DAA/ADA lab Programs

List of Java programs click the required to get it

1.Merge Sort
2.Quick Sort
3.Greedy method
4.Dijkstra's Algorithm
5.Prim's Algorithm
6.Kruskal's Algorithm
7.Dynamic Programming method
8.Floyd's Algorithm
9.Travelling Sales Person problem 

10.Subset

11.Hamiltonian Cycle

Merge Sort

Sort a given set of n integer elements using Merge Sort method and compute its time complexity. Run the program for varied values of n > 5000, and record the time taken to sort. Plot a graph of the time taken versus n on graph sheet. The elements can be read from a file or can be generated using the random number generator. Demonstrate using Java how the divideand - conquer method works along with its time complexity analysis: worst case, average case and best case.

import java.util.Random;
import java.util.Random;
import java.util.Scanner;
public class mergesort 
{
static int max=10000;
void merge( int[] array,int low, int mid,int high)
{
int i=low;
int j=mid+1;
int k=low;
int[]resarray;
resarray=new int[max];
while(i<=mid&&j<=high)
{
if(array[i]<array[j])
{
resarray[k]=array[i];
i++;
k++;
}
else
{
resarray[k]=array[j];
j++;
k++;
}
}
while(i<=mid)
resarray[k++]=array[i++];
while(j<=high)
resarray[k++]=array[j++];
for(int m=low;m<=high;m++)
array[m]=resarray[m];
}
void sort( int[] array,int low,int high)
{
if(low<high)
{
int mid=(low+high)/2;
sort(array,low,mid);
sort(array,mid+1,high);
merge(array,low,mid,high);
}
}
public static void main(String[] args)
{
int[] array;
int i;
System.out.println("Enter the array size");
Scanner sc =new Scanner(System.in);
int n=sc.nextInt();
array= new int[max];
Random generator=new Random();
for( i=0;i<n;i++)
array[i]=generator.nextInt(20);
System.out.println("Array before sorting");
for( i=0;i<n;i++)
System.out.println(array[i]+" ");
long startTime=System.nanoTime();
mergesort m=new mergesort();
m.sort(array,0,n-1);
long stopTime=System.nanoTime();
long elapseTime=(stopTime-startTime);
System.out.println("Time taken to sort array is:"+elapseTime+"nano seconds");
System.out.println("Sorted array is");
for(i=0;i<n;i++)
System.out.println(array[i]);
}
}

Enter the array size
10
Array before sorting
13
9
13
16
13
3
0
6
4
5
Time taken to sort array is:171277nano seconds
Sorted array is
0
3
4
5
6
9
13
13
13
16

Quick Sort

Sort a given set of n integer elements using Quick Sort method and compute its time complexity. Run the program for varied values of n > 5000 and record the time taken to sort. Plot a graph of the time taken versus n on graph sheet. The elements can be read from a file or can be generated using the random number generator. Demonstrate using Java how the divide -and- conquer method works along with its time complexity analysis: worst case, average case and best case.

import java.util.Random;

import java.util.Scanner;
public class quicksort
{
static int max=2000;
int partition (int[] a, int low,int high)
{
int p,i,j,temp;
p=a[low];
i=low+1;
j=high;
while(low<high)
{
while(a[i]<=p&&i<high)
i++;
while(a[j]>p)
j--;
if(i<j)
{
temp=a[i];
a[i]=a[j];
a[j]=temp;
}
else
{
temp=a[low];
a[low]=a[j];
a[j]=temp;
return j;
}
}
return j;
}
void sort(int[] a,int low,int high)
{
if(low<high)
{
int s=partition(a,low,high);
sort(a,low,s-1);
sort(a,s+1,high);
}
}
public static void main(String[] args)
{
// TODO Auto-generated method stub
int[] a;
int i;
System.out.println("Enter the array size");
Scanner sc =new Scanner(System.in);
int n=sc.nextInt();
a= new int[max];
Random generator=new Random();
for( i=0;i<n;i++)
a[i]=generator.nextInt(20);
System.out.println("Array before sorting");
for( i=0;i<n;i++)
System.out.println(a[i]+" ");
long startTime=System.nanoTime();
quicksort m=new quicksort();
m.sort(a,0,n-1);
long stopTime=System.nanoTime();
long elapseTime=(stopTime-startTime);
System.out.println("Time taken to sort array is:"+elapseTime+"nano seconds");
System.out.println("Sorted array is");
for(i=0;i<n;i++)
System.out.println(a[i]);
}
}

Enter the array size
10
Array before sorting
17
17
12
2
10
3
18
15
15
17
Time taken to sort array is:16980 nano seconds
Sorted array is
23
10
12
15
15
17
17
17
18

Greedy Method

Implement in Java, the 0/1 Knapsack problem using Greedy method.

import java.util.Scanner;
public class knapsacgreedy
{
public static void main(String[] args)
{
int i,j=0,max_qty,m,n;
float sum=0,max;
Scanner sc = new Scanner(System.in);
int array[][]=new int[2][20];
System.out.println("Enter no of items");
n=sc.nextInt();
System.out.println("Enter the weights of each items");
for(i=0;i=0)
{
max=0;
for(i=0;imax)
{
max=((float)array[1][i])/((float)array[0][i]);
j=i;
}
}
if(array[0][j]>m)
{
System.out.println("Quantity of item number: " + (j+1) + " added is"+m);
sum+=m*max;
m=-1;
}
else
{
System.out.println("Quantity of item
number: " + (j+1) + " added is " + array[0][j]);
m-=array[0][j];
sum+=(float)array[1][j];
array[1][j]=0;
}
}
System.out.println("The total profit is " + sum);
sc.close();
}
}

Enter no of items
3
Enter the weights of each items
20 25 10
Enter the values of each items
30 40 35
Enter maximum volume of knapsack :
40
Quantity of item number: 3 added is 10
Quantity of item number: 2 added is 25
Quantity of item number: 1 added is 5
The total profit is 82.5

Dijkstra's Algorithm

From a given vertex in a weighted connected graph, find shortest paths to other vertices using Dijkstra's algorithm. Write the program in Java.

import java.util.Scanner;
public class Dijkstra
{
int d[]=new int[10];
int p[]=new int[10];
int visited[]=new int[10];
public void dijk(int[][]a, int s, int n)
{
int u=-1,v,i,j,min;
for(v=0;v<n;v++)
{
d[v]=99;
p[v]=-1;
}
d[s]=0;
for(i=0;i<n;i++)
{
min=99;
for(j=0;j<n;j++)
{
if(d[j]<min&& visited[j]==0)
{
min=d[j];
u=j;
}
}
visited[u]=1;
for(v=0;v<n;v++)
{
if((d[u]+a[u][v]<d[v])&&(u!=v)&&visited[v]==0)
{
d[v]=d[u]+a[u][v];
p[v]=u;
}
}
}
}
void path(int v,int s)
{
if(p[v]!=-1)
path(p[v],s);
if(v!=s)
System.out.print("->"+v+" ");
}
void display(int s,int n)
{
int i;
for(i=0;i<n;i++)
{
if(i!=s)
{
System.out.print(s+" ");
path(i,s);
}
if(i!=s)
System.out.print("="+d[i]+" ");
System.out.println();
}
}
public static void main(String[] args)
{
int a[][]=new int[10][10];
int i,j,n,s;
System.out.println("enter the number of vertices");
Scanner sc = new Scanner(System.in);
n=sc.nextInt();
System.out.println("enter the weighted matrix");
for(i=0;i<n;i++)
for(j=0;j<n;j++)
a[i][j]=sc.nextInt();
System.out.println("enter the source vertex");
s=sc.nextInt();
Dijkstra tr=new Dijkstra();
tr.dijk(a,s,n);
System.out.println("the shortest path between source"+s+"to remaining
vertices are");
tr.display(s,n);
sc.close();
}
}

Output:
enter the number of vertices
5
enter the weighted matrix
0 3 99 7 99
3 0 4 2 99
99 4 0 5 6
5 2 5 0 4
99 99 6 4 0
enter the source vertex
0
the shortest path between source0to remaining vertices are
0 ->1 =3
0 ->1 ->2 =7
0 ->1 ->3 =5
0 ->1 ->3 ->4 =9

Prim's Algorithm

Find Minimum Cost Spanning Tree of a given undirected graph using Prim's algorithm. Implement the program in Java language.

import java.util.Scanner;
public class prims
{
public static void main(String[] args)
{
int w[][]=new int[10][10];
int n,i,j,s,k=0;
int min;
int sum=0;
int u=0,v=0;
int flag=0;
int sol[]=new int[10];
System.out.println("Enter the number of vertices");
Scanner sc=new Scanner(System.in);
n=sc.nextInt();
for(i=1;i<=n;i++)
sol[i]=0;
System.out.println("Enter the weighted graph");
for(i=1;i<=n;i++)
for(j=1;j<=n;j++)
w[i][j]=sc.nextInt();
System.out.println("Enter the source vertex");
s=sc.nextInt();
sol[s]=1;
k=1;
while (k<=n-1)
{
min=99;
for(i=1;i<=n;i++)
for(j=1;j<=n;j++)
if(sol[i]==1&&sol[j]==0)
if(i!=j&&min>w[i][j])
{
min=w[i][j];
u=i;
v=j;
}
sol[v]=1;
sum=sum+min;
k++;
System.out.println(u+"->"+v+"="+min);
}
for(i=1;i<=n;i++)
if(sol[i]==0)
flag=1;
if(flag==1)
System.out.println("No spanning tree");
else
System.out.println("The cost of minimum spanning tree is"+sum);
sc.close();
}
}

Output:
Enter the number of vertices
4
Enter the weighted graph
0 1 3 10
1 0 10 4
3 10 0 2
10 4 2 0
Enter the source vertex
1
1->2=1
1->3=3
3->4=2
The cost of minimum spanning tree is 6

Kruskal's Algorithm

Find Minimum Cost Spanning Tree of a given undirected graph using Kruskal's algorithm Implement the program in Java language.

import java.util.Scanner;
public class kruskal
{
int parent[]=new int[10];
int find(int m)
{
int p=m;
while(parent[p]!=0)
p=parent[p];
return p;
}
void union(int i,int j)
{
if(i<j)
parent[i]=j;
else
parent[j]=i;
}
void krkl(int[][]a, int n)
{
int u=0,v=0,min,k=0,i,j,sum=0;
while(k<n-1)
{
min=99;
for(i=1;i<=n;i++)
for(j=1;j<=n;j++)
if(a[i][j]<min&&i!=j)
{
min=a[i][j];
u=i;
v=j;
}
i=find(u);
j=find(v);
if(i!=j)
{
union(i,j);
System.out.println("("+u+","+v+")"+"="+a[u][v]);
sum=sum+a[u][v];
k++;
}
a[u][v]=a[v][u]=99;
}
System.out.println("The cost of minimum spanning tree = "+sum);
}
public static void main(String[] args)
{
int a[][]=new int[10][10];
int i,j;
System.out.println("Enter the number of vertices of the graph");
Scanner sc=new Scanner(System.in);
int n;
n=sc.nextInt();
System.out.println("Enter the wieghted matrix");
for(i=1;i<=n;i++)
for(j=1;j<=n;j++)
a[i][j]=sc.nextInt();
kruskal k=new kruskal();
k.krkl(a,n);
sc.close();
}
}

Output:
Enter the number of vertices of the graph
4
Enter the wieghted matrix
0 1 3 10
1 0 10 4
3 10 0 2
10 4 2 0
(1,2)=1
(3,4)=2
(1,3)=3
The cost of minimum spanning tree = 6

Dynamic Programming method

Implement in Java, the 0/1 Knapsack problem using Dynamic Programming method

import java.util.Scanner;
public class knapsackDP
{
public void solve(int[] wt, int[] val, int W, int N)
{
int i,j;
int[][] sol = new int[N + 1][W + 1];
for ( i = 0; i <= N; i++)
{
for ( j = 0; j <= W; j++)
{
if(i==0||j==0)
sol[i][j]=0;
else if(wt[i]>j)
sol[i][j]=sol[i-1][j];
else
sol[i][j]=Math.max((sol[i-1][j]), (sol[i - 1][j - wt[i]] + val[i]));
}
}
System.out.println("The optimal solutionis"+sol[N][W]);
int[] selected = new int[N + 1];
for(i=0;i<N+1;i++)
selected[i]=0;
i=N;
j=W;
while (i>0&&j>0)
{
if (sol[i][j] !=sol[i-1][j])
{
selected[i] = 1;
j = j - wt[i];
}
i- -;
}
System.out.println("\nItems selected : ");
for ( i = 1; i < N + 1; i++)
if (selected[i] == 1)
System.out.print(i +" ");
System.out.println();
}
public static void main(String[] args)
{
Scanner scan = new Scanner(System.in);
knapsackDP ks = new knapsackDP();
System.out.println("Enter number of elements ");
int n = scan.nextInt();
int[] wt = new int[n + 1];
int[] val = new int[n + 1];
System.out.println("\nEnter weight for "+ n +" elements");
for (int i = 1; i <= n; i++)
wt[i] = scan.nextInt();
System.out.println("\nEnter value for "+ n +"
elements");
for (int i = 1; i <= n; i++)
val[i] = scan.nextInt();
System.out.println("\nEnter knapsack weight ");
int W = scan.nextInt();
ks.solve(wt, val, W, n);
}
}

Output:
Enter number of elements
3
Enter weight for 3 elements
20 25 10
Enter value for 3 elements
30 40 35
Enter knapsack weight
40
The optimal solution is 75
Items selected :
2 3

Floyd's Algorithm

Write Java programs to Implement All-Pairs Shortest Paths problem using Floyd's algorithm.

import java.util.Scanner;
public class Floyd
{
void flyd(int[][] w,int n)
{
int i,j,k;
for(k=1;k<=n;k++)
for(i=1;i<=n;i++)
for(j=1;j<=n;j++)
w[i][j]=Math.min(w[i][j], w[i][k]+w[k][j]);
}
public static void main(String[] args)
{
int a[][]=new int[10][10];
int n,i,j;
System.out.println("enter the number of vertices");
Scanner sc=new Scanner(System.in);
n=sc.nextInt();
System.out.println("Enter the weighted matrix");
for(i=1;i<=n;i++)
for(j=1;j<=n;j++)
a[i][j]=sc.nextInt();
floyd f=new floyd();
f.flyd(a, n);
System.out.println("The shortest path matrix is");
for(i=1;i<=n;i++)
{
for(j=1;j<=n;j++)
{
System.out.print(a[i][j]+" ");
}
System.out.println();
}
sc.close();
}
}

Output:
enter the number of vertices
4
Enter the weighted matrix
0 99 3 99
2 0 99 99
99 7 0 1
6 99 99 0
The shortest path matrix is
0 10 3 4
2 0 5 6
7 7 0 1
6 16 9 0

Travelling Sales Person problem

Write Java programs to Implement Travelling Sales Person problem using Dynamic programming.

import java.util.Scanner;
class TSPExp
{
int weight[][],n,tour[],finalCost;
final int INF=1000;
TSPExp()
{
Scanner s=new Scanner(System.in);
System.out.println("Enter no. of nodes:=>");
n=s.nextInt();
weight=new int[n][n];
tour=new int[n-1];
for(int i=0;i<n;i++)
{
for(int j=0;j<n;j++)
{
if(i!=j)
{
System.out.print("Enter weight of "+(i+1)+" to "+(j+1)+":=>");
weight[i][j]=s.nextInt();
}
}
}
System.out.println();
System.out.println("Starting node assumed to be node 1.");
eval();
}
public int COST(int currentNode,int inputSet[],int setSize)
{
if(setSize==0)
return weight[currentNode][0];
int min=INF;
int setToBePassedOnToNextCallOfCOST[]=new int[n-1];
for(int i=0;i<setSize;i++)
{
int k=0;//initialise new set
for(int j=0;j<setSize;j++)
{
if(inputSet[i]!=inputSet[j])
setToBePassedOnToNextCallOfCOST[k++]=inputSet[j];
}
int
temp=COST(inputSet[i],setToBePassedOnToNextCallOfCOST,setSize-1);
if((weight[currentNode][inputSet[i]]+temp) <min)
{
min=weight[currentNode][inputSet[i]]+temp;
}
}
return min;
}
public int MIN(int currentNode,int inputSet[],int setSize)
{
if(setSize==0)
return weight[currentNode][0];
int min=INF,minindex=0;
int setToBePassedOnToNextCallOfCOST[]=new int[n-1];
for(int i=0;i<setSize;i++) //considers each node of inputSet
{
int k=0;
for(int j=0;j<setSize;j++)
{
if(inputSet[i]!=inputSet[j])
setToBePassedOnToNextCallOfCOST[k++]=inputSet[j];
}
int
temp=COST(inputSet[i],setToBePassedOnToNextCallOfCOST,setSize-1);
if((weight[currentNode][inputSet[i]]+temp) < min)
{
min=weight[currentNode][inputSet[i]]+temp;
minindex=inputSet[i];
}
}
return minindex;
}
public void eval()
{
int dummySet[]=new int[n-1];
for(int i=1;i<n;i++)
dummySet[i-1]=i;
finalCost=COST(0,dummySet,n-1);
constructTour();
}
public void constructTour()
{
int previousSet[]=new int[n-1];
int nextSet[]=new int[n-2]; for(int i=1;i<n;i++)
previousSet[i-1]=i;
int setSize=n-1;
tour[0]=MIN(0,previousSet,setSize);
for(int i=1;i<n-1;i++)
{
int k=0;
for(int j=0;j<setSize;j++)
{
if(tour[i-1]!=previousSet[j])
nextSet[k++]=previousSet[j];
}
--setSize;
tour[i]=MIN(tour[i-1],nextSet,setSize);
for(int j=0;j<setSize;j++)
previousSet[j]=nextSet[j];
}
display();
}
public void display()
System.out.println();
System.out.print("The tour is 1-");
for(int i=0;i<n-1;i++)
System.out.print((tour[i]+1)+"-");
System.out.print("1");
System.out.println();
System.out.println("The final cost is "+finalCost);
}
}
class TSP
{
public static void main(String args[])
{
TSPExp obj=new TSPExp();
}
}

Output:
Enter no. of nodes:=>
4
Enter weight of 1 to 2:=>2
Enter weight of 1 to 3:=>5
Enter weight of 1 to 4:=>7
Enter weight of 2 to 1:=>2
Enter weight of 2 to 3:=>8
Enter weight of 2 to 4:=>3
Enter weight of 3 to 1:=>5
Enter weight of 3 to 2:=>8
Enter weight of 3 to 4:=>1
Enter weight of 4 to 1:=>7
Enter weight of 4 to 2:=>3
Enter weight of 4 to 3:=>1
Starting node assumed to be node 1.
The tour is 1-2-4-3-1
The final cost is 11

Subset

Design and implement in Java to find a subset of a given set S = {Sl, S2,.....,Sn} of n positive integers whose SUM is equal to a given positive integer d. For example, if S ={1, 2, 5, 6, 8} and d= 9, there are two solutions {1, 2, 6} and {1, 8}. Display a suitable message, if the given problem instance doesn't have a solution.

import java.util.Scanner;
import static java.lang.Math.pow;
public class subSet
{
void subset(int num,int n, int x[])
{
int i;
for(i=1;i<=n;i++)
x[i]=0;
for(i=n;num!=0;i--)
{
x[i]=num%2;
num=num/2;
}
}
public static void main(String[] args)
{
int a[]=new int[10];
int x[]=new int[10];
int n,d,sum,present=0;
int j;
System.out.println("enter the number of elements of set");
Scanner sc=new Scanner(System.in);
n=sc.nextInt();
System.out.println("enter the elements of set");
for(int i=1;i<=n;i++)
a[i]=sc.nextInt();
System.out.println("enter the positive integer sum");
d=sc.nextInt();
if(d>0)
{
for(int i=1;i<=Math.pow(2,n)-1;i++)
{
subSet s=new subSet();
s.subset(i,n,x);
sum=0;
for(j=1;j<=n;j++)
if(x[j]==1)
sum=sum+a[j];
if(d==sum)
{
System.out.print("Subset={");
present=1;
for(j=1;j<=n;j++)
if(x[j]==1)
System.out.print(a[j]+",");
System.out.print("}="+d);
System.out.println();
}
}
}
if(present==0)
System.out.println("Solution does not exists");
}
}

Output:
enter the number of elements of set
5
enter the elements of set
1 2 5 6 8
enter the positive integer sum
9
Subset={1,8,}=9
Subset={1,2,6,}=9

Hamiltonian Cycle

Design and implement in Java to find all Hamiltonian Cycles in a connected undirected Graph G of n vertices using backtracking principle.

import java.util.*;
class Hamiltoniancycle
{
private int adj[][],x[],n;
public Hamiltoniancycle()
{
Scanner src = new Scanner(System.in);
System.out.println("Enter the number of nodes");
n=src.nextInt();
x=new int[n];
x[0]=0;
for (int i=1;i<n; i++)
x[i]=-1;
adj=new int[n][n];
System.out.println("Enter the adjacency matrix");
for (int i=0;i<n; i++)
for (int j=0; j<n; j++)
adj[i][j]=src.nextInt();
}
public void nextValue (int k)
{
int i=0;
while(true)
{
x[k]=x[k]+1;
if (x[k]==n)
x[k]=-1;
if (x[k]==-1)
return;
if (adj[x[k-1]][x[k]]==1)
for (i=0; i<k; i++)
if (x[i]==x[k])
break;
if (i==k)
if (k<n-1 || k==n-1 && adj[x[n-1]][0]==1)
return;
}
}
public void getHCycle(int k)
{
while(true)
{
nextValue(k);
if (x[k]==-1)
return;
if (k==n-1)
{
System.out.println("\nSolution : ");
for (int i=0; i<n; i++)
System.out.print((x[i]+1)+" ");
System.out.println(1);
}
else getHCycle(k+1);
}
}
}
class HamiltoniancycleExp
{
public static void main(String args[])
{
Hamiltoniancycle obj=new Hamiltoniancycle();
obj.getHCycle(1);
}
}

Output:
Enter the number of nodes
6
Enter the adjacency matrix
0 1 1 1 0 0
1 0 1 0 0 1
1 1 0 1 1 0
1 0 1 0 1 0
0 0 1 1 0 1
0 1 0 0 1 0
Solution :
1 2 6 5 3 4 1
Solution :
1 2 6 5 4 3 1
Solution :
1 3 2 6 5 4 1
Solution :
1 3 4 5 6 2 1
Solution :
1 4 3 5 6 2 1
Solution :
1 4 5 6 2 3 1