Kings in tournament What is the difference between king,serf and strong king?
How to prove that if a tournament has no vertex with zero in-degree, then it has at least 3 kings. 
 A: Let $T$ be a tournament. Say that a vertex $u$ is reachable in $k$ steps from a vertex $v$ if there is a path of length $k$ from $v$ to $u$. A vertex $v$ in $T$ is a king if each other vertex of $T$ is reachable from $v$ in at most $2$ steps. A vertex $v$ of $T$ is a serf if $v$ is reachable in at most $2$ steps from each other vertex of $T$. For distinct vertices $v$ and $u$ of $T$ let $b(v,u)$ be the number of paths of length $2$ from $v$ to $u$. (Equivalently, $b(v,u)$ is the number of vertices $w$ distinct from both $u$ and $v$ such that $vwu$ is a path in $T$.) A vertex $v$ of $T$ is a strong king if $b(v,u)>b(u,v)$ for every vertex $u$ such that $uv$ is an edge of $T$. (Note that this does imply that $v$ is a king.)
In game terms, a player $v$ is king if for each other player $u$, either $v$ beats $u$, or there is a player $w$ such that $v$ beats $w$ and $w$ beats $u$. A player $v$ is a serf if for each other player $u$, either $u$ beats $v$, or there is a player $w$ such that $u$ beats $w$ and $w$ beats $v$. A player $v$ is a strong king if whenever some player $u$ beats $v$, $b(v,u)>b(u,v)$.
The theorem that you want to prove is that if $T$ has no vertex with in-degree $0$ (i.e., if no player beats all of the other players), then $T$ has at least $3$ kings, i.e., at least $3$ vertices from which every other vertex is reachable in at most $2$ steps. Here’s an extended HINT:


*

*Show that any vertex of $T$ of maximal out-degree is a king. This shows that $T$ has at least one king.  

*Show that if $v$ is any vertex of $T$, there is a king $u$ such that $uv$ is an edge (i.e., $u$ beats $v$).  

*Show that $T$ cannot have exactly $1$ or exactly $2$ kings.

