## Related questions with answers

Show that in a tournament (defined preceding mentioned theorem) it is always possible to rank the contestants so that the person ranked ith beats the person ranked (i +1)st. (Hint: Use the mentioned theorem.)

Solution

VerifiedA tournament is a directed graph $G = (V,E)$ whose underlying (undirected) graph is $K_n$. We will prove the claim by induction. For $n = 2$ the graph has only one directed edge $(x,y)$. Then we rank $x$ as 1st and $y$ as 2nd. Suppose that the claim holds for some $n \in \mathbb{N}$ and let us prove it for $n+1$. Let $x \in V$ be any vertex. If we remove $x$ from $G$, we get a tournament $G'$ with $n-1$ vertices. By the inducitive assumption, the vertices of $G'$ can be ranked (or renamed) as $1$, $2$, ..., $n$ such that $i$ beats $i+1$ (that means that $G'$ contains the directed edge $(i,i+1)$).

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