Union-Find Algorithm | (Union By Rank and Find by Optimized Path Compression)

Check whether a given graph contains a cycle or not.

Example:

Input: 

Output: Graph contains Cycle.

Input: 

Output: Graph does not contain Cycle.

Prerequisites: Disjoint Set (Or Union-Find), Union By Rank and Path Compression

We have already discussed union-find to detect cycle. Here we discuss find by path compression, where it is slightly modified to work faster than the original method as we are skipping one level each time we are going up the graph. Implementation of find function is iterative, so there is no overhead involved.Time complexity of optimized find function is O(log*(n)), i.e iterated logarithm, which converges to O(1) for repeated calls.

Refer this link for
Proof of log*(n) complexity of Union-Find
Explanation of find function:
Take Example 1 to understand find function:



(1)call find(8) for first time and mappings will be done like this:

It took 3 mappings for find function to get the root of node 8. Mappings are illustrated below:
From node 8, skipped node 7, Reached node 6.
From node 6, skipped node 5, Reached node 4.
From node 4, skipped node 2, Reached node 0.

(2)call find(8) for second time and mappings will be done like this:

It took 2 mappings for find function to get the root of node 8. Mappings are illustrated below:
From node 8, skipped node 5, node 6 and node 7, Reached node 4.
From node 4, skipped node 2, Reached node 0.

(3)call find(8) for third time and mappings will be done like this:

Finally, we see it took only 1 mapping for find function to get the root of node 8. Mappings are illustrated below:
From node 8, skipped node 5, node 6, node 7, node 4, and node 2, Reached node 0.
That is how it converges path from certain mappings to single mapping.

Explanation of example 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.

C++

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// CPP progxrm 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;
}

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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 [tabbyending] Output:

Graph contains Cycle.

Time Complexity(Find) : O(log*(n))
Time Complexity(union) : O(1)



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