README.md

Topological Ordering

Finds a Topological Ordering of vertices in a Directed Acyclic Graph

Problem Statement

Topological Ordering: arranges nodes as v₁, v₂ ..., v_n so that for every edge (v_i, v_j), i < j

• Must be a Directed Graph
• Must NOT contain a cycle

Graph 1

Graph 1 Topological Ordering

Can can be more than 1 topologcal ordering. Algorithm finds 1 if 1 exists

Algorithm

A DAG must have a node with no incoming edges
Pick a node v with no incoming edges
Print v
Calculate a topological ordering on G – {v}
Ends when all nodes are included in the ordering

Keep track of 2 things:

• count[w] = number of incoming edges
• S = set of nodes with no incoming edges

Set-Up

• Scan through the graph to initialize count[] and S
• O(m + n)

Finding the Ordering

• Remove v from S
• Decrement count[w] for all edges from v to w
  add w to S if count[w] hits 0
• O(1) per edge

Runtime

O(m + n)

Usage

Detects cycles if graph is not a DAG and prints an error message
Node names are integers for simplicity of the code & start at 0

• Create a graph as an adjacency list
  ArrayList<ArrayList<Integer>> graph1 = new ArrayList<ArrayList<Integer>>();
  • Add rows for each vertex. graph1.get(u) is a list of nodes representing edges FROM u
  • Orphan nodes are allowed (i.e. a node with no incoming or outgoing edges). It doesn't have to be the last node in the graph, but is probably easier if it is. Just make sure no edges go to that node. (e.g. if node 5 is an orphan, then 5 must not appear in any of the rows of the adjacency list)
  • Nodes with no outgoing edges just need an empty ArrayList
graph1.add(new ArrayList<Integer>());
• Run TopologicalOrdering.findTopologicalOrdering(graph1);

Graph 2

Graph 2 Topological Ordering

Graph 3 (No topological ordering)

Contains a cycle, & no nodes with no incoming edges

Graph 4 (No topological ordering)

Contains a cycle

References

• Kevin Wayne Slides
• University of Washington slides
• Rajesh Rao Slides