Question

Task 3: Grid Graphs and Mazes Part A - Generating Grid Graphs In the lecture we said that we can use Prim's algorithm to create mazes by starting from a regular "grid graph and then finding a spanning tree. Implement a function grid graph (m, n) that takes as input two positive integers, and returns the adjacency matrix of a m by n grid graphie, a graph with ses vertices that, when drawn on a regular m by n grid, has edges exactly between vertices that lie and neighbouring points on the grid. See Figure. 1 for an illustration For example: >>> grid graph(2, 3) [[0, 1, 0, 1, 0, 0]. (1, 0, 1, 0, 1, 0). [0, 1, 0, 0, 0, 1). [1, 0, , 0, 1, 0]. [0, 1, 0, 1, 0, 1). [0, 0, 1, 0, 1, 0]] Hints: Think about appropriate ways to decompose the problem. How can you translate between the id of a vertex i the adjacency matrix and the two grid coordinates? How can you decide whether vertices with indices i and are neighbours in the grid? The module template provides a function print as grid (graph, n) that you can use to visually. For example: >>> grid-graph(2, 3) >>> print_as_grid(grid_graph (2, 3). 3) >>> print_as_grid(grid_graph (3,2), 2) >>> print_as_grid(grid_graph (5, 10), 10) 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

Answer 1

program for degre of a graph

public class Degree
{
   static class Graph
   {
       int v, e;
       int[][] direction;
       Graph(int v, int e) {
           this.v = v;
           this.e = e;
           direction = new int[v][];
           for (int i = 0; i < v; i++)
               direction[i] = new int[v];
       }
   }
   static Graph create(int v, int e)
   {
       Graph G = new Graph(v, e);
       G.direction[0][1] = 1;
       G.direction[0][2] = 1;
       G.direction[0][3] = 1;
       G.direction[1][0] = 1;
       G.direction[1][3] = 1;
       G.direction[2][0] = 1;
       G.direction[2][3] = 1;
       G.direction[3][0] = 1;
       G.direction[3][1] = 1;
       G.direction[3][2] = 1;

       return G;
   }

   static int find(Graph G, int vertex)
   {
       int deg = 0;
       for (int i = 0; i < G.v; i++) {
           if (G.direction[vertex][i] == 1)
               deg++;
       }
       return deg;
   }
   public static void main(String[] args)
   {
       int vers = 4;
       int edgs = 5;
       Graph G = create(vers, edgs);
       int vert = 0;
       int deg = find(G, vert);
       System.out.println(deg);
   }
}

OUTPUT

3
Note : the index of vertex starts from zero

.

Program for existence of path in graph

import java.io.*;
import java.util.*;
import java.util.LinkedList;
public class Graph
{
   private int Vertex;  
   private LinkedList<Integer> adjacent[];
   Graph(int vertex)
   {
       Vertex = vertex;
       adjacent = new LinkedList[vertex];
       for (int i=0; i<vertex; ++i)
           adjacent[i] = new LinkedList();
   }
   void add(int vertex,int wt)
   {
   adjacent[vertex].add(wt);
  
   }
   Boolean isReach(int s, int d)
   {
       LinkedList<Integer>t;
       boolean visit[] = new boolean[Vertex];
       LinkedList<Integer> qu = new LinkedList<Integer>();
       visit[s]=true;
       qu.add(s);
       Iterator<Integer> it;
       while (qu.size()!=0)
       {
           s = qu.poll();

           int n;
           it = adjacent[s].listIterator();
           while (it.hasNext())
           {
               n = it.next();
               if (n==d)
                   return true;
               if (!visit[n])
               {
                   visit[n] = true;
                   qu.add(n);
               }
           }
       }
       return false;
   }
   public static void main(String args[])
   {
       Graph gh = new Graph(4);
       gh.add(0, 1);
       gh.add(0, 2);
       gh.add(1, 2);
       gh.add(2, 0);
       gh.add(2, 3);
       gh.add(3, 3);

       int start = 1;
       int end = 3;
       if (gh.isReach(start, end))
           System.out.println("true");
       else
           System.out.println("false");;
   }
}

OUTPUT

true

NOTE: Add edge as per your need .

Kindly Thumbs Up for the effort .

Thank you .