A videogame tournament has $2012$ participating teams. Two rounds have been played so far. Prove that we can still split the teams into two groups of $1006$ teams each so that no teams of the same group have played each other before. This graph would be a bipartite graph of $1006$ vertices each. However, what do I do for the next step to solve this problem?
2026-03-29 21:40:53.1774820453
A videogame tournament has $2012$ participating teams. Two rounds have been played so far. Prove that we can still split the teams into two groups
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Build a graph, in which the $2012$ vertices are teams and the $2012$ edges are games played in the first two rounds.
Notice your graph is regular of degree $2$. Such graphs consist of disjoint cycles. Now notice it cannot have odd cycles. Suppose a cycle has $2k+1$ vertices, then at least $k+1$ of these belong to one given round, this implies one player played at least $2$ matches in that round. We conclude all cycles are even.
Knowing that the graph consists of disjoint even cycles it is easy to see the vertices can be split into two independent sets of $1006$ vertices.