# Topological Sort

Authors: Benjamin Qi, Nathan Chen

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

Ordering the vertices of a directed acyclic graph such that each vertex is visited before its children.

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 – try your best to solve 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 | ||||
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CSA | interactive, both versions |

## DFS

Resources | ||||
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CPH | example walkthrough | |||

CP2 | code | |||

cp-algo | code |

C++

#include <bits/stdc++.h>using namespace std;#define pb push_backint 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.io.*;import java.util.*;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 =

Python

import sysMAX_N = 10**5sys.setrecursionlimit(MAX_N)def dfs(node):for i in graph[node]:if not visited[i]:visited[i] = True

### Finding a Cycle

Focus Problem – try your best to solve 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 <algorithm>#include <iostream>#include <vector>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 =

Python

import sysMAX_N = 10**5sys.setrecursionlimit(MAX_N)def dfs(node):visited[node], on_stack[node] = True, Truefor u in adj[node]:if on_stack[u]:

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

Java

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

Python

from collections import deque# The code is in a function so it can run faster.def main():N, M = map(int, input().split())adj = [[] for _ in range(N)]for _ in range(M):a, b = map(int, input().split())a -= 1b -= 1

### 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 | ||||
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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 – try your best to solve this problem before continuing!

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

Solution

## Problems

Status | Source | Problem Name | Difficulty | Tags | |
---|---|---|---|---|---|

CSES | Easy | ## Show TagsTopoSort | |||

Kattis | Easy | ## Show TagsTopoSort | |||

Gold | Easy | ## Show TagsTopoSort | |||

CF | Easy | ## Show TagsTopoSort | |||

CF | Easy | ## Show TagsTopoSort | |||

Gold | Normal | ## Show TagsBinary Search, TopoSort | |||

CSES | Hard | ## Show TagsTopoSort |

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