$1.(15$ points.$)$ Suppose that $\backslash ($ G $\backslash )$ is a directed graph. It has an adjacency structure G$.$Adj, a set of edges $\backslash ($ G $.$ E $\backslash ),$ and a set of vertexes $\backslash ($ G $.$ V $\backslash ).$ Each vertex $\backslash ($ v $\backslash$in G $.$ V $\backslash )$ has an attribute $\backslash ($ v $.$ m a r k $\backslash$in$\backslash \{\backslash )$ TRUE, FALSE $\backslash \}.$ It has no other attributes! In particular, it has no color, $\backslash ($ d$,$ f $\backslash ),$ or $\backslash (\backslash$pi $\backslash )$ attributes like Cormen's vertexes do$.$

A vertex $\backslash ($ v $\backslash$in G $.$ V $\backslash )$ is an origin vertex in $\backslash ($ G $\backslash )$ if there is exactly one path from $\backslash ($ v $\backslash )$ to every other vertex in G$.$V$.$ The word origin is meant to suggest that every vertex in $\backslash ($ G $\backslash )$ is reachable starting from $\backslash ($ v $\backslash ).$ Don't bother looking for origin vertex online: the only hits you will get are irrelevant to this question: they deal with parabolas from high$-$school algebra.

Write a procedure IS$-$ORIGIN $\backslash (($G$,$ u$)\backslash )$ that returns TRUE if the vertex $\backslash ($ u $\backslash )$ is an origin in the graph $\backslash ($ G $\backslash ),$ and returns FALSE otherwise. Your procedure must always terminate, even if $\backslash ($ G $\backslash )$ has cycles or self edges. You will lose many points if it uses vertex attributes other than mark.