
This result was used by K. B. Reid and C. Thomassen [23] in order to
characterize the pair of _hereditarily isomorphic_ tournaments, that is, the
pair of tournaments $T$ , $T^{\prime}$ on a set $V$ such that for every subset
$X$ of $V$, the subtournaments $T[X]$ and $T^{\prime}[X]$ are isomorphic. A
relaxed version of this notion is the following. Consider two tournaments $T$
and $T^{\prime}$ on the same vertex set $V$, with $\mid V\mid=n\geq 2$. Let
$k$ be an non negative integer $k$ with $k\leq n$. The tournaments $T$ and
$T^{\prime}$ are _$\\{k\\}$ -hypomorphic_, whenever for every set $X$ of $k$
vertices, the subtournaments $T[X]$ and $T^{\prime}[X]$ are isomorphic. If $T$
and $T^{\prime}$ are $\\{n-k\\}$-hypomorphic, we say that $T$ and $T^{\prime}$
are _$\\{-k\\}$ -hypomorphic_. Let $F$ be a set of integers. The tournaments
$T$ and $T^{\prime}$ are _$F$ -hypomorphic_, if for every $p\in F$, $T$ and
$T^{\prime}$ are $\\{p\\}$-hypomorphic, in particular, if
$F=\\{0,\ldots,k\\}$, we say that $T$ and $T^{\prime}$ are _$(\leq k)$
-hypomorphic_. For example, every two tournaments on the same vertex set with
at least $2$ elements are $(\leq 2)$-hypomorphic. The tournament $T$ is _$F$
-reconstructible_ provided that every tournament $F$-hypomorphic to $T$ is
isomorphic to $T$. This notion was introduced by R. Fraïssé [11] in 1970.

In 1972, G. Lopez ([15],[16]) showed that a tournament, with at least $6$
vertices, is $(\leq 6)$-reconstructible (see also [17]). It follows from a
”Combinatorial Lemma” of M. Pouzet (see Section $2$) that a tournament, with
at least $12$ vertices, is $\\{-6\\}$-reconstructible.

On the other hand, P. K. Stockmeyer [25] showed that the tournaments are not,
in general, $\\{-1\\}$-reconstructible, invalidating so the conjecture of Ulam
[26] for tournaments. Then, M. Pouzet ([1],[2]) proposed the
$\\{-k\\}$-reconstruction problem of tournaments. P. Ille [14] established
that a tournament with at least $11$ vertices is $\\{-5\\}$-reconstructible.
G. Lopez and C. Rauzy ([18],[19]) showed that a tournament with at least $10$
vertices is $\\{-4\\}$-reconstructible. The $\\{-k\\}$-reconstruction problem
of tournaments is still open for $k\in\\{2,3\\}$.

In 1995, Y. Boudabbous and A. Boussaïri [6] studied the $\\{-3\\}$ -recons-
truction of decomposable tournaments, for which they give a partial positive
answer.

In this paper, we find and improve, in a less complicated way, the results of
this study. In order to present our results, we need the Gallai’s
decomposition theorem [12] .

###  1.3 Gallai’s decomposition

Given a tournament $T=(V,A)$, we define on $V$ a binary relation $\mathcal{R}$
as follows: for all $x\in V$, $x\mathcal{R}x$ and for $x\neq y\in V$,
$x\mathcal{R}y$ if there exist two integers $n,\,m\geq 1$ and two sequences
$x_{0}=x,\ldots,x_{n}=y$ and $y_{0}=y,\ldots,y_{m}=x$ of vertices of $T$ such
that $x_{i}\longrightarrow x_{i+1}$ for $i=0,\ldots,n-1$ and
$y_{j}\longrightarrow y_{j+1}$ for $j=0,\ldots,m-1$. Clearly, $\mathcal{R}$ is
an equivalence relation on $V$. The equivalence classes of $\mathcal{R}$ are
called the _strongly connected components_ of $T$. A tournament is then
_strongly connected_ if it has at most one strongly connected component,
otherwise, it is _non-strongly connected_.

The next result is due to J. W. Moon [20].

###### Lemma 1

[20] Given a strongly connected tournament $T=(V,A)$ with $n\geq 3$ vertices,
for every integer $k\in\\{3,\ldots,n\\}$ and for every $x\in V$, there exists
a subset $X$ of $V$ such that $x\in X$, $\mid X\mid=k$ and the subtournament
$T[X]$ is strongly connected.

Let $T$ be a tournament on a set $V$. A partition $\mathcal{P}$ of $V$ is an
_interval partition_ of $T$ if all the elements of $\mathcal{P}$ are intervals
of $T$. It ensues that the elements of $\mathcal{P}$ may be considered as the
vertices of a new tournament, the _quotient_
$T/\mathcal{P}=(\mathcal{P},A/\mathcal{P})$ _of $T$ by $\mathcal{P}$_, defined
in the following way: for any $X\neq Y\in\mathcal{P}$, $(X,Y)\in
A/\mathcal{P}$ if $(x,y)\in A$, for $x\in X$ and $y\in Y$. On another hand, a
subset $X$ of $V$ is a _strong interval_ of $T$ provided that $X$ is an
interval of $T$ and for every interval $Y$ of $T$, if $X\cap Y\neq\emptyset$,
then $X\subset Y$ or $Y\subset X$. Here, for each tournament $T=(V,A)$ with
$\mid V\mid\geq 2$, $\mathcal{P}(T)$ denotes the family of maximal, strong
intervals of $T$, under the inclusion, amongst the strong intervals of $T$
distinct from $V$. Clearly, $\mathcal{P}(T)$ realizes an interval partition of
$T$.

Consider a tournament $H=(V,A)$. For every $x\in V$ is associated the
tournament $T_{x}=(V_{x},A_{x})$ such that the $V_{x}$’s are mutually
disjoint. The lexicographical sum of $T_{x}$’s over $H$ is the tournament $T$
denoted by $H(T_{x};x\in V)$ and defined on the union of $V_{x}$’s as follows:
given $u\in V_{x}$ and $v\in V_{y}$, where $x$, $y\in V$, $(u,v)$ is an arc of
$T$ if either $x=y$ and $(u,v)\in A_{x}$ or $x\neq y$ and $(x,y)\in A$. This
operation consists in fact to replace every vertex $x$ of $V$ by $T_{x}$ so
that $V_{x}$ becomes an interval; we say that the vertex $x$ is _dilated_ by
$T_{x}$. For example, an almost transitive tournament is obtained from a
$3$-cycle by dilating one of its vertices by a transitive tournament.

The Gallai’s decomposition theorem [12] consists in the following examination
of the quotient $T/\mathcal{P}(T)$.

###### Theorem 2

([9],[12]) Let $T$ be a tournament with at least two vertices.

  1. 1.

The tournament $T$ is non-strongly connected if and only if $T/\mathcal{P}(T)$
is transitive. In addition, if $T$ is non-strongly connected, then
$\mathcal{P}(T)$ is the family of the strongly connected components of $T$.

  2. 2.

The tournament $T$ is strongly connected if and only if $T/\mathcal{P}(T)$ is
indecomposable and $\mid\mathcal{P}(T)\mid\geq 3$.

We complete this subsection by the following notation.

###### Notation 3

For every tournament $T$ defined on a vertex set $V$ with at least two
elements, we associate the partition $\widetilde{\mathcal{P}}(T)$ of $V$
defined from $\mathcal{P}(T)$ as follows:

  * •

If $T$ is strongly connected, $\widetilde{\mathcal{P}}(T)=\mathcal{P}(T)$.

  * •

If $T$ is non-strongly connected, a subset $A$ of $V$ belongs to
$\widetilde{\mathcal{P}}(T)$ if and only if either $A\in\mathcal{P}(T)$ and
$\mid A\mid\geq 2$, or $A$ is a maximal union of consecutive vertices of the
transitive tournament $T/\mathcal{P}(T)$ which are singletons.

###  1.4 Statement of the results

In this paper, we begin by proving the following theorem (Section $3$). That
improves the result obtained by K. B. Reid and C. Thomassen [23]. (Note that,
the study of [23] is easly reduced to the decomposable case).

###### Theorem 4

A decomposable tournament which has at least $9$ vertices is $\\{-3\\}$-self
dual if and only if it is either transitive or almost transitive.

Then, we characterize each tournament $\\{-3\\}$-hypomorphic to a decomposable
tournament with at least $9$ vertices (Section $4$). For the statement, we
need additional notation.

###### Notation 5

Consider a set $P$ of non zero integers, an integer $n\geq 6$ and
$q\in\\{2,3\\}$. We denote by:

  * •

$I_{n,P}$, the class of tournaments with $n$ vertices which are
indecomposable, not self dual and $\\{p\\}$-self dual for every $p\in P$;
these tournaments being considered up to an isomorphism.

  * •

$C_{3}(I_{n,P})$ (resp. $O_{q}(I_{n,P})$), the class of tournaments with $n+2$
(resp. $n+q-1$) vertices obtained, from the $3$-cycle $C_{3}$ (resp. from the
transitive tournament $O_{q}$), by dilating one of its vertices by a
tournament belonging to the class $I_{n,P}$; these tournaments being
considered up to an isomorphism.

  * •

For each integer $m\geq 8$, we denote by $\Omega_{m}$ the union
$C_{3}(I_{m-2,\\{-1,-2,-3\\}})\\\ \cup O_{3}(I_{m-2,\\{-1,-2,-3\\}})\cup
O_{2}(I_{m-1,\\{-2,-3\\}})$.

Here is the characterization.

###### Theorem 6

Consider a decomposable tournament $T$ with $n\geq 9$ vertices and let
$T^{\prime}$ be a tournament $\\{-3\\}$-hypomorphic to $T$. Then, we have:

  1. 1.

$\widetilde{\mathcal{P}}(T^{\prime})=\widetilde{\mathcal{P}}(T)$ and one of
the following situations is achieved.

$(a)$

$T$ is almost transitive and $T^{\prime}\sim T$.

$(b)$

$T$ is not almost transitive,
$T/\widetilde{\mathcal{P}}(T)=T^{\prime}/\widetilde{\mathcal{P}}(T)$ and one
of the following situations is achieved.

$(i)$

$T\not\in\Omega_{n}$, for every $X\in\widetilde{\mathcal{P}}(T)$,
$T^{\prime}[X]\sim T[X]$ and $T^{\prime}\sim T$.

$(ii)$

$T\in\Omega_{n}$ and $T^{\prime}\not\sim T$.

  2. 2.

$T$ is not $\\{-3\\}$-reconstructible if and only if $T\in\Omega_{n}$.

We deduce the following results.

###### Corollary 7

Let $T$ be a decomposable tournament on a set $V$, with $\mid V\mid\geq 9$. If
there is no interval $X$ of $T$ such that $T[X]$ is indecomposable and $\mid
V\setminus X\mid\leq 2$, then $T$ is $\\{-3\\}$-reconstructible.

###### Corollary 8

If there exists an integer $n_{0}\geq 7$ for which the indecomposable
tournaments with at least $n_{0}$ vertices are $\\{-3\\}$-reconstructible,
then the tournaments with at least $n_{0}+2$ vertices are
$\\{-3\\}$-reconstructible.

By Theorem 17 (Section $2$) and Corollary 8, we reduce the
$\\{-3\\}$-reconstruction problem of tournaments to the indecomposable case
between a tournament and its dual.

##  2  Preliminary results

In this section, we recall and prove some results which will be used in next
sections.

First, concerning the decomposability, we recall the following notation, lemma
and corollary.

###### Notation 9

Given a tournament $T=(V,A)$, for each subset $X$ of $V$, such that
$\mid\\!X\\!\mid\geq 3$ and $T[X]$ is indecomposable, we associate the
following subsets of $V\setminus X$.

  * •

$Ext(X)$ is the set of $x\in V\setminus X$ such that $T[X\cup\\{x\\}]$ is
indecomposable.

  * •

$[X]$ is the set of $\ x\in V\setminus X\,$ such that $X$ is an interval of
$T[X\cup\\{x\\}]$.

  * •

For every $u\in X$, $X(u)$ is the set of $x\in V\setminus X\ $such that
$\\{u,x\\}$ is an interval of $T[X\cup\\{x\\}]$.

###### Lemma 10

[10] Let $T=(V,A)$ be a tournament and let $X$ be a subset of $V$ such that
$\mid\\!X\\!\mid\geq 3$ and $T[X]$ is indecomposable. The family $Ext(X),[X]$
and $X(u)$ where $u\in X$ constitutes a partition of $V\setminus X$. (Some
elements of this family can be empty).

The next result follows from Lemma 10.

###### Corollary 11

[10] Let $T=(V,A)$ be an indecomposable tournament. If $X$ is a subset of $V$,
such that $\mid\\!X\\!\mid\geq 3$, $\mid\\!V\setminus X\\!\mid\geq 2$ and
$T[X]$ is indecomposable, then there are distinct $x,y\in V\setminus X$ such
that $T[X\cup\\{x,y\\}]$ is indecomposable.

Second, we consider the following remark and notation.

###### Remark 12

i)

Up to an isomorphism, there exist four tournaments with four vertices:
$O_{4}$, $C_{4}$, $\delta^{+}$ and $\delta^{-}$. In addition, both of them is
decomposable.

ii)

Given two $\\{3\\}$-hypomorphic tournaments $T$ and $T^{\prime}$ with the same
vertex set $V$ with $\mid V\mid\geq 4$, $T$ and $T^{\prime}$ are $(\leq
4)$-hypomorphic if and only if for every subset $X$ of $V$, if $T[X]$ or
$T^{\prime}[X]$ is a diamond, then $T^{\prime}[X]\sim T[X]$.

iii)

Consider two $\\{3\\}$-hypomorphic tournaments $T$ and $T^{\prime}$. If $T$ is
without diamonds then $T^{\prime}$ and $T$ are $\\{4\\}$-hypomorphic.

###### Notation 13

Given a tournament $T=(V,A)$, a subset $F$ of $V$ and a tournament $H$, we
denote by $S(T,H;F)=\\{X\subset V;\,F\subset X$ and $T[X]\sim H\\}$ and
$n(T,H;F)=\mid S(T,H;F)\mid$.

Then, we recall the following ”Combinatorial Lemma” of M. Pouzet [21].

###### Lemma 14

[21] Let $p$, $r$ be two positive integers, $E$ be a set of at least $p+r$
elements and $U$, $U^{\prime}$ be two sets of subsets of $p$ elements of $E$.
If for each subset $Q$ of $E$ with $\mid Q\mid=p+r$, the number of the
elements of $U$ which are contained in $Q$ is equal to the number of the
elements of $U^{\prime}$ which are contained in $Q$, then for every finite
subsets $P^{\prime}$ and $Q^{\prime}$ of $E$, such that $P^{\prime}$ is
contained in $Q^{\prime}$ and $Q^{\prime}\setminus P^{\prime}$ has at least
$p+r$ elements, the number of elements of $U$ containing $P^{\prime}$ and
included in $Q^{\prime}$ is equal to the number of elements of $U^{\prime}$
containing $P^{\prime}$ and included in $Q^{\prime}$. In particular, if $E$
has at least $2p+r$ elements, then $U$ and $U^{\prime}$ are equal.

From the Combinatorial Lemma, we have the next corollaries.

###### Corollary 15

[22] Consider positive integers $n$, $p$, $h$ such that $p<n$ and $h\leq n-p$,
a tournament $H$ with $h$ vertices and two tournaments $T$ and $T^{\prime}$
defined on the same vertex set $V$ with $\mid V\mid=n$. If $T$ and
$T^{\prime}$ are $\\{-p\\}$-hypomorphic then for each subset $X$ of $V$ of at
most $p$ elements, $n(T^{\prime},H;X)=n(T,H;X)$.

###### Corollary 16

[21] Consider two tournaments $T=(V,A)$ and $T^{\prime}=(V,A^{\prime})$ and an
integer $p$ such that $0<p<\mid V\mid$. If $T$ and $T^{\prime}$ are
$\\{p\\}$-hypomorphic, then $T$ and $T^{\prime}$ are $\\{q\\}$-hypomorphic for
each $q\in\\{1,\ldots,min(p,\mid V\mid-p)\\}$.
In particular, if $\mid V\mid\geq 6$ and $T$ and $T^{\prime}$ are
$\\{-3\\}$-hypomorphic, then $T$ and $T^{\prime}$ are $(\leq 3)$-hypomorphic.

Now, recall the following theorem, called the ”Inversion Theorem”, which was
obtained by A. Boussaïri, P. Ille, G. Lopez and S. Thomassé ([8],[9]).

###### Theorem 17

([8],[9]) Given an indecomposable tournament $T$ with at least $3$ vertices,
the only tournaments which are $\\{3\\}$-hypomorphic to $T$ are $T$ and
$T^{\ast}$.

The following corollary is a consequence of Theorem 17.

###### Corollary 18

[9] Let $T$ and $T^{\prime}$ be two $\\{3\\}$-hypomorphic tournaments with at
least $3$ verices.

i)

$\mathcal{P}(T)=\mathcal{P}(T^{\prime})$.

ii)

$T$ is strongly connected (resp. indecomposable) if and only if $T^{\prime}$
is strongly connected (resp. indecomposable).

iii)

If $T$ is strongly connected, then the quotients $T^{\prime}/\mathcal{P}(T)$
and $T/\mathcal{P}(T)$ are either equal or dual.

From this corollary, we obtain the following remark.

###### Remark 19

Let $T$ and $T^{\prime}$ be two $\\{3\\}$-hypomorphic tournaments on a set $V$
with $\mid V\mid\geq 3$, and $I$ be a subset of $V$ such that
$T\left[I\right]$ is strongly connected. If $I$ is an interval of $T$, then
$I$ is an interval of $T^{\prime}$.

Given a tournament $T$ on a set $V$ and a subset $I$ of $V$, we denote by
$I^{+}_{T}$ (resp. $I^{-}_{T}$ ) the set of vertices $x\in V\setminus I$ such
that $I\longrightarrow x$ (resp. $x\longrightarrow I$).

We complete this section by the following result.

###### Proposition 20

Let $T$ and $T^{\prime}$ be two tournaments on a set $V$ with
$\left|V\right|\geq 6$, and $I$ be an interval of $T$ such that $\mid
I\mid\geq 3$ and $T\left[I\right]$ is indecomposable.

i)

If $T$ and $T^{\prime}$ are $\\{3,-2\\}$-hypomorphic (resp.
$\\{-3\\}$-hypomorphic) and $\left|V\setminus I\right|\geq 2$ (resp.
$\left|V\setminus I\right|\geq 3$), then $T\left[I\right]\backsim
T^{\prime}\left[I\right]$.

ii)

If $T$ and $T^{\prime}$ are $\\{3,-2\\}$-hypomorphic (resp.
$\\{-3\\}$-hypomorphic) and $\left|V\setminus I\right|\geq 3$ (resp.
$\left|V\setminus I\right|\geq 4$), then
$\left|I^{+}_{T}\right|=\left|I^{+}_{T^{\prime}}\right|$ and
$\left|I^{-}_{T}\right|=\left|I^{-}_{T^{\prime}}\right|$.

iii)

If $T$ and $T^{\prime}$ are $\\{-3\\}$-hypomorphic and $\left|V\setminus
I\right|\geq 4$, then $T\left[I\right]\backsim T^{\prime}\left[I\right]$,
$I^{+}_{T}=I^{+}_{T^{\prime}}$ and $I^{-}_{T}=I^{-}_{T^{\prime}}$.

Proof.

First, note that if $T$ and $T^{\prime}$ are $\\{-3\\}$-hypomorphic, then $T$
and $T^{\prime}$ are $\\{3\\}$-hypomorphic (by Corollary 16). Moreover, as
$T[I]$ is indecomposable (in particular, it is strongly connected because
$\mid I\mid\geq 3$) and $T[I]$ and $T^{\prime}[I]$ are $\\{3\\}$-hypomorphic,
then by Corollary 18, $T^{\prime}[I]$ is indecomposable and by Remark 19, $I$
is an interval of $T^{\prime}$.

i)

Let $a\neq b\in I$ and $J$ be a subset of $V$ containing $\\{a,b\\}$ such that
$T[J]$ is indecomposable and $\mid I\mid=\mid J\mid$ and denote by $H$ the
subtournament $T[I]$. As $I$ is an interval of $T$, then $I\cap J$ is an
interval of $T[J]$. However, $\\{a,b\\}\subset I\cap J$ and $T[J]$ is
indecomposable, then $I\cap J=J$ and hence $I=J$ (because $\mid I\mid=\mid
J\mid$). Thus, $I$ is the only subset $J$ of $V$ containing $\\{a,b\\}$ such
that $T[J]$ is indecomposable and $\mid J\mid=\mid I\mid$. In particular
$S(T,H;\\{a,b\\})=\\{I\\}$ and then $n(T,H;\\{a,b\\})=1$.
By interchanging $T$ and $T^{\prime}$ in the previous result, $I$ is the only
subset $J$ of $V$ containing $\\{a,b\\}$ such that $T^{\prime}[J]$ is
indecomposable and $\mid I\mid=$ $\mid J\mid$. In particular,
$S(T^{\prime},H;\\{a,b\\})\subset\\{I\\}$. Lastly, as $T$ and $T^{\prime}$ are
$\\{-2\\}$-hypomorphic (resp. $\\{-3\\}$-hypomorphic), $\mid V\mid\geq 6$ and
$\mid I\mid\leq$

$\mid V\mid-2$ (resp. $\mid I\mid\leq\mid V\mid-3$), then by Corollary 15,
$n(T,H;\\{a,b\\})=n(T^{\prime},H;\\{a,b\\})$. As $n(T,H;\\{a,b\\})=1$, then
$S(T^{\prime},H;\\{a,b\\})\neq\emptyset$ and so
$S(T^{\prime},H;\\{a,b\\})=\\{I\\}$. Consequently, $T^{\prime}[I]\sim T[I]$.

ii)

Let $a\neq b\in I$ and denote by $H$ a tournament with vertex set
$I\cup\\{u\\}$ (where $u\notin I$) such that $H[I]=T[I]$ and $I\longrightarrow
u$. Clearly, if $I^{+}_{T}\neq\emptyset$, then for each $x\in I^{+}_{T}$,
$I\cup\\{x\\}\in S(T,H;\\{a,b\\})$. Conversely, assume that
$S(T,H;\\{a,b\\})\neq\emptyset$ and consider an element $J$ of
$S(T,H;\\{a,b\\})$. Let $f$ be an isomorphism from $H$ to $T[J]$ and let
$\alpha=f(u)$. As $I$ is the unique non trivial interval of $H$ and
$f(I)=J\setminus\\{\alpha\\}$, then $J\setminus\\{\alpha\\}$ is the unique non
trivial interval of $T[J]$. However, $I\cap J$ is an interval of $T[J]$ and
$\\{a,b\\}\subset I\cap J$, then $I\cap J=J\setminus\\{\alpha\\}$ and hence
$J\setminus\\{\alpha\\}\subset I$. So, $J\setminus\\{\alpha\\}=I$ (because
$\mid I\mid=\mid J\setminus\\{\alpha\\}\mid$). Thus, $J=I\cup\\{\alpha\\}$ and
$\alpha\in I^{+}_{T}$.
We conclude that $S(T,H;\\{a,b\\})=\\{I\cup\\{x\\};\,x\in I^{+}_{T}\\}$ and
hence,
$n(T,H;\\{a,b\\})=\mid I^{+}_{T}\mid$.
As $T^{\prime}[I]\sim T[I]$ (by i)), then by interchanging $T$ and
$T^{\prime}$ in the previous result, we deduce that
$n(T^{\prime},H;\\{a,b\\})=$ $\mid I^{+}_{T^{\prime}}\mid$. Lastly, as $T$ and
$T^{\prime}$ are $\\{-2\\}$-hypomorphic (resp. $\\{-3\\}$-hypomorphic) and
$\mid I\cup\\{u\\}\mid\leq\mid V\mid-2$ (resp. $\mid I\cup\\{u\\}\mid\leq\mid
V\mid-3$), then by Corollary 15, $n(T,H;\\{a,b\\})=n(T^{\prime},H;\\{a,b\\})$.
Therefore, $\mid I^{+}_{T}\mid=\mid I^{+}_{T^{\prime}}\mid$ and hence, $\mid
I^{-}_{T}\mid=\mid I^{-}_{T^{\prime}}\mid$.

iii)

By ii), we have $\left|I^{+}_{T}\right|=\left|I^{+}_{T^{\prime}}\right|$.
Assume now that $I^{+}_{T}\setminus I^{+}_{T^{\prime}}\neq\emptyset$ and let
$x\in I^{+}_{T}\setminus I^{+}_{T^{\prime}}$. As $I$ is an interval of
$T^{\prime}$, then $x\in I^{-}_{T^{\prime}}$. The tournaments $T-x$ and
$T^{\prime}-x$ are $\\{3,-2\\}$-hypomorphic, $\mid V\setminus\\{x\\}\mid\geq
6$, and $\left|(V\setminus\left\\{x\right\\})\setminus I\right|\geq 3$, then
by ii), $\left|I^{+}_{T-x}\right|=\left|I^{+}_{T^{\prime}-x}\right|$. However,
$\left|I^{+}_{T-x}\right|=\left|I^{+}_{T}\right|-1\text{ and
}\left|I^{+}_{T^{\prime}-x}\right|=\left|I^{+}_{T^{\prime}}\right|$;
contradiction. It follows that $I^{+}_{T}\subset I^{+}_{T^{\prime}}$ and then
$I^{+}_{T}=I^{+}_{T^{\prime}}$. By duality, we obtain
$I^{-}_{T}=I^{-}_{T^{\prime}}$.

$\Box$

##  3 Proof of Theorem 4

For the tournaments without diamonds, H. Bouchaala and Y. Boudabbous [5]
established the following result.

###### Proposition 21

[5] Given a tournament $T$ without diamonds and which has at least $9$
vertices, $T$ is $\\{-3\\}$-self dual if and only if it is strongly self dual.

We present now some results concerning tournaments embedding a diamond.

###### Remark 22

A diamond $\delta$ has a unique non trivial interval $I$. Moreover,
$\delta[I]$ is a $3$-cycle.

###### Lemma 23

[4] If $T=(V,A)$ is a tournament embedding a diamond, then each vertex of $T$
is contained in at least one diamond of $T$.

The following proposition was obtained by M. Sghiar in 2004. This result plays
an important role in the proof of Theorem 4.

###### Proposition 24

[24] Let $T$ be a tournament, with at least $8$ vertices, embedding a diamond.
If $T$ has an interval of cardinality $2$, then $T$ is not $\\{-3\\}$-self
dual.

For the proof of this proposition, we need some definitions and notations.
Given a tournament $T=(V,A)$, if $X=\\{a,b,c,d\\}$ is a subset of $V$ such
that $T[X]$ is a diamond and $T[\\{a,b,c\\}]$ is a $3$-cycle, we say that $X$
is a diamond of $T$ of _center_ $d$ and cycle $\\{a,b,c\\}$. Let $x\neq y\in
V$, we denote:

  * •

$\delta_{T,\\{x,y\\}}^{+}$ (resp. $\delta_{T,\\{x,y\\}}^{-}$), the number of
positive (resp. negative) diamonds of $T$ whose cycle contains $\\{x,y\\}$.

  * •

$C_{T,\\{x,y\\}}$, the set of elements $w$ of $V$ such that $T[\\{x,y,w\\}]$
is a $3$-cycle.

  * •

$\delta_{T,\\{x,y,w\\}}^{+}$ (resp. $\delta_{T,\\{x,y,w\\}}^{-}$), the number
of positive (resp. negative) diamonds of $T$ whose cycle is $\\{x,y,w\\}$,
where $w$ is an element of $C_{T,\\{x,y\\}}$.

  * •

$D_{T,\\{x\\}}^{+}(y)$ (resp. $D_{T,\\{x\\}}^{-}(y)$), the number of positive
(resp. negative) diamonds of $T$ passing by $x$ and whose center is $y$.

  * •

$D_{T,\\{x,y\\}}^{+}$ (resp. $D_{T,\\{x,y\\}}^{-}$), the number of positive
(resp. negative) diamonds of $T$ passing by $x$ and $y$.

  * •

$\delta_{T}^{+}(x)$ (resp. $\delta_{T}^{-}(x)$), the number of positive (resp.
negative) diamonds of $T$ whose center is $x$.

###### Lemma 25

[24] Let $T=(V,A)$ be a $\\{-3\\}$-self dual tournament with at least $7$
vertices. If $T$ embeds a diamond, then every vertex of $T$ is the center of
at least one diamond of $T$.

Proof. Suppose for a contradiction that there exists a vertex $x$ of $T$ such
that $\delta^{+}_{T}(x)=\delta^{-}_{T}(x)=0$. From Lemma 23, there exists a
diamond $\sigma$ of $T$ containing $x$. By interchanging $T$ and $T^{\ast}$,
we can assume that $\sigma$ is a negative diamond. Let $y$ be the center of
$\sigma$. So, $D^{-}_{T,\\{x\\}}(y)\neq 0$ and $D^{+}_{T,\\{y\\}}(x)=0$. If
$C_{T,\\{x,y\\}}=\emptyset$, then
$\delta^{+}_{T,\\{x,y\\}}=\delta^{-}_{T,\\{x,y\\}}=0$. If
$C_{T,\\{x,y\\}}\neq\emptyset$, then pick $w\in C_{T,\\{x,y\\}}$ and let
$X=\\{x,y,w\\}$. As $T$ and $T^{\ast}$ are $\\{-3\\}$-hypomorphic and $\mid
V\mid\geq 7$, then from Corollary 15,
$n(T,\delta^{+};X)=n(T^{\ast},\delta^{+};X)$. So,
$n(T,\delta^{+};X)=n(T,\delta^{-};X)$ and hence
$\delta^{+}_{T,X}=\delta^{-}_{T,X}$. However,
$\delta^{+}_{T,\\{x,y\\}}=\sum\limits_{w\in
C_{T,\\{x,y\\}}}\delta_{T,\\{x,y,w\\}}^{+}$ and
$\delta_{T,\\{x,y\\}}^{-}=\sum\limits_{w\in
C_{T,\\{x,y\\}}}\delta_{T,\\{x,y,w\\}}^{-}$. Thus,
$\delta^{+}_{T,\\{x,y\\}}=\delta^{-}_{T,\\{x,y\\}}$. On the other hand, we
have
$D^{-}_{T,\\{x,y\\}}=n(T,\delta^{-};\\{x,y\\})=n(T^{\ast},\delta^{+};\\{x,y\\})$,
$D^{+}_{T,\\{x,y\\}}=n(T,\delta^{+};\\{x,y\\})$ and from Corollary 15,
$n(T,\delta^{+};\\{x,y\\})=n(T^{\ast},\delta^{+};\\{x,y\\})$, hence
$D^{+}_{T,\\{x,y\\}}=D^{-}_{T,\\{x,y\\}}$. However,
$D^{+}_{T,\\{x,y\\}}=D^{+}_{T,\\{y\\}}(x)+\delta^{+}_{T,\\{x,y\\}}$,
$D^{-}_{T,\\{x,y\\}}=D^{-}_{T,\\{x\\}}(y)+\delta^{-}_{T,\\{x,y\\}}$ and
$\delta^{+}_{T,\\{x,y\\}}=\delta^{-}_{T,\\{x,y\\}}$, thus,
$D^{+}_{T,\\{y\\}}(x)=D^{-}_{T,\\{x\\}}(y)$; which contradicts the fact that
$D^{-}_{T,\\{x\\}}(y)\neq 0$ and $D^{+}_{T,\\{y\\}}(x)=0$. $\Box$

###### Lemma 26

[24] Consider a $\\{-2\\}$-self dual (resp. $\\{-3\\}$-self dual) tournament
$T=(V,A)$ with at least $7$ (resp. $8$) vertices and two distinct vertices
$a,b$ of $T$. If $\\{a,b\\}$ is an interval of $T$ then
$\delta^{+}_{T}(a)=\delta^{-}_{T}(a)$.

Proof. Let $H$ be the tournament obtained from one positive diamond by
dilating its center by a tournament with $2$ vertices. Let
$\Delta^{+}_{T}(a)=\\{X\subset V;\,T[X]$ is a positive diamond with center
$a\\}$. Let $X$ be an element of $\Delta^{+}_{T}(a)$. As $\\{a,b\\}$ is an
interval of $T$, so $\\{a,b\\}\cap X$ is an interval of $T[X]$. Then, by
Remark 22, $b\not\in X$ and $\\{a,b\\}$ is an interval of $T[X\cup\\{b\\}]$.
Hence, $X\cup\\{b\\}\in S(T,H;\\{a,b\\})$. Let’s consider the map
$f\,:\,\Delta^{+}_{T}(a)\longrightarrow S(T,H;\\{a,b\\})$ defined by: for each
$X\in\Delta^{+}_{T}(a)$, $f(X)=X\cup\\{b\\}$. Clearly, $f$ is bijective and so
$\delta^{+}_{T}(a)=n(T,H;\\{a,b\\})$. By interchanging $T$ and $T^{\ast}$, we
deduce that
$\delta^{-}_{T}(a)=\delta^{+}_{T^{\ast}}(a)=n(T^{\ast},H;\\{a,b\\})$. On the
other hand, as $T$ and $T^{\ast}$ are $\\{-2\\}$-hypomorphic (resp.
$\\{-3\\}$-hypomorphic) and $\mid V\mid\geq 7$ (resp. $\mid V\mid\geq 8$),
then from Corollary 15, $n(T,H;\\{a,b\\})=n(T^{\ast},H;\\{a,b\\})$. Thus,
$\delta^{+}_{T}(a)=\delta^{-}_{T}(a)$. $\Box$

Proof of Proposition $24$. Assume by contradiction, that $T$ is
$\\{-3\\}$-self dual and has an interval $\\{a,b\\}$ with $a\neq b$. By Lemma
25, $T$ has a diamond $T\left[X\right]$ with center $a$. Assume for example
that $T\left[X\right]$ is a positive diamond. Clearly, $b\notin X$. Consider a
vertex $x\in X\setminus\\{a\\}$. As $T-x$ (resp. $T$) is $\\{-2\\}$-self dual
(resp. $\\{-3\\}$-self dual) and $\\{a,b\\}$ is an interval of $T-x$ (resp.
$T$), then, by Lemma 26, $\delta_{T-x}^{+}(a)=\delta_{T-x}^{-}(a)$ (resp.
$\delta_{T}^{+}(a)=\delta_{T}^{-}(a)$). So,
$0=\delta_{T}^{+}(a)-\delta_{T}^{-}(a)=\delta_{T-x}^{+}(a)+D_{T,\\{x\\}}^{+}(a)-\delta_{T-x}^{-}(a)=D_{T,\\{x\\}}^{+}(a)$;
contradiction.$\Box$

Theorem 4 is an immediate consequence of Proposition 21 and the below
proposition.

###### Proposition 27

Every decomposable tournament with at least $8$ vertices embedding a diamond
is not $\\{-3\\}$-self dual.

Proof. Let $T=(V,A)$ be a decomposable tournament with at least $8$ vertices
embedding a diamond. Assume by contradiction that $T$ is $\\{-3\\}$-self dual.
By Proposition 24, $T$ has no interval of size $2$. Let $I$ be a minimal
(under the inclusion) non trivial interval of $T$. Clearly, the subtournament
$T\left[I\right]$ is indecomposable. And then it is strongly connected.
Firstly, assume that $\left|V\setminus I\right|\geq 4$. Let $z\in V\setminus
I$ and suppose, for example that $z\longrightarrow I$ in $T$. As $T$ and
$T^{\ast}$ are $\\{-3\\}$-hypomorphic, then from Proposition 20,
$z\longrightarrow I$ in $T^{\ast}$; contradiction. Secondly, assume that
$\left|V\setminus I\right|\leq 3$. Let $k=\mid V\setminus I\mid$. We have so
$k\in\\{1,2,3\\}$ and $\mid I\mid\geq 8-k$. As the subtournament $T[I]$ is
strongly connected and $8-k\in\\{5,6,7\\}$, then from Lemma 1, there exists a
subset $X$ of $I$ such that $\mid X\mid=4-k$ and $T[I]-X$ is strongly
connected. Let $Y$ be a subset of $V\setminus I$ such that $\mid Y\mid=k-1$.
Clearly, the subtournament $T-(X\cup Y)$ is not self dual; contradicts the
fact that $\mid X\cup Y\mid=3$. $\Box$

##  4 Proof of Theorem 6

The proof of Theorem 6 is based on the next result.

###### Proposition 28

Let $T$ be a strongly connected and decomposable tournament on a set $V$ with
$\mid V\mid=n\geq 9$, which is not almost transitive. If $T^{\prime}$ is a
tournament $\\{-3\\}$-hypomorphic to $T$, then the following assertions hold.

  1. 1.

$\mathcal{P}(T^{\prime})=\mathcal{P}(T)$ and
$T^{\prime}/\mathcal{P}(T)=T/\mathcal{P}(T)$.

  2. 2.

If there exists $X\in\mathcal{P}(T)$ such that $T^{\prime}[X]\not\sim T[X]$,
then $\mid\mathcal{P}(T)\mid=3$ and $\mid X\mid=n-2$.

  3. 3.

If for each $X\in\mathcal{P}(T)$, $\mid X\mid\leq n-3$, then for each
$X\in\mathcal{P}(T)$, $T^{\prime}[X]\sim T[X]$ and in particular
$T^{\prime}\sim T$.

For the proof of the last proposition, we use the following remark and lemma.

###### Remark 29

Let $T$ and $T^{\prime}$ be two tournaments on a set $V$ with $\mid V\mid\geq
3$ and $\Gamma$ be a common interval partition of $T$ and $T^{\prime}$ such
that $T/\Gamma=T^{\prime}/\Gamma$. Given a non empty subset $A$ of $V$ and let
$\Gamma_{A}=\\{X\cap A;\,X\in\Gamma$ and $X\cap A\neq\emptyset\\}$. Then
$\Gamma_{A}$ is a common interval partition of $T[A]$ and $T^{\prime}[A]$ and
$T[A]/\Gamma_{A}=T^{\prime}[A]/\Gamma_{A}$. Suppose that for each
$Y\in\Gamma_{A}$, there exists an isomorphism $\varphi_{Y}$ from $T[Y]$ onto
$T^{\prime}[Y]$ and consider the map $f\,:\,A\longrightarrow A$ defined by:
for each $x\in A$, $f(x)=\varphi_{Y}(x)$ where $Y$ is the unique element of
$\Gamma_{A}$ such that $x\in Y$. Then, $f$ is an isomorphism from $T[A]$ onto
$T^{\prime}[A]$. In particular, if for each $X\in\Gamma$, $T[X]$ and
$T^{\prime}[X]$ are hereditarily isomorphic, then $T$ and $T^{\prime}$ are
hereditarily isomorphic.

###### Lemma 30

[3] Let $T=(V,A)$ and $T^{\prime}=(V^{\prime},A^{\prime})$ be two isomorphic
tournaments, $f$ be an isomorphism from $T$ onto $T^{\prime}$, $i\in V$ and
$R_{i}$ (resp. $R_{i}^{\prime}$) be a tournament defined on a vertex set
$I_{i}$ (resp. $I_{i}^{\prime}$) disjoint from $V$ (resp. $V^{\prime}$). Let
$R$ (resp. $R^{\prime}$) be the tournament obtained from $T$ (resp.
$T^{\prime}$) by dilating the vertex $i$ (resp. $f(i)$) by $R_{i}$ (resp.
$R_{i}^{\prime}$). Then $R\sim R^{\prime}$ if and only if $R_{i}\sim
R_{i}^{\prime}$.

Note that this lemma is a simple generalization of a result communicated by A.
Boussaïri, and on which $V^{\prime}=V$ and $f=id_{V}$.

Proof of Proposition 28.

  1. 1.

By Corollary 16, $T$ and $T^{\prime}$ are $(\leq 3)$-hypomorphic. So, from
Corollary 18, $\mathcal{P}(T)=\mathcal{P}(T^{\prime})$ and
$T^{\prime}/\mathcal{P}(T)=T/\mathcal{P}(T)$ or
$T^{\prime}/\mathcal{P}(T)=T^{\ast}/\mathcal{P}(T)$.
Assume by contradiction that
$T^{\prime}/\mathcal{P}(T)=T^{\ast}/\mathcal{P}(T)$. In this case, we are
going to show that for every $X\in\mathcal{P}(T)$, $T[X]$ is transitive, and
thus by Remark 29, $T^{\prime}$ is hereditarily isomorphic to $T^{\ast}$.
Since $T^{\prime}$ is $\\{-3\\}$-hypomorphic to $T$, then $T$ is
$\\{-3\\}$-self dual. By Theorem 4, the tournament $T$ is almost transitive;
which contradicts the hypothesis. For that, proceed by contradiction and
consider an element $X$ of $\mathcal{P}(T)$ and a subset
$\\{\alpha,\beta,\gamma\\}$ of $X$ such that $T[\\{\alpha,\beta,\gamma\\}]$ is
a $3$-cycle. As $T$ is strongly connected, there is $a\in V\setminus X$ such
that $X\longrightarrow a$. From Corollary 15,
$n(T,\delta^{+};\\{\alpha,\beta,a\\})=n(T^{\prime},\delta^{+};\\{\alpha,\beta,a\\})$.
Moreover, $n(T,\delta^{+};\\{\alpha,\beta,a\\})\neq 0$, because
$T[\\{\alpha,\beta,\gamma,a\\}]\sim\delta^{+}$, then there exists a subset $K$
of $V$ such that $\\{\alpha,\beta,a\\}\subset K$ and $T^{\prime}[K]$
$\sim\delta^{+}$. Hence, $\mid K\cap X\mid=3$, because otherwise $K\cap X$ is
an interval with two elements of $T^{\prime}[K]$; which contradicts Remark 22.
So, $T^{\prime}[K]$ is written: $(K\cap X)\longleftarrow a$; which contradicts
the fact that $T^{\prime}[K]\sim\delta^{+}$.

  2. 2.

By 1, we have $\mathcal{P}(T^{\prime})=\mathcal{P}(T)$ and
$T^{\prime}/\mathcal{P}(T)=T/\mathcal{P}(T)$. We distinguish the following two
cases.

     * •

If for every $X\in\mathcal{P}(T)$, $\mid X\mid\leq
n-\left|\mathcal{P}(T)\right|-2$.
Let $X\in\mathcal{P}(T)$ and let $H$ be the tournament obtained from
$T/\mathcal{P}(T)$ by dilating the vertex $X$ by $T[X]$. Assume that $\mid
X\mid\geq 2$ and consider a subset $A$ of $X$ with $2$ elements. Consider a
subset $B$ of $V$ containing $X$ such that:
$\forall\,Y\in\mathcal{P}(T)\setminus\\{X\\}$, $\mid B\cap Y\mid=1$. Clearly,
$B\in S(T,H;A)$ and hence $n(T,H;A)\neq\emptyset$. From Corollary 15,
$n(T^{\prime},H;A)=n(T,H;A)$, then $n(T^{\prime},H;A)\neq 0$. So, there exists
a subset $K$ of $V$ such that: $A\subset K$ and $T^{\prime}[K]\sim H$. Then
$\mathcal{P}(T^{\prime}[K])$ have a unique element $J$ non reduced to a
singleton. Clearly, $T^{\prime}[J]\sim T[X]$, in particular, $\mid J\mid=\mid
X\mid$.
Let $P_{1}=\\{Y\in\mathcal{P}(T);\,\mid Y\cap K\mid\geq 2\\}$ and
$P_{2}=\\{Y\in\mathcal{P}(T);\\\ \mid Y\cap K\mid=1\\}$. The set $K$ is the
union of the two disjoint sets $K_{1}=\bigcup\limits_{Y\in P_{1}}K\cap Y$ and
$K_{2}=\bigcup\limits_{Y\in P_{2}}K\cap Y$. As $A\subset X\cap K$, then $X\in
P_{1}$ and so, $P_{2}\subset(\mathcal{P}(T)\setminus\\{X\\})$, in particular,
$\mid P_{2}\mid\leq\mid\mathcal{P}(T)\mid-1$. For all $Y\in P_{1}$, $Y\cap K$
is a non trivial interval of $T^{\prime}[K]$, then $Y\cap K\subset J$, so
$K_{1}\subset J$ and thus $(K\setminus J)\subset K_{2}$.
Thus, $\mid\mathcal{P}(T)\mid-1=\mid K\setminus J\mid\leq\sum\limits_{Y\in
P_{2}}\mid K\cap Y\mid=\mid P_{2}\mid\leq\mid\mathcal{P}(T)\mid-1$. So, $\mid
P_{2}\mid=\mid\mathcal{P}(T)\mid-1$ and then
$P_{2}=\mathcal{P}(T)\setminus\\{X\\}$ and $P_{1}=\\{X\\}$. Thus, $K=(X\cap
K)\cup K_{2}$. So, $\mid X\cap K\mid=\mid K\mid-\mid K_{2}\mid=$ $\mid
K\mid-\mid P_{2}\mid=\mid K\mid-\mid\mathcal{P}(T)\mid+1=\mid J\mid$. As in
addition $X\cap K\subset J$, then, $X\cap K=J$. So, $J\subset X$ and then
$J=X$ because $\mid J\mid=\mid X\mid$. Consequently, $T^{\prime}[X]\sim T[X]$.

     * •

If there exists an element $X$ of $\mathcal{P}(T)$ such that $\mid
X\mid>n-\left|\mathcal{P}(T)\right|-2$.
In this case, it is clear that for every $Y\in\mathcal{P}(T)\setminus\\{X\\}$,
$\mid Y\mid\leq 3$, so $T^{\prime}[Y]$ and $T[Y]$ are hereditarily isomorphic.
Suppose that $\mid V\setminus X\mid\geq 3$. Let $x$ be an element of $X$ and
$B$ be a subset of $V\setminus X$ such that $\mid B\mid=3$. Denote by
$V_{(x,B)}$ the set $(V\setminus(X\cup B))\cup\\{x\\}$ and by $T_{(x,B)}$
(resp. $T_{(x,B)}^{\prime}$) the subtournament $T[V_{(x,B)}]$ (resp.
$T^{\prime}[V_{(x,B)}]$). The tournament $T-B$ (resp. $T^{\prime}-B$) is
obtained from the tournament $T_{(x,B)}$ (resp. $T_{(x,B)}^{\prime}$) by
dilating the vertex $x$ by $T[X]$ (resp. $T^{\prime}[X]$). Moreover, by Remark
29, there exists an isomorphism $g$ from $T_{(x,B)}$ onto $T_{(x,B)}^{\prime}$
such that $g(x)=x$. As in addition, $T-B$ and $T^{\prime}-B$ are isomorphic,
then, from Lemma 30, $T^{\prime}[X]$ is isomorphic to $T[X]$; which permits to
conclude.

  3. 3.

Is a direct consequence of 2.

$\square$

###### Lemma 31

Every strongly connected and decomposable tournament, which has $8$ vertices,
is $\\{-2,-3\\}$-reconstructible.

Proof. Let $H$ be a strongly connected and decomposable tournament defined on
a vertex set $X$ with $\mid X\mid=8$ and let $H^{\prime}$ be a tournament
$\\{-2,-3\\}$-hypomorphic to $H$. The tournaments $H$ and $H^{\prime}$ are
$\\{3\\}$-hypomorphic, by Corollary 16, so they are $\\{3,5,6\\}$-hypomorphic.
Besides, by Corollary 18, $\mathcal{P}(H)=\mathcal{P}(H^{\prime})$, and
$H^{\prime}/\mathcal{P}(H)=H/\mathcal{P}(H)$ or
$H^{\prime}/\mathcal{P}(H)=H^{\ast}/\mathcal{P}(H)$. Let’s put
$Q=\mathcal{P}(H)$, and discuss according to its cardinal.

  * •

If $\mid Q\mid>3$.
If for every $Z\in Q$, $\mid Z\mid\leq 3$. By $(\leq 3)$-hypomorphy,
$H^{\prime}[Z]\sim H[Z];\,\forall Z\in Q$. If $H^{\prime}/Q=H/Q$, clearly
$H^{\prime}\sim H$. Suppose hence that $H^{\prime}/Q=H^{\ast}/Q$. In this
case, $H^{\prime}$ is hereditarily isomorphic to $H^{\ast}$. Then, $H$ is
$\\{-3\\}$-self dual. From Proposition 27, $H$ is without diamonds. By Remark
12, iii), $H$ is $\\{4\\}$-self dual. Thus, $H$ is $\\{4,5,6\\}$-self dual.
So, $H$ is $(\leq 6)$-self dual and it is thus self dual, by the $(\leq
6)$-reconstruction of tournaments with at least $6$ vertices [17]. It follows
that $H^{\prime}\sim H$.
If there exists $Y\in Q$ such that $\mid Y\mid\geq 4$. In this case, as $\mid
X\mid=8$, $\mid Q\mid>3$ and every tournament with $4$ vertices is
decomposable, then $\mid Q\mid=5$, $\mid Y\mid=4$ and for every $Z\in
Q\setminus\\{Y\\}$, $\mid Z\mid=1$.

    * –

If $H^{\prime}/Q=H/Q$. For $z\in X\setminus Y$, as $\mid Y\cup\\{z\\}\mid=\mid
X\mid-3$, then, $H[Y\cup\\{z\\}]\sim H^{\prime}[Y\cup\\{z\\}]$. It follows
that $H^{\prime}[Y]\sim H[Y]$, and then $H^{\prime}\sim H$.

    * –

If $H^{\prime}/Q=H^{\ast}/Q$.

      * *

If $H[Y]$ is not a diamond. In this case, clearly $H^{\prime}[Y]$ is
hereditarily isomorphic to $H^{\ast}[Y]$ and hence $H^{\prime}$ and $H^{\ast}$
are hereditarily isomorphic. As $H^{\prime}$ is $\\{-3\\}$-hypomorphic to $H$,
then $H$ is $\\{-3\\}$-self dual. From Proposition 27, $H$ is then without
diamonds, and it is clearly self dual and thus, $H^{\prime}\sim H$.

      * *

If $H[Y]$ is a diamond. In this case, if $H^{\prime}[Y]$ is not isomorphic to
$H[Y]$, then $H^{\prime}$ is hereditarily isomorphic to $H^{\ast}$ and so, $H$
is $\\{-3\\}$-self dual; which contradicts Proposition 27. So,
$H^{\prime}[Y]\sim H[Y]$. For $z\in X\setminus Y$, it is easy to verify that
$H^{\prime}[Y\cup\\{z\\}]$ is not isomorphic to $H[Y\cup\\{z\\}]$; which
contradicts the $\\{-3\\}$-hypomorphy between $H^{\prime}$ and $H$.

  * •

If $\mid Q\mid=3$. We distinguish the following two sub-cases.

    * –

If there exists an unique $Y\in Q$ such that $\mid Y\mid>1$.
In this case, $Q$ is written: $Q=\\{\\{a\\},\\{b\\},Y\\}$ where $\mid Y\mid=6$
and $a\longrightarrow Y\longrightarrow b$ in $H$. As $\mid Y\mid=6$, we have:
$H^{\prime}[Y]\sim H[Y]$ by $\\{-2\\}$-hypomorphy. Thus, $H^{\prime}\sim H$.

    * –

If there are $Y\neq Z\in Q$ such that $\min(\mid Y\mid,\mid Z\mid)>1$.
In this case, we can write: $Q=\\{Y_{1},Y_{2},Y_{3}\\}$ with: $\mid
Y_{3}\mid\leq\mid Y_{2}\mid\leq$ $\mid Y_{1}\mid$, $\mid Y_{2}\mid\geq 2$ and
$\mid Y_{1}\mid\geq 3$. By considering a subset $A_{1}$ (resp. $A_{2}$) with
$3$ (resp. $2$) elements of $Y_{1}$ (resp. $Y_{2}$) and an element $y_{3}$ of
$Y_{3}$, we see that the isomorphy between $H[A_{1}\cup A_{2}\cup\\{y_{3}\\}]$
and $H^{\prime}[A_{1}\cup A_{2}\cup\\{y_{3}\\}]$ requires that:
$H^{\prime}/Q=H/Q$. So, if $\mid Y_{1}\mid=3$, then for every
$i\in\\{1,2,3\\}$, $H^{\prime}[Y_{i}]\sim H[Y_{i}]$ and then $H^{\prime}\sim
H$. Suppose thus that $\mid Y_{1}\mid\geq 4$. As $\mid X\mid=8$, then $\mid
Y_{1}\mid\in\\{4,5\\}$ and $\mid Y_{2}\mid\leq 3$. By considering an element
$y_{2}$ of $Y_{2}$, we see that the isomorphy between
$H[Y_{1}\cup\\{y_{2}\\}]$ and $H^{\prime}[Y_{1}\cup\\{y_{2}\\}]$ requires the
isomorphy between $H^{\prime}[Y_{1}]$ and $H[Y_{1}]$. It follows that
$H^{\prime}\sim H$.

$\Box$

Proof of Theorem 6. Consider a decomposable tournament $T$ defined on a vertex
set $V$ with $\mid V\mid=n\geq 9$, a tournament $T^{\prime}$
$\\{-3\\}$-hypomorphic to $T$. By Corollary 16, $T^{\prime}$ is $(\leq
3)$-hypomorphic to $T$. So, from Corollary 18,
$\mathcal{P}(T)=\mathcal{P}(T^{\prime})$. In particular, if $T$ is strongly
connected, then $\widetilde{P}(T)=\widetilde{P}(T^{\prime})$.
If $T$ is almost transitive, then clearly $T^{\prime}\sim T$. Let’s suppose
that $T$ is not almost transitive. For the proof, we distinguish the following
two cases.

$\ast$

$T$ is strongly connected.

In this case, from Proposition 28,
$T^{\prime}/\mathcal{P}(T)=T/\mathcal{P}(T)$ and we can suppose that
$\mid\mathcal{P}(T)\mid=3$ and there exists $X\in\mathcal{P}(T)$ such that
$\mid X\mid=n-2$. Let $\mathcal{P}(T)=\\{X,\\{a\\},\\{b\\}\\}$ where
$X\longrightarrow a\longrightarrow b\longrightarrow X$ in $T$. We verify
easily that $T^{\prime}[X]$ and $T[X]$ are $\\{-1,-2,-3\\}$-hypomorphic
(because $T^{\prime}$ and $T$ are $\\{-3\\}$-hypomorphic). We consider the
following three cases.

  * •

$T[X]$ is non-strongly connected.

As the tournaments $T^{\prime}[X]$ and $T[X]$ are $\\{-1\\}$-hypomorphic, then
they are isomorphic, since the non-strongly connected tournaments with at
least $5$ vertices are $\\{-1\\}$-reconstructible [13].

  * •

$T[X]$ is strongly connected and decomposable.

Let $Q=\mathcal{P}(T[X])$. Assume that there exists $Y\in Q$ such that $\mid
Y\mid=$
$\mid X\mid-2$. In this case, we have $\mid Q\mid=3$ and $T[X]/Q$ is a
$3$-cycle. Moreover, as $T^{\prime}[X]$ and $T[X]$ are $(\leq 3)$-hypomorphic,
then by Corollary 18, $\mathcal{P}(T^{\prime}[X])=\mathcal{P}(T[X])$ and
$T^{\prime}[X]/\mathcal{P}(T[X])=T[X]/\mathcal{P}(T[X])$ or
$T^{\prime}[X]/\mathcal{P}(T[X])=T^{\ast}[X]/\mathcal{P}(T[X])$. As in
addition, $T^{\prime}[Y]\sim T[Y]$ (because $T^{\prime}[X]$ and $T[X]$ are
$\\{-2\\}$-hypomorphic), then $T^{\prime}[X]\sim T[X]$. Thus, clearly
$T^{\prime}\sim T$.
Now, suppose that for every $Y\in Q$, $\mid Y\mid<\mid X\mid-2$. In this case,
$T[X]$ is not almost transitive. If $\mid X\mid\geq 9$, as $T^{\prime}[X]$ and
$T[X]$ are $\\{-3\\}$-hypomorphic, then by Proposition 28, $T^{\prime}[X]\sim
T[X]$ and hence, clearly $T^{\prime}\sim T$. If $\mid X\mid=7$. As
$T^{\prime}[X]$ and $T[X]$ are $\\{-1,-2,-3\\}$-hypomorphic, then they are
$(\leq 6)$-hypomorphic, and thus $T^{\prime}[X]\sim T[X]$ (by [17]). If $\mid
X\mid=8$. As $T^{\prime}[X]$ and $T[X]$ are $\\{-2,-3\\}$-hypomorphic, then,
by Lemma 31, $T^{\prime}[X]\sim T[X]$.

  * •

If $T[X]$ is indecomposable.
In this case, from Theorem 17, $T^{\prime}[X]=T[X]$ or
$T^{\prime}[X]=T^{\ast}[X]$ (because $T^{\prime}[X]$ and $T[X]$ are $(\leq
3)$-hypomorphic). If $T^{\prime}[X]=T[X]$, then clearly $T^{\prime}\sim T$. If
$T^{\prime}[X]=T^{\ast}[X]$, then $T[X]$ is $\\{-1,-2,-3\\}$-self dual
(because $T^{\prime}[X]$ and $T[X]$ are $\\{-1,-2,-3\\}$-hypomorphic). We
obtain: If $T[X]$ is self dual, then $T^{\prime}[X]\sim T[X]$ and clearly
$T^{\prime}\sim T$. If $T[X]$ is not self dual, then $T\in
C_{3}(I_{n-2,\\{-1,-2,-3\\}})\subset\Omega_{n}$ and clearly
$T^{\prime}\not\sim T$.

$\ast$

$T$ is non-strongly connected.

Observe that if $T$ is a transitive tournament, then $T^{\prime}$ is also
transitive (by $(\leq 3)$-hypomorphy) and the result is obvious. Suppose then
that $T$ is a non-strongly connected tournament which is not transitive. The
result follows from the following five facts.

Fact 1. Let $X$ be an element of $\widetilde{\mathcal{P}}(T)$ such that $T[X]$
is strongly connected with $\mid X\mid\geq 3$ and let $a$ be an element of
$V\setminus X$. As $\mathcal{P}(T^{\prime})=\mathcal{P}(T)$, then
$T^{\prime}[X]$ is strongly connected, $X\in\widetilde{P}(T^{\prime})$ and we
have:
$a\longrightarrow X$ in $T$ if and only if $a\longrightarrow X$ in
$T^{\prime}$.

Indeed :
Let $\\{\alpha,\beta,\gamma\\}$ be a subset of $X$ such that
$T[\\{\alpha,\beta,\gamma\\}]$ is a $3$-cycle and suppose, for example, that
$X\longrightarrow a$ in $T$. From Corollary 15,
$n(T,\delta^{+};\\{\alpha,\beta,a\\})$
$=n(T^{\prime},\delta^{+};\\{\alpha,\beta,a\\})$. As
$n(T,\delta^{+};\\{\alpha,\beta,a\\})\neq 0$, thus there exists a subset $K$
of $V$ such that $\\{\alpha,\beta,a\\}\subset K$ and
$T^{\prime}[K]\sim\delta^{+}$. Hence, $\mid K\cap X\mid=3$, because otherwise
$K\cap X$ is an interval with two elements of $T^{\prime}[K]$; which
contradicts Remark 22. As $a\in V\setminus X$ and $K\setminus\\{a\\}\subset
X$, then $K\setminus\\{a\\}$ is an interval of $T^{\prime}[K]$ and thus
$T^{\prime}[K]$ is a diamond of center $a$. So, $X\longrightarrow a$ in
$T^{\prime}$.

Fact 2. $\widetilde{\mathcal{P}}(T^{\prime})=\widetilde{\mathcal{P}}(T)$.

Indeed :
Consider an element $Y$ of $\widetilde{\mathcal{P}}(T)$. We distinguish the
following two cases.

  * •

If $T[Y]$ is strongly connected and $\mid Y\mid\geq 3$. In this case,
$Y\in\mathcal{P}(T)$. Then $Y\in\mathcal{P}(T^{\prime})$ and hence,
$Y\in\widetilde{\mathcal{P}}(T^{\prime})$.

  * •

If $T[Y]$ is a transitive. In this case, there exists an element $Z$ of
$\widetilde{\mathcal{P}}(T^{\prime})$ such that $Y\subset Z$, because
otherwise, $Y$ admits a partition $\\{Y_{1},Y_{2}\\}$ such that there is
$K\in\mathcal{P}(T^{\prime})$ with $\mid K\mid\geq 3$, $T^{\prime}[K]$ is
strongly connected and in $T^{\prime}$ we have $Y_{1}\longrightarrow
K\longrightarrow Y_{2}$; which contradicts the Fact 1. While exchanging the
roles of $T$ and $T^{\prime}$, we can hence deduce that $Y=Z$ and then,
$Y\in\widetilde{\mathcal{P}}(T^{\prime})$ .

Fact 3. $T^{\prime}/\widetilde{\mathcal{P}}(T)=T/\widetilde{\mathcal{P}}(T)$.

Indeed:

Proceed by the absurd and suppose that there exist two distinct elements $X$
and $Y$ of $\widetilde{\mathcal{P}}(T)$ such that $X\longrightarrow Y$ in $T$
and $Y\longrightarrow X$ in $T^{\prime}$. From the Fact 1, $T[X]$ and $T[Y]$
are transitive. So, $X$ and $Y$ are not consecutive in
$T/\widetilde{\mathcal{P}}(T)$. Thus, there exists an element $Z$ of
$\widetilde{\mathcal{P}}(T)$ such that $T[Z]$ is strongly connected, $\mid
Z\mid\geq 3$ and $X\longrightarrow Z\longrightarrow Y$ in $T$. So, by Fact 1,
$X\longrightarrow Z\longrightarrow Y$ in $T^{\prime}$ and then
$X\longrightarrow Y$ in $T^{\prime}$; which is absurd.

Fact 4. If $T\not\in\Omega_{n}$, then for all
$X\in\widetilde{\mathcal{P}}(T)$, $T^{\prime}[X]\sim T[X]$.

Indeed:

Suppose that $T\not\in\Omega_{n}$ and consider an element $X$ of
$\widetilde{\mathcal{P}}(T)$. As $T$ and $T^{\prime}$ are $(\leq
3)$-hypomorphic, we can assume that $\mid X\mid\geq 4$ and $T[X]$ is strongly
connected.

We distinguish the following cases.

  * •

If $\mid X\mid\leq n-3$.
Consider $H=T[X]$ and $A\subset X$ such that $\mid A\mid=2$. From Corollary
15, $n(T,H;A)=n(T^{\prime},H;A)$. As $n(T,H;A)\neq 0$, then
$n(T^{\prime},H;A)\neq 0$. So, there exists a subset $K$ of $V$ such that
$A\subset K$ and $T^{\prime}[K]\sim H$. We have then $K=X$, because otherwise,
there exists $Y\in\widetilde{\mathcal{P}}(T)\setminus\\{X\\}$ such that $K\cap
Y\neq\emptyset$ and hence $T^{\prime}[K]$ is non-strongly connected (because
$K\cap X$ is also non empty); which is absurd. So, $T^{\prime}[X]\sim H=T[X]$.

  * •

If $\mid X\mid=n-1$.
Let $\\{a\\}=V\setminus X$ and suppose, for example, that $X\longrightarrow a$
in $T$. As $\mid X\mid-2=n-3$, then $T^{\prime}[X]$ and $T[X]$ are
$\\{-2\\}$-hypomorphic. Now we shall prove that these two tournaments are
$\\{-3\\}$-hypomorphic. For that, consider a subset $A$ of $X$ such that $\mid
A\mid=3$. It is clear that $T^{\prime}-A$ and $T-A$ are isomorphic and in
these two tournaments, we have $(X\setminus A)\longrightarrow a$. So,
$T^{\prime}[X\setminus A]\sim T[X\setminus A]$. Hence, $T^{\prime}[X]$ and
$T[X]$ are $\\{-3\\}$-hypomorphic. So, $T^{\prime}[X]$ and $T[X]$ are
$\\{-2,-3\\}$-hypomorphic.

    * –

If $T[X]$ is decomposable. Let $Q=\mathcal{P}(T[X])$. If there exists $Y\in Q$
such that $\mid Y\mid=\mid X\mid-2$. In this case, $\mid Q\mid=3$ and $T[X]/Q$
is a $3$-cycle. As $\mid Y\mid=n-3$, then $T^{\prime}[Y]\sim T[Y]$. As in
addition, $T^{\prime}[X]$ and $T[X]$ are $(\leq 3)$-hypomorphic, then, by
Corollary 18, $\mathcal{P}(T^{\prime}[X])=Q$ and $T^{\prime}[X]/Q$ $=T[X]/Q$
or $T^{\prime}[X]/Q$ $=T^{\ast}[X]/Q$. So we deduce immediately that
$T^{\prime}[X]\sim T[X]$. Now suppose that for every $Y\in Q$, $\mid
Y\mid<\mid X\mid-2$. In this case, $T[X]$ is not almost transitive.
Distinguish the following two cases. If $\mid X\mid\geq 9$. In this case, by
applying Proposition 28 for the tournaments $T[X]$ and $T^{\prime}[X]$, we
obtain $T^{\prime}[X]\sim T[X]$. If $\mid X\mid=8$. As $T^{\prime}[X]$ and
$T[X]$ are $\\{-2,-3\\}$-hypomorphic, then, by Lemma 31, $T^{\prime}[X]\sim
T[X]$.

    * –

If $T[X]$ is indecomposable. We have $T^{\prime}[X]$ and $T[X]$ are $(\leq
3)$-hypomorphic. So, by Theorem 17, $T^{\prime}[X]=T[X]$ or
$T^{\prime}[X]=T^{\ast}[X]$. Consider then the case where
$T^{\prime}[X]=T^{\ast}[X]$. If $T[X]$ is self dual, then $T^{\prime}[X]\sim
T[X]$. Suppose hence that $T[X]$ is not self dual. As $T[X]$ is
$\\{-2,-3\\}$-self dual (because $T^{\prime}[X]$ and $T[X]$ are
$\\{-2,-3\\}$-hypomorphic), then $T\in
O_{2}(I_{n-1,\\{-2,-3\\}})\subset\Omega_{n}$; which is absurd.

  * •

If $\mid X\mid=n-2$.
Let $\\{a,b\\}=V\setminus X$ and suppose, for example, that $X\longrightarrow
a$ in $T$. As $\mid X\mid-1=n-3$, then $T^{\prime}[X]$ and $T[X]$ are
$\\{-1\\}$-hypomorphic. Now, we shall prove that $T^{\prime}[X]$ and $T[X]$
are $\\{-2,-3\\}$-hypomorphic. For that, for every $i\in\\{2,3\\}$, let
$A_{i}$ be a subset of $X$ such that $\mid A_{i}\mid=i$. We have
$T-(A_{2}\cup\\{a\\})\sim T^{\prime}-(A_{2}\cup\\{a\\})$. From Fact 3,
$(X\setminus A_{2})\longrightarrow b$ in $T$ if and only if $(X\setminus
A_{2})\longrightarrow b$ in $T^{\prime}$. So, $T^{\prime}[X\setminus
A_{2}]\sim T[X\setminus A_{2}]$ and thus, $T^{\prime}[X]$ and $T[X]$ are
$\\{-2\\}$-hypomorphic. Furthermore, as $T^{\prime}-A_{3}\sim T-A_{3}$, then
by Fact 3, we can see that $T^{\prime}[X\setminus A_{3}]\sim T[X\setminus
A_{3}]$. So, $T^{\prime}[X]$ and $T[X]$ are $\\{-1,-2,-3\\}$-hypomorphic.

    * –

If $T[X]$ is decomposable. Pose $Q=\mathcal{P}(T[X])$. If there exists $Y\in
Q$ such that $\mid Y\mid=\mid X\mid-2$. In this case, $\mid Q\mid=3$ and
$T[X]/Q$ is a $3$-cycle. Besides, by $\\{-3\\}$-hypomorphy, the two
subtournaments $T^{\prime}[Y\cup\\{a\\}]$ and $T[Y\cup\\{a\\}]$ are isomorphic
(because $\mid Y\cup\\{a\\}\mid=n-3$). As in addition, $Y\longrightarrow a$ in
both these tournaments, then $T^{\prime}[Y]\sim T[Y]$. As in addition,
$T^{\prime}[X]$ and $T[X]$ are $(\leq 3)$-hypomorphic, then, by Corollary 18,
we can see that $T^{\prime}[X]\sim T[X]$. Now, suppose that for every $Y\in
Q$, $\mid Y\mid<\mid X\mid-2$. In this case, $T[X]$ is not almost transitive.
We distinguish the following three cases.
If $\mid X\mid\geq 9$, we conclude by Proposition 28.
If $\mid X\mid=8$, we conclude by Lemma 31.
If $\mid X\mid=7$, as $T^{\prime}[X]$ and $T[X]$ are
$\\{-1,-2,-3\\}$-hypomorphic, then they are $(\leq 6)$-hypomorphic and hence
$T^{\prime}[X]\sim T[X]$.

    * –

If $T[X]$ is indecomposable. As $T^{\prime}[X]$ and $T[X]$ are $(\leq
3)$-hypomorphic, then from Theorem 17, $T^{\prime}[X]=T[X]$ or
$T^{\prime}[X]=T^{\ast}[X]$. Consider then the case where
$T^{\prime}[X]=T^{\ast}[X]$. If $T[X]$ is self dual, then $T^{\prime}[X]\sim
T[X]$. Suppose now that $T[X]$ is not self dual. As in addition $T[X]$ is
$\\{-1,-2,-3\\}$-self dual (because $T^{\prime}[X]$ and $T[X]$ are
$\\{-1,-2,-3\\}$-hypomorphic), hence the tournament $T$ belongs to
$O_{3}(I_{n-2,\\{-1,-2,-3\\}})$ and then to $\Omega_{n}$; which is absurd.

Fact 5. If $T\in\Omega_{n}$, then $T^{\prime}\not\sim T$.

This fact is an immediate consequence of facts 2 and 3.

$\square$

We shall now prove Corollary 8. Before that let us observe that Corollary 7 is
an immediate consequence of Theorem 6 because each element $T$ of $\Omega_{n}$
(where $n\geq 9$) has an interval $X$ such that $T[X]$ is indecomposable and
$\mid X\mid\in\\{n-1,n-2\\}$.

Proof of Corollary 8. Suppose that there exists an integer $n_{0}\geq 7$ such
that the indecomposable tournaments with at least $n_{0}$ vertices are
$\\{-3\\}$-reconstructible and consider a tournament $T$ with $n\geq n_{0}+2$
vertices. Then, the classes $I_{n-2,\\{-3\\}}$ and $I_{n-1,\\{-3\\}}$ are
empty. So, the classes $I_{n-2,\\{-1,-2,-3\\}}$ and $I_{n-1,\\{-2,-3\\}}$ are
empty. Thus, $\Omega_{n}$ is empty and then $T\notin\Omega_{n}$. By Theorem 6,
the tournament $T$ is then $\\{-3\\}$-reconstructible.

$\square$

_AKNOWLEDGEMENT_
We would like to thank the anonymous referee for his insightful comments.

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