I did this question in a course and it is 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 G 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 $$(s,v),(s,w),(v,w),(v,t),(w,t),(t,s).$$ Suppose the vertices are ordered $s,v,w,t$. Then there is one backwards arc, the $(t,s)$ arc. No ordering of the vertices has zero backwards arcs, and some have more than one. please help and tell me what it means by saying NOT acyclic?