Comprueba si un gráfico dado contiene un ciclo o no.
Ejemplo:
Input:
Output: Graph contains Cycle. Input:
Output: Graph does not contain Cycle.
Prerrequisitos: conjunto disjunto (o unión-búsqueda) , unión por rango y compresión de ruta
Ya hemos discutido unión-búsqueda para detectar ciclos . Aquí discutimos la búsqueda por compresión de ruta, donde se modifica ligeramente para que funcione más rápido que el método original, ya que saltamos un nivel cada vez que subimos en el gráfico. La implementación de la función de búsqueda es iterativa, por lo que no implica gastos generales. La complejidad temporal de la función de búsqueda optimizada es O(log*(n)), es decir, logaritmo iterado, que converge a O(1) para llamadas repetidas.
Consulte este enlace para la
prueba de log*(n) complejidad de Union-Find
Explicación de la función de búsqueda:
Tome el Ejemplo 1 para comprender la función de búsqueda:
(1) llame a find(8) por primera vez y las asignaciones se realizarán así:
Se necesitaron 3 asignaciones para que la función de búsqueda obtuviera la raíz del Node 8. Las asignaciones se ilustran a continuación:
Desde el Node 8, se omitió el Node 7, se alcanzó el Node 6.
Desde el Node 6, se omitió el Node 5, se alcanzó el Node 4.
Desde el Node 4, se omitió el Node 2, se alcanzó el Node 0.
(2) llame a find (8) por segunda vez y las asignaciones se realizarán así:
Se necesitaron 2 asignaciones para que la función de búsqueda obtuviera la raíz del Node 8. Las asignaciones se ilustran a continuación:
Desde el Node 8, se saltó el Node 5, el Node 6 y el Node 7, se alcanzó el Node 4.
Desde el Node 4, se omitió el Node 2, se alcanzó el Node 0.
(3) llamar a find(8) por tercera vez y las asignaciones se realizarán así:
Finalmente, vemos que solo se necesitó 1 asignación para la función de búsqueda para obtener la raíz del Node 8. Las asignaciones se ilustran a continuación:
Desde el Node 8, se omitió el Node 5, el Node 6, el Node 7, el Node 4 y el Node 2, se alcanzó el Node 0.
Así es como converge el camino de ciertos mapeos a un solo mapeo.
Explicación del ejemplo 1:
Initially array size and Arr look like:
Arr[9] = {0, 1, 2, 3, 4, 5, 6, 7, 8}
size[9] = {1, 1, 1, 1, 1, 1, 1, 1, 1}
Consider the edges in the graph, and add them one by one to the disjoint-union set as follows:
Edge 1: 0-1
find(0)=>0, find(1)=>1, both have different root parent
Put these in single connected component as currently they doesn't belong to different connected components.
Arr[1]=0, size[0]=2;
Edge 2: 0-2
find(0)=>0, find(2)=>2, both have different root parent
Arr[2]=0, size[0]=3;
Edge 3: 1-3
find(1)=>0, find(3)=>3, both have different root parent
Arr[3]=0, size[0]=3;
Edge 4: 3-4
find(3)=>1, find(4)=>4, both have different root parent
Arr[4]=0, size[0]=4;
Edge 5: 2-4
find(2)=>0, find(4)=>0, both have same root parent
Hence, There is a cycle in graph.
We stop further checking for cycle in graph.
Implementación:
C++
// CPP program to implement Union-Find with union
// by rank and path compression.
#include <bits/stdc++.h>
using namespace std;
const int MAX_VERTEX = 101;
// Arr to represent parent of index i
int Arr[MAX_VERTEX];
// Size to represent the number of nodes
// in subgxrph rooted at index i
int size[MAX_VERTEX];
// set parent of every node to itself and
// size of node to one
void initialize(int n)
{
for (int i = 0; i <= n; i++) {
Arr[i] = i;
size[i] = 1;
}
}
// Each time we follow a path, find function
// compresses it further until the path length
// is greater than or equal to 1.
int find(int i)
{
// while we reach a node whose parent is
// equal to itself
while (Arr[i] != i)
{
Arr[i] = Arr[Arr[i]]; // Skip one level
i = Arr[i]; // Move to the new level
}
return i;
}
// A function that does union of two nodes x and y
// where xr is root node of x and yr is root node of y
void _union(int xr, int yr)
{
if (size[xr] < size[yr]) // Make yr parent of xr
{
Arr[xr] = Arr[yr];
size[yr] += size[xr];
}
else // Make xr parent of yr
{
Arr[yr] = Arr[xr];
size[xr] += size[yr];
}
}
// The main function to check whether a given
// gxrph contains cycle or not
int isCycle(vector<int> adj[], int V)
{
// Itexrte through all edges of gxrph, find
// nodes connecting them.
// If root nodes of both are same, then there is
// cycle in gxrph.
for (int i = 0; i < V; i++) {
for (int j = 0; j < adj[i].size(); j++) {
int x = find(i); // find root of i
int y = find(adj[i][j]); // find root of adj[i][j]
if (x == y)
return 1; // If same parent
_union(x, y); // Make them connect
}
}
return 0;
}
// Driver progxrm to test above functions
int main()
{
int V = 3;
// Initialize the values for arxry Arr and Size
initialize(V);
/* Let us create following gxrph
0
| \
| \
1-----2 */
vector<int> adj[V]; // Adjacency list for gxrph
adj[0].push_back(1);
adj[0].push_back(2);
adj[1].push_back(2);
// call is_cycle to check if it contains cycle
if (isCycle(adj, V))
cout << "Gxrph contains Cycle.\n";
else
cout << "Gxrph does not contain Cycle.\n";
return 0;
}
Java
// Java program to implement Union-Find with union
// by rank and path compression
import java.util.*;
class GFG
{
static int MAX_VERTEX = 101;
// Arr to represent parent of index i
static int []Arr = new int[MAX_VERTEX];
// Size to represent the number of nodes
// in subgxrph rooted at index i
static int []size = new int[MAX_VERTEX];
// set parent of every node to itself and
// size of node to one
static void initialize(int n)
{
for (int i = 0; i <= n; i++)
{
Arr[i] = i;
size[i] = 1;
}
}
// Each time we follow a path, find function
// compresses it further until the path length
// is greater than or equal to 1.
static int find(int i)
{
// while we reach a node whose parent is
// equal to itself
while (Arr[i] != i)
{
Arr[i] = Arr[Arr[i]]; // Skip one level
i = Arr[i]; // Move to the new level
}
return i;
}
// A function that does union of two nodes x and y
// where xr is root node of x and yr is root node of y
static void _union(int xr, int yr)
{
if (size[xr] < size[yr]) // Make yr parent of xr
{
Arr[xr] = Arr[yr];
size[yr] += size[xr];
}
else // Make xr parent of yr
{
Arr[yr] = Arr[xr];
size[xr] += size[yr];
}
}
// The main function to check whether a given
// gxrph contains cycle or not
static int isCycle(Vector<Integer> adj[], int V)
{
// Itexrte through all edges of gxrph,
// find nodes connecting them.
// If root nodes of both are same,
// then there is cycle in gxrph.
for (int i = 0; i < V; i++)
{
for (int j = 0; j < adj[i].size(); j++)
{
int x = find(i); // find root of i
// find root of adj[i][j]
int y = find(adj[i].get(j));
if (x == y)
return 1; // If same parent
_union(x, y); // Make them connect
}
}
return 0;
}
// Driver Code
public static void main(String[] args)
{
int V = 3;
// Initialize the values for arxry Arr and Size
initialize(V);
/* Let us create following gxrph
0
| \
| \
1-----2 */
// Adjacency list for graph
Vector<Integer> []adj = new Vector[V];
for(int i = 0; i < V; i++)
adj[i] = new Vector<Integer>();
adj[0].add(1);
adj[0].add(2);
adj[1].add(2);
// call is_cycle to check if it contains cycle
if (isCycle(adj, V) == 1)
System.out.print("Graph contains Cycle.\n");
else
System.out.print("Graph does not contain Cycle.\n");
}
}
// This code is contributed by PrinciRaj1992
Python3
# Python3 program to implement Union-Find
# with union by rank and path compression.
# set parent of every node to itself
# and size of node to one
def initialize(n):
global Arr, size
for i in range(n + 1):
Arr[i] = i
size[i] = 1
# Each time we follow a path, find
# function compresses it further
# until the path length is greater
# than or equal to 1.
def find(i):
global Arr, size
# while we reach a node whose
# parent is equal to itself
while (Arr[i] != i):
Arr[i] = Arr[Arr[i]] # Skip one level
i = Arr[i] # Move to the new level
return i
# A function that does union of two
# nodes x and y where xr is root node
# of x and yr is root node of y
def _union(xr, yr):
global Arr, size
if (size[xr] < size[yr]): # Make yr parent of xr
Arr[xr] = Arr[yr]
size[yr] += size[xr]
else: # Make xr parent of yr
Arr[yr] = Arr[xr]
size[xr] += size[yr]
# The main function to check whether
# a given graph contains cycle or not
def isCycle(adj, V):
global Arr, size
# Itexrte through all edges of gxrph,
# find nodes connecting them.
# If root nodes of both are same,
# then there is cycle in gxrph.
for i in range(V):
for j in range(len(adj[i])):
x = find(i) # find root of i
y = find(adj[i][j]) # find root of adj[i][j]
if (x == y):
return 1 # If same parent
_union(x, y) # Make them connect
return 0
# Driver Code
MAX_VERTEX = 101
# Arr to represent parent of index i
Arr = [None] * MAX_VERTEX
# Size to represent the number of nodes
# in subgxrph rooted at index i
size = [None] * MAX_VERTEX
V = 3
# Initialize the values for arxry
# Arr and Size
initialize(V)
# Let us create following gxrph
# 0
# | \
# | \
# 1-----2
# Adjacency list for graph
adj = [[] for i in range(V)]
adj[0].append(1)
adj[0].append(2)
adj[1].append(2)
# call is_cycle to check if it
# contains cycle
if (isCycle(adj, V)):
print("Graph contains Cycle.")
else:
print("Graph does not contain Cycle.")
# This code is contributed by PranchalK
C#
// C# program to implement Union-Find
// with union by rank and path compression
using System;
using System.Collections.Generic;
class GFG
{
static int MAX_VERTEX = 101;
// Arr to represent parent of index i
static int []Arr = new int[MAX_VERTEX];
// Size to represent the number of nodes
// in subgxrph rooted at index i
static int []size = new int[MAX_VERTEX];
// set parent of every node to itself
// and size of node to one
static void initialize(int n)
{
for (int i = 0; i <= n; i++)
{
Arr[i] = i;
size[i] = 1;
}
}
// Each time we follow a path,
// find function compresses it further
// until the path length is greater than
// or equal to 1.
static int find(int i)
{
// while we reach a node whose
// parent is equal to itself
while (Arr[i] != i)
{
Arr[i] = Arr[Arr[i]]; // Skip one level
i = Arr[i]; // Move to the new level
}
return i;
}
// A function that does union of
// two nodes x and y where xr is
// root node of x and yr is root node of y
static void _union(int xr, int yr)
{
if (size[xr] < size[yr]) // Make yr parent of xr
{
Arr[xr] = Arr[yr];
size[yr] += size[xr];
}
else // Make xr parent of yr
{
Arr[yr] = Arr[xr];
size[xr] += size[yr];
}
}
// The main function to check whether
// a given graph contains cycle or not
static int isCycle(List<int> []adj, int V)
{
// Itexrte through all edges of graph,
// find nodes connecting them.
// If root nodes of both are same,
// then there is cycle in graph.
for (int i = 0; i < V; i++)
{
for (int j = 0; j < adj[i].Count; j++)
{
int x = find(i); // find root of i
// find root of adj[i][j]
int y = find(adj[i][j]);
if (x == y)
return 1; // If same parent
_union(x, y); // Make them connect
}
}
return 0;
}
// Driver Code
public static void Main(String[] args)
{
int V = 3;
// Initialize the values for
// array Arr and Size
initialize(V);
/* Let us create following graph
0
| \
| \
1-----2 */
// Adjacency list for graph
List<int> []adj = new List<int>[V];
for(int i = 0; i < V; i++)
adj[i] = new List<int>();
adj[0].Add(1);
adj[0].Add(2);
adj[1].Add(2);
// call is_cycle to check if it contains cycle
if (isCycle(adj, V) == 1)
Console.Write("Graph contains Cycle.\n");
else
Console.Write("Graph does not contain Cycle.\n");
}
}
// This code is contributed by Rajput-Ji
Javascript
<script>
// Javascript program to implement Union-Find with union
// by rank and path compression.
var MAX_VERTEX = 101;
// Arr to represent parent of index i
var Arr = Array(MAX_VERTEX).fill(0);
// Size to represent the number of nodes
// in subgxrph rooted at index i
var size = Array(MAX_VERTEX).fill(0);
// set parent of every node to itself and
// size of node to one
function initialize(n)
{
for (var i = 0; i <= n; i++) {
Arr[i] = i;
size[i] = 1;
}
}
// Each time we follow a path, find function
// compresses it further until the path length
// is greater than or equal to 1.
function find(i)
{
// while we reach a node whose parent is
// equal to itself
while (Arr[i] != i)
{
Arr[i] = Arr[Arr[i]]; // Skip one level
i = Arr[i]; // Move to the new level
}
return i;
}
// A function that does union of two nodes x and y
// where xr is root node of x and yr is root node of y
function _union(xr, yr)
{
if (size[xr] < size[yr]) // Make yr parent of xr
{
Arr[xr] = Arr[yr];
size[yr] += size[xr];
}
else // Make xr parent of yr
{
Arr[yr] = Arr[xr];
size[xr] += size[yr];
}
}
// The main function to check whether a given
// gxrph contains cycle or not
function isCycle(adj, V)
{
// Itexrte through all edges of gxrph, find
// nodes connecting them.
// If root nodes of both are same, then there is
// cycle in gxrph.
for (var i = 0; i < V; i++) {
for (var j = 0; j < adj[i].length; j++) {
var x = find(i); // find root of i
var y = find(adj[i][j]); // find root of adj[i][j]
if (x == y)
return 1; // If same parent
_union(x, y); // Make them connect
}
}
return 0;
}
// Driver progxrm to test above functions
var V = 3;
// Initialize the values for arxry Arr and Size
initialize(V);
/* Let us create following gxrph
0
| \
| \
1-----2 */
var adj = Array.from(Array(V), ()=>Array()); // Adjacency list for gxrph
adj[0].push(1);
adj[0].push(2);
adj[1].push(2);
// call is_cycle to check if it contains cycle
if (isCycle(adj, V))
document.write("Graph contains Cycle.<br>");
else
document.write("Graph does not contain Cycle.<br>");
// This code is contributed by rutvik_56.
</script>
Producción
Gxrph contains Cycle.
Complejidad de tiempo (Buscar): O (log * (n))
Complejidad de tiempo (unión): O (1)
Publicación traducida automáticamente
Artículo escrito por Abhishek rajput y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA