(Optional) Introduction to Functional Graphs

Authors: Siyong Huang, Benjamin Qi, Andrew Wang

Directed graphs in which every vertex has exactly one outgoing edge.

Focus Problem – read through this problem before continuing!

It's easy to solve the above problem in O(N2)\mathcal{O}(N^2) time. We'll solve it in O(N)\mathcal{O}(N).


In a functional graph, each node has exactly one out-edge. This is also commonly referred to as a successor graph.


You can think of every connected component of a functional graph as a rooted tree plus an additional edge going out of the root.

Floyd's Algorithm

Floyd's Algorithm, also commonly referred to as the Tortoise and Hare Algorithm, is capable of detecting cycles in a functional graph in O(N)\mathcal{O}(N) time and O(1)\mathcal{O}(1) memory (not counting the graph itself).

Example - Cooperative Game

Focus Problem – read through this problem before continuing!

Hint 1

Hint 2



Solution - Badge

Solution 1


int N;
vi P,ans;
bool in_cycle;
void gen(int x) {
if (ans[x] != -2) {
if (ans[x] == -1) ans[x] = x, in_cycle = 1; // found a cycle!
ans[x] = -1; gen(P[x]);


import java.util.*;
public class badge
static InputReader in = new InputReader(;
static PrintWriter out = new PrintWriter(System.out);
public static final int MN = 1010;
public static int N;
public static int[] p = new int[MN], f = new int[MN];

This code generates the answer independently for each connected component. Note that it uses 0-indexing, not 1-indexing.

Try simulating the algorithm on the following directed graph in CSAcademy's Graph Editor.

0 1
1 2
2 3
3 4
4 2
5 6
6 1
  • On the first step, we make the following recursive calls: gen(0) -> gen(1) -> gen(2) -> gen(3) -> gen(4) -> gen(2), at which point we stop since ans[2] = -1. Since we have reached 2 for the second time, we know that 2 is part of a cycle and ans[2] = 2. Similarly, ans[3] = 3 and ans[4] = 4 since they are part of the cycle. On the other hand, ans[0] = ans[1] = 2 since neither of them are part of the cycle.

  • Later, we make the following recursive calls when we start at vertex 5: gen(5) -> gen(6) -> gen(1). We already know that ans[1] = 2, so ans[5] = ans[6] = 2 as well.

Solution 2

go(x) generates answers for all vertices in the connected component containing x. Requires reverse adjacency lists (radj).


int N;
vi P,ans;
V<vi> radj;
void fill_radj(int x) {
trav(t,radj[x]) if (ans[t] == -1) {
ans[t] = ans[x];

Count Cycles

The following sample code counts the number of cycles in such a graph. The "stack" contains nodes that can reach the current node. If the current node points to a node v on the stack (on_stack[v] is true), then we know that a cycle has been created. However, if the current node points to a node v that has been previously visited but is not on the stack, then we know that the current chain of nodes points into a cycle that has already been considered.


//Each node points to next_node[node]
bool visited[MAXN], on_stack[MAXN];
int number_of_cycles = 0, next_node[MAXN];
void dfs(int n)
visited[n] = on_stack[n] = true;
int u = next_node[n];


//source: Mayank Kumar
import java.util.*;
public class badge2{
static boolean[] visited=new boolean[MAXN],onStack=new boolean[MAXN];
static int numberOfCycles=0;
static int[] nextNode=new int[MAXN];
public static void main(String[] args) throws IOException{
//Take in input
for(int i=1;i!=N;++i)

KK-th Successor

As described briefly in CPH 16.3, can be found in O(logK)\mathcal{O}(\log K) time using binary jumping. See the Platinum module for details.


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External Sol

Additional problems involving functional graphs can be found in the Tree DP and Binary Jumping modules.

Module Progress:

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