Quiz-1 for DCET 2025 (Introduction to Python)

 1. Introduction to Python

https://docs.google.com/forms/d/e/1FAIpQLSc3LFrMaswLg6mltJ3NN5QGbALA9Uc53KWPp4oV5HN4SaM0Iw/viewform?usp=sharing&ouid=105493060103049146790

Design and implement C/C++ Program to 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.

 #include <stdio.h>

#include <stdlib.h>

#include <time.h>

// Function to merge two subarrays

void merge(int arr[], int left, int mid, int right) {

    int i, j, k;

    int n1 = mid - left + 1;

    int n2 = right - mid;

    int L[n1], R[n2];

    for (i = 0; i < n1; i++)

        L[i] = arr[left + i];

    for (j = 0; j < n2; j++)

        R[j] = arr[mid + 1 + j];

    i = 0;

    j = 0;

    k = left;

    while (i < n1 && j < n2) {

        if (L[i] <= R[j]) {

            arr[k] = L[i];

            i++;

        } else {

            arr[k] = R[j];

            j++;

        }

        k++;

    }

    while (i < n1) {

        arr[k] = L[i];

        i++;

        k++;

    }

    while (j < n2) {

        arr[k] = R[j];

        j++;

        k++;

    }

}

// Merge Sort function

void mergeSort(int arr[], int left, int right) {

    if (left < right) {

        int mid = left + (right - left) / 2;

        mergeSort(arr, left, mid);

        mergeSort(arr, mid + 1, right);

        merge(arr, left, mid, right);

    }

}

int main() {

    int n;

    printf("Enter number of elements (n > 5000): ");

    scanf("%d", &n);

    if (n <= 5000) {

        printf("Please enter n greater than 5000.\n");

        return 1;

    }

    int *arr = (int *)malloc(n * sizeof(int));

    if (arr == NULL) {

        printf("Memory allocation failed.\n");

        return 1;

    }

    // Generating random numbers

    srand(time(0));

    for (int i = 0; i < n; i++) {

        arr[i] = rand() % 100000; // Random numbers between 0 and 99999

    }

    clock_t start, end;

    double cpu_time_used;

    start = clock();

    mergeSort(arr, 0, n - 1);

    end = clock();

    cpu_time_used = ((double)(end - start)) / CLOCKS_PER_SEC;

    printf("Time taken to sort %d elements: %f seconds\n", n, cpu_time_used);

    free(arr);

    return 0;

}

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GCD of two numbers and its application...

The greatest common divisor (gcd) of two numbers is the largest positive integer that divides both numbers without leaving a remainder. The gcd can be found using the Euclidean algorithm

GCD is useful in cases when you want different amounts of things to be arranged in the same number of order. 

For example there are 32 soldiers and 48 bandsman and during the parade you want them to march in the same number of rows. 

So , you calculate the HCF which is 8 and thus you can make 8 rows each for each group.

Write C++ program to find GCD two numbers a and b using  user defined function. i.e. GCD(a,b) using Euclids algorithm.

 #include <iostream>

using namespace std;

int gcd(int a, int b) {

   if (b == 0)

   return a;

   return gcd(b, a % b);

}

int main() {

   int a , b;

   cout<<"Enter the values of a and b: "<<endl;

   cin>>a>>b;

   cout<<"GCD of "<< a <<" and "<< b <<" is "<< gcd(a, b);

   return 0;

}

Output:

Enter the values of a and b: 

10

20

GCD of 10 and 20 is 10

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Module 5: L2: Design and implement the presence of Hamiltonian Cycle in an undirected Graph G of n vertices.

 

Design and implement the presence of Hamiltonian Cycle in an undirected Graph G of n vertices.

 

importjava.util.*;

 

 

classHamiltoniancycle

 

{

privateintadj[][],x[],n;

publicHamiltoniancycle()

{

Scanner src = newScanner(System.in);

System.out.println("Enter the number of nodes");

n=src.nextInt();

x=new int[n];

x[0]=0;

for(inti=1;i<n; i++)

x[i]=-1;

adj=new int[n][n];

System.out.println("Enter the adjacency matrix"); for (inti=0;i<n; i++)

for(intj=0; j<n; j++)

adj[i][j]=src.nextInt();

}

 

public void nextValue (intk)

 

{

inti=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(intk)

{

while(true)

{

nextValue(k);

if(x[k]==-1)

return;

if(k==n-1)

{

System.out.println("\nSolution : ");

for(inti=0; i<n; i++)

System.out.print((x[i]+1)+" ");

System.out.println(1);

}

else

getHCycle(k+1);

}

 

}

}

 

 

classHamiltoniancycleExp

{

public static void main(String args[])

{

Hamiltoniancycleobj=newHamiltoniancycle(); 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 :					4	1
1	2	6	5	3
Solution :					3	1
1	2	6	5	4
Solution :					4	1
1	3	2	6	5
Solution :					2	1
1	3	4	5	6
Solution :					2	1
1	4	3	5	6
Solution :					3	1
1	4	5	6	2

Module-5: L1 Subset Problem :Java Program 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

 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(intnum,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) {

 

//                    TODO Auto-generated method stub int a[]=new int[10];

 

int x[]=new int[10]; intn,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