Suppose you have an undirected graph G=(V,E) and you want to determine if you ca
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Suppose you have an undirected graph G=(V,E) and you want to determine if you can assign two colors (blue and red) to the vertices such that adjacent vertices are different colors. This is the graph Two-Color problem. If the assignment of two colors is possible, then a 2-coloring is a function C: V -> {blue, red} such that C(u) C(v) for every edge (u,v) E. Note: a graph can have more than one 2- coloring. Give an O(V + E) algorithm to determine the 2-coloring of a graph if one exists or terminate with the message that the graph is not Two-Colorable. Assume that the input graph G=(V,E) is represented using adjacency lists. (a) Give a verbal description of the algorithm and provide detailed pseudocode. (b) Analyze the running time.
4. Suppose you have an undirected graph G (V,E) and you want to determine if you can assign two colors (blue and red) to the vertices such that adjacent vertices are different colors. This is the graph Two-Color problem. If the assignment of two colors is possible, then a 2-coloring is a function C: V {blue, red) such that C(u) C(v) for every edge (u,v) e E. Note: a graph can have more than one 2- coloring. Give an o (V E) algorithm to determine the 2-coloring of a graph if one exists or terminate with the message that the graph is not Two-Colorable. Assume that the input graph G (V,E) is represented using adjacency lists (a) Give a verbal description of the algorithm and provide detailed pseudocode. (b) Analyze the running timeExplanation / Answer
Wheather a graph is 2 colorable is not is equal to checking of wheather a graph is biparttite or not.
A Bipartite Graph is a graph whose vertices can be divided into two independent sets, U and V such that every edge (u, v) either connects a vertex from U to V or a vertex from V to U. In other words, for every edge (u, v), either u belongs to U and v to V, or u belongs to V and v to U. We can also say that there is no edge that connects vertices of same set.
A bipartite graph is possible if the graph coloring is possible using two colors such that vertices in a set are colored with the same color. Note that it is possible to color a cycle graph with even cycle using two colors.
Algorithm to check if a graph is Bipartite:
One approach is to check whether the graph is 2-colorable or not using backtracking algorithm m coloring problem.
Following is a simple algorithm to find out whether a given graph is Birpartite or not using Breadth First Search (BFS).
1. Assign RED color to the source vertex (putting into set U).
2. Color all the neighbors with BLUE color (putting into set V).
3. Color all neighbor’s neighbor with RED color (putting into set U).
4. This way, assign color to all vertices such that it satisfies all the constraints of m way coloring problem where m = 2.
5. While assigning colors, if we find a neighbor which is colored with same color as current vertex, then the graph cannot be colored with 2 vertices (or graph is not Bipartite)
Here is the Java Code for this algorithm
// Java program to find out whether a given graph is Bipartite or not
import java.util.*;
import java.lang.*;
import java.io.*;
class Bipartite
{
final static int V = 4; // No. of Vertices
// This function returns true if graph G[V][V] is Bipartite, else false
boolean isBipartite(int G[][],int src)
{
// Create a color array to store colors assigned to all veritces.
// Vertex number is used as index in this array. The value '-1'
// of colorArr[i] is used to indicate that no color is assigned
// to vertex 'i'. The value 1 is used to indicate first color
// is assigned and value 0 indicates second color is assigned.
int colorArr[] = new int[V];
for (int i=0; i<V; ++i)
colorArr[i] = -1;
// Assign first color to source
colorArr[src] = 1;
// Create a queue (FIFO) of vertex numbers and enqueue
// source vertex for BFS traversal
LinkedList<Integer>q = new LinkedList<Integer>();
q.add(src);
// Run while there are vertices in queue (Similar to BFS)
while (q.size() != 0)
{
// Dequeue a vertex from queue
int u = q.poll();
// Find all non-colored adjacent vertices
for (int v=0; v<V; ++v)
{
// An edge from u to v exists and destination v is
// not colored
if (G[u][v]==1 && colorArr[v]==-1)
{
// Assign alternate color to this adjacent v of u
colorArr[v] = 1-colorArr[u];
q.add(v);
}
// An edge from u to v exists and destination v is
// colored with same color as u
else if (G[u][v]==1 && colorArr[v]==colorArr[u])
return false;
}
}
// If we reach here, then all adjacent vertices can
// be colored with alternate color
return true;
}
// Driver program to test above function
public static void main (String[] args)
{
int G[][] = {{0, 1, 0, 1},
{1, 0, 1, 0},
{0, 1, 0, 1},
{1, 0, 1, 0}
};
Bipartite b = new Bipartite();
if (b.isBipartite(G, 0))
System.out.println("Yes");
else
System.out.println("No");
}
}
Output:
The above algorithm works only if the graph is strongly connected. In above code, we always start with source 0 and assume that vertices are visited from it. One important observation is a graph with no edges is also Bipiartite. Note that the Bipartite condition says all edges should be from one set to another.
We can extend the above code to handle cases when a graph is not connected. The idea is repeatedly call above method for all not yet visited vertices.
Running Time:
Time Complexity of the above approach is same as that Breadth First Search. In above implementation is O(V^2) where V is number of vertices. If graph is represented using adjacency list, then the complexity becomes O(V+E).
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