Question

Intro to problem:

I am given recipes how to craft items. Recipe is in format : {element that is being crafter}: {list of elements, that is needed}. Before I can craft element x, I need to know how to craft elements it’s made of. So I want to find in what order do I have to learn recipes.

For valid input, like following everything works:

// Input:
{ 
    "F1: F2 F3 F4", "F5: F6 F4", "F6: F7 F8 F4", "F2: F3 F8 F4", "F8: F4",
    "F9: F4", "F7: F4", "F10: F7 F4", "F11: F4", "F4:", "F3: F6"
}
// Output:
[F4, F7, F8, F6, F3, F2, F1, F5, F9, F10, F11]

The problem is, that task is more complex. Time to time I have some recipes missing or thy are invalid. Example of invalid input: { "F1: F2", "F2: F1" }.

Code example:

mp contains recipe name as key and elements as value, labels are unique mp keys and result will contain answer. I’m looking for a way to return empty result if infinite loop is met.

private void getArray(HashMap<String, ArrayList<String>> mp,
        ArrayList<String> result, ArrayList<String> labels) {
    for (String a : labels) {
        if (mp.get(a) != null)
            for (String label : mp.get(a))
                getArray(mp, result, label);
        if (!result.contains(a))
            result.add(a);
    }
}

private void getArray(HashMap<String, ArrayList<String>> mp,
        ArrayList<String> result, String label) {
    if (result.contains(label))
        return;
    if (mp.get(label) == null) {
        result.add(label);
        return;
    }
    for (String l : mp.get(label))
        getArray(mp, result, l);
    if (!result.contains(label))
        result.add(label);
}

Edit

Problem solved.

For any Google’s that stumble up this, this is what I came up with:

/** <p>
 * <b>Topological sort</b> solves a problem of - finding a linear ordering
 * of the vertices of <i>V</i> such that for each edge <i>(i, j) ∈ E</i>,
 * vertex <i>i</i> is to the left of vertex <i>j</i>. (Skiena 2008, p. 481)
 * </p>
 * 
 * <p>
 * Method is derived from of <a
 * href="http://en.wikipedia.org/wiki/Topological_sort#Algorithms" > Kahn's
 * pseudo code</a> and traverses over vertices as they are returned by input
 * map. Leaf nodes can have null or empty values. This method assumes, that
 * input is valid DAG, so if cyclic dependency is detected, error is thrown.
 * tSortFix is a fix to remove self dependencies and add missing leaf nodes.
 * </p>
 * 
 * <pre>
 * // For input with elements:
 * { F1=[F2, F3, F4], F10=[F7, F4], F11=[F4], F2=[F3, F8, F4], F3=[F6], 
 *   F4=null, F5=[F6, F4], F6=[F7, F8, F4], F7=[F4], F8=[F4], F9=[F4]}
 *   
 * // Output based on input map type: 
 * HashMap: [F4, F11, F8, F9, F7, F10, F6, F5, F3, F2, F1]
 * TreeMap: [F4, F11, F7, F8, F9, F10, F6, F3, F5, F2, F1]
 * </pre>
 * 
 * @param g
 *            <a href="http://en.wikipedia.org/wiki/Directed_acyclic_graph"
 *            > Directed Acyclic Graph</a>, where vertices are stored as
 *            {@link java.util.HashMap HashMap} elements.
 * 
 * @return Linear ordering of input nodes.
 * @throws Exception
 *             Thrown when cyclic dependency is detected, error message also
 *             contains elements in cycle.
 * 
 */
public static <T> ArrayList<T> tSort(java.util.Map<T, ArrayList<T>> g)
        throws Exception
/**
 * @param L
 *            Answer.
 * @param S
 *            Not visited leaf vertices.
 * @param V
 *            Visited vertices.
 * @param P
 *            Defined vertices.
 * @param n
 *            Current element.
 */
{
    java.util.ArrayList<T> L = new ArrayList<T>(g.size());
    java.util.Queue<T> S = new java.util.concurrent.LinkedBlockingDeque<T>();
    java.util.HashSet<T> V = new java.util.HashSet<T>(), 
    P = new java.util.HashSet<T>();
    P.addAll(g.keySet());
    T n;

    // Find leaf nodes.
    for (T t : P)
        if (g.get(t) == null || g.get(t).isEmpty())
            S.add(t);

    // Visit all leaf nodes. Build result from vertices, that are visited
    // for the first time. Add vertices to not visited leaf vertices S, if
    // it contains current element n an all of it's values are visited.
    while (!S.isEmpty()) {
        if (V.add(n = S.poll()))
            L.add(n);
        for (T t : g.keySet())
            if (g.get(t) != null && !g.get(t).isEmpty() && !V.contains(t)
                    && V.containsAll(g.get(t)))
                S.add(t);
    }

    // Return result.
    if (L.containsAll(P))
        return L;

    // Throw exception.
    StringBuilder sb = new StringBuilder(
            "\nInvalid DAG: a cyclic dependency detected :\n");
    for (T t : P)
        if (!L.contains(t))
            sb.append(t).append(" ");
    throw new Exception(sb.append("\n").toString());
}

/**
 * Method removes self dependencies and adds missing leaf nodes.
 * 
 * @param g
 *            <a href="http://en.wikipedia.org/wiki/Directed_acyclic_graph"
 *            > Directed Acyclic Graph</a>, where vertices are stored as
 *            {@link java.util.HashMap HashMap} elements.
 */
public static <T> void tSortFix(java.util.Map<T, ArrayList<T>> g) {
    java.util.ArrayList<T> tmp;
    java.util.HashSet<T> P = new java.util.HashSet<T>();
    P.addAll(g.keySet());

    for (T t : P)
        if (g.get(t) != null || !g.get(t).isEmpty()) {
            (tmp = g.get(t)).remove(t);
            for (T m : tmp)
                if (!P.contains(m))
                    g.put(m, new ArrayList<T>(0));
        }
}

Answer 1

The problem you are solving is known as topological sort. Kahn’s algorithm solves the problem while also detecting invalid input (that is, containing cycles).