Let $n\geq 3$ be an integer, set $[n] = \{1,\ldots,n\}$ and let $G=([n],E)$ be an undirected connected [bridgeless](https://en.wikipedia.org/wiki/Bridge_(graph_theory)) graph. Is there an orientation (explanation below) of $G$ such that for all $a\neq b\in [n]$ there is a directed path leading from $a$ to $b$?




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An *orientation* of $G=([n],E)$ is a directed graph $G\_d=([n], E\_d)$ with $E\_d \subseteq [n]\times[n]$ where for every $\{x,y\} \in E$ exactly one of $\big\{(x,y),(y,x)\big\}$ is a member of $E\_d$.



Orientations in connected bridgeless graphs
Yes. This is known as [Robbins' theorem](https://en.wikipedia.org/wiki/Robbins%27_theorem).


