10.4.1. Topological Sort¶

Assume that we need to schedule a series of tasks, such as classes or construction jobs, where we cannot start one task until after its prerequisites are completed. We wish to organize the tasks into a linear order that allows us to complete them one at a time without violating any prerequisites. We can model the problem using a DAG. The graph is directed because one task is a prerequisite of another – the vertices have a directed relationship. It is acyclic because a cycle would indicate a conflicting series of prerequisites that could not be completed without violating at least one prerequisite. The process of laying out the vertices of a DAG in a linear order to meet the prerequisite rules is called a topological sort.

Figure 10.5.1 illustrates the problem. An acceptable topological sort for this example is J1, J2, J3, J4, J5, J6, J7. However, other orders are also acceptable, such as J1, J3, J2, J6, J4, J5, J7.

10.4.2. Depth-first solution¶

A topological sort may be found by performing a DFS on the graph. When a vertex is visited, no action is taken (i.e., function PreVisit does nothing). When the recursion pops back to that vertex, function PostVisit adds the vertex to a stack. In the end, the stack is returned to the caller.

The reason that we use a stack is that this algorithm produces the vertices in reverse topological order. And if we pop the elements in the stack, they will be popped in topological order.

So the DFS algorithm yields a topological sort in reverse order. It does not matter where the sort starts, as long as all vertices are visited in the end. Here is implementation for the DFS-based algorithm.

static <V> Stack<V> topsortDFS(Graph<V> G) {
    Set<V> visited = new MapSet<V>();
    Stack<V> sortedVertices = new LinkedStack<V>();
    for (V v : G.vertices()) {
        if (!visited.contains(v))
            topsortHelperDFS(G, v, sortedVertices, visited);
    }
    return sortedVertices;
}

static <V> void topsortHelperDFS(Graph<V> G, V v, Stack<V> sortedVertices, Set<V> visited) {
    if (!visited.contains(v)) {
        visited.add(v);
        for (Edge<V> edge : G.outgoingEdges(v)) {
            V w = edge.end;
            topsortHelperDFS(G, w, sortedVertices, visited);
        }
        sortedVertices.push(v); // PostVisit
    }
}

Using this algorithm starting at J1 and visiting adjacent neighbors in alphabetic order, vertices of the graph in Figure 10.5.1 are pushed to the stack in the order J7, J5, J4, J6, J2, J3, J1. Popping them one by one yields the topological sort J1, J3, J2, J6, J4, J5, J7.

Here is another example.

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10.4.3. Queue-based Solution (optional)¶

We can implement topological sort using a queue instead of recursion, as follows.

First visit all edges, counting the number of edges that lead to each vertex (i.e., count the number of prerequisites for each vertex). All vertices with no prerequisites are placed on the queue. We then begin processing the queue. When vertex \(v\) is taken off of the queue, it is added to a list containing the topological order, and all neighbors of \(v\) (that is, all vertices that have \(v\) as a prerequisite) have their counts decremented by one. Place on the queue any neighbor whose count becomes zero. If the queue becomes empty without having added all vertices to the final list, then the graph contains a cycle (i.e., there is no possible ordering for the tasks that does not violate some prerequisite). The order in which the vertices are added to the final list is the correct one, so if traverse the final list we will get the elements in topological order. Applying the queue version of topological sort to the graph of Figure 10.5.1 produces J1, J2, J3, J6, J4, J5, J7. Here is an implementation for the algorithm.

static <V> Queue<V> topsortBFS(Graph<V> G) {
    // Initialize the prerequisite counts
    Map<V, Integer> Count = new SeparateChainingHashMap<>();
    for (V v : G.vertices())
        Count.put(v, 0);
    for (V v : G.vertices()) {
        for (Edge<V> edge : G.outgoingEdges(v))
            // Add one to v's prereq count
            Count.put(edge.end, Count.get(edge.end) + 1);
    }

    // Initialize the queue
    Queue<V> Q = new LinkedQueue<>();
    for (V v : G.vertices()) {
        // V has no prerequisites
        if (Count.get(v) == 0)
            Q.enqueue(v);
    }

    // Process the vertices
    Queue<V> sortedVertices = new LinkedQueue<V>();
    while (!Q.isEmpty()) {
        V v = Q.dequeue();
        // PreVisit for vertex v
        sortedVertices.enqueue(v);
        for (Edge<V> edge : G.outgoingEdges(v)) {
            Count.put(edge.end, Count.get(edge.end) - 1);
            if (Count.get(edge.end) == 0)
                Q.enqueue(edge.end);
        }
    }
    return sortedVertices;
}

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