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How to construct a k-regular graph?
A match can only be played by exactly two different players. I would've thought this was easier. Simple combinatorics show that the number of matches played would have to be $\frac{1}{2}p\cdot m$ so that either $p$ or $m$ must be even. But is that enough? If either $p$ or $m$ is even (and of course, $m \lt p$ ), does that guarantee I can arrange matches between different players such that each of the $p$ players plays exactly $m$ different matches (that is, there are no rematches)? I will rephrase the question more rigorously:
Let $p, m \in \mathbb N$. Define $I_p = \{ n \in \mathbb N ; n \leq p \} = \{1, 2, ..., p\} \text{ }$ to be the set of players and
$\begin{eqnarray*} (I_p, 2) &=& \{ P\subset I_p ; |P|=2 \}\\ &=&\{\{a,b\} \subset I_p; a\neq b\} \end{eqnarray*}$ the set of all subsets of $I_p$ with two elements to be the set of matches between players.
We will say a set $S \subset (I_p, 2)$ is a valid matchup of $m$ matches per each of $p$ players if $\begin{eqnarray*} f:&I_p& \rightarrow \mathbb N \\ &n&\mapsto |\text{ }\{\text{ }\{n,a\} \in S\ \text{ }\}\text{ }| \end{eqnarray*}$
is the constant function $f \equiv m\text{ }.$ For which choice of $p, m$ is there a valid matchup of $m$ matches per each of $p$ players?
In terms of graph theory, it is equivalent to asking: for which natural numbers $p, m$ is there a $m$-regular simple graph on $p$ vertices?