What is Kruskals Minimum Spanning Tree Algorithm?
Given a connected and undirected graph, a spanning tree of that graph is a subgraph that is a tree and connects all the vertices together. A single graph can have many different spanning trees. A minimum spanning tree (MST) or minimum weight spanning tree for a weighted, connected and undirected graph is a spanning tree with weight less than or equal to the weight of every other spanning tree. The weight of a spanning tree is the sum of weights given to each edge of the spanning tree.
How many edges does a minimum spanning tree has?
A minimum spanning tree has (V – 1) edges where V is the number of vertices in the given graph.
What are the applications of Minimum Spanning Tree?
See this for applications of MST.
Below are the steps for finding MST using Kruskal’s algorithm
Sort all the edges in non-decreasing order of their weight.
Pick the smallest edge. Check if it forms a cycle with the spanning tree
formed so far. If cycle is not formed, include this edge. Else, discard it.
Repeat step#2 until there are (V-1) edges in the spanning tree.
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The step#2 uses Union-Find algorithm to detect cycle. So we recommend to read following post as a prerequisite.
Union-Find Algorithm | Set 1 (Detect Cycle in a Graph)
Union-Find Algorithm | Set 2 (Union By Rank and Path Compression)
The algorithm is a Greedy Algorithm. The Greedy Choice is to pick the smallest weight edge that does not cause a cycle in the MST constructed so far. Let us understand it with an example: Consider the below input graph.

The graph contains 9 vertices and 14 edges. So, the minimum spanning tree formed will be having (9 – 1) = 8 edges.
After sorting:
Weight Src Dest
1 7 6
2 8 2
2 6 5
4 0 1
4 2 5
6 8 6
7 2 3
7 7 8
8 0 7
8 1 2
9 3 4
10 5 4
11 1 7
14 3 5
Now pick all edges one by one from sorted list of edges
1. Pick edge 7-6: No cycle is formed, include it.

2. Pick edge 8-2: No cycle is formed, include it.

3. Pick edge 6-5: No cycle is formed, include it.

4. Pick edge 0-1: No cycle is formed, include it.

5. Pick edge 2-5: No cycle is formed, include it.

6. Pick edge 8-6: Since including this edge results in cycle, discard it.
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7. Pick edge 2-3: No cycle is formed, include it.

8. Pick edge 7-8: Since including this edge results in cycle, discard it.
9. Pick edge 0-7: No cycle is formed, include it.

10. Pick edge 1-2: Since including this edge results in cycle, discard it.
11. Pick edge 3-4: No cycle is formed, include it.

Since the number of edges included equals (V – 1), the algorithm stops here.
Recommended:
Please try your approach on {IDE} first, before moving on to the solution.
C++
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
struct Edge
{
int src, dest, weight;
};
struct Graph
{
int V, E;
struct Edge* edge;
};
struct Graph* createGraph(int V, int E)
{
struct Graph* graph = (struct Graph*) malloc( sizeof(struct Graph) );
graph->V = V;
graph->E = E;
graph->edge = (struct Edge*) malloc( graph->E * sizeof( struct Edge ) );
return graph;
}
struct subset
{
int parent;
int rank;
};
int find(struct subset subsets[], int i)
{
if (subsets[i].parent != i)
subsets[i].parent = find(subsets, subsets[i].parent);
return subsets[i].parent;
}
void Union(struct subset subsets[], int x, int y)
{
int xroot = find(subsets, x);
int yroot = find(subsets, y);
if (subsets[xroot].rank < subsets[yroot].rank)
subsets[xroot].parent = yroot;
else if (subsets[xroot].rank > subsets[yroot].rank)
subsets[yroot].parent = xroot;
else
{
subsets[yroot].parent = xroot;
subsets[xroot].rank++;
}
}
int myComp(const void* a, const void* b)
{
struct Edge* a1 = (struct Edge*)a;
struct Edge* b1 = (struct Edge*)b;
return a1->weight > b1->weight;
}
void KruskalMST(struct Graph* graph)
{
int V = graph->V;
struct Edge result[V];
int e = 0;
int i = 0;
qsort(graph->edge, graph->E, sizeof(graph->edge[0]), myComp);
struct subset *subsets =
(struct subset*) malloc( V * sizeof(struct subset) );
for (int v = 0; v < V; ++v)
{
subsets[v].parent = v;
subsets[v].rank = 0;
}
while (e < V - 1)
{
struct Edge next_edge = graph->edge[i++];
int x = find(subsets, next_edge.src);
int y = find(subsets, next_edge.dest);
if (x != y)
{
result[e++] = next_edge;
Union(subsets, x, y);
}
}
printf("Following are the edges in the constructed MST\n");
for (i = 0; i < e; ++i)
printf("%d -- %d == %d\n", result[i].src, result[i].dest,
result[i].weight);
return;
}
int main()
{
int V = 4;
int E = 5;
struct Graph* graph = createGraph(V, E);
graph->edge[0].src = 0;
graph->edge[0].dest = 1;
graph->edge[0].weight = 10;
graph->edge[1].src = 0;
graph->edge[1].dest = 2;
graph->edge[1].weight = 6;
graph->edge[2].src = 0;
graph->edge[2].dest = 3;
graph->edge[2].weight = 5;
graph->edge[3].src = 1;
graph->edge[3].dest = 3;
graph->edge[3].weight = 15;
graph->edge[4].src = 2;
graph->edge[4].dest = 3;
graph->edge[4].weight = 4;
KruskalMST(graph);
return 0;
}
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JAVA
import java.util.*;
import java.lang.*;
import java.io.*;
class Graph
{
class Edge implements Comparable<Edge>
{
int src, dest, weight;
public int compareTo(Edge compareEdge)
{
return this.weight-compareEdge.weight;
}
};
class subset
{
int parent, rank;
};
int V, E;
Edge edge[];
Graph(int v, int e)
{
V = v;
E = e;
edge = new Edge[E];
for (int i=0; i<e; ++i)
edge[i] = new Edge();
}
int find(subset subsets[], int i)
{
if (subsets[i].parent != i)
subsets[i].parent = find(subsets, subsets[i].parent);
return subsets[i].parent;
}
void Union(subset subsets[], int x, int y)
{
int xroot = find(subsets, x);
int yroot = find(subsets, y);
if (subsets[xroot].rank < subsets[yroot].rank)
subsets[xroot].parent = yroot;
else if (subsets[xroot].rank > subsets[yroot].rank)
subsets[yroot].parent = xroot;
else
{
subsets[yroot].parent = xroot;
subsets[xroot].rank++;
}
}
void KruskalMST()
{
Edge result[] = new Edge[V];
int e = 0;
int i = 0;
for (i=0; i<V; ++i)
result[i] = new Edge();
Arrays.sort(edge);
subset subsets[] = new subset[V];
for(i=0; i<V; ++i)
subsets[i]=new subset();
for (int v = 0; v < V; ++v)
{
subsets[v].parent = v;
subsets[v].rank = 0;
}
i = 0;
while (e < V - 1)
{
Edge next_edge = new Edge();
next_edge = edge[i++];
int x = find(subsets, next_edge.src);
int y = find(subsets, next_edge.dest);
if (x != y)
{
result[e++] = next_edge;
Union(subsets, x, y);
}
}
System.out.println("Following are the edges in the constructed MST");
for (i = 0; i < e; ++i)
System.out.println(result[i].src+" -- "+result[i].dest+" == "+
result[i].weight);
}
public static void main (String[] args)
{
int V = 4;
int E = 5;
Graph graph = new Graph(V, E);
graph.edge[0].src = 0;
graph.edge[0].dest = 1;
graph.edge[0].weight = 10;
graph.edge[1].src = 0;
graph.edge[1].dest = 2;
graph.edge[1].weight = 6;
graph.edge[2].src = 0;
graph.edge[2].dest = 3;
graph.edge[2].weight = 5;
graph.edge[3].src = 1;
graph.edge[3].dest = 3;
graph.edge[3].weight = 15;
graph.edge[4].src = 2;
graph.edge[4].dest = 3;
graph.edge[4].weight = 4;
graph.KruskalMST();
}
}
Python
Python
from collections import defaultdict
class Graph:
def __init__(self,vertices):
self.V= vertices
self.graph = []
def addEdge(self,u,v,w):
self.graph.append([u,v,w])
def find(self, parent, i):
if parent[i] == i:
return i
return self.find(parent, parent[i])
def union(self, parent, rank, x, y):
xroot = self.find(parent, x)
yroot = self.find(parent, y)
if rank[xroot] < rank[yroot]:
parent[xroot] = yroot
elif rank[xroot] > rank[yroot]:
parent[yroot] = xroot
else :
parent[yroot] = xroot
rank[xroot] += 1
def KruskalMST(self):
result =[]
i = 0
e = 0
self.graph = sorted(self.graph,key=lambda item: item[2])
parent = [] ; rank = []
for node in range(self.V):
parent.append(node)
rank.append(0)
while e < self.V -1 :
u,v,w = self.graph[i]
i = i + 1
x = self.find(parent, u)
y = self.find(parent ,v)
if x != y:
e = e + 1
result.append([u,v,w])
self.union(parent, rank, x, y)
print "Following are the edges in the constructed MST"
for u,v,weight in result:
print ("%d -- %d == %d" % (u,v,weight))
g = Graph(4)
g.addEdge(0, 1, 10)
g.addEdge(0, 2, 6)
g.addEdge(0, 3, 5)
g.addEdge(1, 3, 15)
g.addEdge(2, 3, 4)
g.KruskalMST()
Following are the edges in the constructed MST
2 -- 3 == 4
0 -- 3 == 5
0 -- 1 == 10
Time Complexity: O(ElogE) or O(ElogV). Sorting of edges takes O(ELogE) time. After sorting, we iterate through all edges and apply find-union algorithm. The find and union operations can take atmost O(LogV) time. So overall complexity is O(ELogE + ELogV) time. The value of E can be atmost O(V2), so O(LogV) are O(LogE) same. Therefore, overall time complexity is O(ElogE) or O(ElogV)