2 corrected spelling

Let me also stay state an explicit criterion of the existence of orientation with out-degrees at most $d$: any induced subgraph on some, say, $k$ vertices contains at most $dk$ edges. This is clearly necessary, and the proof that it is sufficient is not hard: orient edges arbitrarily and consider the following procedure.

If out-degree of some vertex $a$ is at least $d+1$, then consider the set of vertices $x$, for which there exist oriented path from $a$. If all out-degrees of such vertices are at least $d$, then the set of them contradicts to our assumption. If deree of $x$ is less then $d$, then invert all edges on the path from $a$ to $x$.

Repeating this stuff we kill all high (more then $d$) out-degree after a fnite number of steps.

1

Let me also stay explicit criterion of the existence of orientation with out-degrees at most $d$: any induced subgraph on some, say, $k$ vertices contains at most $dk$ edges. This is clearly necessary, and the proof that it is sufficient is not hard: orient edges arbitrarily and consider the following procedure.

If out-degree of some vertex $a$ is at least $d+1$, then consider the set of vertices $x$, for which there exist oriented path from $a$. If all out-degrees of such vertices are at least $d$, then the set of them contradicts to our assumption. If deree of $x$ is less then $d$, then invert all edges on the path from $a$ to $x$.

Repeating this stuff we kill all high (more then $d$) out-degree after a fnite number of steps.