TST

Let $f(n)$ be the answer to the problem. We claim that $f(n)=\lceil \log n\rceil$ for all $n\ge 1$. The proof is divided into showing that this quantity is both an upper and a lower bound.

Lower Bound. We show that every tournament on $n$ vertices has directed-edge-chromatic number at least $\lceil \log n\rceil$. We prove this by strong induction on $n$. As our base case, if $n=1$, then clearly $f(1)\ge 0=\lceil \log 1\rceil$. Now, for $n\ge 2$, assume by the strong inductive hypothesis that $f(k)\ge \lceil \log k\rceil$ for all $k<n$. Consider any proper directed-edge-coloring of an $n$-vertex directed graph (vertex set $V$) with $f(n)\ge 1$ colors. Pick an arbitrary color, and let $U_1$ be the set of vertices $u$ for which there exists an edge $\overrightarrow{uv}$ of that color, and let $U_2$ be the set of vertices $u$ for which there exists an edge of the form $\overrightarrow{vu}$. Clearly since the coloring is proper $U_1\cap U_2=\varnothing$. Thus, there exists $i\in \{1,2\}$ such that $U_i$ has size at most $n/2$. Notice that $V\setminus U_i$ has size at least $n/2$ and also has no edges of the originally chosen color. Thus, the graph induced by $V\setminus U_i$ has a proper directed-edge-coloring of $f(n)-1$ colors. By the inductive hypothesis, this implies that $$f(n)-1\ge f(|V\setminus U_i|)\ge \lceil \log |V\setminus U_i|\rceil \ge \lceil \log (n/2)\rceil =\lceil \log n\rceil -1.$$ Therefore, $f(n)\ge \lceil \log n\rceil$.

Upper Bound. We exhibit a proper directed-edge-coloring with $\lceil \log n\rceil$ colors of a tournament with $n$ vertices. Consider the tournament with vertices $0,\dots ,n-1$ such that there is a directed edge from $i$ to $j$ if and only if $i>j$. Label our colors $0,\dots ,\lceil \log n\rceil -1$. Color the edge $\overrightarrow{ij}$ the largest $k$ such that $i$ has a $1$ and $j$ has a $0$ in the $2^k$th place of their binary representations. Such a $k$ always exists and is at most $\lceil \log n\rceil -1$ because $i>j$ and $n-1$ requires at most $\lceil \log n\rceil$ digits in its binary representation. Assume for sake of contradiction that this is not a proper directed-edge-coloring. Then, there would exist edge $\overrightarrow{ij}$ and $\overrightarrow{jk}$ of color $l$. Thus, $j$ would have both $0$ and $1$ in the $2^l$th place of its binary representation, contradiction. Hence, this exhibited coloring is proper so $f(n)\le \lceil \log n\rceil$.

Thus, the minimal directed-edge-chromatic number among tournaments on $n$ vertices is $\lceil \log n\rceil$.