# 14.4. Topological Sort¶

## 14.4.1. 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 14.4.1 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.

### 14.4.1.1. 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`

prints the vertex.
This 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 void topsortDFS(Graph G) {
int v;
for (v=0; v<G.nodeCount(); v++) {
G.setValue(v, null); // Initialize
}
for (v=0; v<G.nodeCount(); v++) {
if (G.getValue(v) != VISITED) {
tophelp(G, v);
}
}
}
static void tophelp(Graph G, int v) {
G.setValue(v, VISITED);
int[] nList = G.neighbors(v);
for (int i=0; i< nList.length; i++) {
if (G.getValue(nList[i]) != VISITED) {
tophelp(G, nList[i]);
}
}
printout(v);
}
```

Using this algorithm starting at J1 and visiting adjacent neighbors in alphabetic order, vertices of the graph in Figure 14.4.1 are printed out in the order J7, J5, J4, J6, J2, J3, J1. Reversing this yields the topological sort J1, J3, J2, J6, J4, J5, J7.

Here is another example.

### 14.4.1.2. Queue-based Solution¶

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 printed, 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 printing all of the vertices, then the graph contains a cycle (i.e., there is no possible ordering for the tasks that does not violate some prerequisite). The printed order for the vertices of the graph in Applying the queue version of topological sort to the graph of Figure 14.4.1 produces J1, J2, J3, J6, J4, J5, J7. Here is an implementation for the algorithm.

Here is the code to implement the queue-based topological sort:

```
static void topsortBFS(Graph G) { // Topological sort: Queue
Queue Q = new LQueue(G.nodeCount());
int[] Count = new int[G.nodeCount()];
int[] nList;
int v;
for (v=0; v<G.nodeCount(); v++) { Count[v] = 0; } // Initialize
for (v=0; v<G.nodeCount(); v++) { // Process every edge
nList = G.neighbors(v);
for (int i=0; i< nList.length; i++) {
Count[nList[i]]++; // Add to v's prereq count
}
}
for (v=0; v<G.nodeCount(); v++) { // Initialize Queue
if (Count[v] == 0) { // V has no prerequisites
Q.enqueue(v);
}
}
while (Q.length() > 0) { // Process the vertices
v = (Integer)Q.dequeue();
printout(v); // PreVisit for Vertex V
nList = G.neighbors(v);
for (int i=0; i< nList.length; i++) {
Count[nList[i]]--; // One less prerequisite
if (Count[nList[i]] == 0) { // This vertex is now free
Q.enqueue(nList[i]);
}
}
}
}
```

The inverse problem of determining whether a proposed node ordering is a valid topological sort of the graph can be solved with an algorithm nearly identical to the queue-based topological sort algorithm. First process the graph to generate the count array with the incoming degree of each node. Assuming that the proposed ordering has a length of \(n\), move through the nodes of the proposed ordering in order from the beginning. For each node \(v\), check that it’s count is zero. Then decrement the count by one for each neighbor reachable by \(v\). If all nodes have a count of zero when they are visited in this order, then this is a valid topological sort.