One of the things I generally say about myself is that I love learning. I can spend hours upon hours reading papers and algorithms to better understand a topic. Some of these topics are stand alone segments that I can understand in one sitting. Sometimes, however, there is a need to read up on some preliminary work in order to fully understand a concept.

Lets say that I was interested in organizing this information into a new course. The order I present these topics is very important. Knowing which topics depend on one another allows me to use the topological sorting algorithm to determine an ordering for the topics that respects the preliminary work.

The input for the topoligical sorting algorithm is a Directed Acyclic Graph (DAG). This is a set of relationships between pairs of topics, where if topic 1 must be understood before topic 2, we would add the relationship (topic 1, topic 2) to the graph. DAGs can be visualized by a set of nodes (points) representing the topics. Relationships like the one above (topic 1, topic 2) can then represented by a directed arc originating at topic 1 and flowing in the direction of topic 2. We say that the graph is “Acyclic” because there cannot be a cycle in the topic preliminaries. This amounts to us saying that a topic cannot be a prerequisite for itself. An example of a DAG is shown in the image above.

With the topics represented as a DAG, the topologial ordering algorithm works by searching the set of nodes for the one with no arcs coming into it. This node (or these nodes is multiple are present) represents the topic that can be covered next without losing understanding of the material. Such a node is guaranteed to exist by the acyclic property of the DAG. Once the node is selected, we can remove this node as well as all arcs that originate at this node from the DAG. The algorithm then repeats the procedure of searching for a nod with no arcs coming into it. This process repeats until there are no remaining nodes from which to choose.

Now lets see how the topological sort algorithm works on the graph above. We will first need to count the in-degree (the number of arcs coming into) each node.

Node | Indegree

—————-

0 | 2

1 | 2

2 | 0

3 | 2

4 | 2

5 | 2

6 | 0

7 | 2

8 | 3

Node to be removed (i.e. node with the minimum indegree): Node 2.

Arcs connected to node 2: (2, 5), (2, 3)

Resulting Indegree Count:

Node | Indegree

—————-

0 | 2

1 | 2

3 | 1

4 | 2

5 | 1

6 | 0

7 | 2

8 | 3

Node to be removed: Node 6:

Arcs connected to node 6: (6, 1), (6, 3), (6, 4), (6, 5), (6, 7), (6, 8)

Resulting Indegree Count:

Node | Indegree

—————-

0 | 2

1 | 1

3 | 0

4 | 1

5 | 0

7 | 1

8 | 2

Node to be removed: Node 3

Arcs connected to node 3: (3, 0), (3, 8)

Resulting Indegree Count:

Node | Indegree

—————-

0 | 1

1 | 1

4 | 1

5 | 0

7 | 1

8 | 1

Node to be removed: Node 5

Arcs connected to node 5: (5, 0), (5, 8)

Resulting Indegree Count:

Node | Indegree

—————-

0 | 0

1 | 1

4 | 1

7 | 1

8 | 0

Node to be removed: Node 0:

Arcs connected to node 0: (0, 1), (0, 4)

Resulting Indegree Count:

Node | Indegree

—————-

1 | 0

4 | 0

7 | 1

8 | 0

Node to be removed: Node 1

Arcs connected to node 1: none

Resulting Indegree Count:

Node | Indegree

—————-

4 | 0

7 | 1

8 | 0

Node to be removed: Node 4

Arcs connected to node 4: none

Resulting Indegree Count:

Node | Indegree

—————-

7 | 1

8 | 0

Node to be removed: Node 8

Arcs connected to node 8: (8, 7)

Resulting Indegree Count:

Node | Indegree

—————-

7 | 0

Node to be removed: Node 7

Arcs connected to node 7: none

Resulting Indegree Count:

Node | Indegree

—————-

Since there are no nodes remaining, we have arrived at a topological ordering. Going through this iteration, we can see that we arrived at the ordering (2, 6, 3, 5, 0, 1, 4, 8, 7). There were several occasions where there were multiple nodes with indegree of 0 and we could have selected an alternative node. This would have given us a different topological ordering of the nodes, but it would still be valid.

There are more learning opportunities and an interactive demonstration of the algorithm at Topological Sort Examples at LEARNINGlover.

]Any advantage to this over a DFS where you mark “exit times” and sort in descending order of those?