Laboratory Manual
PROGRAMING LABORATORY
Second Year - Information Technology
Teaching Scheme
Examination Scheme
Theory : ——
Term Work: 50 Marks
Practical : 04 Hrs/Week
Practical : 50 Marks
Oral :
——
Prepared By
Prof. S. S. Kare
Department of Information Technology
Vidya Pratishthan’s College of Engineering
Baramati – 413133, Dist- Pune (M.S.)
INDIA
June 2013
Table of Contents
1 Set
1.1
1.2
1.3
1.4
1.5
1.6
1.7
operations using array
Problem Statement . . . . . . . . . .
Pre Lab . . . . . . . . . . . . . . . .
Hardware and Software Requirement
Theory . . . . . . . . . . . . . . . . .
Procedure . . . . . . . . . . . . . . .
Post Lab . . . . . . . . . . . . . . . .
Viva Questions . . . . . . . . . . . .
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1
1
1
1
2
2
5
5
2 String operations
2.1 Problem Statement . . . . . . . . . .
2.2 Pre Lab . . . . . . . . . . . . . . . .
2.3 Hardware and Software Requirement
2.4 Theory . . . . . . . . . . . . . . . . .
2.5 Procedure . . . . . . . . . . . . . . .
2.6 Post Lab . . . . . . . . . . . . . . . .
2.7 Viva Questions . . . . . . . . . . . .
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6
6
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7
7
15
15
Structure manipulation
3.1 Problem Statement . . . . . . . . . .
3.2 Pre Lab . . . . . . . . . . . . . . . .
3.3 Hardware and Software Requirement
3.4 Theory . . . . . . . . . . . . . . . . .
3.4.1 Introduction to C structure .
3.5 Procedure . . . . . . . . . . . . . . .
3.6 Post Lab . . . . . . . . . . . . . . . .
3.7 Viva Questions . . . . . . . . . . . .
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16
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16
17
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20
24
25
4 Searching and Sorting I
4.1 Problem Statement . . . . . . . . . .
4.2 Pre Lab . . . . . . . . . . . . . . . .
4.3 Hardware and Software Requirement
4.4 Theory . . . . . . . . . . . . . . . . .
4.5 Procedure . . . . . . . . . . . . . . .
4.6 Post Lab . . . . . . . . . . . . . . . .
4.7 Viva Questions . . . . . . . . . . . .
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26
26
26
26
27
27
30
30
5 Matrix Operations
5.1 Problem Statement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.2 Pre Lab . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.3 Hardware and Software Requirement . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
31
31
31
31
3
i
TABLE OF CONTENTS
5.4
5.5
5.6
5.7
Theory . . . . .
Procedure . . .
Post Lab . . . .
Viva Questions
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TABLE OF CONTENTS
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32
32
32
32
6 Quick Sort
6.1 Problem Statement . . . . . . . . . .
6.2 Pre Lab . . . . . . . . . . . . . . . .
6.3 Hardware and Software Requirement
6.4 Theory . . . . . . . . . . . . . . . . .
6.5 Procedure . . . . . . . . . . . . . . .
6.6 Post Lab . . . . . . . . . . . . . . . .
6.7 Viva Questions . . . . . . . . . . . .
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33
33
33
33
34
34
35
35
7 Sparse Matrix
7.1 Problem Statement . . . . . . . . . .
7.2 Pre Lab . . . . . . . . . . . . . . . .
7.3 Hardware and Software Requirement
7.4 Theory . . . . . . . . . . . . . . . . .
7.5 Procedure . . . . . . . . . . . . . . .
7.6 Post Lab . . . . . . . . . . . . . . . .
7.7 Viva Questions . . . . . . . . . . . .
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36
36
36
36
37
37
38
38
8 Singly Linked List
8.1 Problem Statement . . . . . . . . . .
8.2 Pre Lab . . . . . . . . . . . . . . . .
8.3 Hardware and Software Requirement
8.4 Theory . . . . . . . . . . . . . . . . .
8.5 Procedure . . . . . . . . . . . . . . .
8.6 Post Lab . . . . . . . . . . . . . . . .
8.7 Viva Questions . . . . . . . . . . . .
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39
39
39
39
40
40
43
43
9 Implementation of DLL
9.1 Problem Statement . . . . . . . . . .
9.2 Pre Lab . . . . . . . . . . . . . . . .
9.3 Hardware and Software Requirement
9.4 Theory . . . . . . . . . . . . . . . . .
9.5 Procedure . . . . . . . . . . . . . . .
9.6 Post Lab . . . . . . . . . . . . . . . .
9.7 Viva Questions . . . . . . . . . . . .
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44
44
44
44
45
45
46
46
10 Polynomial Using Circular Linked-List
10.1 Problem statement . . . . . . . . . . . .
10.2 Pre Lab . . . . . . . . . . . . . . . . . .
10.3 Hardware and Software Requirement . .
10.4 Theory . . . . . . . . . . . . . . . . . . .
10.5 Post Lab . . . . . . . . . . . . . . . . . .
10.6 Viva Questions . . . . . . . . . . . . . .
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48
48
48
48
49
51
51
References
52
ii
List of Tables
iii
List of Figures
8.1
SLL . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
40
9.1
9.2
DLL . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
DLL1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
45
45
10.1 Circular list . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
49
iv
Assignment 1
Set operations using array
1.1
Problem Statement
Write a program to perform Set operations using array.
• Union
• Intersection
• Difference
• Symmetric Difference
1.2
Pre Lab
• Concept of Linked-List
• Concepts Of structures
1.3
Hardware and Software Requirement
• Hardware requirement :
– Intel pentium III/M 1.4 GHz, or Pentium IV 1.4 GHz minimum
– Microsoft Windows XP/2000
– 512 MB RAM (1 GB+ recommended)
– 900 MB free hard disk space
1
1.4. THEORY
Set operations using array
• Software Requirement
– Turbo c
1.4
Theory
1. Union: The Union of two sets A and B in the set of all those elements that are present in set A
or in set B in both.
Example: A= {1, 2}
B= {1, 2, 5}
A∪B = {1, 2, 5}
2. Intersection: The intersection of two sets A and B is the set of all those elements that are in set
A and set B. i.e. Common elements of set A and set B.
Example: A= {1, 2}
B= {1, 2, 3}
A∩B = {1, 2}
3. Difference: The elements that are presented in first set but not in second the set is called difference
of first set from second set.
Example: A={ 1,2,3,4}
B={2,3,5}
A-B = {1,4}
4. Symmetric Difference: The Symmetric Difference of the two sets A and B denoted by A⊕B is
defined as
A⊕B = (A∪B) - (A∩B) Example: A= {a, b,e,g}
B ={d,e,f,g} A⊕B = {a,b,d,f}
1.5
Procedure
I.Pseudo code for union of two sets :
int uni(int a[],int b[],int u[],int m,int n)
{
for(i=0 to m)
{
c[i]=a[i];
// copy A set as it is to C set
}
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1.5. PROCEDURE
Set operations using array
for(j=0 to n)
{
f=0;
for(k=0 to m)
{
if(c[k]==b[j])
{
f=1;
break;
}
}
// end of inner for loop
if(f==0)
{
c[m+z]=b[j];
z++;
}
}
//end of outer for loop
Print C Array & return val of m+z .
}
// end of function.
II.Pseudo code for intersection of two sets :
int inter(int a[],int b[],int v[],int m,int n)
{
for(i=0 to m)
{
for(j=0 to n)
{
if(a[i]==b[j])
// check for common element
{
c[k]=a[i];
k++;
}
}
// end of inner for loop
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1.5. PROCEDURE
}
// end of outer for loop
Print
}
Set operations using array
C array & return val. of
k.
//end of function.
III.Pseudo code for difference of two sets :
void diff (int a[],int b[],int c[],int m,int n)
{
k=0,flag=0;
for(i=0 to m)
{
flag=0;
for(j=0 to n)
{
if(a[i]==b[j])
{
flag=1;
break;
}
}
// end of inner for loop
if(flag==0)
{
c[k]=a[i];
k++;
}
}
// end of outer
Print
}
for loop
C array.
// end of function.
IV.Pseudo code for symmetric difference of two sets :
For finding symmetric difference, we have to call difference function once again,
but now we have to pass U array, V array & their size as a parameter.
i.e. diff(U,V,U1,V1)
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1.6. POST LAB
Set operations using array
where U1= size of U array
V1= size of V array.
1.6
Post Lab
Union, intersection , difference and symmetric difference of two sets.
1.7
Viva Questions
1. How set elements can be sotred in computer memory?
2. How symmetric difference is calculated between two sets?
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Assignment 2
String operations
2.1
Problem Statement
Perform following String operations with and without pointers to arrays (without using the library
functions)
1. Substring
2. Palindrome
3. Compare
4. Copy
5. Reverse
2.2
Pre Lab
• Concept of Linked-List
• Concepts Of structures
2.3
Hardware and Software Requirement
• Hardware requirement :
– Intel pentium III/M 1.4 GHz, or Pentium IV 1.4 GHz minimum
– Microsoft Windows XP/2000
– 512 MB RAM (1 GB+ recommended)
6
2.4. THEORY
String operations
– 900 MB free hard disk space
• Software Requirement
– Turbo c
2.4
Theory
In computing, a C string is a character sequence stored as a one-dimensional character array and
terminated with a null character
Basic String Handling Functions:All the string handling functions are prototyped in : #include <string.h>
2.5
Procedure
I.Pseudo code for string length: //String Length//
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2.5. PROCEDURE
String operations
void length(char a[])
{
int i=0;
while(a[i]!=’\0’)
{
i++;
// increment Val of i, up to null character.
}
Print i.
} // end of function.
II.Pseudo code for string reverse: //String Reverse//
void reverse(char a[])
{
int i=0;
while(a[i]!=’\0’)
{
i++;
}
i--;
for(j=i to 0 )
{
printf("%c",a[j]);
}
}
// end of function.
III.Pseudo code for string copy: //String Copy//
void copy(char a[],char b[])
{
int i=0;
while(b[i]!=’\0’)
{
a[i]=b[i];
// copy b array to a array
i++;
}
a[i]=’\0’;
// add null character after copying b array to a array.
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2.5. PROCEDURE
String operations
for(i=0;a[i]!=’\0’;i++)
{
printf("%c",a[i]);
}
}
//end of function
IV. Pseudo code for string concatenate: //String Concatenate//
void concat(char a[],char b[])
{
int i=0,j=0;
while(a[i]!=’\0’)
{
c[i]=a[i];
i++;
}
while(b[j]!=’\0’)
{
c[i+j]=b[j];
j++;
}
c[i+j]=’\0’;
for(i=0;c[i]!=’\0’;i++)
{
printf("%c",c[i]);
}
}
// end of function.
V. Pseudo code for string palindrome: //Palindrome//
void palim(char a[])
{
int i=0,j=0,f=0;
while(a[j]!=’\0’)
{
j++;
}
j--;
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2.5. PROCEDURE
String operations
while(a[i]!=’\0’)
{
if(a[i]!=a[j])
{
f=1;
}
i++;
j--;
}
if(f==0)
{
printf("\n String is palindrome");
}
Else
{
printf("\n String is not palindrome");
}
}
}
// end of function.
VI.Pseudo code for Substring :
void sub(char a[])
{
printf("\n Enter starting number");
scanf("%d",&n1);
printf("\n Enter
Ending number");
scanf("%d",&n2);
for (i=n1;i<=n2;i++)
{
Print a[i]
}
}
// end of function.
VII.Pseudo code for String compare:
//Comparision//
void comp(char a[],char b[])
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2.5. PROCEDURE
String operations
{
int i=0,j=0,f=0;
while (a[i]!=’\0’ || b[j]!=’\0’)
{
if(a[i]!=b[j])
{
f=1;
// if any character is not same, set flag to 1.
break;
}
i++;
j++;
}
if (f==0)
{
print Strings are same.
}
Else
print Strings are not same.
}
// end of function.
VIII.Pseudo code for String length using pointer:
Let input is given as a string, p is a pointer as, Set p=&a[0];
//String Length//
void length(char *p)
{
int i=0;
while(*p!=’\0’)
{
p++;
i++;
}
print Val of i.
}
// end of function.
IX.Pseudo code for String reverse using pointer:
Let input is given as a string, p is a pointer as, Set p=&a[0];
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2.5. PROCEDURE
String operations
//String Reverse//
void reverse(char *p)
{
int i=0,j;
while(*p!=’\0’)
{
p++;
i++;
}
p--;
i--;
for(j=i;j>=0;j--)
{
printf("%c",*p);
p--;
}
}
// end of function.
X. Pseudo code for String concatenate using pointer: Let input is given as a string & b string,
p & q are pointers as, Set p=&a[0] , q=&b[0];
//String Concatenate//
void concat(char *p,char *q)
{
char c[100],*r;
r=&c[0];
while(*p!=’\0’)
{
*r=*p;
r++;
p++;
}
while(*q!=’\0’)
{
*r=*q;
r++;
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2.5. PROCEDURE
String operations
q++;
}
*r=’\0’;
printf("%s",c);
}
XI. Pseudo code for String palindrome using pointer: Let input is given as a string, p is a
pointer as, Set p=&a[0];
//Palindrome//
void palim(char *p)
{
int f=0;
char *r;
*r=*p;
while(*r!=’\0’)
{
r++;
// move pointer r up to last character, & set p to first character.
}
r--;
while(*p!=’\0’)
{
if(*p!=*r)
{
f=1;
}
p++;
r--;
}
if(f==0)
{
printf("\n String is palimdrome");
}
Else
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2.5. PROCEDURE
String operations
{
printf("\n String is not palimdrome");
}
}
// end of function.
XII.Pseudo code for finding sbstring using pointer: Let input is given as a string, p is a
pointer as, Set p=&a[0];
//SUBSTRING//
void sub(char *a)
{
int n1,n2,i;
printf("\n Enter starting no. of string:");
scanf("%d",&n1);
printf("\n Enter ending no.of string:");
scanf("%d",&n2);
a=&a[n1];
for(i=n1;i<=n2;i++)
{
printf("%c",*p);
p++;
}
}
// end of function.
XIII.Pseudo code for String comparison using pointer: Let input is given as a string & b
string, p & q are pointers as, Set p=&a[0] , q=&b[0];
//Comparison//
void comp(char *p,char *q)
{
int i=0,j=0,f=0;
while (*p!=’\0’ || *q!=’\0’)
{
if(*p!=*q)
{
f=1;
break;
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2.6. POST LAB
String operations
}
p++;
q++;
}
if (f==0)
{
printf("\n String are same");
}
Else
printf("\n String are not same");
}
2.6
// end of function.
Post Lab
String operations like length, palindrome of string, comparison,copy ,reverse, substring etc.
2.7
Viva Questions
1. What is a mean by palindrome of a string ?
2. Define string ?
3. Where string comparison operation can be used ?
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Assignment 3
Structure manipulation
3.1
Problem Statement
Create a database using array of structures and perform following operations on it:
1. Create database
2. Display database
3. Add record
4. Search record
5. Modify a record
6. delete a record
3.2
Pre Lab
• Concepts Of structures
• Concept of Array
3.3
Hardware and Software Requirement
• Hardware requirement :
– Intel pentium III/M 1.4 GHz, or Pentium IV 1.4 GHz minimum
– Microsoft Windows XP/2000
16
3.4. THEORY
Structure manipulation
– 512 MB RAM (1 GB+ recommended)
– 900 MB free hard disk space
• Software Requirement
– Ubuntu 10.04
3.4
Theory
3.4.1
Introduction to C structure
In some programming contexts, you need to access multiple data types under a single name for easy
manipulation; for example you want to refer to address with multiple data like house number, street,
zip code, country C supports structure which allows you to wrap one or more variables with different
data types. A structure can contain any valid data types like int, char, float even arrays and other
structures. Each variable in structure is called a structure member.
Defining structure:- To define a structure, you can use struct keyword. Here is the common syntax
of structure definition:
struct struct_name
{
structure_member
};
The name of structure is followed the rule of variable name. Here is an example of defining address
structure:
struct address
{
unsigned int house_number;
char street_name[50];
int zip_code;
char country[50];
};
It contains house number as an positive integer, street name as a string, zip code as an integer and
country as a string.
Declaring structure:The above example only defines an address structure without create any instance of it. To create or
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3.4. THEORY
Structure manipulation
declare a structure you have two ways:
The first way is declare a structure follows by structure definition like this :
struct struct_name
{
structure_member;
...
} instance_1,instance_2 instance_n;
In the second way you can declare your structure instance at a different location in your source code
after structure definition. Here is structure declaration syntax :
struct struct_name instance_1 instance_2 instance_n;
Complex structure:Structure can contains arrays or other structures so it is sometimes called complex structure. For
example address structure is a complex structure. We can define a complex structure which contains
address structure as follows:
struct customer
{
char name[50];
structure address billing_addr;
structure address shipping_addr;
};
accessing structure member:- To access structure members we can use dot operator (.) between
structure name and structure member name as follows:
structure name. structure member
For example to access street name of structure address we do as follows:
struct address billing_addr;
billing_addr.country = "US";
If the structure contains another structure, we can use dot operator to access nested structure and use
dot operator again to access variables of nested structure.
struct customer jack;
jack.billing_addr.country = "US";
Initializing structure:- C programming language treats a structure as a custom data type therefore
you can initialize a structure like a variable. Here is an example of initialize product structure:
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3.4. THEORY
Structure manipulation
struct product
{
char name[50];
double price;
} book = { "C programming language",40.5};
In above example, we define product structure, then we declare and initialize book structure with its
name and price.
Structure and pointer:- A structure can contain pointers as structure members and we can create
a pointer to a structure as follows:
struct invoice
{
char* code;
char date[20];
};
struct address billing_addr;
struct address *pa = &billing_addr;
Shorthand structure with typedef keyword:- To make your source code more concise, you can
use typedef keyword to create a synonym for a structure. This is an example of using typedef keyword
to define address structure so when you want to create an instance of it you can omit the keyword struct
typedef struct
{
unsigned int house_number;
char street_name[50];
int zip_code;
char country[50];
} address;
address billing_addr;
address shipping_addr;
In assignment there is Structure manipulation (for any database like Employee or Bank database) with
and without pointers to structures.
Here we declare structure as,
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3.5. PROCEDURE
Structure manipulation
struct Employee
{
int id;
char name[100];
int basic;
}emp[100];
3.5
Procedure
I. Pseudo code for create:
void create()
{
int i=0;
Do
{
printf("\nEnter Employee id,name,basic:");
scanf("%d%s%d",&emp[i].id,emp[i].name,&emp[i].basic);
printf("\n Add more record");
ans=getche();
i++;
count++;
}while(ans==’y’);
}
// end of function.
II.Pseudo code for display:
void display()
{
for(i=0 to count )
{
printf("\nEmployee Id:%d\t\tEmployee Name:%s\t\tEmployee
Basic:%d",emp[i].id,emp[i].nam
}
}
// end of function.
III.Pseudo code for search:
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3.5. PROCEDURE
Structure manipulation
void search()
{
printf("\n Enter the id u want to search");
scanf("%d",&key);
for(i=0 to count )
{
if(key==emp[i].id)
{
printf("\n %d Id is present ",key);
break;
}
// end of if.
}
// end of for.
if(i==count)
printf("\nId is not present in record");
}
// end of function.
IV.Pseudo code for modify:
Void modify()
{
printf("\n Enter the id u modify");
scanf("%d",&key);
for(i=0 to count )
{
if(key==emp[i].id)
{
printf("\n Enter the Id to b modified ");
scanf("%d",&emp[i].id);
printf("\n Enter the name to b modified ");
scanf("%s",emp[i].name);
printf("\n Enter the basic salary to b modified ");
scanf("%d",&emp[i].basic);
}
}
// end of if.
// end of for.
}
// end of function.
V.Pseudo code for delete:
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3.5. PROCEDURE
Structure manipulation
void del()
{
int f=0,key,i,j=0;
printf("\n Enter the id u want delete");
scanf("%d",&key);
for(i=0 to count )
{
if(key==emp[i].id)
{
for( j=I to count - 1
)
{
emp[j].id=emp[j+1].id;
emp[j].basic=emp[j+1].basic;
strcpy(emp[j].name,emp[j+1].name);
}
// end of for.
Count--;
}
// end of if.
}
// end of outer for.
if(i==count)
{
printf("\n record not present");
}
give call to display function.
} // end of function.
VI. Pseudo code for create with pointer:
void create()
{
char ans;
p=&emp[0];
Do
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3.5. PROCEDURE
Structure manipulation
{
printf("\nEnter Employee id,name,basic:");
scanf("%d%s%d",&p->id,p->name,&p->basic);
p++;
count++;
printf("\n Add more record");
ans=getche();
}while(ans==’y’);
}
// end of function.
VII.Pseudo code for display with pointer:
void display()
{
int i;
p=&emp[0];
for(i=0;i<count;i++)
{
printf("\nEmployee Id:%d\t\tEmployee Name:%s\t\tEmployee
Basic:%d",p->i
p++;
}
}
// end of for.
// end of function.
VIII.Pseudo code for search with pointer:
void search()
{
int key,i;
p=&emp[0];
printf("\n Enter the id u want to search");
scanf("%d",&key);
for(i=0;i<count;i++)
{
if(key==p->id)
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3.6. POST LAB
Structure manipulation
{
printf("\n %d Id is present ",key);
break;
}
// end of if.
p++;
}
// end of for.
if(i==count)
printf("\nId is not present in record");
}
// end of function.
IX.Pseudo code for modify with pointer:
void modify()
{
int key,i;
p=&emp[0];
printf("\n Enter the id u modify");
scanf("%d",&key);
for(i=0;i<count;i++)
{
if(key==p->id)
{
printf("\n Enter the Id to b modified ");
scanf("%d",&p->id);
printf("\n Enter the name to b modified ");
scanf("%s",p->name);
printf("\n Enter the basic salary to b modified ");
scanf("%d",&p->basic);
}
// end of if.
p++;
}
// end of for.
}
3.6
// end of function.
Post Lab
Creation of database using array of structures, addition of record, deletion of record and modification
of record
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3.7. VIVA QUESTIONS
3.7
Structure manipulation
Viva Questions
1. What is a structure?
2. How structure size is calculated?
3. what is difference between structure and union?
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Assignment 4
Searching and Sorting I
4.1
Problem Statement
Accept student information (e.g. Roll No, Name, Percentage etc.)
1. Display the data in descending order of Percentage (Bubble Sort)
2. display the data in descending order of percentage (selection sort)
3. Search for particular string using binary search
4.2
Pre Lab
• Concept Array
• Concepts Of structures
4.3
Hardware and Software Requirement
• Hardware requirement :
– Intel pentium III/M 1.4 GHz, or Pentium IV 1.4 GHz minimum
– Microsoft Windows XP/2000
– 512 MB RAM (1 GB+ recommended)
– 900 MB free hard disk space
• Software Requirement
– Ubuntu 10.04
26
4.4. THEORY
4.4
Searching and Sorting I
Theory
1. Bubble Sort: The basic idea underlying the bubble sort is to compare the adjacent elements and
moves the largest element to the bottom of the list.
2. Selection Sort: A selection sort is one in which we successively select the smallest element, second
smallest element, third smallest element, and so on, and place them into their proper sorted
positions.
3. Binary Search: In binary searching we compare the item with the element in the center of the list
and then restrict our attention to only the first or second half of the list, depending on whether
the ’item’ comes before or after the central one. If the ’item’ is in the first half of the list, the
procedure is repeated on the first of the list. In this way at each step we reduce the length of the
list to be searched by half. And this process continues until either the required ’item’ is found or
the search intervals become empty. E.g. target item = 123.
4.5
Procedure
I. SelectionSort (int a [], int n)
{
int i,j,temp;
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4.5. PROCEDURE
Searching and Sorting I
for (i=0; i<n-1;i++)
{
for (j=i+1;j<n;j++)
{
if (a[i]> a[j])
{
temp =a[i];
a [i]=a[j];
a [j]= temp;
}
}
}
}
II.BubbleSort (int a [], int n)
{
int i,j,t,flag;
flag=1;
for (i=0; i<n && flag= =1;i++)
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4.5. PROCEDURE
Searching and Sorting I
{
flag=0;
for (j=0;j<n-i-1;j++)
{
if (a[j]> a[j+1])
{
flag=1;
t=a[j];
a [j]=a[j+1];
a[j+1]= t;
}
}
}
}
III.Binary Search (int a [], int n, int item)
int mid, lower, upper;
lower= 0;
upper= n-1;
while (lower<= upper)
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4.6. POST LAB
Searching and Sorting I
{
mid = (lower+ upper)/2;
if (a[mid] = = item)
return (mid);
else if ( a[mid]> item)
upper =mid -1;
else
lower = mid +1;
}
return (-1);
}
4.6
Post Lab
Sorting of database using various sorting techniques.
4.7
Viva Questions
1. What is mean by sorting ?
2. What is time complexity ?
3. Write the complexity of bubble sort and selection sort.
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Assignment 5
Matrix Operations
5.1
Problem Statement
Implement following Matrix operations:
1. Addition with pointers to arrays
2. Multiplication without pointers to arrays
3. Transpose with pointers to arrays
4. Saddle point without pointers
5.2
Pre Lab
• Concept of Linked-List
• Concepts Of structures
5.3
Hardware and Software Requirement
• Hardware requirement :
– Intel pentium III/M 1.4 GHz, or Pentium IV 1.4 GHz minimum
– Microsoft Windows XP/2000
– 512 MB RAM (1 GB+ recommended)
– 900 MB free hard disk space
31
5.4. THEORY
Matrix Operations
• Software Requirement
– Ubuntu 10.04
5.4
Theory
5.5
Procedure
5.6
Post Lab
Addition, Multiplication and Evaluation of polynomial using circular linked list.
5.7
Viva Questions
1. a
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Assignment 6
Quick Sort
6.1
Problem Statement
Write a program to implement quick sort on a list of numbers.
6.2
Pre Lab
• Concept of Array
• Concepts Of quick sorting technique
6.3
Hardware and Software Requirement
• Hardware requirement :
– Intel pentium III/M 1.4 GHz, or Pentium IV 1.4 GHz minimum
– Microsoft Windows XP/2000
– 512 MB RAM (1 GB+ recommended)
– 900 MB free hard disk space
• Software Requirement
– Ubuntu 10.04
33
6.4. THEORY
6.4
Quick Sort
Theory
Quicksort sorts by employing a divide and conquer strategy to divide a list into two sub-lists. The steps
are:
1. Pick an element, called a pivot, from the list.
2. Reorder the list so that all elements with values less than the pivot come before the pivot, while all
elements with values greater than the pivot come after it (equal values can go either way). After this
partitioning, the pivot is in its final position. This is called the partition operation.
3. Recursively sort the sub-list of lesser elements and the sub-list of greater elements.
6.5
Procedure
I.void quicksort(int list[],int m,int n)
int key,i,j,k;
if( m < n)
{
k = choose_pivot(m,n);
swap(&list[m],&list[k]);
key = list[m];
i = m+1;
j = n;
while(i <= j)
{
while((i <= n) && (list[i] <= key))
i++;
while((j >= m) && (list[j] > key))
j--;
if( i < j)
swap(&list[i],&list[j]);
}
// swap two elements
swap(&list[m],&list[j]);
// recursively sort the lesser list
quicksort(list,m,j-1);
quicksort(list,j+1,n);
}
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6.6. POST LAB
Quick Sort
}
II.void printlist(int list[],int n)
int i;
for(i=0;i<n;i++)
printf("%d\t",list[i]);
}
6.6
Post Lab
Sorting of numbers in ascending or descending order
6.7
Viva Questions
1. What is a time complexity of quick sort.
2. What is a recursive function.
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Assignment 7
Sparse Matrix
7.1
Problem Statement
Represent sparse matrix using two dimensional array or structure and perform simple and fast transpose.
7.2
Pre Lab
• Concept of two dimensional arrays
• Concepts Of Structures
7.3
Hardware and Software Requirement
• Hardware requirement :
– Intel pentium III/M 1.4 GHz, or Pentium IV 1.4 GHz minimum
– Microsoft Windows XP/2000
– 512 MB RAM (1 GB+ recommended)
– 900 MB free hard disk space
• Software Requirement
– Turbo c
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7.4. THEORY
Sparse Matrix
7.4
Theory
7.5
Procedure
I.Simple Transpose Algorithm: Procedure TRANSPOSE (A,B)
// A is a matrix represented in sparse form
// B is set to be its transpose
1 (m,n,t)
(A(0,1),A(0,2),A(0,3))
2 (B(0,1),B(0,2),B(0,3))
3 if t
4 q
(n,m,t)
0 then return
1
5 for col
6 for p
1 to n do
1 to t do
7 if A(p,2) = col
8 then [(B(q,1),B(q,2),B(q,3))
9
(A(p,2),A(p,1),A(p,3))
10 q
q+1]
11 end
12 end
13 end of TRANSPOSE
Time complexity of the algorithm is O(nt).
This time complexity becomes O(n2m) when t is of the order nm. This is worse than the O(nm) time
using arrays.
II.Fast Transpose Algorithm: Procedure FAST-TRANSPOSE (A,B) //A is an array representing a
sparse matrix m n with t nonzero elements. The transpose is stored in B using only O(t+n) operations
//
declare S(1:n),T(1:n);
1 (m,n,t)
(A(0,1),A(0,2), A(0,3))
2 (B(0,1),B(0,2), B(0,3))
3 if t
(n,m,t)
0 then return
4 for i
1 to n do S(i)
5 for i
1 to t do
6 S(A(i,2))
0 end
S(A(i,2)) +1
7 end
8 T(1)
1
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7.6. POST LAB
9 for i
1 to n do
10 T(i)
T(i-1) +S(i-1)
Sparse Matrix
11 end
12 for i
13 j
1 to t do
A(i,2)
14 (B(T(j),1), B(T(j),2) ,B(T(j),3))
15 (A(i,2), A(i,1), A(i,3))
16 T(j)
T(j) + 1
17 end
18 end FAST_TRANSPOSE
Time complexity of fast transpose algorithm is O(n+t).
This computing time becomes O(nm) when t is of the order of nm.
7.6
Post Lab
Simple and fast transpose operations.
7.7
Viva Questions
1. How fast transpose is calculated ?
2. What is a sparse matrix .
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Assignment 8
Singly Linked List
8.1
Problem Statement
Problem Statement: Create a singly linked list with options:
a. insert (at front, at end, in the middle),
b. delete (at front, at end, in the middle),
c. Display,
d. Display Reverse
8.2
Pre Lab
• Concept of Linked-List
• Concepts Of structures
8.3
Hardware and Software Requirement
• Hardware requirement :
– Intel pentium III/M 1.4 GHz, or Pentium IV 1.4 GHz minimum
– Microsoft Windows XP/2000
– 512 MB RAM (1 GB+ recommended)
– 900 MB free hard disk space
• Software Requirement
– Ubuntu 10.04
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8.4. THEORY
8.4
Singly Linked List
Theory
Declare a structure to represent a node in the SLL
typedef struct link
int data;
struct link next;
node;
Create the head node
head = (node) malloc (sizeof(node));
head→data=-99;
head→next=NULL;
Declare the menu
Read the choice & switch according to that
Pass the head node as a parameter to each function call
Figure 8.1: SLL
8.5
Procedure
I.Pseudo code for insert a node:
void insert(node *head)
{
node *t,*temp;
t=(node*)malloc(sizeof(node));
printf("\nEnter the new node data:");
scanf("%d",&t->data);
t->next=NULL;
temp=head;
while(temp->next!=NULL)
{
temp=temp->next;
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8.5. PROCEDURE
Singly Linked List
}
temp->next=t;
}
// end of function.
II.Pseudo code for display :
void display(node *head)
{
node *temp;
temp=head->next;
while(temp!=NULL)
{
printf("\t%d",temp->data);
temp=temp->next;
}
}
// end of function.
III.Pseudo code for delete a node:
void del(node *head)
{
node *temp,*t;
printf("\nEnter the element which is to be delete");
scanf("%d",&c);
temp=head;
while(temp->next!=NULL)
{
if(temp->next->data==c)
{
t=temp->next;
temp->next=temp->next->next;
free(t);
printf("\nData is deleted");
}
Else
{
temp=temp->next;
}
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8.5. PROCEDURE
Singly Linked List
}
display(head);
}
// end of function.
IV.Pseudo code for reverse:
void reverse(node *head)
{
node *p,*q,*r;
p=head->next;
q=NULL;
while(p!=NULL)
{
r=q;
q=p;
p=p->next;
q->next=r;
}
head->next= q;
} // end of function.
V.Pseudo code for revert:
void revert(node *head)
{
node *p,*q,*r;
p=head->next;
q=NULL;
while(p!=NULL)
{
r=q;
q=p;
p=p->next;
q->next=r;
}
head->next=q;
display(head);
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8.6. POST LAB
Singly Linked List
} // end of function.
8.6
Post Lab
Operations on single link list: addition, deletion, reverse and display
8.7
Viva Questions
1. What is the difference between SLL and DLL?
2. What are the applications of SLL.
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Assignment 9
Implementation of DLL
9.1
Problem Statement
Implement DLL and perform following operations:
• Insert
• Delete
• Modify
• Forward
• Backward
9.2
Pre Lab
• Concept of Linked-List
• Concepts Of structures
9.3
Hardware and Software Requirement
• Hardware requirement :
– Intel pentium III/M 1.4 GHz, or Pentium IV 1.4 GHz minimum
– Microsoft Windows XP/2000
– 512 MB RAM (1 GB+ recommended)
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9.4. THEORY
Implementation of DLL
– 900 MB free hard disk space
• Software Requirement
– Ubuntu 10.04
9.4
Theory
In computer science, a doubly-linked list is a linked data structure that consists of a set of data records,
each having two special link fields that contain references to the previous and to the next record in the
sequence. It can be viewed as two singly-linked lists formed from the same data items, in two opposite
orders.
Figure 9.1: DLL
The two links allow walking along the list in either direction with equal ease. Compared to a singlylinked list, modifying a doubly-linked list usually requires changing more pointers, but is sometimes
simpler because there is no need to keep track of the address of the previous node. In computer science,
a doubly-linked list is a linked data structure that consists of a set of data records, each having two
special link fields that contain references to the previous and to the next record in the sequence. It can
be viewed as two singly-linked lists formed from the same data items, in two opposite orders.
Figure 9.2: DLL1
The two links allow walking along the list in either direction with equal ease. Compared to a singlylinked list, modifying a doubly-linked list usually requires changing more pointers, but is sometimes
simpler because there is no need to keep track of the address of the previous node
9.5
Procedure
I.Insert a Node:
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9.6. POST LAB
Implementation of DLL
function insertAfter(List list, Node node, Node newNode)
newNode.prev := node
newNode.next := node.next
if node.next == null
list.lastNode := newNode
Else
node.next.prev := newNode
node.next := newNode
function insertBefore(List list, Node node, Node newNode)
newNode.prev := node.prev
newNode.next := node
if node.prev is null
list.firstNode := newNode
Else
node.prev.next := newNode
node.prev
:= newNode
II.Remove Node:
function remove(List list, Node node)
if node.prev == null
list.firstNode := node.next
Else
node.prev.next := node.next
if node.next == null
list.lastNode := node.prev
Else
node.next.prev := node.prev
destroy node
9.6
Post Lab
Insertion ,deletion, modification and display DLL
9.7
Viva Questions
1. What is a difference between DLL and SLL?
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9.7. VIVA QUESTIONS
Implementation of DLL
2. What is a node structure for DLL ?
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Assignment 10
Polynomial Using Circular
Linked-List
10.1
Problem statement
Represent single variable polynomial as a circular linked list. Accept the terms in the polynomial in
any order, i.e. not necessarily in the decreasing order of exponent. Sort while creating polynomial
in the decreasing order of exponent and write a menu driven program to perform display, addition,
multiplication and evaluation.
10.2
Pre Lab
• concept of Linked-List
• Concepts Of structures
10.3
Hardware and Software Requirement
• Hardware requirement :
– Intel pentium III/M 1.4 GHz, or Pentium IV 1.4 GHz minimum
– Microsoft Windows XP/2000
– 512 MB RAM (1 GB+ recommended) 900 MB free hard disk space
– SVGA or higher-resolution monitor (XGA recommended)
• Software Requirement
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10.4. THEORY
Polynomial Using Circular Linked-List
– Turbo c
10.4
Theory
Circular lists:
In some situations, a circular list is needed in which nodes form a ring. Here each node has a successor.
An example of such a situation is when several processes are using the same computer resource for
the same amount of time, and we have to assure that no process accesses the resource before all other
processes did. A process is may be a program in execution. The processes are put on a circular list, and
control passes to next one when any of the nodes (processes) is serviced.
A circular list is accessible through one of its nodes, usually called the tail. Note that unlike standard
linked lists, there is no node with NULL pointer in a circularly linked list.
Figure 10.1: Circular list
Structure used for polynomial representation using CLL is,
typedef struct node
{
int coeff;
int exp;
struct node *next;
}node;
Algorithm:
Steps for creating polynomial:
1. create new node head give its link to itself. i.e. head -¿next=head. (first node on list)
2. Enter coefficient and exponent for new term .insert this term in a list by using insert function.
3. Repeate step 2 till you want to add terms in polynomial.
4. Stop.
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10.4. THEORY
Polynomial Using Circular Linked-List
Steps for inserting node in the list:
1. while inserting new term on polynomial check all the nodes previously present on list. If exponent
of new term is matching with exponent of any node(temp) on list then just add the coefficient of
new term in the coefficient of temp.and don’t create new node.
2. If coefficient of new term is not matching with any node present on list then Then create new node
with new coeff and expo .
3. Add new node at the end of list.
4. return starting ponter of list to create function.
5. stop.
Steps for addition of two polynomials:
1. head1 is starting pointer for polynomial 1 and head2 is starting pointer for polynomial 2.
2. Create a node head3 as a starting node of third polynomial.
3. consider first term from first polynomials (term1 ) an first term from second polynomial. (term2)
If exponents are equal then add coeff and keep expo same and add this insert this node in list 3
(third Poly) by using insert function.
If expo of term1 is greater then insert term1 in list3 else copy term 2 in List3.
4. Repeate this for all the terms from poly1 and poly2 .
5. stop.
Steps for multiplication of two polynomials:
Initially I=0 , j=0
1. Consider ith term from list1 multiply coeff of that term with jth term in list 2 add coeff of these
two terms .Now with this coeff and exponent add new term in list three by using insert function.
2. consider next term in list2 (j+1)
3. Repeate step1 and 2 for all terms in list 2.
4. consider next term in list 1.( i+1)
5. Repeat step 2 3 and 4 for all terms in list 1.
Steps for evaluation of a polynomial:
J=0 ,ans=0;
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10.5. POST LAB
Polynomial Using Circular Linked-List
1. Enter value of x;
2. Consider Jth term of list1 and multiply its coefficient with power of list1 consider its coeff is cf
and exponent is ex.
3. Now multiply cf with xex add this answer in previous ans. Increment j
4. Repeat step 2 and 3 for all the terms in list1.
5. return final answer to main function
6. Stop.
10.5
Post Lab
Addition, Multiplication and Evaluation of polynomial using circular linked list.
10.6
Viva Questions
1. What is Circular Linked-List?
2. What is a Difference between CLL, DLL and SLL?
3. What are the advantages of CLL over SLL?
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References
[1] B Forouzan, ”Data Structures:A pseudo code approach with C”, Cenage Learning,.
[2] Yashwant Kanetkar, ”Let us C” and ”Pointers in C”, BPB Publication.
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