Нахождение цикла в ориентированном графе, какова цель метода проверки в этом коде?

Question

Код, предоставленный этим источником, обнаруживает циклические ориентированные графы с использованием алгоритмов DFS. Код (из источника):

 /******************************************************************************
 *  Compilation:  javac DirectedCycle.java
 *  Execution:    java DirectedCycle input.txt
 *  Dependencies: Digraph.java Stack.java StdOut.java In.java
 *  Data files:   https://algs4.cs.princeton.edu/42digraph/tinyDG.txt
 *                https://algs4.cs.princeton.edu/42digraph/tinyDAG.txt
 *
 *  Finds a directed cycle in a digraph.
 *  Runs in O(E + V) time.
 *
 *  % java DirectedCycle tinyDG.txt 
 *  Directed cycle: 3 5 4 3 
 *
 *  %  java DirectedCycle tinyDAG.txt 
 *  No directed cycle
 *
 ******************************************************************************/

/**
 *  The {@code DirectedCycle} class represents a data type for 
 *  determining whether a digraph has a directed cycle.
 *  The <em>hasCycle</em> operation determines whether the digraph has
 *  a simple directed cycle and, if so, the <em>cycle</em> operation
 *  returns one.
 *  <p>
 *  This implementation uses depth-first search.
 *  The constructor takes time proportional to <em>V</em> + <em>E</em>
 *  (in the worst case),
 *  where <em>V</em> is the number of vertices and <em>E</em> is the number of edges.
 *  Afterwards, the <em>hasCycle</em> operation takes constant time;
 *  the <em>cycle</em> operation takes time proportional
 *  to the length of the cycle.
 *  <p>
 *  See {@link Topological} to compute a topological order if the
 *  digraph is acyclic.
 *  <p>
 *  For additional documentation,
 *  see <a href="https://algs4.cs.princeton.edu/42digraph">Section 4.2</a> of
 *  <i>Algorithms, 4th Edition</i> by Robert Sedgewick and Kevin Wayne.
 *
 *  @author Robert Sedgewick
 *  @author Kevin Wayne
 */
public class DirectedCycle {
    private boolean[] marked;        // marked[v] = has vertex v been marked?
    private int[] edgeTo;            // edgeTo[v] = previous vertex on path to v
    private boolean[] onStack;       // onStack[v] = is vertex on the stack?
    private Stack<Integer> cycle;    // directed cycle (or null if no such cycle)

    /**
     * Determines whether the digraph {@code G} has a directed cycle and, if so,
     * finds such a cycle.
     * @param G the digraph
     */
    public DirectedCycle(Digraph G) {
        marked  = new boolean[G.V()];
        onStack = new boolean[G.V()];
        edgeTo  = new int[G.V()];
        for (int v = 0; v < G.V(); v++)
            if (!marked[v] && cycle == null) dfs(G, v);
    }

    // check that algorithm computes either the topological order or finds a directed cycle
    private void dfs(Digraph G, int v) {
        onStack[v] = true;
        marked[v] = true;
        for (int w : G.adj(v)) {

            // short circuit if directed cycle found
            if (cycle != null) return;

            // found new vertex, so recur
            else if (!marked[w]) {
                edgeTo[w] = v;
                dfs(G, w);
            }

            // trace back directed cycle
            else if (onStack[w]) {
                cycle = new Stack<Integer>();
                for (int x = v; x != w; x = edgeTo[x]) {
                    cycle.push(x);
                }
                cycle.push(w);
                cycle.push(v);
                assert check();
            }
        }
        onStack[v] = false;
    }

    /**
     * Does the digraph have a directed cycle?
     * @return {@code true} if the digraph has a directed cycle, {@code false} otherwise
     */
    public boolean hasCycle() {
        return cycle != null;
    }

    /**
     * Returns a directed cycle if the digraph has a directed cycle, and {@code null} otherwise.
     * @return a directed cycle (as an iterable) if the digraph has a directed cycle,
     *    and {@code null} otherwise
     */
    public Iterable<Integer> cycle() {
        return cycle;
    }


    // certify that digraph has a directed cycle if it reports one
    private boolean check() {

        if (hasCycle()) {
            // verify cycle
            int first = -1, last = -1;
            for (int v : cycle()) {
                if (first == -1) first = v;
                last = v;
            }
            if (first != last) {
                System.err.printf("cycle begins with %d and ends with %d\n", first, last);
                return false;
            }
        }


        return true;
    }

    /**
     * Unit tests the {@code DirectedCycle} data type.
     *
     * @param args the command-line arguments
     */
    public static void main(String[] args) {
        In in = new In(args[0]);
        Digraph G = new Digraph(in);

        DirectedCycle finder = new DirectedCycle(G);
        if (finder.hasCycle()) {
            StdOut.print("Directed cycle: ");
            for (int v : finder.cycle()) {
                StdOut.print(v + " ");
            }
            StdOut.println();
        }

        else {
            StdOut.println("No directed cycle");
        }
        StdOut.println();
    }

}

Я не совсем понимаю цель check Funtion. Как видно из последнего условного утверждения в dfs функция, что одна и та же вершина всегда добавляется в верхней и нижней части cycle стек, поэтому метод проверки всегда должен возвращать true независимо от того, является ли цикл действительным или нет.

Я что-то упустил или это check функция избыточна в этом коде? Благодарю.