Consider our algorithm for computing a topological ordering that is based on depth$-$first search $($i$.$e$.,$ NOT the "straightforward solution"$).$ Suppose we run this algorithm on a graph that is NOT directed acyclic. Obviously it won't compute a topological order $($since none exist$).$

Does it compute an ordering that minimizes the number of edges that go backward?

For example, consider the four$-$node graph with the six directed edges $.$ Suppose the vertices are ordered $.$ Then there is one backwards arc, the arc. No ordering of the vertices has zero backwards arcs, and some have more than one.