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Topological Sort

Authors: Benjamin Qi, Nathan Chen

Contributors: Michael Cao, Andi Qu, Andrew Wang, Qi Wang, Maggie Liu

An ordering of vertices in a directed acyclic graph that ensures that a node is visited before every node it has a directed edge to.

Prerequisites

To review, a directed graph consists of edges that can only be traversed in one direction. Additionally, an acyclic graph defines a graph which does not contain cycles, meaning you are unable to traverse across one or more edges and return to the node you started on. Putting these definitions together, a directed acyclic graph, sometimes abbreviated as DAG, is a graph which has edges which can only be traversed in one direction and does not contain cycles.

Topological Sort

Focus Problem – read through this problem before continuing!

A topological sort of a directed acyclic graph is a linear ordering of its vertices such that for every directed edge $u\to v$ from vertex $u$ to vertex $v$, $u$ comes before $v$ in the ordering.

There are two common ways to topologically sort, one involving DFS and the other involving BFS.

Resources
CSA

interactive, both versions

DFS

Resources
CPH

example walkthrough

CP2

code

cp-algo

code

C++

#include <bits/stdc++.h>using namespace std;
#define pb push_back
int N; // Number of nodesvector<int> graph[100000], top_sort; // Assume that this graph is a DAGbool visited[100000];
void dfs(int node) {

Java

import java.util.*;import java.io.*;
public class CourseSchedule {	static List<Integer>[] graph;	static List<Integer> topo = new ArrayList<Integer>();	static int N;	static boolean[] visited;	public static void main(String[] args) throws Exception {		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

Finding a Cycle

Focus Problem – read through this problem before continuing!

We can modify the algorithm above to return a directed cycle in the case where a topological sort does not exist. To find the cycle, we add each node we visit onto the stack until we detect a node already on the stack.

For example, suppose that our stack currently consists of $s_1,s_2,\ldots,s_i$ and we then visit $u=s_j$ for some $j\le i$. Then $s_j\to s_{j+1}\to \cdots\to s_i\to s_j$ is a cycle. We can reconstruct the cycle without explicitly storing the stack by marking $u$ as not part of the stack and recursively backtracking until $u$ is reached again.

C++

#include <vector>#include <iostream>#include <algorithm>using namespace std;
bool visited[(int)1e5 + 5], on_stack[(int)1e5 + 5];vector<int> adj[(int)1e5 + 5];vector<int> cycle;int N, M;bool dfs(int n) {

Java

import java.io.*;import java.util.*;
public class cycle {	static List<Integer>[] adj;	static boolean[] visited, onStack;	static List<Integer> cycle;	static int N, M;	public static void main(String[] args) throws IOException {		BufferedReader sc = new BufferedReader(new InputStreamReader(System.in));

Warning!

This code assumes that there are no self-loops.

BFS

The BFS version is known as Kahn's Algorithm.

C++

Course Schedule Solution

#include <iostream>#include <vector>#include <queue>using namespace std;
const int MAX_N = 100000;
int main(){	int n, m;

Java

public class TopoSort {	static int[] inDegree;	static List<Integer>[] edge; // adjacency list
static int N; // number of nodes
static void topological_sort() {		Queue<Integer> q = new LinkedList<>();		for (int i = 0; i < N; i++) {			if (inDegree[i] == 0) {

Pro Tip

If the length of the array containing the end order does not equal the number of elements that need to be sorted, then there is a cycle in the graph.

Optional

We can also use Kahn's algorithm to extract the lexicographically minimum topological sort by breaking ties lexographically.

Although the above code does not do this, one can simply replace the queue with a priority_queue to implement this extension.

Dynamic Programming

Resources
CPH

One useful property of directed acyclic graphs is, as the name suggests, that no cycles exist. If we consider each node in the graph as a state, we can perform dynamic programming on the graph if we process the states in an order that guarantees for every edge $u\to v$ that $u$ is processed before $v$. Fortunately, this is the exact definition of a topological sort!

Focus Problem – read through this problem before continuing!

In this task, we must find the longest path in a DAG.

Solution

Problems

StatusSourceProblem NameDifficultyTags
CSESEasy
Show TagsTopoSort
KattisEasy
Show TagsTopoSort
GoldEasy
Show TagsTopoSort
CFEasy
Show TagsTopoSort
CFEasy
Show TagsTopological Sort
GoldNormal
Show TagsBinary Search, TopoSort
CSESHard
Show TagsTopoSort

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