text:
\begin{document}
\maketitle
\begin{abstract}
In this paper,
we completely characterize the multipartite tournaments whose competition graphs are complete.
\end{abstract}
\noindent
{\it Keywords.} multipartite tournament; competition graph; complete graph.

\noindent
{{{\it 2010 Mathematics Subject Classification.} 05C20, 05C75}}

\section{Introduction}

Given a digraph $D$,
$N^+_D(x)$ and $N^-_D(x)$ denote the sets of out-neighbors and in-neighbors, respectively, of a vertex $x$ in $D$.
When no confusion is likely, we omit $D$ in $N^+_{D}(x)$ and $N^+_{D}(x)$ to just write $N^+(x)$ and $N^-(x)$, respectively.

The \emph{competition graph} $C(D)$ of a digraph $D$
is the (simple undirected) graph $G$ defined by
$V(G)=V(D)$ and $E(G)=\{uv \mid u,v \in V(D), u \neq v,
N_D^+(u) \cap N_D^+(v) \neq \emptyset \}$.
Competition graphs arose in
connection with an application in ecology (see \cite{Cohen})
and also have applications in coding,
radio transmission, and modeling of complex economic systems.
Early literature of the study on competition graphs is summarized
in the survey papers by Kim~\cite{kim1993competition} and Lundgren~\cite{lundgren1989food}.

For a digraph $D$,
the \emph{underlying graph} of $D$
is the graph $G$
such that $V(G)=V(D)$ and $E(G)=\{ uv \mid (u,v) \in A(D) \}$.
An \emph{orientation} of a graph $G$
is a digraph having no directed $2$-cycles, no loops, and no multiple arcs
whose underlying graph  is $G$.
A \emph{tournament} is an orientation of a complete graph.
A \emph{$k$-partite tournament} is  an orientation of a complete $k$-partite graph for some positive integer $k \geq 2$.
If a digraph is a $k$-partite tournament for some integer $k \geq 2$, then it is called a \emph{multipartite tournament}.
The competition graphs of tournaments and those of bipartite tournaments
have been actively studied

(see \cite{cho2002domination}, \cite{choi20171},  \cite{eoh2019niche},
  \cite{eoh2020m},
 \cite{factor2007domination},
  \cite{fisher2003domination},
  \cite{fisher1998domination}, and \cite{kim2016competition}
for papers related to this topic).

It seems to be a natural shift to take
a look at the competition graphs of
$k$-partite tournaments for an integer $k \geq 3$.
Even if the competition graph of a bipartite tournament is disconnected (see Proposition 1.1 of \cite{kim2016competition}), that of a $k$-partite tournament may be connected as long as $k$ is an integer greater than or equal to $3$.

We began our research on competition graphs of multipartite tournaments with answering a fundamental question ``For each integer $k \geq 3$, which $k$-partite tournaments do have complete graphs as their competition graphs?".
After a rather long struggle, we finally managed to figure out the sizes of partite sets of multipartite tournaments whose competition graphs are complete.

In Section~\ref{sec:preliminaries}, we develop tools to be used frequently in settling our question.
In the remaining sections, we characterize $k$-partite tournaments whose competition graphs are complete for an integer $k \geq 3$. Especially, Section~\ref{sec:6-partite} deals the case $k \geq 6$; Section~\ref{sec:3-partite} treats the case $k=3$; Section~\ref{sec:4,5-partite} takes care of the case $k=4$ and the case $k=5$.

\section{Preliminaries} \label{sec:preliminaries}
In this section, we develop tools to be used frequently in characterizing $k$-partite tournaments whose competition graphs are complete for an integer $k \geq 3$.

\begin{Lem} \label{lem:two-partite-set-complete}
Suppose that $D$ is a multipartite tournament whose competition graph is complete.
Then the out-neighbors of each vertex are included in at least two partite sets of $D$.
\end{Lem}
\begin{proof}
If there exists a vertex $v$ whose out-neighbors are included in one partite set of $D$, then $v$ and its out-neighbors cannot have a common out-neighbor and so $v$ cannot be adjacent to its out-neighbors in $C(D)$, which is a contradiction.
\end{proof}

The following is immediate consequence of Lemma~\ref{lem:two-partite-set-complete}.
\begin{Cor} {\label{cor:no-biparte-complete}}
There is no bipartite tournament whose competition graph is complete.
\end{Cor}

\begin{Prop} \label{prop:two-partite-out-neighbor-2-2}
Suppose that $D$ is a  multipartite tournament whose competition graph is complete.
If the out-neighbors of a vertex $v$ are included in exactly two partite sets $U$ and $V$ of $D$,
then $|N^+(v) \cap U | \geq 2$ and $|N^+(v) \cap V | \geq 2$.
\end{Prop}

\begin{proof}
Suppose that there exists a vertex $v$ whose out-neighbors are included in exactly two partite sets $U$ and $V$ of $D$.
To reach a contradiction, suppose that $|N^+(v) \cap U | < 2$ or $|N^+(v) \cap V | < 2$.

Without loss of generality, we may assume $|N^+(v) \cap U | < 2$.
Then $N^+(v) \cap U = \{u\}$ for some vertex $u$ in $D$.
Since $C(D)$ is complete, each vertex in $N^+(v) \cap V$ is adjacent to $v$ in $C(D)$.
Since $N^+(v) \subseteq U \cup V$,
a common out-neighbor of $v$ and the vertices in $N^+(v) \cap V$ belong to $N^+(v) \cap U=\{u\}$.
Therefore $u$ is a common out-neighbor of $v$ and the vertices in $N^+(v) \cap V$.
Since $u$ and $v$ are adjacent in $C(D)$, $u$ and $v$ have a common out-neighbor $w$ in $D$.

By the way, since $u \in U$ and $N^+(v) \subseteq U \cup V$, $w \in N^+(v) \cap V$, which is impossible by our assumption that no directed $2$-cycle is allowed in a digraph.
Hence $|N^+(v) \cap U | \geq 2$ and so the statement is true.
\end{proof}

\begin{Cor} \label{cor:outdegree-at-least-complete}
Suppose that $D$ is a multipartite tournament whose competition graph is complete.
Then each vertex has outdegree at least $3$.
\end{Cor}
\begin{proof}
Take a vertex $v$ in $D$.
Then the out-neighbors of $v$ belong to at least two partite sets of $D$ by Lemma~\ref{lem:two-partite-set-complete}.
If the out-neighbors of $v$ belong to exactly two partite sets, then $v$ has outdegree at least $4$ by Proposition~\ref{prop:two-partite-out-neighbor-2-2}.
If the out-neighbors of $v$ belong to at least three partite sets, then it is obvious that $v$ has outdegree at least $3$.
Therefore $v$ has outdegree at least $3$.
Since $v$ was arbitrarily chosen,
the statement is true.
\end{proof}

\begin{Cor} \label{cor:n.b.of-vertices-outdegree3}
Suppose that $D$ is a multipartite tournament whose competition graph is complete. Then there exist at least $\max \{4|V(D)|-|A(D)|,0\}$ vertices of outdegree $3$ in $D$.
\end{Cor}

\begin{proof}
Let $l$ be the number of vertices of outdegree $3$.
Since each vertex in $D$ has outdegree at least $3$ by Corollary~\ref{cor:outdegree-at-least-complete},
\[4(|V(D)|-l)+3l \leq |A(D)|.\]
Therefore $4|V(D)|-|A(D)| \leq l$.
Thus the statement is true.
\end{proof}

\begin{Lem} \label{lem:complete-outdegree-3}
Suppose that $D$ is a multipartite tournament whose competition graph is complete.
If a vertex $u$ has outdegree $3$ in $D$,
then its out-neighbors form a directed cycle.
\end{Lem}

\begin{proof}
Each pair of vertices has a common out-neighbor in $D$ since $C(D)$ is complete.
Suppose that a vertex $u$ has outdegree $3$ in $D$.
Let $N^+(u)=\{v_1,v_2,v_3\}$.
Then, since $u$ and $v_i$ are adjacent in $C(D)$, $u$ and $v_i$ have a common out-neighbor in $\{v_1, v_2, v_3\} \setminus \{v_i\}$ for each $1\leq i \leq 3$.
Therefore $v_1$, $v_2$, and $v_3$ form a directed cycle in $D$.
\end{proof}

The following lemma is immediately true by the definition of the competition graph.
\begin{Lem} \label{lem:subdigraph}
Let $D$ be a digraph and $D'$ be a subdigraph of $D$.
Then the competition graph of $D'$ is a subgraph of the competition graph of $D$.
\end{Lem}

\begin{Lem} \label{lem:adding-vertex}
Suppose that $D$ is a digraph with at least two vertices whose competition graph is complete.
Let $D'$ be a digraph with the vertex set $V(D)\cup\{v\}$ where $v$ is not a vertex of $D$.
If  $A(D)\subset A(D')$ and $N^+_D(u) \subseteq N^+_{D'}(v)$ for some vertex $u$ in $D$,
then the competition graph of $D'$ is complete.
\end{Lem}

\begin{proof}
Suppose that $A(D)\subset A(D')$ and $N^+_D(u) \subseteq N^+_{D'}(v)$ for some vertex $u$ in $D$.
Take two vertices $x$ and $y$ in $D'$.
If $x \neq v $ and $y \neq v$, then $x$ and $y$ are adjacent in $C(D')$ by Lemma~\ref{lem:subdigraph}. By symmetry, now we suppose $x=v$.
Then $y \neq v$, so $y \in V(D)$.
Since $|V(D)| \geq 2$ and $C(D)$ is complete, $N^+_D(y) \cap N^+_D(u) \neq \emptyset$.
Then $N^+_{D'}(x) \cap N^+_{D'}(y)=N^+_{D'}(v) \cap N^+_{D'}(y) \supseteq N^+_D(u) \cap N^+_D(y) \neq \emptyset$ and so $x$ and $y$ are adjacent in $C(D')$.
\end{proof}

\begin{Lem} \label{lem:making-multi-complete}
Let $k$ and $l$ be positive integers with $l \geq k \geq 3$; $n_1,\ldots, n_k$ be positive integers such that $n_1 \geq \cdots \geq n_k$;  $n'_1, \ldots ,n'_l$ be positive integers such that
$n'_1 \geq \cdots \geq n'_l$, $n'_1 \geq n_1$,

$n'_2 \geq n_2,\ldots$, and $n'_k \geq n_k$.
If $D$ is an orientation of $K_{n_1,\ldots,n_k}$ whose competition graph is complete,
then there exists an orientation $D'$ of $K_{n'_1,\ldots,n'_l}$ whose competition graph is complete.

\end{Lem}
\begin{proof}
Suppose that  $D$ is an orientation of $K_{n_1,\ldots,n_k}$ whose competition graph is complete.
Let $V_1,V_2,\ldots,V_k$ be partite sets of $D$  satisfying $|V_i|=n_i$ for each $1 \leq i \leq k$.
Then we construct an orientation of $K_{n'_1,n_2,\ldots,n_k}$ whose competition graph is complete in the following way.
If $n'_1 = n_1$, then we take $D$ as a desired orientation.
Suppose $n'_1 > n_1$.
We add a new vertex $v$ to $V_1$ so that
\begin{equation} \label{eq:lem:making-multi-complete}
 A(D)\subseteq A(D_1) \quad \text{and} \quad  N^+_D(u) \subseteq N^+_{D_1}(v)
 \end{equation}
for some vertex $u$ in $V_1$.

Then $C(D_1)$ is complete by Lemma~\ref{lem:adding-vertex}.
We may repeat this process until we obtain a desired orientation $D_{n'_1-n_1}$.
Inductively, we obtain an orientation $D_{t}$ of $K_{n'_1,\ldots,n'_k}$ whose competition graph is complete where $t=(n'_1+\cdots+n'_k)-(n_1+\cdots+n_k)$.
If $l=k$, then we are done.
Suppose $l > k$.
Then we construct a $(k+1)$-partite tournament $D_{t+1}$ by adding a new vertex $w$ to $D_t$ so that $V_{k+1}:=\{w\}$ is a partite set of $D_{t+1}$,
\[ A(D_t)\subseteq A(D_{t+1}), \quad \text{and} \quad N^+_{D_t}(u) \subseteq N^+_{D_{t+1}}(w).  \]
where $u$ is the vertex satisfying the second containment in \eqref{eq:lem:making-multi-complete}.
Then $C(D_{t+1})$ is complete by Lemma~\ref{lem:adding-vertex}.
By applying a similar argument for obtaining a digraph $D_{t+n'_{k+1}}$, we may show that there exists an orientation of $K_{n'_1,\ldots,n'_{k+1}}$ whose competition graph is complete.
We may repeat this process until we obtain
an orientation of $K_{n'_1,\ldots,n'_l}$ whose competition graph is complete.
Therefore the statement is true.
\end{proof}

\begin{Lem} \label{lem:orientation-8-vertices}
If $D$ is a $k$-partite tournament with $8$ vertices whose competition graph is complete for some $k\in \{5,6\}$ and has at least two vertices of outdegree at least $4$,
then $D$ is an orientation of $K_{2,2,1,1,1,1}$
in which there exist exactly two vertices of outdegree at least $4$.
\end{Lem}

\begin{proof}
Suppose that there exists a $k$-partite tournament $D$ with $8$ vertices whose competition graph is complete for some $k\in \{5,6\}$ and which has at least two vertices of outdegree at least $4$.
Suppose $k=5$.
It is easy to check that the numbers of arcs in $K_{4,1,1,1,1},K_{3,2,1,1,1}$, and $K_{2,2,2,1,1}$ are $22,24$, and $25$, respectively.
Therefore $|A(D)|$ becomes maximum when $D$ is an orientation of $K_{2,2,2,1,1}$, so $|A(D)| \leq 25$.
By Corollary~\ref{cor:n.b.of-vertices-outdegree3}, there exist at least $\max \{4|V(D)|-|A(D)|,0\}$ vertices of outdegree $3$ in $D$, so at least $7$ vertices has outdegree $3$ in $D$, which is a contradiction.
Therefore $k=6$ and $D$ is an orientation of $K_{3,1,1,1,1,1}$ or $K_{2,2,1,1,1,1}$.
If $D$ is an orientation of $K_{3,1,1,1,1,1}$, then $|A(D)|=25$ and so, by the same reason, we reach a contradiction.
Thus $D$ is an orientation of $K_{2,2,1,1,1,1}$.
By the way, $D$ has $8$ vertices and $26$ arcs, so $4|V(D)|-|A(D)| =6$.
	Then there exist at least $6$ vertices of outdegree $3$ by Corollary~\ref{cor:n.b.of-vertices-outdegree3}.
Therefore $D$ has exactly two vertices of outdegree at least $4$.
\end{proof}

\begin{Prop} \label{prop:complete-condition-outdegree-3}
Suppose that $D$ is a $k$-partite tournament whose competition graph is complete for some integer $k \in \{4,5,6\}$ and there exists a vertex $u$ of outdegree $3$.
Then $D$ contains a subdigraph isomorphic to the digraph $\tilde{D}$ in Figure~\ref{fig:subdigraph-degree3} and $|V(D)| \geq 9$. In particular, if $k=4$, then $|V(D)| \geq 10 $.
\end{Prop}

\begin{proof}
Each pair of vertices has a common out-neighbor in $D$ since $C(D)$ is complete.
Let $N^+(u)=\{v_1,v_2,v_3\}$.
Without loss of generality, we may assume that $C:=v_1 \to v_2 \to v_3 \to v_1 $ is a directed cycle of $D$ by Lemma~\ref{lem:complete-outdegree-3}.
Let $w_i$ be a common out-neighbor of $v_i$ and $v_{i+1}$ for each $1 \leq i \leq 3$ (identify $v_4$ with $v_1$).
If $w_j=w_k$ for some distinct $j,k \in \{1,2,3\}$,
then $N^+(u)=\{v_1,v_2,v_3\} \subseteq N^-(w_j)$ and so $w_j$ does not share a common out-neighbor with $u$, which is a contradiction.
Therefore $w_1$, $w_2$, and $w_3$ are all distinct.
Thus, so far, we have a subdigraph $\tilde{D}$ of $D$ with the vertex set $\{u,v_1,v_2,v_3,w_1,w_2,w_3\}$ given in Figure~\ref{fig:subdigraph-degree3}.

Suppose, to the contrary, that $|V(D)|=7$.
Then $V(D)=V(\tilde{D})$ and so $v_i$ has outdegree $3$ in $D$ for each $1\leq i \leq 3$.
Therefore the out-neighbors of $v_i$ form a directed $3$-cycle in $D$ for each $1\leq i \leq 3$ by Lemma~\ref{lem:complete-outdegree-3}.
Since $N^+(v_3)=\{v_1,w_2,w_3\}$ and $\{(v_1,w_3),(w_2,v_1)\} \subset A(D)$,
$(w_3,w_2) \in A(D)$.
By arguing similarly for $N^+(v_2)$ and $N^+(v_1)$,
we may show that $D_1$ given in Figure~\ref{fig:subdigraph-degree3-example} is a subdigraph of $D$.

\begin{figure}
\begin{center}
\begin{tikzpicture}[x=1.0cm, y=1.0cm]
   \tikzset{->-/.style={decoration={
  markings,
  mark=at position #1 with {\arrow{>}}},postaction={decorate}}}

   \vertex (u) at (0,0) [label=above:$u$]{};
   \vertex (v2) at (2,0) [label=left:$v_2$]{};
   \vertex (v3) at (2,2) [label=above:$v_3$]{};
   \vertex (v1) at (2,-2) [label=below:$v_1$]{};
   \vertex (w2) at (4,0) [label=right:$w_2$]{};
    \vertex (w3) at (4,2) [label=above:$w_3$]{};
    \vertex (w1) at (4,-2) [label=below:$w_1$]{};
   \path
(u) edge [->-=0.97,thick] (v1)
(u) edge [->-=0.97,thick] (v3)
(u) edge [->-=0.97, bend left=30, thick] (v2)
(v1) edge [->-=0.97,bend left=15,thick] (v2)
(v3) edge [->-=0.97,thick] (w3)
(v2) edge [->-=0.97,thick] (w1)
(v2) edge [->-=0.97,thick] (w2)
(v3) edge [->-=0.97,thick] (w2)
(v1) edge [->-=0.97,thick] (w3)
(v1) edge [->-=0.97,thick] (w1)
(w3) edge [->-=0.97,thick] (v2)
(w2) edge [->-=0.97,thick] (v1)
(w1) edge [->-=0.97,thick] (v3)
   (v2) edge [->-=0.97,bend left=15,thick] (v3)
   (v3) edge [->-=0.97,bend left=20,thick] (v1)
   ;
   ;
\draw (3, -3) node{$\tilde{D}$};
\end{tikzpicture}
 \end{center}
  \caption{The subdigraph $\tilde{D}$ obtained in the proof of Proposition~\ref{prop:complete-condition-outdegree-3}}
  \label{fig:subdigraph-degree3}
\end{figure}

\begin{figure}
\begin{center}
\begin{tikzpicture}[x=1.0cm, y=1.0cm]
 \tikzset{->-/.style={decoration={
  markings,
  mark=at position #1 with {\arrow{>}}},postaction={decorate}}}
   \vertex (u) at (0,0) [label=above:$u$]{};
   \vertex (v2) at (2,0) [label=left:$v_2$]{};
   \vertex (v3) at (2,2) [label=above:$v_3$]{};
   \vertex (v1) at (2,-2) [label=below:$v_1$]{};
   \vertex (w2) at (4,0) [label=right:$w_2$]{};
    \vertex (w3) at (4,2) [label=above:$w_3$]{};
    \vertex (w1) at (4,-2) [label=below:$w_1$]{};
   \path

(u) edge [->-=0.97,thick] (v1)
(u) edge [->-=0.97,thick] (v3)
(u) edge [->-=0.97, bend left=30, thick] (v2)
(v1) edge [->-=0.97,bend left=15,thick] (v2)
(v3) edge [->-=0.97,thick] (w3)
(v2) edge [->-=0.97,thick] (w1)
(v2) edge [->-=0.97,thick] (w2)
(v3) edge [->-=0.97,thick] (w2)
(v1) edge [->-=0.97,thick] (w3)
(v1) edge [->-=0.97,thick] (w1)
(w3) edge [->-=0.97,thick] (v2)
(w2) edge [->-=0.97,thick] (v1)
(w1) edge [->-=0.97,thick] (v3)
   (v2) edge [->-=0.97,bend left=15,thick] (v3)
   (v3) edge [->-=0.97,bend left=20,thick] (v1)
(w1) edge [->-=0.97,bend right=40,thick] (w3)
    (w2) edge [->-=0.97,thick] (w1)
    (w3) edge [->-=0.97,thick] (w2)
	;
	;
\draw (3, -3) node{$D_1$};
\end{tikzpicture}
\hspace{2em}
\begin{tikzpicture}[x=1.0cm, y=1.0cm]
 \tikzset{->-/.style={decoration={
  markings,
  mark=at position #1 with {\arrow{>}}},postaction={decorate}}}
   \vertex (u) at (0,0) [label=above:$u$]{};
   \vertex (v2) at (2,0) [label=left:$v_2$]{};
   \vertex (v3) at (2,2) [label=above:$v_3$]{};
   \vertex (v1) at (2,-2) [label=below:$v_1$]{};
   \vertex (w2) at (4,0) [label=right:$w_2$]{};
    \vertex (w3) at (4,2) [label=above:$w_3$]{};
    \vertex (w1) at (4,-2) [label=below:$w_1$]{};
     \vertex (x) at (0.6,-1.5) [label=above:$x$]{};
   \path

(u) edge [->-=0.97,thick] (v1)
(u) edge [->-=0.97,thick] (v3)
(u) edge [->-=0.97, bend left=30, thick] (v2)
(v1) edge [->-=0.97,bend left=15,thick] (v2)
(v3) edge [->-=0.97,thick] (w3)
(v2) edge [->-=0.97,thick] (w1)
(v2) edge [->-=0.97,thick] (w2)
(v3) edge [->-=0.97,thick] (w2)
(v1) edge [->-=0.97,thick] (w3)
(v1) edge [->-=0.97,thick] (w1)
(w3) edge [->-=0.97,thick] (v2)
(w2) edge [->-=0.97,thick] (v1)
(w1) edge [->-=0.97,thick] (v3)
   (v2) edge [->-=0.97,bend left=15,thick] (v3)
   (v3) edge [->-=0.97,bend left=20,thick] (v1)
(w1) edge [->-=0.97,bend right=40,thick] (w3)
(x) edge [->-=0.97,thick] (v1)
	;
	;
\draw (3, -3) node{$D_2$};
\end{tikzpicture}
 \end{center}
  \caption{The subdigraphs $D_1$ and $D_2$ considered in the proof of Proposition~\ref{prop:complete-condition-outdegree-3}}

  \label{fig:subdigraph-degree3-example}
\end{figure}

	It is easy to see from $D_1$ that $u$ is the only possible common out-neighbor of each pair of $w_1$, $w_2$, and $w_3$ in $D$.
	Therefore
	$N^-(u)=\{w_1,w_2,w_3\}$ and so $D$ is a $7$-partite tournament, which is a contradiction to the hypothesis.
Thus $|V(D)| \geq 8$.

To reach a contradiction, suppose that $|V(D)|=8$.
Then  $V(D)=V(\tilde{D}) \cup \{x\}$ for some vertex $x$ in $D$ and

 \begin{equation}
\label{eq:prop:complete-condition-outdegree-3-0}
|N^+(v_i)|=3 \mbox{ or } 4 \end{equation} for each $1\leq i \leq 3$.
Since $x$ and $u$ must be adjacent and $N^+(u)=\{v_1,v_2,v_3\}$, one of $v_1$, $v_2$, $v_3$ is a common out-neighbor of $u$ and $x$.
Therefore $ |N^+(v_j)|=3$
for some $j \in \{1,2,3\}$.
Without loss of generality, we may assume \[|N^+(v_1)|=3.\] Then \[N^+(v_1)=\{v_2,w_1,w_3\}\]
and
\[(x,v_1)\in A(D).\]
By Lemma~\ref{lem:complete-outdegree-3}, the out-neighbors of $v_1$ form a directed cycle.
Therefore
$\{v_1,v_2,v_3,w_1,w_3\}$ forms a $5$-tournament, so \[k \geq 5.\]
By the way, since $(w_3,v_2)$ and $(v_2,w_1)$ are arcs of $D$, \[(w_1,w_3)\in A(D)\]
(see the digraph $D_2$ given in Figure~\ref{fig:subdigraph-degree3-example} for an illustration).
If $w_2$ is a common out-neighbor of $w_1$ and $w_3$, then, since $N^+(v_1)=\{v_2,w_1,w_3\}$, $v_1$ and $w_2$ cannot have a common out-neighbor, which is a contradiction.
Therefore \begin{equation}
\label{eq:prop:complete-condition-outdegree-3-2}N^+(w_2) \cap \{w_1,w_3\} \neq \emptyset. \end{equation}

 We first claim that $\{v_1,v_2,v_3,w_1,w_2,w_3\}$ forms a tournament in $D$.
  Since $D_2$ is a subgraph of $D$, we need to show that $\{w_1,w_2,w_3\}$ forms a tournament in $D$.
 		As we have shown that $\{v_1,v_2,v_3,w_1,w_3\}$ is a tournament in $D$, it remains to show that $w_2$ is adjacent to $w_1$ and $w_3$ in $D$.

 	Suppose, to the contrary, that there is no arc between $w_1$ and $w_2$.
Then $(w_2,w_3) \in A(D)$ by~\eqref{eq:prop:complete-condition-outdegree-3-2}.
Then the vertices $v_1$, $w_2$, $w_3$ cannot form a directed cycle. Yet,  $v_1$, $w_2$, $w_3$ are out-neighbors of $v_3$, so $|N^+(v_3)|=4$ by Lemma~\ref{lem:complete-outdegree-3} and \eqref{eq:prop:complete-condition-outdegree-3-0}.
Since $x$ is the only possible new out-neighbor of $v_3$ in $D$,

$N^+(v_3)=\{v_1,w_2,w_3,x\}$.
Since $x$ is the only possible common out-neighbor of $w_2$ and $v_2$, $N^+(w_2) \cap N^+(v_2)=\{x\}$.
Thus $N^+(v_2)=\{x,v_3,w_1,w_2\}$ and $\{v_1,w_3,x\} \subseteq N^+(w_2)$.
Since $v_2$ and $v_3$ have outdegree $4$, $D$ is an orientation of $K_{2,2,1,1,1,1}$ by Lemma~\ref{lem:orientation-8-vertices}.
Then $\{w_1,w_2\}$ forms a partite set of $D$.
Since $u$ has outdegree $3$ in $D$, $(w_2,u) \in A(D)$ and so
$\{u,v_1,w_3,x \} \subseteq N^+(w_2)$.
Then $v_2$, $v_3$, and $w_2$ has outdegree at least $4$, which contradicts Lemma~\ref{lem:orientation-8-vertices}.
Thus there is an arc between $w_1$ and $w_2$.

 Now we suppose, to the contrary, that there is no arc between $w_2$ and $w_3$.
	Then $v_1$, $w_2$, $w_3$ cannot form a directed cycle. Since they are out-neighbors of $v_3$, $|N^+(v_3)|=4$ by Lemma~\ref{lem:complete-outdegree-3} and \eqref{eq:prop:complete-condition-outdegree-3-0} and so $N^+(v_3)=\{v_1,w_2,w_3,x\}$.
Since there is no arc between $w_2$ and $w_3$, there is an arc $(w_2,w_1)$ in $D$ by \eqref{eq:prop:complete-condition-outdegree-3-2}.
For the same reason, $x$ is the only possible common out-neighbor of $w_3$ and $v_2$, so $x \in N^+(w_3) \cap N^+(v_2)$.
Thus $v_2$ has outdegree $4$ by~\eqref{eq:prop:complete-condition-outdegree-3-0}.
	Since $v_3$ also has outdegree  $4$,
	$D$ is an orientation of $K_{2,2,1,1,1,1}$ by Lemma~\ref{lem:orientation-8-vertices}.
Thus $\{w_2,w_3\}$ is a partite set of $D$.
Since $N^+(v_1)=\{v_2,w_1,w_3\}$ and $x \in N^+(w_3) \cap N^+(v_2)$,
$(x,w_1)$ must be an arc of $D$ in order for
$v_1$ and $x$ to be adjacent.
Since $u$ is the only possible common out-neighbor of $x$ and $w_1$, there exist arcs $(x,u)$ and $(w_1,u)$ in $D$.
Then $\{x,u,v_1,v_2,v_3,w_1\}$ forms a tournament and we reach a contradiction to the fact that $D$ is an orientation of $K_{2,2,1,1,1,1}$ with $\{w_2,w_3\}$ as a partite set of $D$.
Therefore $\{v_1,v_2,v_3,w_1,w_2,w_3\}$ forms a tournament as we claimed.
Thus $k =6$ and $u$ and $x$ belong to a partite set of size at least $2$. Furthermore, since $N^+(u)=\{v_1,v_2,v_3\}$, $u$ cannot form a partite set with $v_1$, $v_2$, or $v_3$ and so $u$ and exactly one of $w_1$, $w_2$, and $w_3$ belong to the same partite set.

Suppose, to the contrary, that $(w_3,w_2)\in A(D)$.
Then $(w_2,w_1) \in A(D)$ by~\eqref{eq:prop:complete-condition-outdegree-3-2}.
Therefore $w_1\to w_3 \to w_2 \to w_1$ forms a directed cycle.
Then, for each pair of $w_1$, $w_2$, and $w_3$, $x$ and $u$ are its only possible common out-neighbors.
Since $u$ and one of $w_1$, $w_2$, and $w_3$ belong to the same partite set, exactly one pair of $w_1$, $w_2$, and $w_3$ can prey on $u$.
Then the other two pair of  $w_1$, $w_2$, and $w_3$ prey on $x$.
Therefore $\{w_1,w_2,w_3 \} \subseteq N^-(x)$. Thus $x$ and exactly one of $v_2$ and $v_3$ belong to the same partite set (recall that we assumed $(x,v_1) \in A(D)$) and so $|N^+(x) \cap \{v_1,v_2,v_3\} | \leq 2$. Hence, for some $j \in \{1,2,3\}$, a common out-neighbor of $x$ and $v_j$  is contained in $\{u,w_1,w_2,w_3\}$.
Since $\{w_1,w_2,w_3 \} \subseteq N^-(x)$, $u$ must be a common out-neighbor of $x$ and $v_j$, which contradicts the fact $  N^+(u)= \{v_1,v_2,v_3\}$.
Therefore $(w_3,w_2)\notin A(D)$ and so
 \[(w_2,w_3)\in A(D).\]
 Thus $N^+(w_3) \subseteq \{u,v_2,x\}$ and so, by Corollary~\ref{cor:outdegree-at-least-complete}, $N^+(w_3)=\{u,v_2,x\}$.
Then $x$ is the only possible common out-neighbor of each pair of $v_3$ and $w_3$, and $v_2$ and $w_3$.
 Therefore $N^+(v_3)=\{v_1,w_2,w_3,x\}$ and $N^+(v_2)=\{v_3,w_1,w_2,x\}$
 by~\eqref{eq:prop:complete-condition-outdegree-3-0}.
Thus $D$ is an orientation of $K_{2,2,1,1,1,1}$ by Lemma~\ref{lem:orientation-8-vertices}.
Moreover, since $N^+(v_1)=\{v_2,w_3,w_1\}$,
$w_1$ is the only possible common out-neighbor of $x$ and $v_1$ and so
$(x,w_1) \in A(D)$.
Then $u$ must be a common out-neighbor of $w_1$ and $w_3$.
Therefore $\{w_2,u\}$ and $\{x,v_1\}$ are the partite sets of size $2$ in $D$.
Then, since $\{v_2,v_3\} \subset N^-(x)$ and $N^+(u)=\{v_1,v_2,v_3\}$,
$u$ and $x$ has no common out-neighbor, which is a contradiction.

Therefore we have shown that $|V(D)| \neq 8$ and so $|V(D)| \geq 9$.

To show the ``particular" part, suppose $k=4$.
Let $V_1,V_2,V_3,V_4$ be the partite sets of $D$.
By the above argument, $u$, $v_1$, $v_2$, and $v_3$ belong to distinct partite sets.
Without loss of generality, we may assume that $u \in V_1$, $v_1 \in V_2$, $ v_2 \in V_3$, and $v_3 \in V_4$.
Let $y_i$ be a common out-neighbor of $v_i$ and $w_i$ in $D$ for each $1\leq i \leq 3$.
If $y_1=y_2=y_3$, then $\{v_1,v_2,v_3\} \subseteq N^-(y_1)$, which implies that $u$ and $y_1$ do not share a common out-neighbor, and we reach a contradiction.
Therefore at least two of $y_1$, $y_2$, and $y_3$ are distinct.
By the above argument, $\{u,w_1,w_2,w_3\} \subseteq V_1$.
Therefore $y_i$ cannot be $w_j$ for each $1\leq i,j \leq 3$.
Suppose, to the contrary, that $|V(D)|=9$.
Then exactly two of $y_1$, $y_2$, and $y_3$ are the same.

Without loss of generality, we may assume $y_1=y_2$ and $y_1 \neq y_3$.
Neither $v_1$ nor $v_2$ is a common out-neighbor of $y_1$ and $u$.
Thus $v_3$ must be a common out-neighbor of $y_1$ and $u$.
Yet, $y_1$ is a common out-neighbor of $v_1$, $v_2$, $w_1$, and $w_2$, so $y_1 \in V_4$ and we reach a contradiction.
Thus $|V(D)| \geq 10$.
\end{proof}

\section{$k$-partite tournaments whose competition graphs are complete for $k \geq 6$}\label{sec:6-partite}
\
In this section, we characterize $k$-partite tournaments whose competition graphs are complete for an integer $k \geq 6$.

\begin{Prop} \label{prop:k>=7-complete}
Let $k$ be a positive integer greater than or equal to $7$.
For positive integers $n_1,n_2,\ldots,n_k$,
there exists an orientation of $K_{n_1,\ldots,n_k}$ whose competition graph is complete.
\end{Prop}

\begin{proof}
The digraph in Figure~\ref{fig:7-tournament-complete} is an orientation of $K_{1,1,1,1,1,1,1}$ whose competition graph is complete.
Therefore the statement is true
by Lemma~\ref{lem:making-multi-complete}.
\end{proof}

\begin{figure}
\begin{center}
 \begin{tikzpicture}[auto,thick]

       \tikzstyle{player}=[minimum size=5pt,inner sep=0pt,outer sep=0pt,draw,circle]

    \tikzstyle{player1}=[minimum size=2pt,inner sep=0pt,outer sep=0pt,fill,color=black, circle]
    \tikzstyle{source}=[minimum size=5pt,inner sep=0pt,outer sep=0pt,ball color=black, circle]
    \tikzstyle{arc}=[minimum size=5pt,inner sep=1pt,outer sep=1pt, font=\footnotesize]
    \path (90:1.5cm)     node [player]  (v0)
    {};
    \path (40:1.5cm)     node [player]  (v1)  {};
    \path (-10:1.5cm)     node [player]  (v2){};
    \path (-60:1.5cm)     node [player]  (v3){};
 \path (140:1.5cm)     node [player]  (v6){};
  \path (190:1.5cm)     node [player]  (v5){};
    \path (240:1.5cm)     node [player]  (v4){};

   \draw[black,thick,-stealth] (v0) - +(v1);
 \draw[black,thick,-stealth] (v0) - +(v2);
  \draw[black,thick,-stealth] (v0) - +(v4);

  \draw[black,thick,-stealth] (v1) - +(v2);
  \draw[black,thick,-stealth] (v1) - +(v3);
  \draw[black,thick,-stealth] (v1) - +(v5);

  \draw[black,thick,-stealth] (v2) - +(v3);
  \draw[black,thick,-stealth] (v2) - +(v4);
  \draw[black,thick,-stealth] (v2) - +(v6);

  \draw[black,thick,-stealth] (v3) - +(v4);
  \draw[black,thick,-stealth] (v3) - +(v0);
  \draw[black,thick,-stealth] (v3) - +(v5);

  \draw[black,thick,-stealth] (v4) - +(v1);
  \draw[black,thick,-stealth] (v4) - +(v5);
  \draw[black,thick,-stealth] (v4) - +(v6);

  \draw[black,thick,-stealth] (v5) - +(v2);
  \draw[black,thick,-stealth] (v5) - +(v0);
  \draw[black,thick,-stealth] (v5) - +(v6);

  \draw[black,thick,-stealth] (v6) - +(v1);
  \draw[black,thick,-stealth] (v6) - +(v3);
  \draw[black,thick,-stealth] (v6) - +(v0);

    \end{tikzpicture}
 \end{center}
 \caption{An orientation of $K_{1,1,1,1,1,1,1}$}
\label{fig:7-tournament-complete}
\end{figure}
We now completely characterize sizes of partite sets of a $k$-partite tournament whose competition graph is complete when $k=2$ or $k \geq 7$
by Corollary~\ref{cor:no-biparte-complete} and Proposition~\ref{prop:k>=7-complete}.
Accordingly, it remains to study $k$-partite tournaments whose competition graphs are complete for $ 3 \leq k \leq 6$.
Especially, in the rest of this section, we characterize $6$-partite tournaments whose competition graphs are complete.

The following lemma is obviously true by the definition of competition graph.

\begin{Lem} \label{lem:deleting-vertex-complete}
Let $D$ be a nontrivial digraph whose competition graph is complete.
If a vertex $v$ has indegree at most $1$ in $D$, then $C(D-v)$ is complete.
\end{Lem}

\begin{Prop} \label{prop:indegree-condition-complete}
Let $D$ be a $k$-partite tournament  for some positive integer $k\geq 3$ whose competition graph is complete with the partite sets $V_1, \ldots, V_k$.
Then
there exists a $k$-partite tournament $D^*$ with the partite sets $V_1, \ldots, V_k$
such that $C(D^*)$ is complete and each vertex in $D^*$ has indegree at least $2$.
\end{Prop}

\begin{proof}
Suppose that there exists a vertex $v_1$ of indegree at most $1$ in $V_{t_1}$ for some $t_1 \in \{1,\ldots,k\}$.
Let $D_1=D-v_1$.
Then $C(D_1)$ is complete by Lemma~\ref{lem:deleting-vertex-complete}.
By Corollary~\ref{cor:no-biparte-complete},
$D_1$ is not a bipartite tournament.
Suppose that there exists a vertex $v_{2}$ of indegree at most $1$ in $V_{t_2}$ for some $t_2 \in \{1,\ldots,k\}$ in $D_1$.
Let $D_2=D_1-v_{2}$.
Therefore $C(D_2)$ is complete by Lemma~\ref{lem:deleting-vertex-complete} and so, by Corollary~\ref{cor:no-biparte-complete}, $D_2$ is not a bipartite tournament.
We keep repeating this process.
Since $D$ has a finite number of vertices, this process terminates to produce digraphs $D_1, D_2, \ldots, D_l$ the competition graph of each of which is complete and none of which is a bipartite tournament.
Since $C(D_l)$ is complete, the number of partite sets in $D_l$ is at least $3$.

The fact that the process ended with $D_l$ implies that each vertex in $D_l$ has indegree at least $2$.
As some of partite sets of $D_l$ are proper subsets of corresponding partite sets of $D$,
we need to add vertices to obtain a desired $k$-partite tournament.
Let $X$ be the partite set of $D_{l-1}$ to which $v_l$ belongs.
Then $X \subseteq V_{t_l}$.
In the following, we construct a multipartite tournament $D^*_{l-1}$ from $D_l$ such that
$V(D_{l-1})=V(D^*_{l-1})$, $D_{l-1}$ and $D^*_{l-1}$ have the identical partite sets, and $C(D^*_{l-1})$ is complete.
We consider two cases for $X$.

{\it Case 1}. $X=\{v_l\}$.
We take a vertex $v'$ in $D_l$.
Then $v'$ has indegree at least $2$.
Now we add $v_l$ to $D_l$ so that $\{v_l\}$ is a partite set of $D^*_{l-1}$, $v_l$ takes the out-neighbors and the in-neighbors
of $v'$ as its out-neighbors and in-neighbors, respectively, and the remaining out-neighbors and in-neighbors of $v_l$ are arbitrarily taken.
Then the indegree of $v_l$ in $D^*_{l-1}$ is at least $2$.
Moreover,
\[V(D_{l})\cup \{v_l\}=V(D^*_{l-1}), \quad A(D_l) \subset A(D^*_{l-1}), \quad  \text{and} \quad  N^+_{D_l}(v') \subset N^+_{D^*_{l-1}}(v_{l}). \]

{\it Case 2}. $\{v_l\}  \varsubsetneq X$.
Then there exists a vertex $v'$ distinct from $v_{l}$ in $X$.
Since $D_l=D_{l-1}-v_{l}$, $v'$ is a vertex of $D_l$.
Now we add $v_l$ to the partite set of $D_l$ where $v'$ belongs so that
$\{v_{l},v'\}$ is involved in a partite set of $D^*_{l-1}$, $v_l$ takes the out-neighbors and the in-neighbors
of $v'$ as its out-neighbors and in-neighbors, respectively.
Then the indegree of $v_l$ in $D^*_{l-1}$ is at least $2$ since the indegree of $v'$ is at least $2$ in $D_l$.
Moreover,
\[V(D_{l})\cup \{v_l\}=V(D^*_{l-1}), \quad A(D_l) \subset A(D^*_{l-1}), \quad \text{and} \quad  N^+_{D_l}(v') \subseteq N^+_{D^*_{l-1}}(v_{l}). \]
In both cases, $C(D^*_{l-1})$ is complete by Lemma~\ref{lem:adding-vertex}.

Now we add $v_{l-1}$ to $D^*_{l-1}$
and apply an argument similar to the above one to obtain $D^*_{l-2}$ whose competition graph is complete and each vertex in which has indegree at least $2$.
We may repeat this process until we obtain a $k$-partite tournament
$D^*_0$ whose competition graph is complete and each vertex in which has indegree at least $2$.
Since we added $v_i$ to the partite set of $D^*_i$ which is included in $V_{t_i}$ for each $1\leq i \leq l$,
it is true that the partite sets of $D^*_0$ are the same as $D$.
Thus $D^*_0$ is a desired $k$-partite tournament.
\end{proof}

\begin{Prop} \label{prop:K_411111-no}
There is no orientation of $K_{4,1,1,1,1,1}$ whose competition graph is complete.
\end{Prop}

\begin{proof}
Suppose, to the contrary, that there exists an orientation of $K_{4,1,1,1,1,1}$ whose competition graph is complete.
Then, by Proposition~\ref{prop:indegree-condition-complete},  there exists an orientation $D$ of $K_{4,1,1,1,1,1}$ whose competition graph is complete and each vertex in which has indegree at least $2$.
Let $V_1,\ldots,V_6$ be the partite sets of $D$ with $|V_1|=4$.

By Corollary~\ref{cor:outdegree-at-least-complete},
each vertex has outdegree at least $3$ in $D$.
Then, since each vertex has indegree at least $2$ in $D$,

\begin{equation} \label{equation:prop:K_411111-no}
|N^+(v)|=3 \quad \text{and} \quad |N^-(v)|=2
\end{equation}
for each vertex $v$ in $V_1$.

By Corollary~\ref{cor:n.b.of-vertices-outdegree3}, there exist at least $\max \{4|V(D)|-|A(D)|,0\}$ vertices of outdegree $3$ in $D$.
Since $4|V(D)|-|A(D)|=6$,
there exist at least $6$ vertices of outdegree $3$.
Thus at least two vertices of outdegree $3$ do not belong to $V_1$.
Let $u$ be a vertex of outdegree $3$ which is not in $V_1$.
Without loss of generality, we may assume $u \in V_2$.

Let $N^+(u)=\{v_1,v_2,v_3\}$.
By Lemma~\ref{lem:complete-outdegree-3},
$N^+(u)$ forms a directed cycle in $D$ and
we may assume $v_1 \to v_2 \to v_3 \to v_1$.
Since each out-neighbor of $u$ has indegree at least $3$ by Proposition~\ref{prop:complete-condition-outdegree-3},
$N^+(u) \cap V_1 = \emptyset$ by~\eqref{equation:prop:K_411111-no}.
Therefore we may assume that $V_3=\{v_1\}$, $V_4=\{v_2\}$, $V_5=\{v_3\}$, $V_6=\{x\}$, and $v_1$ is a common out-neighbor of $x$ and $u$.
Then $\{u,v_3,x\} \subseteq N^-(v_1)$.
Let $w_1$ be a common out-neighbor of $v_1$ and $v_2$.
Then $w_1 \in V_1$.
Therefore, by~\eqref{equation:prop:K_411111-no}, $N^-(w_1)=\{v_1,v_2\}$ and $N^+(w_1)=\{u,v_3,x\}$.
Thus  $N^+(w_1)\subseteq N^-(v_1)$
and so $w_1$ and $v_1$ have no common out-neighbor, which is a contradiction.
\end{proof}

Let $D$ be a digraph with the vertex set $\{v_1,v_2,\ldots,v_n\}$ and $A=(a_{ij})$ be the {\it adjacency matrix }of $D$ such that \begin{displaymath}
a_{ij} = \left\{ \begin{array}{ll}
1 & \textrm{if there is an arc $(v_i,v_j)$ in $D$,}\\
0 & \textrm{otherwise.}\
\end{array} \right.
\end{displaymath}

Now we are ready to introduce one of our main theorems.
\begin{Thm} \label{thm:complete-6-partite}
Let $n_1,\ldots,n_6$ be positive integers such that $n_1 \geq \cdots \geq n_6$.
There exists an orientation $D$ of $K_{n_1,n_2,\ldots,n_6}$ whose competition graph is complete if and only if one of the following holds: (a) $n_1\geq5$ and $n_2=1$; (b) $n_1 \geq 3$, $n_2 \geq 2$, and $n_3=1$; (c) $n_3\geq 2$.
\end{Thm}

\begin{proof}
To show the ``only if" part, suppose that there exists an orientation $D$ of $K_{n_1,n_2,\ldots,n_6}$ such that $C(D)$ is complete.
We suppose $n_3=1$.

{\it Case 1}. $n_2=1$.
If $n_1 \leq 4$, then
 there exists an orientation of $K_{4,1,1,1,1,1}$ whose competition graph is complete by Lemma~\ref{lem:making-multi-complete}, which contradicts Proposition~\ref{prop:K_411111-no}.
Therefore $n_1 \geq 5$.

{\it Case 2}. $n_2 \geq 2$.
Then $n_1 \geq 2$.
Suppose, to the contrary, that $n_1 = 2$.
Then $n_2=2$, so $D$ is an orientation of $K_{2,2,1,1,1,1}$.
Therefore $4|V(D)|-|A(D)|=6$.
By Corollary~\ref{cor:n.b.of-vertices-outdegree3},
there exists a vertex of outdegree $3$ in $D$.

Therefore $|V(D)| \geq 9$ by Proposition~\ref{prop:complete-condition-outdegree-3}, which is a contradiction.
Thus $n_1 \geq 3$.
Hence the ``only if" part is true.

Now we show the ``if" part.
Let $D_\alpha$, $D_\beta$, $D_\gamma$ be the digraphs whose adjacency matrix are $A_1$, $A_2$, and $A_3$, respectively, given in Figure~\ref{fig:6-partite-complete}.
It is easy to check that
 the inner product of each pair of rows in each matrix is nonzero, so the competition graphs of $D_\alpha$, $D_\beta$, and $D_\gamma$ are complete.
By applying Lemma~\ref{lem:making-multi-complete} to $D_\alpha$, $D_\beta$, and $D_\gamma$,
we may obtain orientations $D'_\alpha$, $D'_\beta$, and $D'_\gamma$ of $K_{n_1,n_2,n_3,n_4,n_5,n_6}$ whose competition graphs are complete for (a) $n_1\geq5$ and $n_2=1$; (b) $n_1 \geq 3$, $n_2 \geq 2$; (c) $n_3\geq 2$, respectively.
Therefore we have shown that the ``if" part is true.
\end{proof}

\begin{figure}
  \centering
    \begin{equation*}
A_1=
\left(\begin{array}{*{10}c}
 0 & 0 & 0 & 0 & 0 &
  0 & 1 & 0 & 1 & 1  \\
  0 & 0 & 0 & 0 & 0 &
  0 & 1 & 1 & 0 & 1  \\
  0 & 0 & 0 & 0 & 0 &
  1 & 0 & 1 & 1 & 0  \\
  0 & 0 & 0 & 0 & 0 &
  1 & 0 & 1 & 0 & 1 \\
  0 & 0 & 0 & 0 & 0 &
  1 & 1 & 0 & 1 & 0 \\

 1 & 1 & 0 & 0 & 0 &
  0 & 1 & 1 & 0 & 0  \\

 0 & 0 & 1 & 1 & 0 &
  0 & 0 & 1 & 1 & 0  \\

 1 & 0 & 0 & 0 & 1 &
  0 & 0 & 0 & 1 & 1  \\

 0 & 1 & 0 & 1 & 0 &
  1& 0 & 0 & 0 & 1  \\

 0 & 0 & 1 & 0 & 1 &
  1 & 1 & 0 & 0 & 0  \\
   \end{array}
    \right)
    \end{equation*}
    \begin{equation*}
A_2=
\left(\begin{array}{*{9}c}
 0 & 0 & 0 &
 1 & 0 &
 1 & 0 & 1 & 0   \\
 0 & 0 & 0 &
 0 & 1 &
 1 & 1 & 0 & 1   \\
 0 & 0 & 0 &
 0 & 1 &
 1 & 0 & 1 & 1   \\

 0 & 1 & 1 &
 0 & 0 &
 0 & 0 & 1 & 1   \\
 1 & 0 & 0 &
 0 & 0 &
 1 & 0 & 0 & 1   \\

 0 & 0 & 0 &
 1 & 0 &
 0 & 1 & 0 & 1   \\
 1 & 0 & 1 &
 1 & 1 &
 0 & 0 & 0 & 0   \\
 0 & 1 & 0 &
 0 & 1 &
 1 & 1 & 0 & 0   \\
 1 & 0 & 0 &
 0 & 0 &
 0 & 1 & 1 & 0   \\
   \end{array}
    \right)
    \end{equation*}
    \begin{equation*}
    A_3=
\left(\begin{array}{*{9}c}
 0 & 0 & 0 & 0 & 0 & 0
 & 1 & 1 & 1   \\
 0 & 0 & 1 & 0 & 1 & 1
 & 0 & 1 & 0   \\
 1 & 0 & 0 & 0 & 1 & 0
 & 1 & 1 & 1   \\
 1 & 1 & 0 & 0 & 1 & 0
 & 0 & 0 & 1   \\
 1 & 0 & 0 & 0 & 0 & 0
 & 1 & 1 & 1   \\
 1 & 0 & 1 & 1 & 0 & 0
 & 1 & 0 & 0   \\
 0 & 1 & 0 & 1 & 0 & 0
 & 0 & 1 & 0   \\
 0 & 0 & 0 & 1 & 0 & 1
 & 0 & 0 & 1   \\
 0 & 1 & 0 & 0 & 0 & 1
 & 1 & 0 & 0   \\
   \end{array}
    \right)
    \end{equation*}
      \caption{The adjacency matrices $A_1$, $A_2$, and $A_3$ which are orientations of $K_{5,1,1,1,1,1}$, $K_{3,2,1,1,1,1}$, and $K_{2,2,2,1,1,1}$, respectively, in the proof of Theorem~\ref{thm:complete-6-partite}}
      \label{fig:6-partite-complete}
    \end{figure}

\section{Tripartite tournaments whose competition graphs are complete}\label{sec:3-partite}

In this section, we characterize tripartite tournaments whose competition graphs are complete.

    \begin{Prop}\label{thm4} Let $D$ be an orientation of $K_{n_1, n_2, n_3}$ whose competition graph is complete for some positive integers $n_1, n_2$, and $n_3$. Then $n_i \ge 4$ for each $1\leq i \leq 3$. \end{Prop}
        \begin{proof}
        Let $V_1, V_2,$ and $V_3$ be the partite sets of $D$ with $|V_i|=n_i$ for each $1\leq i \leq 3$.
        Suppose, to the contrary, that $n_j\leq 3$ for some $j \in \{1,2,3\}$.
        Without loss of generality, we may assume that $n_1 \leq 3$.
        Take $v_1 \in V_1$.
        By Lemma~\ref{lem:two-partite-set-complete},
        the out-neighbors of each vertex in $D$ are included in at least two partite sets.
        Then, since $D$ is tripartite tournament, the out-neighbors of each vertex in $D$ are included in exactly two partite sets.
        Thus, by Proposition~\ref{prop:two-partite-out-neighbor-2-2}, there are four vertices $u_1, u_2, w_1,$ and $w_2$ such that $\{u_1, u_2\} \subseteq N^+(v_1) \cap V_2$ and $\{w_1, w_2\} \subseteq N^+(v_1) \cap V_3$.
        By the same proposition, there are two vertices $v_2$ and $v_3$ in $V_1$ such that $\{v_2, v_3\} \subseteq N^+(u_1) \cap V_1$.
        Since $(v_1, u_1) \in A(D)$, $v_2$ and $v_3$ are distinct from $v_1$.
        Since $n_1 \leq 3$, $n_1=3$. Then
         $V_1 = \{v_1, v_2, v_3\}$.
        Therefore, by Proposition~\ref{prop:two-partite-out-neighbor-2-2}, $N^+(v) \cap V_1 =\{v_2,v_3\}$ for each vertex $v$ in $N^+(v_1)$.
        This implies that each out-neighbor of $v_1$ has $v_2$ as its out-neighbor.
        Therefore $v_1$ and $v_2$ cannot have a common out-neighbor, which is a contradiction.

        \end{proof}

    \begin{Lem}\label{lem6} If $D$ is an orientation of $K_{4,4,4}$ with partite sets $V_1, V_2$, and $V_3$ whose competition graph is complete, then, for distinct $i,j \in \{1,2,3\}$ and each $u \in V_i$, $|N^+(u) \cap V_j|=|N^-(u) \cap V_j|=2$. \end{Lem}
        \begin{proof}
       Suppose that there exists an orientation $D$ of $K_{4,4,4}$ with partite sets $V_1, V_2$, and $V_3$ whose competition graph is complete.
         Take distinct $i$ and $j$ in $\{1,2,3\}$.
        Then there are exactly $16$ arcs between $V_i$ and $V_j$.
        On the other hand, by Proposition~\ref{prop:two-partite-out-neighbor-2-2}, for each $u \in V_i$ and $v \in V_j$,
        \begin{equation*}
        |N^+(u) \cap V_j| \geq2 \quad \text{and} \quad |N^+(v) \cap V_i| \geq2.
        \end{equation*}
        Therefore
        \begin{equation*}
        16=\sum_{u \in V_i} |N^+(u) \cap V_j| + \sum_{v \in V_j} |N^+(v) \cap V_i|\geq 16.
        \end{equation*}
        and so $|N^+(u) \cap V_j|=|N^+(v) \cap V_i|=2$ for each $u \in V_i$ and $v \in V_j$.
        Hence $|N^+(u) \cap V_j|=|N^-(u) \cap V_j|=2$ for each $u \in V_i$.
        \end{proof}

    \begin{Lem}\label{lem7} If $D$ is an orientation of $K_{4,4,4}$ with the partite sets $V_1, V_2$, and $V_3$ whose competition graph is complete, then, for some distinct $i$ and $j$ in $\{1,2,3\}$, there is a pair of vertices $x$ and $y$ in $V_i$ such that $N^+(x) \cap V_j=N^+(y) \cap V_j$. \end{Lem}
        \begin{proof}
        Suppose that there exists an orientation $D$ of $K_{4,4,4}$ with the partite sets $V_1, V_2$, and $V_3$ whose competition graph is complete.
        Suppose, to the contrary, that, for distinct $i,j \in \{1,2,3\}$,

        \begin{equation}\label{lem7:eq1} N^+(u) \cap V_j \neq N^+(v) \cap V_j
        \end{equation}
        for any pair of vertices $u$ and $v$ in $V_i$.
        Fix $i \in \{1,2,3\}$ and $u, v \in V_i$.
        Let $w$ and $z$ be the remaining vertices in $V_i$.
        Since $C(D)$ is complete, $u$ and $v$ have a common out-neighbor in $V_j$ for some $j \in \{1,2,3\} \setminus \{i\}$.
        By Lemma~\ref{lem6} and \eqref{lem7:eq1}, $N^+(u) \cap V_j = \{v_1,v_2\}$ and
        \begin{equation}\label{lem7:eq2}
        N^+(v) \cap V_j = \{v_1,v_3\}
        \end{equation}
        for distinct vertices $v_1,v_2$, and $v_3$ in $V_j$.
        Then, by Lemma~\ref{lem6}, $N^+(v_1) \cap V_i= \{w,z\}$.
        Let $v_4$ be the remaining vertex in $V_j$.
        Then, by Lemma~\ref{lem6} again,
        $N^-(u) \cap V_j =\{v_3,v_4\}$ and
        $N^-(v) \cap V_j=\{v_2,v_4\}$.
        Therefore $N^-(v_4) \cap V_i = \{w,z\}$ by the same lemma and so $N^+(v_4) \cap V_i = \{u,v\}$.
  Thus $v_1$ and $v_4$ cannot have a common out-neighbor in $V_i$.
        Hence they have a common out-neighbor in $V_k$ for $k \in \{1,2,3\} \setminus \{i,j\}$.
        By Lemma~\ref{lem6} and \eqref{lem7:eq1} again, $N^+(v_1) \cap V_k = \{w_1,w_2\}$ and $N^+(v_4) \cap V_k = \{w_1,w_3\}$ for distinct vertices $w_1,w_2$, and $w_3$ in $V_k$.
        Let $w_4$ be the remaining vertex in $V_k$.
        Then, by Lemma~\ref{lem6},
        \[N^-(v_1) \cap V_k = \{w_3,w_4\} \quad \text{and} \quad N^-(v_4) \cap V_k = \{w_2,w_4\}.\]
        Meanwhile we note that $w_2$ and $u$ have a common out-neighbor in $V_j$.
        Since $N^+(u)\cap V_j=\{v_1,v_2\}$ and $N^+(v_1) \cap V_k = \{w_1,w_2\}$, $v_2$ is a common out-neighbor of $w_2$ and $u$.
        Then $N^+(w_2) \cap V_j = \{v_2,v_4\}$ by Lemma~\ref{lem6}.
        Thus $w_2$ and $v$ cannot have a common out-neighbor in $V_j$ by \eqref{lem7:eq2}.
        Since $w_2$ and $v$ belong to $V_k$ and $V_i$, respectively, they cannot have a common out-neighbor in $D$ and we reach a contradiction.
        \end{proof}

    \begin{Thm}\label{thm8}
     There is no orientation of $K_{4,4,4}$ whose competition graph is complete.
  \end{Thm}
        \begin{proof}
          Suppose, to the contrary, that there exists an orientation $D$ of $K_{4,4,4}$ whose competition graph is complete. Let $V_1,V_2$, and $V_3$ be the partite sets of $D$.

        Then, by Lemmas~\ref{lem6} and~\ref{lem7}, for some distinct $i$ and $j$ in $\{1,2,3\}$, there is a pair of vertices $u_1$ and $u_2$ in $V_i$ such that $N_D^+(u_1) \cap V_j = N_D^+(u_2) \cap V_j = \{v_1,v_2\}$ for some vertices $v_1$ and $v_2$ in $V_j$. Without loss of generality, we may assume that $i=1$ and $j=2$.
        Let $u_3$ and $u_4$ (resp.\ $v_3$ and $v_4$) be the remaining vertices in $V_1$ (resp.\ $V_2$).
        Then, by Lemma~\ref{lem6},
        \[N_D^+(u_3) \cap V_2 = N_D^+(u_4) \cap V_2 = \{v_3,v_4\},\]
        \[N_D^+(v_1) \cap V_1 = N_D^+(v_2) \cap V_1 = \{u_3,u_4\},\]
        \[N_D^+(v_3) \cap V_1 = N_D^+(v_4) \cap V_1 = \{u_1,u_2\}\]
        (see Figure~\ref{thm8:fig1} for an illustration).
\begin{figure}
\begin{center}
\begin{tikzpicture}[auto,thick, scale=0.9]
    \tikzstyle{player}=[minimum size=5pt,inner sep=0pt,outer sep=0pt,draw,circle]
    \tikzstyle{source}=[minimum size=5pt,inner sep=0pt,outer sep=0pt,ball color=black, circle]
    \tikzstyle{arc}=[minimum size=5pt,inner sep=1pt,outer sep=1pt, font=\footnotesize]
    \tikzset{->-/.style={decoration={
  markings,
  mark=at position #1 with {\arrow{>}}},postaction={decorate}}}

    \draw (180:3.2cm) node (name) {$V_1$};
    \draw (0:3.2cm) node (name) {$V_2$};

    \draw (270:3.2cm) node (name) {the arcs from $V_1$ to $V_2$};

    \path (-2,1.5)    node [player,label=left:$u_1$]  (u1) {};
    \path (-2,0.5)   node [player,label=left:$u_2$]  (u2) {};
    \path (-2,-0.5)   node [player,label=left:$u_3$]  (u3) {};
    \path (-2,-1.5)   node [player,label=left:$u_4$]  (u4) {};
    \path (2,1.5)   node [player,label=right:$v_1$]  (v1) {};
    \path (2,0.5)   node [player,label=right:$v_2$]  (v2) {};
    \path (2,-0.5)   node [player,label=right:$v_3$]  (v3) {};
    \path (2,-1.5)   node [player,label=right:$v_4$]  (v4) {};

    \draw[black,thick,->-=0.85] (u1) -- (v1);
    \draw[black,thick,->-=0.85] (u1) - + (v2);
    \draw[black,thick,->-=0.85] (u2) - + (v1);
    \draw[black,thick,->-=0.85] (u2) - + (v2);
    \draw[black,thick,->-=0.85] (u3) - + (v3);
    \draw[black,thick,->-=0.85] (u3) - + (v4);
    \draw[black,thick,->-=0.85] (u4) - + (v3);
    \draw[black,thick,->-=0.85] (u4) - + (v4);

\end{tikzpicture}
\hspace{2em}
\begin{tikzpicture}[auto,thick, scale=0.9]
    \tikzstyle{player}=[minimum size=5pt,inner sep=0pt,outer sep=0pt,draw,circle]
    \tikzstyle{source}=[minimum size=5pt,inner sep=0pt,outer sep=0pt,ball color=black, circle]
    \tikzstyle{arc}=[minimum size=5pt,inner sep=1pt,outer sep=1pt, font=\footnotesize]
    \tikzset{->-/.style={decoration={
  markings,
  mark=at position #1 with {\arrow{>}}},postaction={decorate}}}

    \draw (180:3.2cm) node (name) {$V_1$};
    \draw (0:3.2cm) node (name) {$V_2$};

    \draw (270:3.2cm) node (name) {the arcs from $V_2$ to $V_1$};

    \path (-2,1.5)    node [player,label=left:$u_1$]  (u1) {};
    \path (-2,0.5)   node [player,label=left:$u_2$]  (u2) {};
    \path (-2,-0.5)   node [player,label=left:$u_3$]  (u3) {};
    \path (-2,-1.5)   node [player,label=left:$u_4$]  (u4) {};
    \path (2,1.5)   node [player,label=right:$v_1$]  (v1) {};
    \path (2,0.5)   node [player,label=right:$v_2$]  (v2) {};
    \path (2,-0.5)   node [player,label=right:$v_3$]  (v3) {};
    \path (2,-1.5)   node [player,label=right:$v_4$]  (v4) {};

    \draw[black,thick,->-=0.85] (v1) -- (u3);
    \draw[black,thick,->-=0.85] (v1) - + (u4);
    \draw[black,thick,->-=0.85] (v2) - + (u3);
    \draw[black,thick,->-=0.85] (v2) - + (u4);
    \draw[black,thick,->-=0.85] (v3) - + (u1);
    \draw[black,thick,->-=0.85] (v3) - + (u2);
    \draw[black,thick,->-=0.85] (v4) - + (u1);
    \draw[black,thick,->-=0.85] (v4) - + (u2);

\end{tikzpicture}
\caption{The arcs between $V_1$ and $V_2$}\label{thm8:fig1}
\end{center}
\end{figure}
        Therefore each of the following pairs does not have a common out-neighbor in $V_2$: $\{u_1,u_3\}$; $\{u_1,u_4\}$; $\{u_2,u_3\}$; $\{u_2,u_4\}$.
        In addition, each of the following pairs does not have a common out-neighbor in $V_1$: $\{v_1,v_3\}$; $\{v_1,v_4\}$; $\{v_2,v_3\}$; $\{v_2,v_4\}$. Then each of these pairs has a common out-neighbor in $V_3$.
        Let $w_1,w_2,w_3$, and $w_4$ be the common out-neighbors of $\{u_1,u_3\},\{u_1,u_4\},\{u_2,u_3\}$, and $\{u_2,u_4\}$, respectively.
        Then, by Lemma~\ref{lem6}, $w_i \neq w_j$ for distinct $i,j \in \{1,2,3,4\}$ and so $V_3=\{w_1,w_2,w_3,w_4\}$.
        Without loss of generality, we may assume that $w_1,w_2,w_3$, and $w_4$ are the common out-neighbors of $\{v_1,v_3\},\{v_1,v_4\},\{v_2,v_3\}$, and $\{v_2,v_4\}$, respectively.
        Then, by Lemma~\ref{lem6}, $w_1$ and $w_2$ are out-neighbors of $u_1$ and $w_3$ and $w_4$ are out-neighbors of $v_2$ in $V_3$, so $u_1$ and $v_2$ do not have a common out-neighbor in $D$, which is a contradiction.
        \end{proof}

Now we are ready to state one of our main theorems.

\begin{Thm} \label{thm:complete-3-partite}
Let $n_1$, $n_2$, and $n_3$ be positive integers such that $n_1 \geq n_2 \geq n_3$.
There exists an orientation $D$ of $K_{n_1,n_2,n_3}$ whose competition graph is complete if and only if $n_1\geq5$ and $n_3 \geq 4$.
\end{Thm}

\begin{proof}
To show the ``only if" part, suppose that
$D$ is an orientation of $K_{n_1,n_2,n_3}$ whose competition graph is complete.
Then, by Proposition~\ref{thm4},

$n_i \geq 4$ for each $1\leq i \leq 3$.
If $n_1=4$, then $n_2=n_3=4$, which contradicts Theorem~\ref{thm8}.

Therefore $n_1 \geq 5$ and so the ``only if" part is true.

Now we show the ``if" part.
Let $D_\alpha$ be the digraph whose adjacency matrix is $A_4$ given in Figure~\ref{fig:3-partite-complete}.

It is easy to check that $D_\alpha$ is an orientation of $K_{5,4,4}$ and
the inner product of each pair of rows in each matrix is nonzero, so the competition graph of $D_\alpha$ is complete.
If $n_1\geq5$ and $n_3\geq 4$, then, by applying Lemma~\ref{lem:making-multi-complete} to $D_\alpha$,
we obtain an orientation $D'_\alpha$ of $K_{n_1,n_2,n_3}$ whose competition graph is complete.
\end{proof}
        \begin{figure}
    \centering
    \begin{equation*}
A_4=
\left(\begin{array}{*{13}c}
 0&0&0&0&0&0&1&0&1&0&1&1&0\\
 0&0&0&0&0&0&1&1&0&1&0&0&1\\
 0&0&0&0&0&1&0&0&1&1&0&0&1\\
 0&0&0&0&0&1&0&1&0&0&1&0&1\\
 0&0&0&0&0&1&0&1&0&1&0&1&0\\
 1&1&0&0&0&0&0&0&0&1&1&0&0\\
 0&0&1&1&1&0&0&0&0&1&1&0&0\\
 1&0&1&0&0&0&0&0&0&0&0&1&1\\
 0&1&0&1&1&0&0&0&0&0&0&1&1\\
 1&0&0&1&0&0&0&1&1&0&0&0&0\\
 0&1&1&0&1&0&0&1&1&0&0&0&0\\
 0&1&1&1&0&1&1&0&0&0&0&0&0\\
 1&0&0&0&1&1&1&0&0&0&0&0&0
\\
     \end{array}
    \right)
    \end{equation*}
  \caption{
  The adjacency matrix $A_4$ which is an orientation of $K_{5,4,4}$ in the proof of Theorem~\ref{thm:complete-3-partite}
  }\label{fig:3-partite-complete}
\end{figure}

\section{$k$-partite tournaments whose competition graphs are complete for the cases $k=4$ and $k=5$}\label{sec:4,5-partite}

In this section, we characterize $k$-partite tournaments whose competition graphs are complete for the cases $k=4$ and $k=5$.

\begin{Prop} \label{prop:n.b.of.size.1}
Let $D$ be a $k$-partite tournament whose competition graph is complete for some integer $k \in \{4,5\}$.
Then the number of partite sets having size $1$ is at most $k-3$.
\end{Prop}

\begin{proof}
Let $V_1, V_2, \ldots, V_k$ be the partite sets of $D$.

We may assume that $x \in V_1$, $y \in V_2$, and $z \in V_3$.

We suppose $k=4$.
To reach a contradiction, suppose that there are at least $2$ partite sets of size $1$.
Without loss of generality, we may assume $|V_1|=|V_2|=1$.
Then $V_1=\{x\}$ and $V_2=\{y\}$.
Without loss of generality, we may assume $z$ is a common out-neighbor of $x$ and $y$.
Then $N^+(z) \subseteq V_4$, which contradicts Lemma~\ref{lem:two-partite-set-complete}.
Therefore $D$ has at most $1$ partite set of size $1$.

Suppose $k=5$.
To reach a contradiction, suppose that there are at least $3$ partite sets of size $1$.
Without loss of generality, we may assume $|V_1|=|V_2|=|V_3|=1$.
Then $V_1=\{x\}$, $V_2=\{y\}$, and $V_3=\{z\}$.
Suppose that $x$ and $y$ have a common out-neighbor $w$ in $V_4\cup V_5$.
Without loss of generality, we may assume $w\in V_4$.
Then $N^+(w) \subseteq V_3 \cup V_5$.
By Lemma~\ref{lem:two-partite-set-complete},
$N^+(w) \cap V_3 \neq \emptyset $ and $N^+(w) \cap V_5 \neq \emptyset $.
However, $N^+(w) \cap V_3=\{z\}$, which contradicts Proposition~\ref{prop:two-partite-out-neighbor-2-2}.

Thus $w \notin V_4\cup V_5$ and so $w=z$.
By symmetry, the only possible common out-neighbor of $y$ and $z$ is $x$.
Since $z\in N^+(x)$, $x$ cannot be an out-neighbor of $z$ and we reach a contradiction.

Therefore $D$ has at most $2$ partite sets having size $1$.
\end{proof}

By Proposition~\ref{prop:complete-condition-outdegree-3},
the out-neighbors of a vertex of outdegree $3$ in a $k$-partite tournament whose competition graph is complete for some $4 \leq k\leq 5$ form a directed cycle and we have the following lemma.
\begin{Lem} \label{lem:degree3-partite-set-sizes}

Let $D$ be a $k$-partite tournament whose competition graph is complete for some $4 \leq k \leq 5$.
Suppose that a vertex $u$ has outdegree $3$.
If $N^+(u)\subseteq U \cup V \cup W$ for distinct partite sets $U$, $V$, and $W$ of $D$,
then $|U|+ |V|+ |W| \leq |V(D)|-4$.
\end{Lem}

\begin{proof}
Suppose that $N^+(u)\subseteq U \cup V \cup W$ for distinct partite sets $U$, $V$, and $W$ of $D$.
Since $u$ has outdegree $3$,
by Proposition~\ref{prop:complete-condition-outdegree-3},
$D$ contains a subdigraph isomorphic to $\tilde{D}$ given in Figure~\ref{fig:subdigraph-degree3}.
We may assume that the subdigraph is $\tilde{D}$ itself including labels.

We may assume $v_1 \in U$, $v_2 \in V$, and $v_3 \in W$.
Then $\{u,w_1,w_2,w_3\} \cap (U\cup V \cup W) = \emptyset$.
Thus $|V(D) \setminus (U\cup V \cup W)| \geq 4$ and so $|U|+ |V|+ |W|=|U\cup V \cup W| \leq |V(D)| -4$.
\end{proof}

\begin{Cor} \label{cor:no-or-3,3,2,2}
There is no orientation of $K_{3,3,2,2}$ and $K_{3,3,3,1}$  whose competition graphs are complete.
\end{Cor}
\begin{proof}
Suppose, to the contrary, that there exists an orientation $D$ of $K_{3,3,2,2}$ or $K_{3,3,3,1}$ whose competition graph is complete. Then $|A(D)| < 40$.
If each vertex in $D$ has outdegree at least $4$, then $|A(D)| \geq 40$, which is a contradiction.
Therefore there exists a vertex $u$ of outdegree $3$, then, the out-neighbors of $u$ belong to three distinct partite sets $U$, $V$, and $W$ by Lemma~\ref{lem:complete-outdegree-3} and, by Lemma~\ref{lem:degree3-partite-set-sizes},
$|U|+|V|+|W| \leq |V(D)| - 4=6$, which is impossible.
\end{proof}

\begin{Lem} \label{lem:complete-3-partite}
Let $n_1$, $n_2$, and $n_3$ be positive integers such that $n_1 \geq n_2 \geq n_3$.
If $D$ is an orientation of $K_{n_1,n_2,n_3,1}$ whose competition graph is complete,
then $n_3 \geq 3$.
\end{Lem}
\begin{proof}
Suppose that there exists an orientation $D$ of $K_{n_1,n_2,n_3,1}$ whose competition graph is complete.
Then $n_3 \geq 2$ by Proposition~\ref{prop:n.b.of.size.1}.
Suppose, to the contrary, that $n_3=2$.
Let $V_1, \ldots, V_4$ be the partite sets of $D$ satisfying $|V_1|=n_1$, $|V_2|=n_2$, $|V_3|=2$, and $|V_4|=1$.
Let $V_3=\{x_1,x_2\}$ and $y$ be a common out-neighbor of $x_1$ and $x_2$.
Then $V_3 \cap N^+(y) = \emptyset$, so, by Lemma~\ref{lem:two-partite-set-complete}, $N^+(y)$ is included in exactly two partite sets.
If $y \in V_1\cup V_2$, then
$|N^+(y) \cap V_4 |=1$, which contradicts Proposition~\ref{prop:two-partite-out-neighbor-2-2}.
Therefore $y \notin V_1\cup V_2$ and so
$y \in V_4$. Thus $V_4=\{y\}$.
Hence $N^+(y) \subseteq V_1 \cup V_2$.
Take a vertex $u$ in $N^+(y)$.
Then $N^+(u) \subseteq V_1 \cup V_3$ or $V_2 \cup V_3$.
Therefore $N^+(u)$ is included in exactly two partite sets by Lemma~\ref{lem:two-partite-set-complete}.
Since $|V_3|=2$,
 $N^+(u)\cap V_3=V_3$, that is, $u$ is a out-neighbor of neither $x_1$ nor $x_2$, by Proposition~\ref{prop:two-partite-out-neighbor-2-2}.
Since $u$ was arbitrarily chosen in $N^+(y)$,
any out-neighbor of $y$ is a out-neighbor of neither $x_1$ nor $x_2$.

Thus $x_1$ and $y$ have no common out-neighbor in $D$, which is a contradiction.
Hence $n_3 \neq 2$ and so $n_3 \geq 3$.
\end{proof}

Now we are ready to identify the sizes of partite sets of $4$-partite tournaments whose competition graphs are complete.
\begin{Thm} \label{thm:complete-4-partite}
Let $n_1,\ldots,n_4$ be positive integers such that $n_1 \geq \cdots \geq n_4$.
There exists an orientation $D$ of $K_{n_1,n_2,n_3,n_4}$ whose competition graph is complete if and only if one of the following holds:

(a) $n_1 \geq 4$, $n_3 \geq 3$, and $n_4 = 1$;
(b)  $n_1 \geq 4$, $n_3 =2$, and $n_4 =2$;
(c) $ n_3 \geq 3$ and $n_4 \geq 2$.

\end{Thm}

\begin{proof}
To show the ``only if" part,
suppose that $D$ is an orientation of $K_{n_1,n_2,n_3,n_4}$ whose competition graph is complete.

{\it Case 1}. $n_4=1$.
Then $n_3 \geq 3$ by Lemma~\ref{lem:complete-3-partite}, so $n_1 \geq 3$.
If $n_1 =3$, then $n_1=n_2=n_3=3$ and so $D$ is an orientation of $K_{3,3,3,1}$, which contradicts Corollary~\ref{cor:no-or-3,3,2,2}.
Therefore $n_1 \geq 4$.

{\it Case 2}. $n_4 \geq 2$.
Then $n_3 \geq 2$ and so (c) holds.
Suppose $n_3=2$. Then $n_4=2$.
If $n_1=3$, then, by applying Lemma~\ref{lem:making-multi-complete} to $D$, we obtain an orientation $D^*$ of $K_{3,3,2,2}$ whose competition graph is complete, which contradicts Corollary~\ref{cor:no-or-3,3,2,2}.
Therefore $n_1 \geq 4$.
Thus the ``only if" part is true.

Now we show the ``if" part.
Let $D_\alpha$, $D_\beta$, $D_\gamma$ be the digraphs whose adjacency matrices are $A_5$, $A_6$, and $A_7$, respectively, given in Figure~\ref{fig:4-partite-complete}.
It is easy to check that $D_\alpha$, $D_\beta$, and $D_\gamma$ are orientations of $K_{4,3,3,1}$, $K_{4,2,2,2}$, and $K_{3,3,3,2}$, respectively, and the inner product of each pair of rows in each matrix is nonzero, so the competition graphs of $D_\alpha$, $D_\beta$, and $D_\gamma$ are complete.
By applying Lemma~\ref{lem:making-multi-complete} to $D_\alpha$, $D_\beta$, and $D_\gamma$,
we may obtain orientations $D'_\alpha$, $D'_\beta$, and $D'_\gamma$ of $K_{n_1,n_2,n_3,n_4}$ whose competition graphs are complete for (a) $n_1 \geq 4$, $n_3 \geq 3$, and $n_4 \geq 1$; (b) $n_1 \geq 4$, $n_3 =2$, and $n_4 =2$;
(c) $n_3 \geq 3$ and $n_4 \geq 2$, respectively.
Therefore we have shown that the ``if" part is true.
\end{proof}

   \begin{figure}
  \centering
    \begin{equation*}
A_5=
\left(\begin{array}{*{11}c}
  0& 0& 0& 0& 1& 0& 0& 1& 0& 0& 1\\
  0& 0& 0& 0& 0& 1& 0& 0& 1& 0& 1\\
  0& 0& 0& 0& 1& 0& 1& 0& 1& 1& 0\\
  0& 0& 0& 0& 0& 1& 1& 1& 0& 1& 0\\
  0& 1& 0& 1& 0& 0& 0& 1& 1& 0& 0\\
  1& 0& 1& 0& 0& 0& 0& 1& 1& 0& 0\\
  1& 1& 0& 0& 0& 0& 0& 0& 0& 1& 1\\
  0& 1& 1& 0& 0& 0& 1& 0& 0& 0& 1\\
  1& 0& 0& 1& 0& 0& 1& 0& 0& 0& 1\\
  1& 1& 0& 0& 1& 1& 0& 0& 0& 0& 0\\
  0& 0& 1& 1& 1& 1& 0& 0& 0& 1& 0\\
     \end{array}
        \right)
    \end{equation*}
    \begin{equation*}
    A_6=
\left(\begin{array}{*{10}c}
 0 & 0 & 0 & 0 &
 1 & 0
 & 1 & 0
 & 1 & 0   \\
 0 & 0 & 0 & 0 &
 0 & 1
 & 0 & 1
 & 1 & 0   \\
  0 & 0 & 0 & 0 &
 1 & 0
 & 0 & 1
 & 0 & 1   \\
  0 & 0 & 0 & 0 &
 0 & 1
 & 1 & 0
 & 0 & 1   \\
   0 & 1 & 0 & 1 &
 0 & 0
 & 1 & 1
 & 0 & 0   \\
   1 & 0 & 1 & 0 &
 0 & 0
 & 1 & 1
 & 0 & 0   \\
 0 & 1 & 1 & 0 &
 0 & 0
 & 0 & 0
 & 1 & 1   \\
 1 & 0 & 0 & 1 &
 0 & 0
 & 0 & 0
 & 1 & 1   \\
 0 & 0 & 1 & 1 &
 1 & 1
 & 0 & 0
 & 0 & 0   \\
 1& 1 & 0 & 0 &
 1 & 1
 & 0 & 0
 & 0 & 0   \\
   \end{array}
    \right)
    \end{equation*}
     \begin{equation*}
 A_7=
\left(\begin{array}{*{11}c}
0&0&0&1&0&0&0&0&1&1&1\\
0&0&0&0&1&0&0&1&0&1&1\\
0&0&0&1&0&1&1&0&1&0&1\\
0&1&0&0&0&0&0&1&1&1&0\\
1&0&1&0&0&0&0&1&0&0&1\\
1&1&0&0&0&0&1&0&0&1&0\\
1&1&0&1&1&0&0&0&0&0&0\\
1&0&1&0&0&1&0&0&0&1&0\\
0&1&0&0&1&1&0&0&0&0&1\\
0&0&1&0&1&0&1&0&1&0&0\\
0&0&0&1&0&1&1&1&0&0&0\\
  \end{array}
    \right)
    \end{equation*}
      \caption{The adjacency matrices $A_5$, $A_6$, $A_7$ which are orientations of $K_{4,3,3,1}$, $K_{4,2,2,2}$, and $K_{3,3,3,2}$ in the proof of Theorem~\ref{thm:complete-4-partite}.}
      \label{fig:4-partite-complete}
    \end{figure}

In the following, we study $5$-partite tournaments whose competition graphs are complete.

\begin{Thm} \label{thm:no-orientation-5-partite-32211}
There is no orientation of $K_{3,2,2,1,1}$
whose competition graph is complete.
\end{Thm}

\begin{proof}
Suppose, to the contrary, that there exists an orientation $D$ of $K_{3,2,2,1,1}$
whose competition graph is complete.
Let $V_1,\ldots,V_5$ be the partite sets of $D$ with $|V_1|=3$, $|V_2|=|V_3|=2$, and $|V_4|=|V_5|=1$.
Since $4|V(D)|-|A(D)|=5$,
\begin{itemize}
\item[($\dagger$)]
there exist at least $5$ vertices of outdegree $3$ in $D$
\end{itemize}
 by Corollary~\ref{cor:n.b.of-vertices-outdegree3}.
Take a vertex $u$ of outdegree $3$.
Then $D$ contains a subdigraph containing $u$ isomorphic to $\tilde{D}$ given in Figure~\ref{fig:subdigraph-degree3} by Proposition~\ref{prop:complete-condition-outdegree-3}.
We may assume that the subdigraph is $\tilde{D}$ itself including labels.
For each $i=1,2,3$,
since $w_i$ is adjacent to each of $v_1$, $v_2$ and $v_3$ in $D$,

\begin{itemize}
\item[($\S$)] $w_i$ cannot belong to a partite set containing an out-neighbor of $u$.
\end{itemize}
By Lemma~\ref{lem:complete-outdegree-3},
the out-neighbors
$v_1,v_2$, and $v_3$ of $u$ belong to three distinct partite sets $U$, $V$, and $W$.
By Lemma~\ref{lem:degree3-partite-set-sizes},
$|U|+|V|+|W| \leq |V(D)| - 4=5$.
Therefore
\begin{equation}
\label{eq:thm:no-orientation-5-partite-32211-0-0}
|N^+(u) \cap V_1|=|N^+(u) \cap V_4|= |N^+(u) \cap V_5|=1
 \end{equation}
 or
 \begin{equation}
 \label{eq:thm:no-orientation-5-partite-32211-0-1}
  |N^+(u) \cap V_i|=|N^+(u) \cap V_j|=|N^+(u) \cap
  V_k|=1
  \end{equation}
 for $2 \leq i < j < k \leq 5$.
We first show that each vertex in $V_4 \cup V_5$ has outdegree at least $4$.

Suppose, to the contrary, that
 $V_4 \cup V_5$ contains a vertex of outdegree at most $3$. Then, by Corollary~\ref{cor:outdegree-at-least-complete}, the vertex has outdegree $3$.
We may regard it as $u$ since $u$ is a vertex of outdegree $3$ arbitrarily chosen.
Without loss of generality, we may assume $u \in V_5$.
Then $N^+(u) \cap V_5 =\emptyset$.
Therefore \eqref{eq:thm:no-orientation-5-partite-32211-0-0} cannot happen and so
\eqref{eq:thm:no-orientation-5-partite-32211-0-1} holds.
Thus, without loss of generality, we may assume that $v_1 \in V_2$, $v_2 \in V_3$, and $v_3 \in V_4$.

By ($\S$),
$V_1=\{w_1,w_2,w_3\}$.
Let $V_2=\{v_1,x_1\}$ and $V_3=\{v_2,x_2\}$.
Then \[N^-(u)=\{w_1,w_2,w_3,x_1,x_2\}.\]

Since $x_1$ is the only possible common out-neighbor of each of pairs $\{v_2,w_2\}$ and $\{v_2,w_3\}$,
\begin{equation} \label{eq:prop:no-orientation-5-partite-32211-1}
\{v_2,w_2,w_3\} \subseteq N^-(x_1).\end{equation}
In addition, $x_2$ is the only possible common out-neighbor of each of pairs $\{v_1,w_1\}$ and $\{v_1,w_2\}$.
Therefore $\{v_1,w_1,w_2\} \subseteq N^-(x_2).$
By the way, $x_1$ and $x_2$ are the only possible common out-neighbors of $v_3$ and $w_3$.
If $x_2$ is a common out-neighbor of $v_3$ and $w_3$,
then $\{v_1,v_3,w_1,w_2,w_3\} \subseteq N^-(x_2)$ and so $N^+(x_2)\subseteq \{u,x_1\}$, which contradicts Corollary~\ref{cor:outdegree-at-least-complete}.
Therefore $x_1$ is a common out-neighbor of $v_3$ and $w_3$.
Then $\{v_2,v_3,w_2,w_3\} \subseteq N^-(x_1)$ by~\eqref{eq:prop:no-orientation-5-partite-32211-1}, so
 $N^+(x_1) \subseteq \{u,w_1,x_2\}$.
 Thus $N^+(x_1) = \{u,w_1,x_2\}$ by Corollary~\ref{cor:outdegree-at-least-complete}.
However, since $\{w_1 ,x_2\} \subset N^-(u)$,
$N^+(x_1)$ cannot form a directed cycle, which contradicts Lemma~\ref{lem:complete-outdegree-3}.
Hence $u \notin V_4 \cup V_5$ and we reach a contradiction.

Therefore
\begin{equation} \label{eq:prop:no-orientation-5-partite-32211-2}
|N^+(v)| \geq 4
\end{equation}
for each vertex $v$ in $V_4 \cup V_5$.

Now we show that each of $V_2$ and $V_3$ has exactly one vertex of outdegree $3$, which implies that each vertex of $V_1$ has outdegree $3$.

Since $D$ has at least $5$ vertices of outdegree $3$ by ($\dagger$), $V_2 \cup V_3$ has at least $2$ vertices of outdegree $3$ by~\eqref{eq:prop:no-orientation-5-partite-32211-2}.
Take a vertex of outdegree $3$ in $ V_2 \cup V_3$. Then we may regard it as $u$.

Take a vertex $v$ of outdegree $3$ distinct from $u$ in $V_2 \cup V_3$.
Then, $N^+(x) \cap V_2 =\emptyset$ or $N^+(x) \cap V_3 =\emptyset$ for each vertex $x$ in $\{u,v\}$, so, by~\eqref{eq:thm:no-orientation-5-partite-32211-0-0} and \eqref{eq:thm:no-orientation-5-partite-32211-0-1},
\[ |N^+(u) \cap V_i|=|N^+(u) \cap V_4|=|N^+(u) \cap V_5|=1 \]
for some $i \in \{1,2,3\}$
and
\[ |N^+(v) \cap V_{j}|=|N^+(v) \cap V_{4}|=|N^+(v) \cap V_{5}|=1 \]
for some $j \in \{1,2,3\}$.
Thus, since $|V_4|=|V_5|=1$,
the vertices in $V_4 \cup V_5$ are common out-neighbors of $u$ and $v$ and we may assume that
\[ N^+(v)=\{v'_1,v_2,v_3\}\]
for some vertex $v'_1$ in $D$, $V_4=\{v_2\}$, and $V_5=\{v_3\}$ by symmetry.

Since $(v_2,v_3)\in A(D)$,
\[ (v'_1,v_2) \in A(D).\]
by Lemma~\ref{lem:complete-outdegree-3}.
Therefore
\begin{equation} \label{eq:prop:no-orientation-5-partite-32211-001}
\{v,v'_1\} \subseteq N^-(v_2).
\end{equation}
To reach a contradiction, we suppose that $u$ and $v$ are contained in the same partite set.

Without loss of generality, we may assume $\{u,v\}\subseteq V_2$, Then $V_2=\{u,v\}$.
Suppose $v_1 =v'_1$. Then $N^+(u)=N^+(v)$ and $N^-(u)=N^-(v)$.
Therefore any pair of vertices having $v$ as a common out-neighbor has $u$ as a common out-neighbor.
Then, since $C(D)$ is complete,
$C(D-u)$ is complete.
However, $D-u$ is an orientation of $K_{3,1,2,1,1}$, which contradicts Proposition~\ref{prop:n.b.of.size.1}.
Therefore $v_1 \neq v'_1$.
Thus \[N^+(u) \cap N^+(v)=\{v_2,v_3\}.\]
If $v_1 \in V_1$, then
$N^+(u)\subset V_1 \cup V_4 \cup V_5$ and so, by ($\S$), $v=w_i$ for some $i \in \{1,2,3\}$, which contradicts $\{v_2,v_3\}\subseteq N^+(v)$.
Therefore $v_1 \notin V_1$ and so $v_1 \in V_3$.
Thus $N^+(u)\subset V_3\cup V_4 \cup V_5$ and so $V_1=\{w_1,w_2,w_3\}$ by ($\S$).
We may show that, by applying the same argument to $v'_1$, $v'_1 \notin V_1$.
Then $v'_1 \in V_3$, so
$\{v_1,v'_1 \} \subseteq  V_3$.
Therefore $V_3=\{v_1,v'_1 \} $.
We know from $D_1$ that $\{u,v_1,w_3\} \subseteq N^-(v_2) $.
Moreover, $\{v,v'_1\} \subseteq N^-(v_2)$ by~\eqref{eq:prop:no-orientation-5-partite-32211-001}.
Thus
 $\{u,v,v_1,v'_1,w_3 \} \subseteq N^-(v_2)$ and so $N^+(v_2) \subseteq \{v_3, w_1,w_2\}$.
Hence $N^+(v_2) =\{v_3, w_1,w_2\}$ by Corollary~\ref{cor:outdegree-at-least-complete}.
However, $w_1$ and $w_2$ belong to the same partite set $V_1$, which contradicts Lemma~\ref{lem:complete-outdegree-3}.
Therefore $u$ and $v$ belong to the distinct partite sets.
Since $u$ and $v$ were vertices of outdegree $3$ arbitrarily chosen, each of $V_2$ and $V_3$ has at most one vertex of outdegree $3$.
By the way, $V_2 \cup V_3$ has at least $2$ vertices of outdegree $3$, so
we may conclude that
each of $V_2$ and $V_3$ has exactly one vertex of outdegree $3$.
Thus each vertex of $V_1$ has outdegree $3$ by ($\dagger$).

Without loss of generality,
we may assume that $u \in V_2$, $v \in V_3$, and
\[(u,v) \in A(D). \]
Then $v_1=v$.
Therefore
$ |N^+(u) \cap V_3|=|N^+(u) \cap V_4|=|N^+(u) \cap
  V_5|=1 $ by~\eqref{eq:thm:no-orientation-5-partite-32211-0-1}.

If $w_2$, which is a common out-neighbor of $v_2$ and $v_3$, is contained in $V_1$,
then $N^+(w_2) \subseteq V_2 \cup V_3$ and so, by Proposition~\ref{prop:two-partite-out-neighbor-2-2}, $w_2$ has outdegree at least $4$, which is a contradiction to the fact that each vertex of $V_1$ has outdegree $3$.
Therefore $w_2 \in V_2$ by ($\S$).

Then
\[V_2=\{u,w_2\}.\]
Thus $\{w_1,w_3 \} \subset V_1$ by ($\S$) and so each of $w_1$ and $w_3$ has outdegree $3$.
Let
\[V_1=\{w_1,w_3,z\} \quad \text{and} \quad V_3=\{v_1,y\}\] for some vertices $y$ and $z$ in $D$.
 We know from $\tilde{D}$ given in Figure~\ref{fig:subdigraph-degree3} that $N^+(w_1) \cap \{v_1,v_2,v_3\}=\{v_3\}$ and $N^+(w_3)\cap \{v_1,v_2,v_3\}=\{v_2\}$.
Since each of $w_1$ and $w_3$ has outdegree $3$, the out-neighbors of $w_i$ belong to distinct partite sets by Lemma~\ref{lem:complete-outdegree-3} for $i=1,3$. By recalling that $N^+(u)=\{v_1,v_2,v_3\}$, we may conclude that  $N^+(w_1)=\{u,v_3,y\}$ and $N^+(w_3)=\{u,v_2,y\}$.
Since $(u,v_2) \in A(D)$ and $(u,v_3)\in A(D)$, $(v_2,y)\in A(D)$ and $(v_3,y)\in A(D)$ by the same lemma.
Therefore $\{v_2,v_3,w_1,w_3\} \subseteq N^-(y)$ and so $N^+(y) \subseteq \{u,w_2,z\}$.
Thus $N^+(y)= \{u,w_2,z\}$ by Corollary~\ref{cor:outdegree-at-least-complete}.
However, there is no arc between $u$ and $w_2$ and so $N^+(y)$ cannot form a directed cycle, which contradicts Lemma~\ref{lem:complete-outdegree-3}.
\end{proof}

\begin{figure}
  \centering
    \begin{equation*}
     A_8=
\left(\begin{array}{*{10}c}
  0& 0& 0& 1& 0& 0& 0& 1& 1& 1\\
  0& 0& 0& 0& 1& 1& 0& 1& 0& 1\\
  0& 0& 0& 1& 0& 1& 1& 0& 1& 0\\
  0& 1& 0& 0& 0& 0& 0& 0& 1& 1\\
  1& 0& 1& 0& 0& 0& 0& 1& 1& 0\\
  1& 0& 0& 0& 0& 0& 1& 0& 0& 1\\
  1& 1& 0& 1& 1& 0& 0& 0& 0& 0\\
  0& 0& 1& 1& 0& 1& 0& 0& 0& 1\\
  0& 1& 0& 0& 0& 1& 1& 1& 0& 0\\
  0& 0& 1& 0& 1& 0& 1& 0& 1& 0
   \end{array}
    \right)
    \end{equation*}
  \begin{equation*}
    A_9=
\left(\begin{array}{*{9}c}
 0 & 0 & 1 &0 &1 &0 &0 &0 &1 \\
  0 & 0 & 0 &1 &0 &1 &0 &0 &1 \\
  0&1&0&0&0&0&1&0&1\\
  1&0&0&0&0&0 &0 &1 &1\\
  0&1 &1 &1 &0 &0 &0 &1 &0\\
  1 &0 &1 &1 &0 &0 &1 &0 &0\\
  1 &1 &0 &1 &1 &0 &0 &0 &0\\
  1 &1 &1 &0 &0 &1 &0 &0 &0\\
  0 &0 &0 &0 &1 &1 &1 &1 &0\\
   \end{array}
    \right)
    \end{equation*}
      \caption{The adjacency matrices $A_8$ and $A_9$ which are orientations of $K_{3,3,2,1,1}$, $K_{2,2,2,2,1}$ respectively, in the proof of Theorem~\ref{thm:complete-5-partite}.}
      \label{fig:5-partite-complete}
    \end{figure}

\begin{Lem} \label{lem:dividing-partite-set}
Let $k$ be a positive integer with $k \geq 3$; $n_1, \ldots, n_k$,$n'_k,n'_{k+1}$ be positive integers such that $n_k=n'_k+n'_{k+1}$.
If $D$ is an orientation of $K_{n_1,\ldots,n_{k-1},n_k}$ whose competition graph is complete,
then there exists an orientation $D'$ of $K_{n_1,\ldots,n_{k-1},n'_k,n'_{k+1}}$ whose competition graph is complete.
\end{Lem}

\begin{proof}
Suppose that $D$ is an orientation of $K_{n_1,\ldots,n_{k-1},n_k}$ whose competition graph is complete.
Let $V_1,\ldots,V_k$ be partite sets of $D$ satisfying $|V_i|=n_i$ for each $1\leq i \leq k$.
Out of $D$, we construct a $(k+1)$-partite tournament $D'$ whose competition graph is complete in the following way.
We take an $n'_k$-subset $V'_k$ of $V_k$ and let $V'_{k+1}=V_k\setminus V'_k$.
Then $|V'_{k+1}|=n'_{k+1}$.
We keep all the arcs in $D$ so that $A(D) \subset A(D')$.
For each pair $\{x,y\}$ with $x \in V'_k$ and $y\in V'_{k+1}$, we randomly choose one of $(x,y)$ and $(y,x)$ to be an arc of $D'$.
From the fact that $C(D)$ is complete, $V(D)=V(D')$, and $A(D) \subset  A(D')$,
we may conclude that $C(D')$ is complete and so the statement is true.
\end{proof}

\begin{Thm} \label{thm:complete-5-partite}
Let $n_1,n_2,\ldots,n_5$ be positive integers such that $n_1 \geq n_2 \geq \cdots \geq n_5$.
There exists an orientation $D$ of $K_{n_1,n_2,\ldots,n_5}$ whose competition graph is complete if and only if one of the following holds:   (a) $n_1=3$, $n_2=3$, $n_3 \geq 2$, $n_4=1$, and $n_5=1$;
(b) $n_1\geq 4$, $n_2 \geq n_3 \geq 2$, and $n_4 =n_5=1$;
(c) $n_4 \geq 2$.
\end{Thm}

\begin{proof}
To show the ``only if'' part, suppose that there exists an orientation $D$ of $K_{n_1,n_2,\ldots,n_5}$ such that $C(D)$ is complete.
By Proposition~\ref{prop:n.b.of.size.1},
 \[n_3 \geq 2.\]

If $n_4 \geq 2$, then (c) holds.
Now suppose $n_4=1$.
 Then $n_5=1$.
 Suppose, to the contrary, that $n_1=2$.
 Then $D$ is an orientation of $K_{2,2,2,1,1}$.
Since $4|V(D)|-|A(D)|=7 > 0$, $D$ has a vertex of outdegree 3 by Corollary~\ref{cor:n.b.of-vertices-outdegree3}.
Therefore $|V(D)| \geq 9$ by Proposition~\ref{prop:complete-condition-outdegree-3}, which is impossible.
Thus \[n_1 \geq 3.\]
If $n_1 \geq 4$, then (b) holds.
Now suppose $n_1=3$.
Then $n_2 \leq 3$.
If $n_2=2$, then $n_3=2$ and so $D$ is an orientation of $K_{3,2,2,1,1}$, which contradicts Theorem~\ref{thm:no-orientation-5-partite-32211}.
Therefore
$n_2 = 1$ or $3$.
Then, since $n_2 \geq n_3 \geq 2$,
$n_2 =3$ and (a) holds.
Thus we have shown the ``only if" part.

Now we show the ``if" part.
Let $D_\alpha$, $D_\gamma$ be the digraphs whose adjacency matrix are $A_8$ and $A_9$, respectively, given in Figure~\ref{fig:5-partite-complete}.
Then $D_\alpha$ and $D_\gamma$ are orientations of $K_{3,3,2,1,1}$ and $K_{2,2,2,2,1}$, respectively.

Let $D^*_\beta$ be the digraph whose adjacency matrix is $A_6$ given in Figure~\ref{fig:4-partite-complete}.
It is easy to check that the inner product of each pair of rows in each matrix is nonzero, so the competition graphs of $D_\alpha$, $D_\gamma$, and $D^*_\beta$ are complete.
By Lemma~\ref{lem:dividing-partite-set}, we obtain an orientation $D_\beta$ of $K_{4,2,2,1,1}$ whose competition graph is complete from $D^*_\beta$.
By applying Lemma~\ref{lem:making-multi-complete} to $D_\alpha$, $D_\beta$, and $D_\gamma$,
we may obtain orientations $D'_\alpha$, $D'_\beta$ , and $D'_\gamma$ of $K_{n_1,n_2,\ldots,n_5}$ whose competition graphs are complete for (a) $n_1=3$, $n_2=3$, $n_3 \geq 2$, $n_4=1$, and $n_5=1$;
(b) $n_1\geq 4$, $n_2 \geq n_3 \geq 2$, $n_4 =1$, and $n_5=1$;
(c) $n_4 \geq 2$, respectively.
Therefore we have shown that the ``if" part is true.
\end{proof}
By Corollary~\ref{cor:no-biparte-complete},
there is no complete graph that is the competition graph of a bipartite tournament.
For an integer $k \geq 3$,
Proposition~\ref{prop:k>=7-complete}, and Theorems~\ref{thm:complete-3-partite},~\ref{thm:complete-4-partite},~\ref{thm:complete-5-partite},~\ref{thm:complete-6-partite} may be summarized in the aspect of the number of vertices of a complete graph which is the competition graph of a $k$-partite tournament as follows.

\begin{Thm} \label{Thm:summary:K_n-k-partite}
A complete graph $K_n$ is the competition graph of a $k$-partite tournament for some integer $k\geq 3$ if and only if
\[
 \begin{cases}
    n \geq 13  & \mbox{ if $k=3$;}   \\
  n \geq 10 & \mbox{ if $k=4$;}
\\
 n \geq 9 & \mbox{ if $k=5$;}
 \\
 n \geq 9 &  \mbox{ if $k=6$;}
  \\
 n \geq k  & \mbox{ if $k\geq 7$.}
  \end{cases}
 \]
\end{Thm}

\begin{proof}
For an integer $k\geq 3$, suppose that
a complete graph $K_n$ is the competition graph of a $k$-partite tournament which is an orientation of $K_{n_1,n_2,\ldots,n_k}$.
Then $n_1 + n_2 + \cdots + n_k=n$.
If $k=3$, then $\sum_{i=1}^{3}n_i \geq 13$ by Theorem~\ref{thm:complete-3-partite}.
If $k=4$, then $\sum_{i=1}^{4}n_i \geq 10$ by Theorem~\ref{thm:complete-4-partite}.
If $k=5$, then $\sum_{i=1}^{5}n_i \geq 9$ by Theorem~\ref{thm:complete-5-partite}.
If $k=6$, then $\sum_{i=1}^{6}n_i \geq 9$ by Theorem~\ref{thm:complete-6-partite}.
If $k\geq 7$, then $\sum_{i=1}^{k}n_i \geq k$ by Proposition~\ref{prop:k>=7-complete}.
Each of the above theorems also guarantees the existence of a $k$-partite tournament whose competition graph is complete for the corresponding $k$, so the ``if" part is true.
\end{proof}

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