I want to show that for every integer $n$ there exists a tournament graph with $n$ vertices where all the in-degrees are different, and all the out-degrees are different. My first thought was induction but I have been trying to form a result with no effect. Any suggestions?
2026-03-28 01:48:22.1774662502
A tournament graph where all the in-degrees and out-degrees are different.
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Why doesn't induction work? By assumption we have such graph on $n$ vertices. Now add another vertex which is source (it has outdegree = $n$) in new graph and we are done.