Depar tment Of Comp uter Science...Smt. Padmavatibai Raghavendraroa Deshpande Pikalihal Government First Grade College, Mudgal -584125 Affiliated under Gulbarga University Depar tment

Smt. Padmavatibai Raghavendraroa Deshpande Pikalihal

Government First Grade College, Mudgal -584125

Affiliated under

Gulbarga University

Department Of

Computer Science

B.Sc. 1st

Semester Lab Manual (CBCS Syllabus)

OBJECT ORIENTED PROGRAMMING Using C++ Lab

INDEX

OBJECT ORIENTED PROGRAMMING USING C++ Lab

Sl.

No Name of the Program

1

Program to read the ballots and count the votes cast for each candidate using an array variable count. In case, a number read is outside the range 1 to 5, the ballot should be considered as a ‘spoilt ballot’, and the program should also count the number of spoilt ballots.

2 Program to find the factorial of a number using ‘return by reference’ concept.

3 Program to find the largest of three numbers using Inline functions.

4 Program to read a matrix of size m x n from the keyboard and display the same on the screen. Make the row parameter of the matrix as a default argument.

5 Write a main that calls both the functions. Use the concept of

function overloading

6 Program in C++ to swap two variables of data types integer, floating

point number and character type using function overloading

7 Write a C++ program to create a class Sample. Include the following

class members

8 Write a C++ program using class to demonstrate Nesting of member

functions

9 The time object shows the time in hh:mm:ss format. Design a program to add 2 time objects using friend function.

10 Program to find the volume of a cube, cylinder, rectangle using

constructors overloading.

1) An election is contested by five candidates. The candidates are numbered 1 to 5

and the voting is done by marking the candidate number on the ballot paper.

Write a program to read the ballots and count the votes cast for each candidate

using an array variable count. In case, a number read is outside the range 1 to 5,

the ballot should be considered as a ‘spoilt ballot’, and the program should also

count the number of spoilt ballots.

#include<iostream>

usingnamespace std;

main()

{

int a[5]={0,0,0,0,0};

int n,vote,sb=0,i;

cout<<"Enter the number of votes\n";

cin>>n;

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

{

cout<<"Enter your vote : ";

cin>>vote;

switch(vote)

{

case 1:a[0]++;

break;

case 2:a[1]++;

break;

case 3:a[2]++;

break;

case 4:a[3]++;

break;

case 5:a[4]++;

break;

default:sb++;

}

}

cout<<"\n\nResults of Election are as follows\n";

for(i=0;i<5;i++)

cout<<"\ncandidate "<<i+1<<" : "<<a[i];

cout<<"\nSpoiled Ballots : "<<sb;

}

2) Write a program to find the factorial of a number using ‘return by reference’ concept.

#include<iostream>

usingnamespace std;

int&factorial(int,int&f);

main()

{

int n,fact=1;

cout<<"enter a number\n";

cin>>n;

fact=factorial(n,fact);

cout<<n<<"! = "<<fact;

}

int&factorial(int n,int&f)

{

int i;

if(n==0)

{

f=0;

return f;

}

else

{

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

f=f*i;

return f;

}

}

3) Write a program to find the largest of three numbers using Inline functions.

#include<iostream>

usingnamespace std;

inlineint largest(int a,int b,int c)

{

return((a>b)?((a>c)?a:c):((b>c)?b:c));

}

main()

{

int a,b,c;

cout<<"enter values for a,b,c : ";

cin>>a>>b>>c;

cout<<"Largest = "<<largest(a,b,c);

}

4) Write a program to read a matrix of size m x n from the keyboard and display the

same on the screen. Make the row parameter of the matrix as a default argument.

#include<iostream>

usingnamespace std;

main()

{

void matrix(int n,int m=3);

int m,n;

cout<<"Enter the no of cols : ";

cin>>n;

matrix(n);

}

void matrix(int n,int m)

{

int a[20][20];

int i,j;

cout<<"enter the elements of the matrix \n";

for(i=0;i<m;i++)

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

cin>>a[i][j];

cout<<"\n Matrix \n";

for(i=0;i<m;i++)

{

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

cout<<a[i][j]<<"\t";

cout<<"\n";

}

}

5) Write a main that calls both the functions. Use the concept of function

overloading.

#include<iostream>

usingnamespace std;

double power(double m,int n=2);

int power(int m,int n=2);

main()

{

int n;

double m;

cout<<"enter m and n : ";

cin>>m>>n;

cout<<m<<" to the power of "<<n<<" = "<<power(m,n);

cout<<"\n";

{

int m;

cout<<"enter m ";

cin>>m;

cout<<m<<" to the power of 2 = "<<power(m);

}

}

double power(double m,int n)

{

double p=1,i;

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

p=p*m;

cout<<"fun 1\n";

return p;

}

int power(int m,int n)

{

int p=1,i;

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

p=p*m;

cout<<"fun2\n";

return p;

6) Write a program in C++ to swap two variables of data types integer,

floating point number and character type using function overloading.

#include<iostream>

usingnamespace std;

void swap(int a,int b)

{

int t;

t=a;

a=b;

b=t;

cout<<"\nAfter swapping integer values\n a = "<<a<<" b = "<<b;

}

void swap(float a,float b)

{

float t;

t=a;

a=b;

b=t;

cout<<"\nAfter swapping floating point values\n a = "<<a<<" b = "<<b;

}

void swap(char a,char b)

{

char t;

t=a;

a=b;

b=t;

cout<<"\nAfter swapping character type values\n a = "<<a<<" b = "<<b;

}

main()

{

int a,b;

float c,d;

char m,n;

cout<<"enter 2 integer values \n";

cin>>a>>b;

cout<<"enter 2 floating point values \n";

cin>>c>>d;

cout<<"enter 2 character values \n";

cin>>m>>n;

swap(a,b);

swap(c,d);

swap(m,n);

}

7. Write a C++ program to create a class Sample. Include the following

class members.

Read() – a private member function to read the value for any private

variable of type integer

Update() – a public member function to access the private member

function

Write()—a public member function to display the value of private data

member

#include<iostream>

usingnamespace std;

class sample

{

private: int n;

void read(int a)

{

n=a;

}

public: void update(int a)

{

read(a);

}

void write()

{

cout<<"\nPrivate member n : "<<n;

}

};

main()

{

int x;

cout<<"enter a value : ";

cin>>x;

sample s;

s.update(x);

s.write();

}

8. Write a C++ program using class to demonstrate Nesting of member

functions.

#include<iostream>

usingnamespace std;

class Nest

{

int m1;

int m2;

public: void getdata(int a,int b)

{

m1=a;

m2=b;

display();

}

void display()

{

cout<<"m1 : "<<m1;

cout<<"\tm2 : "<<m2;

}

};

int main()

{

Nest n;

n.getdata(95,99);

}

9. The time object shows the time in hh:mm:ss format. Design a program

to add 2 time objects using friend function.

#include<iostream>

usingnamespace std;

class time

{

int h;

int m;

int s;

public: void getdata(int a,int b,int c)

{

h=a;

m=b;

s=c;

}

void disp()

{

cout<<h<<":"<<m<<":"<<s;

}

friend time add(time t1,time t2);

};

time add(time t1,time t2)

{

time t;

int hh,mm,ss;

t.s=t1.s+t2.s;

ss=t.s/60;

t.s=t.s%60;

t.m=ss+t1.m+t2.m;

mm=t.m/60;

t.m=t.m%60;

t.h=mm+t1.h+t2.h;

if(t.h>=24)

{

t.h=t.h-24;

}

return t;

}

main()

{

int a,b,c;

time t1,t2,t3;

cout<<"\nenter 3 values(hh:mm:ss)";

cin>>a>>b>>c;

t1.getdata(a,b,c);

cout<<"\nenter 3 values(hh:mm:ss)";

cin>>a>>b>>c;

t2.getdata(a,b,c);

cout<<"\n\nEntered time values are\n";

t1.disp();

cout<<"\t";

t2.disp();

cout<<"\n\nSum of two time objects = ";

t3=add(t1,t2);

t3.disp();

}

10. Write a program to find the volume of a cube, cylinder, rectangle using

constructors overloading.

#include<iostream>

usingnamespace std;

class volume

{

float v;

public: volume(float s)

{

v=s*s*s;

cout<<"\nVolume of cube of side "<<s<<" = "<<v;

}

volume(float l,float b,float h)

{

v=l*b*h;

cout<<"\nVolume of rectangle of length,breadth & height ";

cout<<l<<b<<h<<" = "<<v;

}

volume(float r,float h)

{

v=(22/7)*r*r*h;

cout<<"\nVolume of cylinder with radius & height ";

cout<<r<<h<<" = "<<v;

}

};

main()

{

float s,l,b,h,r;

cout<<"\nenter length of side of a cube : ";

cin>>s;

volume v1(s);

cout<<"\nenter length,breadth & height of rectangle : ";

cin>>l>>b>>h;

volume v2(l,b,h);

cout<<"\nenter radius & height of the cylinder : ";

cin>>r>>h;

volume v3(r,h);

}



Affiliated under

Gulbarga University

Department Of

Computer Science

B.Sc. 2nd Semester Lab Manual (CBCS Syllabus)

Data Structure using C++ Lab

INDEX

Data Structure using C++ Lab

Sl. No Name of the Program

1 Program to insert and delete an element in an array

2 Program to perform linear search using array;

3 Program to perform the Binary Search operations in array;

4 Program to perform STACK OPERATIONS

5 Program to perform operations on QUEUE.

6 Program to demonstrate a concept of Linked list.

7 Program to perform TRAVERSE a tree in PREORDER

8 Program to TRAVERSE a tree in INORDER

9 Program to TRAVERSE a tree in POSTORDER

10 Program to calculate POSTFIX expression

11 Program to perform SELECTION sort

12 Program to perform BUBBLE sort

13 Program to perform INSERTION sort

14 Program to perform RADIX sort

//1.Program to insert and delete an element in an array

#include<iostream.h>

#include<conio.h>

#include<stdio.h>

#include<process.h>

class arrays

{

private: int a[10],n,i,k,item,ch,j;

public: void opers();

};

void arrays:: opers()

{

cout<<"\n\n OUTPUT FOR ARRAY OPERATIONS \n\n";

cout<<"enter the number of elements you want:\n";

cin>>n;

cout<<"enter the "<<n<<" elements:\n";

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

cin>>a[i];

do

{

cout<<"\n enter the choice: \n 1.insert \n 2.delete \n 3.exit\n";

cin>>ch;

switch(ch)

{

case 1:

cout<<"\n enter the item to be inserted \n";

cin>>item;

cout<<"\n enter the position to insert the item\n";

cin>>k;

j=n;

while(j>=k)

{

a[j+1]=a[j];

j--;

}

a[k]=item;

n++;

cout<<"After insertion the array\n";

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

cout<<"\t"<<a[i];

break;

case 2:

if(n==0)

cout<<"The array is empty,deletion is not possible\n";

else

{

cout<<"enter the position of the item to be deleted\n";

cin>>k;

item=a[k];

j=k;

while(j<=n-1)

{

a[j]=a[j+1];

j++;

}

n--;

if(n==0)

cout<<"\n now the array is empty\n";

else

cout<<"\n After deletion the array is:\n";

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

cout<<"\t"<<a[i];

}

break;

default:

cout<<"the code is mismatch try again!\n";

break;

}

}

while(ch!=3);

}

void main()

{

clrscr();

arrays ar;

ar.opers();

getch();

}

OUTPUT FOR ARRAY OPERATIONS

enter the number of elements you want:

4

enter the 4 elements:

12 34 56 67

enter the choice:

1.insert

2.delete

3.exit

1

enter the item to be inserted

23

enter the position to insert the item

3

After insertion the array

12 34 23 56 78

enter the choice:

1.insert

2.delete

3.exit

2

enter the position of the item to be deleted

3

After deletion the array is:

12 34 56 78

enter the choice:

1.insert

2.delete

3.exit

3

the code is mismatch try again!

//2.Program to perform linear search using array;


#include<conio.h>

class linear

{

private: int i,n,item,a[10],loc;

public: void search();

};

void linear:: search()

{ cout<<"OUT PUT FOR LINEAR SEARCH"<<endl;

cout<<"Enter no of elements"<<endl;

cin>>n;

cout<<"Enter the elements"<<endl;

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

cin>>a[i];

cout<<"Enter the item to search"<<endl;

cin>>item;

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

if(a[i]==item)

{

loc=i;

cout<<"Search is sucessful at location "<<loc;

}

if(loc!=item)

cout<<"\n Search is unsucessful\n";

}

void main()

{

clrscr();

linear ln;

ln.search();

getch();

}

OUT PUT FOR LINEAR SEARCH

Enter no of elements

4

Enter the elements

12 34 45 67

Enter the item to search

67

Search is successful at location 3

//3.program to perform the binary search operations in array;


#include<conio.h>

class binsear

{

private: int a[20],st,i,j,temp,end,n,mid,loc,item;

public: void search();

};

void binsear:: search( )

{

cout<<"\n OutPut of binary search\n";

cout<<"enter number of elements:\n";

cin>>n;

cout<<"enter the elemens of array:\n";

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

cin>>a[i];

cout<<"enter the elements to search:\n";

cin>>item;

st=1;

end=n;

mid = ((st+end)/2);

while((st<=end)&&(a[mid]!=item))

{

if(item<a[mid])

end=mid-1;

else

st=mid+1;

mid=((st+end)/2);

}

if(a[mid]==item)

cout<<"search is successful at location:"<<mid;

else

cout<<"search is unsucessful \n";

}

void main( )

{

binsear br;

clrscr();

br.search();

getch();

}

OUTPUT OF BINARY SEARCH

enter number of elements:

5

enter the elemens of array:

12 34 56 67 23

enter the elements to search:

67

search is successful at location:3

//4. Program to perform STACK OPERATIONS


#include<conio.h>

#include<process.h>

class stacks

{

private: int top,st[10],item,i,ch;

public: void operations();

};

void stacks::operations()

{

top=0;

cout<<"output for stack operations:\n";

cout<<"\n enter the choice:\n 1.push \n 2.pop \n 3.exit\n";

cin>>ch;

while(ch!=3)

{

switch(ch)

{

case 1: cout<<"enter the item to insert in stack:\n";

cin>>item;

st[++top] = item;

break;

case 2: --top;

if(top<=0)

{

cout<<"empty stack";

top=0;

}

break;

case 3:exit(1);

default: cout<<"illegal entry\n";

}

cout<<"\n After stack operation\n";

for(i=1;i<=top;i++)

cout<<"\n"<<st[i];

cout<<"\n enter the choice 1.push 2.pop 3.exit\n";

cin>>ch;

}

}

void main()

{

stacks st;

clrscr();

st.operations();

getch();

}

OUTPUT FOR STACK OPERATIONS:

enter the choice:

1.push

2.pop

3.exit

1

enter the item to insert in stack:

23

After stack operation

23

enter the choice 1.push 2.pop 3.exit

1


24


23

24


1


25


23

24

25


2


23

24


3

//5.Program to peform operations on QUEUE.


#include<conio.h>

#define maxq 10

class queue

{

private: int rear,front,q[10],ch,i,item;

public:

void calc();

};

void queue::calc()

{

front=1;

rear=0;

cout<<"Select your choice"<<endl;

cout<<"1.Insert\n 2.Delete\n 3.Exit"<<endl;

cout<<"Enter your choice"<<endl;

cin>>ch;

do

{

switch(ch)

{

case 1:{

if(rear==maxq)

cout<<"queue is full"<<endl;

else

cout<<"Enter item to be inserted"<<endl;

cin>>item;

q[++rear]=item;

}

break;

case 2:{

if(front>=rear)

cout<<"queue is empty"<<endl;

else

item=q[front++];

}

break;

default:{

cout<<"illegal entry"<<endl;

}

break;

}

cout<<"After queue operation"<<endl;

for(i=front;i<=rear;i++)

cout<<q[i]<<endl;

cout<<"enter choice"<<endl;

cout<<"1.Insert 2.Delete 3.Exit"<<endl;

cin>>ch;

}

while(ch!=3);

}

void main()

{

clrscr();

queue q;

q.calc();

getch();

}

OUT PUT FOR QUEUES:

enter choice:

1.insert 2.delete 3.exit

1

Enter item to be inserted

12

enter choice:

1.insert 2.delete 3.exit

1


After queue operation

12

13

enter choice

1.Insert 2.Delete 3.Exit

1


14


12

13

14

enter choice


2


13

14

enter choice


3

//6. Program to demonstrate a concept of Linked list.


#include<malloc.h>

#include<conio.h>

struct list

{

int info;

struct list*next;

};

typedef struct list node;

node *first,*ptr,*start;

void main()

{

clrscr();

first=(node*)malloc(sizeof(node));

cout<<"Enter the elements at end enter 0"<<endl;

cin>>first->info;

ptr=start=first;

while(first->info!=0)

{

first=(node*)malloc(sizeof(node));

cin>>first->info;

ptr->next=first;

ptr=first;

}

ptr->next=0;

cout<<"List after creation"<<endl;

while(start->next!=0)

{

cout<<start->info<<"->";

start=start->next;

}

cout<<"Null";

getch();

}

OUTPUT:-

Enter the elements at end enter 0

12

14

35

67

99

0

List after creation

12->14->35->67->99->Null

//7. Program to perform TRAVERSE a tree in PREORDER


#include<conio.h>

#include<stdio.h>

#include<alloc.h>

#define null 0

struct tree

{

int info;

struct tree *left,*right;

};

typedef struct tree node;

node * root;

void preorder(node *p)

{

if(p!=null)

{

cout<<p->info;

preorder(p->left);

preorder(p->right);

}

}

void read1(node **p2)

{

node *p1;

int item;

p1=(node *)malloc(sizeof(node));

cin>>item;

if(item!=0)

{

p1->info=item;

p1->left=null;

p1->right=null;

*p2=p1;

}

if(item!=0)

{

cout<<"the left child of is:\n"<<p1->info;

read1(&p1->left);

cout<<"the right child of is:\n"<<p1->info;

read1(&p1->right);

}

}

void main()

{

root=null;

clrscr();

cout<<"\n\nOUTPUT FOR PREORDER TRAVERSAL\n\n";

cout<<"enter the root or 0 to quit\n";

read1(&root);

cout<<"\nThe preorder traversal of the tree is:\n";

preorder(root);

getch();

}

OUTPUT FOR PREORDER TRAVERSAL

enter the root or 0 to quit

24

the left child of is:

24 6


6 0

the right child of is:

6 0


24 5


5 8


8 0


8 0


5 0

The preorder traversal of the tree is:

24658

//8. Program to TRAVERSE a tree in INORDER


#include<conio.h>

#include<stdio.h>

#include<alloc.h>

#define null 0

struct tree

{

int info;


};


node *root;

void inorder(node *p)

{

if(p!=null)

{

inorder(p->left);

cout<<p->info;

inorder(p->right);

}

}


{

node *p1;

int item;


cin>>item;

if(item!=0)

{

p1->info=item;

p1->left=null;

p1->right=null;

*p2=p1;

}

if(item!=0)

{


read1(&p1->left);


read1(&p1->right);

}

}

void main()

{

root=null;

clrscr();

cout<<"\n\nOUTPUT FOR INORDER TRAVERSAL\n\n";


read1(&root);

cout<<"\n The inorder traversal of the tree is:\n";

inorder(root);

getch();

}

OUTPUT FOR INORDER TRAVERSAL


24


24 12


12 0


12 43


43 67


67 0


67 0


43 0


24 0

The inorder traversal of the tree is:

12674324

//9. Program to TRAVERSE a tree in POSTORDER


#include<conio.h>

#include<stdio.h>

#include<alloc.h>

#define null 0

struct tree

{

int info;


};


node *root;

void postorder(node *p)

{

if(p!=null)

{

postorder(p->left);

postorder(p->right);

cout<<p->info;

}

}


{

node *p1;

int item;


cin>>item;

if(item!=0)

{

p1->info=item;

p1->left=null;

p1->right=null;

*p2=p1;

}

if(item!=0)

{


read1(&p1->left);


read1(&p1->right);

}

}

void main()

{

root=null;

clrscr();

cout<<"\n\nOUTPUT FOR POSTORDER TRAVERSAL\n\n";


read1(&root);

cout<<"\nThe inorder traversal of the tree is:\n";

postorder(root);

getch();

}

OUT PUT FOR POST ORDER TRAVERSAL


24


24 3


3 0


3 6


6 0


6 0


24 5


5 9


9 0


9 0


5 0

The inorder traversal of the tree is:

639524

//10.Program to calculate POSTFIX expression


#include<conio.h>

#include<stdio.h>

#include<process.h>

class postfix

{

private: char item;

float top,op1,op2,value,st[20];

public: void pstfxopers();

};

void postfix::pstfxopers()

{

cout<<"\n\n OUTPUT FOR POSTFIX EXPRESSION \n\n";

cout<<"\n enter the posfix expression at end enter ) \n";

cin>>item;

top=0;

while(item!=')')

{

if((item>='0') && (item<='9'))

st[++top]=item-'0';

else

{

op1=st[top--];

op2=st[top--];

switch(item)

{

case '+': value=op2+op1;

break;

case '-': value=op2-op1;

break;

case '*': value=op2*op1;

break;

case '/': value=op2/op1;

break;

default:

{

cout<<"invalid entry\n";

}

}

st[++top] = value;

}

fflush(stdin);

cin>>item;

}

cout<<"\n value of the entered postfix exp:\n"<<value;

}

void main()

{

clrscr();

postfix p;

p.pstfxopers();

getch();

}

OUTPUT FOR POSTFIX EXPRESSION

enter the posfix expression at end enter )

23*7+45*9

)

value of the entered postfix exp:

20

//11. Program to perform SELECTION sort


#include<conio.h>

class selsort

{

private: int a[20],k,n,i,j,loc,min,temp;

public: void sort();

};

void selsort::sort()

{

cout<<"\n OUT PUT FOR SELECTION SORT\n";

cout<<"enter the number of elements:\n";

cin>>n;

cout<<"enter the " <<n<< " elements\n";

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

cin>>a[i];

for(j=1;j<=n;j++)

{

min=a[j];

loc=j;

for(k=j+1;k<=n;k++)

if(min>a[k])

{

min = a[k];

loc=k;

temp=a[j];

a[j]=a[loc];

a[loc]=temp;

}

}

cout<<"sorted elements are: \n";

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

cout<<"\n"<<a[i];

}

void main()

{

selsort srt;

clrscr();

srt.sort();

getch();

}

OUT PUT FOR SELECTION SORT

enter the number of elements:

5

enter the 5 elements

12 24 16 89 87

sorted elements are:

12

16

24

87

89

//12. Program to perform BUBBLE sort


#include<conio.h>

class bublesrt

{

private: int i,j,n,a[20],temp;


};

void bublesrt:: sort()

{

cout<<" OUTPUT FOR BUBBLE SORT\n";

cout<<"enter the number of elements to sort:\n";

cin>>n;

cout<<"enter "<< n <<"elements in an array:\n";

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

cin>>a[i];

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

for(j=1;j<=n-i;j++)

if(a[j]>a[j+1])

{

temp=a[j];

a[j]=a[j+1];

a[j+1]=temp;

}

cout<<"sorted elements are:\n";

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

cout<<"\n"<<a[i];

}

void main()

{

bublesrt br;

clrscr();

br.sort();

getch();

}

OUTPUT FOR BUBBLE SORT

enter the number of elements to sort:

5

enter 5elements in an array:

12 45 34 67 40


12

34

40

45

67

// 13.Program to perform INSERTION sort


#include<conio.h>

class insersort

{

private: int a[10],ptr,i,n,k,temp;


};

void insersort::sort()

{

cout<<"\n OUT PUT FOR INSERTION SORT\n";

cout<<"enter the number of elements:\n";

cin>>n;

cout<<"enter the "<<n<<" elements:\n";

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

cin>>a[i];

for(i=2;i<=n;i++)

{

temp=a[i];

ptr=i-1;

while(temp<a[ptr])

{

a[ptr+1]=a[ptr];

ptr--;

}

a[ptr+1]=temp;

}

cout<<"sorted elements are:\n";

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

cout<<"\n"<<a[i];

}

void main()

{

insersort isrt;

clrscr();

isrt.sort();

getch();

}

OUT PUT FOR INSERTION SORT

enter the number of elements:

5

enter the 5 elements:

12 45 24 78 99


12

24

45

78

99

// 14.Program to perform RADIX sort


#include<conio.h>

class radixsrt

{

private: int c[20],a[20],d,i,j,k,l,m,n,rem,temp,temp1;

public: void radixsort();

};

void radixsrt:: radixsort()

{

cout<<"\n OUT PUT FOR RADIX SORT\n";

cout<<"enter the value of n:\n";

cin>>n;

cout<<"enter the no.of digits:\n";

cin>>d;

cout<<"enter the "<<d<<" numbers:\n";

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

cin>>a[i];

temp1=1;

for(l=1;l<=d;l++)

{

i=1;

for(j=0;j<=9;j++)

{

for(k=1;k<=n;k++)

{

temp=a[k]/temp1;

rem=temp%10;

if(rem==j)

{

c[i]=a[k];

i++;

}

}

}

}

temp1=temp1*10;

cout<<" at pass "<<l<<endl;

for(m=1;m<=n;m++)

{

a[m]=c[m];

cout<<"\n"<<a[m];

}

}

void main()

{

radixsrt rsrt;

clrscr();

rsrt.radixsort();

getch();

}

OUT PUT FOR RADIX SORT

enter the value of n:

4

enter the no.of digits:

2

enter the 2 numbers:

-25

50

-10

60

at pass 3

50

-10

60

40



Affiliated under

Gulbarga University

Department Of

Computer Science

B.Sc. 3rd

Semester Lab Manual (CBCS Syllabus)

Numerical Computing Lab

INDEX

OBJECT ORIENTED PROGRAMMING

Sl. No Name of the Program

1 C++ program for bisection method

2 C++ program for secant method

3 C++ program for Newton Raphson metthod

4

C++ program for fixed point iteration equation,

x=cosx

5 C++ Program for Newton Gregory Forward

Interpolation Formula

6 C++ program for trapezoidal rule

7 C++ program for picard's method

8 C++ program for Taylor's series

9 C++ program for Euler's Method dy/dx=1+xy,y(0)=2 compute y(0.1) and y(0.2) with h=0, correct up to five decimal places.

10 C++ program for Simpson's 1/3 rd rule

11 C++ program for Simpson's 3/8 th rule

12 C++ program for Runge-Kutta 4th order method to solve differential equation dy/dx=x-y/2.

1. C++ program for bisection method


#include<conio.h>

#include<cmath>

#include<iomanip>

using namespace std;

double f(double x);

double f(double x)

{

double a=pow(x,3)-x-11.0;

return a;

}

int main()

{

cout.precision(4);

cout.setf(ios::fixed);

double a,b,c,e,fa,fb,fc;

a:cout<<"enter the initial guesses:\n";

cout<<"a=";

cin>>a;

cout<<"b=";

cin>>b;

cout<<"\n enter the degree of accuracy desired"<<endl;

cin>>e;

if(f(a)*f(b)>0)

{

cout<<"please enter a different initial guess"<<endl;

goto a;

}

else

{

while(fabs(a-b)>=e)

{

c=(a+b)/2.0;

fa=f(a);

fb=f(b);

fc=f(c);

cout<<"a="<<a<<" "<<"b="<<b<<" "<<"c="<<c<<"fc="<<fc<<endl;

if (fc==0)

{

cout<<"the root of the equetion is"<<c;

break;

}

if(fa*fc>0)

{

a=c;

}

else if(fa*fc<0)

{

b=c;

}

}

}

cout<<"the root of the equation is"<<c;

getch();

}

OUTPUT:

enter the initial guesses:

a=1

b=3

enter the degree of accuracy desired

0.0001

a=1.0000 b=3.0000 c=2.0000fc=-5.0000

a=2.0000 b=3.0000 c=2.5000fc=2.1250

a=2.0000 b=2.5000 c=2.2500fc=-1.8594

a=2.2500 b=2.5000 c=2.3750fc=0.0215

a=2.2500 b=2.3750 c=2.3125fc=-0.9460

a=2.3125 b=2.3750 c=2.3438fc=-0.4691

a=2.3438 b=2.3750 c=2.3594fc=-0.2256

a=2.3594 b=2.3750 c=2.3672fc=-0.1025

a=2.3672 b=2.3750 c=2.3711fc=-0.0406

a=2.3711 b=2.3750 c=2.3730fc=-0.0096

a=2.3730 b=2.3750 c=2.3740fc=0.0059

a=2.3730 b=2.3740 c=2.3735fc=-0.0018

a=2.3735 b=2.3740 c=2.3738fc=0.0021

a=2.3735 b=2.3738 c=2.3737fc=0.0001

a=2.3735 b=2.3737 c=2.3736fc=-0.0009

the root of the equation is2.3736

2. C++ program for secant method


#include<conio.h>

#include<iomanip>

#include<cmath>


double f(double x);

double f(double x)

{

double a=pow(x,3)-x-11.0;

return a;

}

int main()

{

cout.precision(4);


double a,b,c,e;

cout<<"enter the initial guess\na=";

cin>>b;

cout<<"b=";

cin>>c;

cout<<"enter the degree of accuracy\n";

cin>>e;

do

{

a=b;

b=c;

c=b-(b-a)/(f(b)-f(a))*f(b);

if (f(c)==0)

{

cout<<"\n the root of the equation is"<<c;

return 0;

}

}

while(abs(c-b)>=e);

cout<<"\n the root of the equation is"<<c;

getch ();

}

OUTPUT:

enter the initial guess

a=9

b=10

enter the degree of accuracy

0.0001


3. C++ program for Newton Raphson metthod


#include<conio.h>

#include<cmath>

#include<iomanip>


double f(double x);

double f(double x)

{

double a=pow(x,3.0)-x-11.0;

return a;

}

double fprime(double x);

double fprime(double x)

{

double b=3*pow(x,2.0)-1.0;

return b;

}

int main()

{

double x,x1,e,fx,fx1;

cout.precision(4);


cout<<"enter the initial guess:n";

cin>>x1;

cout<<"enter the desired accuracy:n";

cin>>e;

fx=f(x);

fx1=fprime(x);

cout<<"x{i}"<<" "<<"x{i+1}"<<" "<<"|x{i+1}-x{i}|"<<endl;

do

{

x=x1;

fx=f(x);

fx1=fprime(x);

x1=x-(fx/fx1);

cout<<x<<" "<<x1<<" "<<abs(x1-x)<<endl;

}

while (fabs(x1-x)>=e);

cout<<"the root of the equation is"<<x1<<endl;

getch ();

}

OUTPUT:

enter the initial guess:n6

enter the desired accuracy:n0.00001

x{i} x{i+1} |x{i+1}-x{i}|

6.0000 4.1402 1.8598

4.1402 3.0330 1.1072

3.0330 2.5116 0.5214

2.5116 2.3815 0.1301

2.3815 2.3737 0.0078

2.3737 2.3736 0.0000

2.3736 2.3736 0.0000


4. C++ program for fixed point iteration equation, x=cosx


#include<conio.h>

#include<cmath>


double f(double x)

{

return cos(x);

}

int main()

{

double p,p0=1,eps=0.001;

int i=1,n=1000;

while(i<=n)

{

p=f(p0);

if (fabs(p-p0)<eps)

{

cout<<p<<endl;

break;

}

cout<<"iteration"<<i<<":p="<<p<<endl;

i++;

p0=p;

cout<<"the solution is"<<p<<endl;

if(i>n)

{

cout<<"solution not found(method diverges)"<<endl;

break;

}

}

cout<<"the approximation solution is x="<<p<<"in the iteration"<<i-1<<endl;

getch();

}

OUTPUT:

iteration1:p=0.540302

the solution is0.540302































0.73876

the approximation solution is x=0.73876in the iteration16

5. C++ Program for Newton Gregory Forward Interpolation Formula


#include<conio.h>

#define maxN 100

#define order_of_diff 4


main()

{

float arr_x[maxN+1],arr_y[maxN+1];

float numerator=1.0;

float denomenator=1.0;

float x,y,p,h,diff_table[maxN+1][order_of_diff+1];

int i,j,k,n;

cout<<"enter the value of n\n";

cin>>n;

cout<<"enter the value of x& y\n";

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

cin>>arr_x[i]>>arr_y[i];

cout<<"enter the value of x at which value of y is to be calculated";

cin>>x;

h=arr_x[1]-arr_x[0];

for(i=0;i<n-1;i++)

diff_table[i][1]=arr_y[i+1]-arr_y[i];

for(j=2;j<=order_of_diff;j++)

for(i=0;i<n-j;i++)

diff_table[i][j]=diff_table[i+1][j-1]-diff_table[i][j-1];

i=0;

while(!(arr_x[i]>x))

i++;

i--;

p=(x-arr_x[i])/h;

y=arr_y[i];

for(k=1;k<=order_of_diff;k++)

{

numerator*=p-k+1;

denomenator*=k;

y+=(numerator/denomenator)*diff_table[i][k];

}

cout<<"when x="<<x<<"y="<<y;

getch();

OUTPUT:

enter the value of n

5

enter the value of x& y

45 114.84

50 96.16

55 83.32

60 74.48

65 68.48

enter the value of x at which value of y is to be calculated

46

when x=46y=110.526

6.C++ program for trapezoidal rule

#include<iostream.h> #include<conio.h> #include<cmath> using namespace std; float f(float(x)) { return(pow(x,3)+pow(x,2)-(4*x)-5); } float g(float(x)) { return(3*pow(x,2)+2*x-4); } float h(float(x)) { return(6*x+4); } int main() { long double a,b,d,i,n,I=0,J=0,A,K=0,E=0; cout<<"given f(x)=x^3+2x^2-4x-5"<<endl; cout<<"enter lower limit"<<endl; cin>>a; cout<<"enter upper limit"<<endl; cin>>b; cout<<"enter the number of intervals:"<<endl; cin>>n; d=(b-a)/n; for(i=0;i<=n;i++) { I=I+f(a+(i*d)); } for(i=1;i<n;i++) { J=J+f(a+(i*d)); } A=(d/2)*(I+J); cout<<"the value of integral under the limits:"<<endl; cout<<A<<endl; for(i=0;i<=n;i++) { K=K+a+(i*d); } E=-((b-a)*d*d*(h(K/n)/12)); cout<<"the total error is:"<<endl; cout<<E<<endl;

getch(); } OUTPUT:

given f(x)=x^3+2x^2-4x-5 enter lower limit 0 enter upper limit 4 enter the number of intervals: 100 the value of integral under the limits: 33.3408 the total error is: -0.00859733

7.C++ program for picard's method

#include<iostream.h> #include<math.h> #include<conio.h> using namespace std; #define y1(x) (1+(x)+pow(x,2)/2) #define y2(x) (1+(x)+pow(x,2)/2+pow(x,3)/3+pow(x,4)/8) #define y3(x) (1+(x)+pow(x,2)/2+pow(x,3)/3+pow(x,4)/8+pow(x,5)/15+pow(x,6)/48) int main() { double start_value,end_value,allowed_error,temp; double y1[30],y2[30],y3[30]; int count; cout<<"\nstart value:\t"; cin>>start_value; cout<<"\nend value:\t"; cin>>end_value; cout<<"\nallowed error:\t"; cin>>allowed_error; for(temp=start_value,count=0;temp<=end_value;temp=temp+allowed_error,count++) { y1[count]=y1(temp); y2[count]=y2(temp); y3[count]=y3(temp); } cout<<"\nx"; for(temp=start_value;temp<=end_value;temp=temp+allowed_error) { cout<<"\t"<<temp; } cout<<"\n\ny(1)"; for(temp=start_value,count=0;temp<=end_value;temp=temp+allowed_error,count++) { cout<<"\t"<<y1[count]; } cout<<"\n\ny(2)"; for(temp=start_value,count=0;temp<=end_value;temp=temp+allowed_error,count++) { cout<<"\t"<<y2[count]; } cout<<"\n\ny(3)";

for(temp=start_value,count=0;temp<=end_value;temp=temp+allowed_error,count++)

{

cout<<"\t"<<y3[count];

}

getch();

OUTPUT:

start value: 0

end value: 3

allowed error: 0.4

x 0 0.4 0.8 1.2 1.6 2 2.4 2.8

y(1) 1 1.48 2.12 2.92 3.88 5 6.28 7.72

y(2) 1 1.50453 2.34187 3.7552 6.06453 9.66667 15.0352 22.7205

y(3) 1 1.5053 2.36917 3.9833 7.11311 13.1333 24.3249 44.2335

8.C++ program for Taylor's series


#include<cmath>

#include<iomanip>

#include<conio.h>


int main()

{

double x0=0,y0=1,h=0.1,y1,y2,y3,y4,y;

y1=3*x0+y0*y0;

y2=3+2*y0*y1;

y3=2*y1*y1+2*y0*y2;

y4=6*y1*y2+2*y0*y3;

y=y0+(y1*h)+(y2*pow(h,2))/2+(y3*pow(h,3))/6+(y4*pow(h,4))/24;

cout<<"the value of y when x=0.1 is"<<setprecision(5)<<fixed<<y<<endl;

getch();

}

OUTPUT:

the value of y when x=0.1 is1.12722

9.C++ program for Euler's Method dy/dx=1+xy,y(0)=2 compute y(0.1) and y(0.2) with

h=0, correct up to five decimal places.

#include<iostream.h> #include<conio.h> #include<iomanip> #include<cmath> using namespace std; double df(double x,double y) { double a=1+x*y; return a; } int main() { int n; double x0,y0,x,y,h; cout.precision(5); cout.setf(ios::fixed); cout<<"\n enter the initial values of x and y respectively:\n"; cin>>x0>>y0; cout<<"\n for what values of xdo you want to find the value of y\n"; cin>>x; cout<<"\n enter the width of the sub interval(h):\n"; cin>>h; cout<<"x"<<setw(19)<<"y"<<setw(19)<<"dy/dx"<<setw(16)<<"y_new\n"; cout<<"_____________\n"; while(fabs(x-x0)>0.0000001) { y=y0+(h*df(x0,y0)); cout<<x0<<setw(16)<<y0<<setw(16)<<df(x0,y0)<<setw(16)<<y<<endl; y0=y; x0=x0+h; } cout<<x0<<setw(16)<<y<<endl; cout<<"the approximate value of y at x="<<x<<"is"<<y<<endl; getch(); }

OUTPUT:

enter the initial values of x and y respectively: 0 2 for what values of xdo you want to find the value of y 0.1 enter the width of the sub interval(h): 0.1 x y dy/dx y_new _____________ 0.00000 2.00000 1.00000 2.10000 0.10000 2.10000 the approximate value of y at x=0.10000is2.10000

10.C++ program for Simpson's 1/3 rd rule


#include<cmath>

#include<conio.h>


double f(double x)

{

double a=1/(1+x*x);

return a;

}

int main()

{

cout.precision(4);


int n,i;

double a,b,c,h,sum=0,integral;

cout<<"\n enter the limits of integration,\n\n initial limit,a=";

cin>>a;

cout<<"\n final limit,b=";

cin>>b;

cout<<"\n enter the no of subintervals(it should be multiple of 3),\nn=";

cin>>n;

double x[n+1],y[n+1];

h=(b-a)/n;

for(i=0;i<n+1;i++)

{

x[i]=a+i*h;

y[i]=f(x[i]);

}

for(i=1;i<n;i+=2)

{

sum=sum+4.0*y[i];

}

for(i=2;i<n-1;i+=2)

{

sum=sum+2.0*y[i];

}

integral=h/3.0*(y[0]+y[n]+sum);

cout<<"\n the define integral is"<<integral<<"\n"<<endl;

getch();

}

OUTPUT:

enter the limits of integration,

initial limit,a=0

final limit,b=6

enter the no of subintervals(it should be multiple of 3),

n=6

the define integral is1.3662

11.C++ program for simpson's 3/8th rule.


#include<conio.h>

#include<cmath>


double f(double x)

{

double a=1/(1+x*x);

return a;

}

int main()

{

cout.precision(4);


int n,i;

double a,b,c,h,sum=0,integral;

cout<<"\n enter the limits of integration,\n\n initial limit,a=";

cin>>a;

cout<<"\n final limit,b=";

cin>>b;

cout<<"\n enter the no of sub intervals(it should be multiple of 3),\nn=";

cin>>n;

double x[n+1],y[n+1];

h=(b-a)/n;

for(i=0;i<n+1;i++)

{

x[i]=a+i*h;

y[i]=f(x[i]);

}

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

{

if(i%3==0)

sum=sum+2*y[i];

else

sum=sum+3*y[i];

}

integral=3*h/8*(y[0]+y[n]+sum);

cout<<"\n the definite integral is"<<integral<<"\n"<<endl;

getch();

}

OUTPUT:

enter the limits of integration,

initial limit,a=0

final limit,b=6

enter the no of sub intervals(it should be multiple of 3),

n=6

the definite integral is1.3571

12.C++ program for Runge-Kutta 4th order method to solve differential equation

dy/dx=x-y/2.

#include<iostream.h> #include<conio.h> using namespace std; float dydx(float x,float y) { return((x-y)/2); } float rungekutta(float x0,float y0,float x,float h) { int n=(int)((x-x0)/h); float k1,k2,k3,k4,k5,k; float y=y0; for(int i=1;i<=n;i++) { k1=h*dydx(x0,y); k2=h*dydx(x0+0.5*h,y+0.5*k1); k3=h*dydx(x0+0.5*h,y+0.5*k2); k4=h*dydx(x0+h,y+k3); y=y+(1.0/6.0)*(k1+2*k2+2*k3+k4); x0=x0+h; } return y; } int main() { float x0=0,y=1,x=2,h=0.2; cout<<"\n the value of y at"<<x<<"is:"<<rungekutta(x0,y,x,h); getch(); } OUTPUT:

the value of y at2is:1.01971

Documents

Depar tment Of Comp uter Science...Smt. Padmavatibai Raghavendraroa Deshpande Pikalihal Government First Grade College, Mudgal -584125 Affiliated under Gulbarga University Depar tment