# A directed bichromatic tournament

Consider a tournament with $$n$$ vertices. Each edge of the directed complete graph (tournament) is colored red or blue. Prove there exist a vertex for which there exists a monochromatic path from it to any other vertex.

My method: Let $$S=\{1,2,...,n\}$$ where they are the vertices. Use induction on n. By inductive hypothesis, WLOG x is connected to $$S\ \{X,y\}$$ via a monochromatic path without edges from $$y$$. Note $$y$$ is unique or I am done. So every vertex is not connected to exactly one via a monochromatic path. 1 would be connected to some $$v$$ which connects the vertex 1 was assumed to be disconnected to. Qed. Is this right? Are there other methods?

• Mar 3, 2020 at 8:46
• No; mine contains bichronatic conditions. My sol is wrong. Mar 3, 2020 at 8:48
• If it's directed and bichromatic then there may not be such a vertex. Let the vertices be named $1,2,3,4$, with all edges going from higher numbers to lower. Let 43 be red, let the other edges be blue. The only path in the graph that respects the directions and visits each vertex is 4321, and it's not monochromatic. Mar 3, 2020 at 8:56
• @GerryMyerson I read the statement as "there is a vertex $x$ such that for every vertex $y$ there is a monochromatic path from $x$ to $y$."
– bof
Mar 3, 2020 at 11:34

Theorem. Let $$T$$ be a (finite) tournament, and suppose each edge of $$T$$ is colored red or blue. Then there is a vertex $$x$$ such that, for every vertex $$y$$, there is a monochromatic (directed) path from $$x$$ to $$y$$.
Proof. Assume for a contradiction that $$T$$ is a minimal counterexample. For each vertex $$x$$ of $$T$$ we can choose a vertex $$f(x)\ne x$$ such that there is a monochromatic path from $$f(x)$$ to every vertex of $$T-x$$ but there is no monochromatic path from $$f(x)$$ to $$x$$; of course this implies that there is a directed edge $$x\to f(x)$$. Since $$T$$ is finite, there is a cycle $$x_1,\dots,x_n,x_1$$ such that $$f(x_i)=x_{i+1}$$ for $$i=1,2,\dots,n-1$$, and $$f(x_n)=x_1$$.
If the edges $$x_1x_2,x_2x_3,\dots,x_nx_1$$ are all the same color, we get an obvious contradiction, since there is then a monochromatic path from $$x_2$$ to $$x_1$$. Therefore the cycle must contain two consecutive edges of different colors. Without loss of generality, suppose $$x_1x_2$$ is red and $$x_2x_3$$ is blue. Now there is a monochromatic path from $$x_3$$ to $$x_1$$. If there is a red path from $$x_3$$ to $$x_1$$, then there is a red path from $$x_3$$ to $$x_2$$; if there is a blue path from $$x_3$$ to $$x_1$$, then there is a blue path from $$x_2$$ to $$x_1$$; in either case we have arrived at a contradiction.