Basics of Data structure using C describing basics concepts

CHAPTER 2 1
Structures (records)
struct {
char name[10];
int age;
float salary;
} person;
strcpy(person.name, “james”);
person.age=10;
person.salary=35000;

CHAPTER 2 2
Create structure data type
typedef struct human_being {
char name[10];
int age;
float salary;
};
or
typedef struct {
char name[10];
int age;
float salary
} human_being;
human_being person1, person2;

CHAPTER 2 3
Unions
Similar to struct, but only one field is active.
Example: Add fields for male and female.
typedef struct sex_type {
enum tag_field {female, male} sex;
union {
int children;
int beard;
} u;
};
typedef struct human_being {
char name[10];
int age;
float salary;
date dob;
sex_type sex_info;
}
human_being person1, person2;
person1.sex_info.sex=male;
person1.sex_info.u.beard=FALSE;

CHAPTER 2 4
Self-Referential Structures
One or more of its components is a pointer to itself.
typedef struct list {
char data;
list *link;
}
list item1, item2, item3;
item1.data=‘a’;
item2.data=‘b’;
item3.data=‘c’;
item1.link=item2.link=item3.link=NULL;
Construct a list with three nodes
item1.link=&item2;
item2.link=&item3;
malloc: obtain a node
a b c

CHAPTER 2 5
Ordered List Examples
 (MONDAY, TUEDSAY, WEDNESDAY,
THURSDAY, FRIDAY, SATURDAYY,
SUNDAY)
 (2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen,
King,
Ace)
 (1941, 1942, 1943, 1944, 1945)
 (a1, a2, a3, …, an-1, an)
ordered (linear) list: (item1, item2, item3, …, itemn)

CHAPTER 2 6
Operations on Ordered List
 Find the length, n , of the list.
 Read the items from left to right (or right to left).
 Retrieve the i’th element.
 Store a new value into the i’th position.
 Insert a new element at the position i , causing
elements numbered i, i+1, …, n to become numbered
i+1, i+2, …, n+1
 Delete the element at position i , causing elements
numbered i+1, …, n to become numbered i, i+1, …,
n-1 array (sequential mapping)? (1)~(4) O (5)~(6) X

Arrays
7
Outline
1. Introduction
2. Arrays
3. Declaring Arrays
4. Examples Using Arrays
5. Representation of arrays in memory
6. Dynamically allocated arrays
7. Array operations
1. Traversing
2. Inserting
3. Deleting
4. Searching
5. Sorting
8. Multidimensional arrays
9. Case Study: polynomials and sparse matrices

Introduction
• Arrays
– Structures of related data items
– Static entity – same size throughout program
– Dynamic data structures can also be created
8

Arrays
9
A linear array is a list of a finite number of homogeneous
data elements (that is data elements of the same type) such
that:
1. The elements of the array are referenced respectively by
an index of n consecutive numbers
2. The elements of the array are stored respectively in
successive memory locations

Arrays
• Array
– Group of consecutive memory locations
– Same name and type
• To refer to an element, specify
– Array name
– Position number
• Format:
arrayname[ position number ]
– First element at position 0
– n element array named c:
• c[ 0 ], c[ 1 ]...c[ n – 1 ]
10
Name of array
(Note that all
elements of this
array have the
same name, c)
Position number
of the element
within array c
c[6]
-45
6
0
72
1543
-89
0
62
-3
1
6453
78
c[0]
c[1]
c[2]
c[3]
c[11]
c[10]
c[9]
c[8]
c[7]
c[5]
c[4]

Arrays
• Array elements are like normal variables
c[ 0 ] = 3;
printf( "%d", c[ 0 ] );
– Perform operations in subscript. If x equals 3
c[ 5 - 2 ] == c[ 3 ] == c[ x ]
11

Declaring Arrays
• When declaring arrays, specify
– Type
– Name
– Number of elements
arrayType arrayName[ numberOfElements ];
– Examples:
int c[ 10 ];
float myArray[ 3284 ];
• Declaring multiple arrays of same type
– Format similar to regular variables
– Example:
int b[ 100 ], x[ 27 ];
12

Examples Using Arrays
• Initializers
int n[ 5 ] = { 1, 2, 3, 4, 5 };
– If not enough initializers, rightmost elements become
0
int n[ 5 ] = { 0 }
• All elements 0
– If too many a syntax error
– C arrays have no bounds checking
• If size omitted, initializers determine it
int n[ ] = { 1, 2, 3, 4, 5 };
– 5 initializers, therefore 5 element array
13

Length of the array
The length of an array is given by the number of elements
stored in it.
The general formula to calculate the length of an array is
Length = upper_bound – lower_bound + 1
where upper_bound - is the index of the last element
lower_bound - is the index of the first element
14

15
Ex: Let Age[5] be an array of integers such that
Age[0] = 2, Age[1] = 5, Age[2] = 3, Age[3] = 1, Age[4] = 7
Show the memory representation of the array and calculate
its length.

ACCESSING THE ELEMENTS OF AN ARRAY
• To access all the elements, we must use a loop.
• subscript must be an integral value or an expression
that evaluates to an integral value.
16

Calculating the Address of an Array Elements
17
Address of data element, A[k] = BA(A) + w(k – lower_bound)
A -is the array
K - is the index of the element of which we have to calculate the
address,
BA – is the base address of the array A,
w - is the size of one element in memory, Ex: size of int is 2.
Given an array int marks[] = {99,67,78,56,88,90,34,85}, calculate the
address of marks[4] if the base address = 1000.
marks[4] = 1000 + 2(4 – 0)
= 1000 + 2(4) = 1008

Example-2 calculating address
An automobile company uses an array AUTO to record the
number of auto-mobiles sold each year from 1932 to 1984.
suppose Auto appears in memory at location 200 and w=4
words. Find the address of the array element for year k=1965
18
The address of the array element for the year k=1965 can
be obtained :
Loc(Auto[1932]) = Base (Auto)+w (1965 - lower bound)
= 200 + 4 (1965-1932)= 332

20
Input values for the element from the key board
Assign values to individual elements

OPERATIONS ON ARRAYS
21
There are a number of operations that can be preformed
on arrays.
1. Traversing an array
2. Inserting an element in an array
3. Searching an element in an array
4. Deleting an element from an array
5. Merging two arrays
6. Sorting an array in ascending or descending order

27
Algorithm to Insert an Element in the Middle of an Array

28
Algorithm to delete an element from the middle of an array

31
• Design, develop and implement a program to merge two
sorted arrays
• Design, develop and implement a program to interchange
the largest and smallest number in an array.

Dynamically allocated arrays
35

 2000 Prentice Hall, Inc.
All rights reserved.
1. Initialize array
2. Loop
3. Print
37
1 /*
2 Histogram printing program */
3 #include <stdio.h>
4 #define SIZE 10
5
6 int main()
7 {
8 int n[ SIZE ] = { 19, 3, 15, 7, 11, 9, 13, 5, 17, 1 };
9 int i, j;
10
11 printf( "%s%13s%17sn", "Element", "Value", "Histogram" );
12
13 for ( i = 0; i <= SIZE - 1; i++ ) {
14 printf( "%7d%13d ", i, n[ i ]) ;
15
16 for ( j = 1; j <= n[ i ]; j++ ) /* print one bar */
17 printf( "%c", '*' );
18
19 printf( "n" );
20 }
21
22 return 0;
23 }

Program Output
38
Element Value Histogram
0 19 *******************
1 3 ***
2 15 ***************
3 7 *******
4 11 ***********
5 9 *********
6 13 *************
7 5 *****
8 17 *****************
9 1 *

Examples Using Arrays
• Character arrays
– String “first” is really a static array of characters
– Character arrays can be initialized using string literals
char string1[] = "first";
• Null character '0' terminates strings
• string1 actually has 6 elements
– It is equivalent to
char string1[] = { 'f', 'i', 'r', 's', 't', '0' };
– Can access individual characters
string1[ 3 ] is character ‘s’
– Array name is address of array, so & not needed for scanf
scanf( "%s", string2 );
• Reads characters until whitespace encountered
• Can write beyond end of array, be careful
39

1. Initialize strings
2. Print strings
2.1 Define loop
2.2 Print characters
individually
2.3 Input string
3. Print string
Program Output
40
2 /*Treating character arrays as strings */
4
5 int main()
6 {
7 char string1[ 20 ], string2[] = "string literal";
8 int i;
9
10 printf(" Enter a string: ");
11 scanf( "%s", string1 );
12 printf( "string1 is: %snstring2: is %sn"
13 "string1 with spaces between characters is:n",
14 string1, string2 );
15
16 for ( i = 0; string1[ i ] != '0'; i++ )
17 printf( "%c ", string1[ i ] );
18
19 printf( "n" );
20 return 0;
21 }
Enter a string: Hello there
string1 is: Hello
string2 is: string literal
string1 with spaces between characters is:
H e l l o

Passing Arrays to Functions
• Passing arrays
– To pass an array argument to a function, specify the name
of the array without any brackets
int myArray[ 24 ];
myFunction( myArray, 24 );
• Array size usually passed to function
– Arrays passed call-by-reference
– Name of array is address of first element
– Function knows where the array is stored
• Modifies original memory locations
• Passing array elements
– Passed by call-by-value
– Pass subscripted name (i.e., myArray[ 3 ]) to function
41

Passing Arrays to Functions
• Function prototype
void modifyArray( int b[], int
arraySize );
– Parameter names optional in prototype
• int b[] could be written int []
• int arraySize could be simply int
42

1. Function definitions
2. Pass array to a
function
2.1 Pass array element
to a function
3. Print
43
1 /*
2 Passing arrays and individual array elements to functions */
4 #define SIZE 5
5
6 void modifyArray( int [], int ); /* appears strange */
7 void modifyElement( int );
8
9 int main()
10 {
11 int a[ SIZE ] = { 0, 1, 2, 3, 4 }, i;
12
13 printf( "Effects of passing entire array call "
14 "by reference:nnThe values of the "
15 "original array are:n" );
16
17 for ( i = 0; i <= SIZE - 1; i++ )
18 printf( "%3d", a[ i ] );
19
20 printf( "n" );
21 modifyArray( a, SIZE ); /* passed call by reference */
22 printf( "The values of the modified array are:n" );
23
24 for ( i = 0; i <= SIZE - 1; i++ )
25 printf( "%3d", a[ i ] );
26
27 printf( "nnnEffects of passing array element call "
28 "by value:nnThe value of a[3] is %dn", a[ 3 ] );
29 modifyElement( a[ 3 ] );
30 printf( "The value of a[ 3 ] is %dn", a[ 3 ] );
31 return 0;
32 }
Entire arrays passed call-by-
reference, and can be modified
Array elements passed call-by-
value, and cannot be modified

3.1 Function
definitions
Program Output
44
33
34 void modifyArray( int b[], int size )
35 {
36 int j;
37
38 for ( j = 0; j <= size - 1; j++ )
39 b[ j ] *= 2;
40 }
41
42 void modifyElement( int e )
43 {
44 printf( "Value in modifyElement is %dn", e *= 2 );
45 }
Effects of passing entire array call by reference:
The values of the original array are:
0 1 2 3 4
The values of the modified array are:
0 2 4 6 8
Effects of passing array element call by value:
The value of a[3] is 6
Value in modifyElement is 12
The value of a[3] is 6

Sorting Arrays
• Sorting data
– Important computing application
– Virtually every organization must sort some data
• Bubble sort (sinking sort)
– Several passes through the array
– Successive pairs of elements are compared
• If increasing order (or identical ), no change
• If decreasing order, elements exchanged
– Repeat
• Example:
– original: 3 4 2 6 7
– pass 1: 3 2 4 6 7
– pass 2: 2 3 4 6 7
– Small elements "bubble" to the top
45

Case Study: Computing Mean, Median
and Mode Using Arrays
• Mean – average
• Median – number in middle of sorted list
– 1, 2, 3, 4, 5
– 3 is the median
• Mode – number that occurs most often
– 1, 1, 1, 2, 3, 3, 4, 5
– 1 is the mode
46

1. Function prototypes
1.1 Initialize array
2. Call functions mean,
median, and mode
47
1 /*
2 This program introduces the topic of survey data analysis.
3 It computes the mean, median, and mode of the data */
5 #define SIZE 99
6
7 void mean( const int [] );
8 void median( int [] );
9 void mode( int [], const int [] ) ;
10 void bubbleSort( int [] );
11 void printArray( const int [] );
12
13 int main()
14 {
15 int frequency[ 10 ] = { 0 };
16 int response[ SIZE ] =
17 { 6, 7, 8, 9, 8, 7, 8, 9, 8, 9,
18 7, 8, 9, 5, 9, 8, 7, 8, 7, 8,
19 6, 7, 8, 9, 3, 9, 8, 7, 8, 7,
20 7, 8, 9, 8, 9, 8, 9, 7, 8, 9,
21 6, 7, 8, 7, 8, 7, 9, 8, 9, 2,
22 7, 8, 9, 8, 9, 8, 9, 7, 5, 3,
23 5, 6, 7, 2, 5, 3, 9, 4, 6, 4,
24 7, 8, 9, 6, 8, 7, 8, 9, 7, 8,
25 7, 4, 4, 2, 5, 3, 8, 7, 5, 6,
26 4, 5, 6, 1, 6, 5, 7, 8, 7 };
27
28 mean( response );
29 median( response );
30 mode( frequency, response );
31 return 0;
32 }

3. Define function
mean
3.1 Define function
median
3.1.1 Sort Array
3.1.2 Print middle
element
48
33
34 void mean( const int answer[] )
35 {
36 int j, total = 0;
37
38 printf( "%sn%sn%sn", "********", " Mean", "********" );
39
40 for ( j = 0; j <= SIZE - 1; j++ )
41 total += answer[ j ];
42
43 printf( "The mean is the average value of the datan"
44 "items. The mean is equal to the total ofn"
45 "all the data items divided by the numbern"
46 "of data items ( %d ). The mean value forn"
47 "this run is: %d / %d = %.4fnn",
48 SIZE, total, SIZE, ( double ) total / SIZE );
49 }
50
51 void median( int answer[] )
52 {
53 printf( "n%sn%sn%sn%s",
54 "********", " Median", "********",
55 "The unsorted array of responses is" );
56
57 printArray( answer );
58 bubbleSort( answer );
59 printf( "nnThe sorted array is" );
60 printArray( answer );
61 printf( "nnThe median is element %d ofn"
62 "the sorted %d element array.n"
63 "For this run the median is %dnn",
64 SIZE / 2, SIZE, answer[ SIZE / 2 ] );

49
65 }
66
67 void mode( int freq[], const int answer[] )
68 {
69 int rating, j, h, largest = 0, modeValue = 0;
70
71 printf( "n%sn%sn%sn",
72 "********", " Mode", "********" );
73
74 for ( rating = 1; rating <= 9; rating++ )
75 freq[ rating ] = 0;
76
77 for ( j = 0; j <= SIZE - 1; j++ )
78 ++freq[ answer[ j ] ];
79
80 printf( "%s%11s%19snn%54sn%54snn",
81 "Response", "Frequency", "Histogram",
82 "1 1 2 2", "5 0 5 0 5" );
83
84 for ( rating = 1; rating <= 9; rating++ ) {
85 printf( "%8d%11d ", rating, freq[ rating ] );
86
87 if ( freq[ rating ] > largest ) {
88 largest = freq[ rating ];
89 modeValue = rating;
90 }
91
92 for ( h = 1; h <= freq[ rating ]; h++ )
93 printf( "*" );
94
3.2 Define function
mode
3.2.1 Increase
frequency[]
depending on
response[]
Notice how the subscript in
frequency[] is the value of an
element in response[]
(answer[])
Print stars depending on value of
frequency[]

3.3 Define bubbleSort
3.3 Define printArray
50
95 printf( "n" );
96 }
97
98 printf( "The mode is the most frequent value.n"
99 "For this run the mode is %d which occurred"
100 " %d times.n", modeValue, largest );
101}
102
103 void bubbleSort( int a[] )
104 {
105 int pass, j, hold;
106
107 for ( pass = 1; pass <= SIZE - 1; pass++ )
108
109 for ( j = 0; j <= SIZE - 2; j++ )
110
111 if ( a[ j ] > a[ j + 1 ] ) {
112 hold = a[ j ];
113 a[ j ] = a[ j + 1 ];
114 a[ j + 1 ] = hold;
115 }
116 }
117
118 void printArray( const int a[] )
119 {
120 int j;
121
122 for ( j = 0; j <= SIZE - 1; j++ ) {
123
124 if ( j % 20 == 0 )
125 printf( "n" );
Bubble sort: if elements out of order,
swap them.

Program Output
51
126
127 printf( "%2d", a[ j ] );
128 }
129 }
********
Mean
********
The mean is the average value of the data
items. The mean is equal to the total of
all the data items divided by the number
of data items (99). The mean value for
this run is: 681 / 99 = 6.8788
********
Median
********
The unsorted array of responses is
7 8 9 8 7 8 9 8 9 7 8 9 5 9 8 7 8 7 8
6 7 8 9 3 9 8 7 8 7 7 8 9 8 9 8 9 7 8 9
6 7 8 7 8 7 9 8 9 2 7 8 9 8 9 8 9 7 5 3
5 6 7 2 5 3 9 4 6 4 7 8 9 6 8 7 8 9 7 8
7 4 4 2 5 3 8 7 5 6 4 5 6 1 6 5 7 8 7
The sorted array is
1 2 2 2 3 3 3 3 4 4 4 4 4 5 5 5 5 5 5 5
5 6 6 6 6 6 6 6 6 6 7 7 7 7 7 7 7 7 7 7
7 7 7 7 7 7 7 7 7 7 7 7 7 8 8 8 8 8 8 8
8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8
9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9
The median is element 49 of
the sorted 99 element array.
For this run the median is 7

Program Output
52
********
Mode
********
Response Frequency Histogram
1 1 2 2
5 0 5 0 5
1 1 *
2 3 ***
3 4 ****
4 5 *****
5 8 ********
6 9 *********
7 23 ***********************
8 27 ***************************
9 19 *******************
The mode is the most frequent value.
For this run the mode is 8 which occurred 27 times.

Searching Arrays: Linear Search and
Binary Search
• Search an array for a key value
• Linear search
– Simple
– Compare each element of array with key value
– Useful for small and unsorted arrays
53

Searching Arrays: Linear Search and
Binary Search
• Binary search
– For sorted arrays
– Compares middle element with key
• If equal, match found
• If key < middle, looks in first half of array
• If key > middle, looks in last half
• Repeat
– Very fast; at most n steps, where 2n > number of
elements
• 30 element array takes at most 5 steps
– 25 > 30 so at most 5 steps
54
5

Multiple-Subscripted Arrays
• Multiple subscripted arrays
– Tables with rows and columns (m by n array)
– Like matrices: specify row, then column
55
Row 0
Row 1
Row 2
Column 0 Column 1 Column 2 Column 3
a[ 0 ][ 0 ]
a[ 1 ][ 0 ]
a[ 2 ][ 0 ]
a[ 0 ][ 1 ]
a[ 1 ][ 1 ]
a[ 2 ][ 1 ]
a[ 0 ][ 2 ]
a[ 1 ][ 2 ]
a[ 2 ][ 2 ]
a[ 0 ][ 3 ]
a[ 1 ][ 3 ]
a[ 2 ][ 3 ]
Row subscript
Array name
Column subscript

Multiple-Subscripted Arrays
• Initialization
– int b[ 2 ][ 2 ] = { { 1, 2 }, { 3, 4 } };
– Initializers grouped by row in braces
– If not enough, unspecified elements set to zero
int b[ 2 ][ 2 ] = { { 1 }, { 3, 4 } };
• Referencing elements
– Specify row, then column
printf( "%d", b[ 0 ][ 1 ] );
56
1 2
3 4
1 0
3 4

1. Initialize variables
1.1 Define functions to
take double scripted
arrays
1.2 Initialize
studentgrades[][]
2. Call functions
minimum, maximum,
and average
1 /*
2 Double-subscripted array example */
4 #define STUDENTS 3
5 #define EXAMS 4
6
7 int minimum( const int [][ EXAMS ], int, int );
8 int maximum( const int [][ EXAMS ], int, int );
9 double average( const int [], int );
10 void printArray( const int [][ EXAMS ], int, int );
11
12 int main()
13 {
14 int student;
15 const int studentGrades[ STUDENTS ][ EXAMS ] =
16 { { 77, 68, 86, 73 },
17 { 96, 87, 89, 78 },
18 { 70, 90, 86, 81 } };
19
20 printf( "The array is:n" );
21 printArray( studentGrades, STUDENTS, EXAMS );
22 printf( "nnLowest grade: %dnHighest grade: %dn",
23 minimum( studentGrades, STUDENTS, EXAMS ),
24 maximum( studentGrades, STUDENTS, EXAMS ) );
25
26 for ( student = 0; student <= STUDENTS - 1; student++ )
27 printf( "The average grade for student %d is %.2fn",
28 student,
29 average( studentGrades[ student ], EXAMS ) );
30
31 return 0;
32 }
Each row is a particular student,
each column is the grades on the
exam.

3. Define functions
33
34 /* Find the minimum grade */
35 int minimum( const int grades[][ EXAMS ],
36 int pupils, int tests )
37 {
38 int i, j, lowGrade = 100;
39
40 for ( i = 0; i <= pupils - 1; i++ )
41 for ( j = 0; j <= tests - 1; j++ )
42 if ( grades[ i ][ j ] < lowGrade )
43 lowGrade = grades[ i ][ j ];
44
45 return lowGrade;
46 }
47
48 /* Find the maximum grade */
49 int maximum( const int grades[][ EXAMS ],
51 {
52 int i, j, highGrade = 0;
53
54 for ( i = 0; i <= pupils - 1; i++ )
55 for ( j = 0; j <= tests - 1; j++ )
56 if ( grades[ i ][ j ] > highGrade )
57 highGrade = grades[ i ][ j ];
58
59 return highGrade;
60 }
61
62 /* Determine the average grade for a particular exam */
63 double average( const int setOfGrades[], int tests )
64 {

3. Define functions
65 int i, total = 0;
66
67 for ( i = 0; i <= tests - 1; i++ )
68 total += setOfGrades[ i ];
69
70 return ( double ) total / tests;
71 }
72
73 /* Print the array */
74 void printArray( const int grades[][ EXAMS ],
76 {
77 int i, j;
78
79 printf( " [0] [1] [2] [3]" );
80
81 for ( i = 0; i <= pupils - 1; i++ ) {
82 printf( "nstudentGrades[%d] ", i );
83
84 for ( j = 0; j <= tests - 1; j++ )
85 printf( "%-5d", grades[ i ][ j ] );
86 }
87 }

Program Output
60
The array is:
[0] [1] [2] [3]
studentGrades[0] 77 68 86 73
Lowest grade: 68
Highest grade: 96
The average grade for student 0 is 76.00

CHAPTER 2 61
Structure Polynomial is
objects: ; a set of ordered pairs of <ei,ai>
where ai in Coefficients and ei in Exponents, ei are integers >= 0
functions:
for all poly, poly1, poly2  Polynomial, coef Coefficients, expon
Exponents
Polynomial Zero( ) ::= return the polynomial,
p(x) = 0
Boolean IsZero(poly) ::= if (poly) return FALSE
else return TRUE
Coefficient Coef(poly, expon) ::= if (expon  poly) return its
coefficient else return Zero
Exponent Lead_Exp(poly) ::= return the largest exponent in
poly
Polynomial Attach(poly,coef, expon) ::= if (expon  poly) return error
else return the polynomial poly
with the term <coef, expon>
inserted
n
e
n
e
x
a
x
a
x
p 

 ...
)
( 1
1
Polynomials A(X)=3X20+2X5+4, B(X)=X4+10X3+3X2+1 ADT for Polynomial

62
Polynomial Remove(poly, expon) ::= if (expon  poly) return the
polynomial poly with the
term whose exponent is
expon deleted
else return error
Polynomial SingleMult(poly, coef, expon) ::= return the polynomial
poly • coef • xexpon
Polynomial Add(poly1, poly2) ::= return the polynomial
poly1 +poly2
Polynomial Mult(poly1, poly2) ::= return the polynomial
poly1 • poly2
*Structure 2.2:Abstract data type Polynomial (p.61)
End Polynomial

63
/* d =a + b, where a, b, and d are polynomials */
d = Zero( )
while (! IsZero(a) && ! IsZero(b)) do {
switch COMPARE (Lead_Exp(a), Lead_Exp(b)) {
case -1: d =
Attach(d, Coef (b, Lead_Exp(b)), Lead_Exp(b));
b = Remove(b, Lead_Exp(b));
break;
case 0: sum = Coef (a, Lead_Exp (a)) + Coef ( b, Lead_Exp(b));
if (sum) {
Attach (d, sum, Lead_Exp(a));
a = Remove(a , Lead_Exp(a));
b = Remove(b , Lead_Exp(b));
}
break;
Polynomial Addition
#define MAX_DEGREE 101
typedef struct {
int degree;
float coef[MAX_DEGREE];
} polynomial;
data structure 1:

CHAPTER 2 64
case 1: d =
Attach(d, Coef (a, Lead_Exp(a)), Lead_Exp(a));
a = Remove(a, Lead_Exp(a));
}
}
insert any remaining terms of a or b into d
*Program 2.4 :Initial version of padd function(p.62)
advantage: easy implementation
disadvantage: waste space when sparse

CHAPTER 2 65
Data structure 2: use one global array to store all polynomials
A(X)=2X1000+1
B(X)=X4+10X3+3X2+1
2 1 1 10 3 1
1000 0 4 3 2 0
coef
exp
starta finisha startb finishb avail
0 1 2 3 4 5 6
*Figure 2.2: Array representation of two polynomials
(p.63)
specification representation
poly <start, finish>
A <0,1>
B <2,5>

CHAPTER 2 66
MAX_TERMS 100 /* size of terms array */
typedef struct {
float coef;
int expon;
} polynomial;
polynomial terms[MAX_TERMS];
int avail = 0;
*(p.62)
storage requirements: start, finish, 2*(finish-start+1)
nonparse: twice as much as (1)
when all the items are nonzero

CHAPTER 2 67
void padd (int starta, int finisha, int startb, int finishb,
int * startd, int *finishd)
{
/* add A(x) and B(x) to obtain D(x) */
float coefficient;
*startd = avail;
while (starta <= finisha && startb <= finishb)
switch (COMPARE(terms[starta].expon,
terms[startb].expon)) {
case -1: /* a expon < b expon */
attach(terms[startb].coef, terms[startb].expon);
startb++
break;
Add two polynomials: D = A + B

CHAPTER 2 68
case 0: /* equal exponents */
coefficient = terms[starta].coef +
terms[startb].coef;
if (coefficient)
attach (coefficient, terms[starta].expon);
starta++;
startb++;
break;
case 1: /* a expon > b expon */
attach(terms[starta].coef, terms[starta].expon);
starta++;
}

CHAPTER 2 69
/* add in remaining terms of A(x) */
for( ; starta <= finisha; starta++)
attach(terms[starta].coef, terms[starta].expon);
/* add in remaining terms of B(x) */
for( ; startb <= finishb; startb++)
attach(terms[startb].coef, terms[startb].expon);
*finishd =avail -1;
}
*Program 2.5: Function to add two polynomial (p.64)
Analysis: O(n+m)
where n (m) is the number of nonzeros in A(B).

CHAPTER 2 70
void attach(float coefficient, int exponent)
{
/* add a new term to the polynomial */
if (avail >= MAX_TERMS) {
fprintf(stderr, “Too many terms in the polynomialn”);
exit(1);
}
terms[avail].coef = coefficient;
terms[avail++].expon = exponent;
}
*Program 2.6:Function to add anew term (p.65)
Problem: Compaction is required
when polynomials that are no longer needed.
(data movement takes time.)

Sparse matrix
71
• A sparse matrix is a matrix in which most of the
elements are zero.
• By contrast, if most of the elements are nonzero,
then the matrix is considered dense.
• The number of zero-valued elements divided by the
total number of elements is called the sparsity of the
matrix (which is equal to 1 minus the density of the
matrix).

Check matrix is sparse or not
72

CHAPTER 2 73






















0
0
0
28
0
0
0
0
0
0
0
91
0
0
0
0
0
0
0
0
6
0
0
0
0
0
0
3
11
0
15
0
22
0
0
15
col1 col2 col3 col4 col5 col6
row0
row1
row2
row3
row4
row5
*Figure 2.3: Two matrices
8/36
6*6
5*3
15/15
Sparse Matrix
sparse matrix
data structure?

CHAPTER 2 75
Structure Sparse_Matrix is
objects: a set of triples, <row, column, value>, where row
and column are integers and form a unique combination,
and
value comes from the set item.
functions:
for all a, b  Sparse_Matrix, x  item, i, j, max_col,
max_row  index
Sparse_Marix Create(max_row, max_col) ::=
return a Sparse_matrix that can hold up to
max_items = max _row  max_col and
whose maximum row size is max_row and
whose maximum column size is max_col.
SPARSE MATRIX ABSTRACT DATA TYPE

CHAPTER 2 76
Sparse_Matrix Transpose(a) ::=
return the matrix produced by interchanging
the row and column value of every triple.
Sparse_Matrix Add(a, b) ::=
if the dimensions of a and b are the same
return the matrix produced by adding
corresponding items, namely those with
identical row and column values.
else return error
Sparse_Matrix Multiply(a, b) ::=
if number of columns in a equals number of
rows in b
return the matrix d produced by multiplying
a by b according to the formula: d [i] [j] =
(a[i][k]•b[k][j]) where d (i, j) is the (i,j)th
element
else return error.
* Structure 2.3: Abstract data type Sparse-Matrix (p.68)

CHAPTER 2 77
row col value row col value
a[0] 6 6 8 b[0] 6 6 8
[1] 0 0 15 [1] 0 0 15
[2] 0 3 22 [2] 0 4 91
[3] 0 5 -15 [3] 1 1 11
[4] 1 1 11 [4] 2 1 3
[5] 1 2 3 [5] 2 5 28
[6] 2 3 -6 [6] 3 0 22
[7] 4 0 91 [7] 3 2 -6
[8] 5 2 28 [8] 5 0 -15
(a) (b)
*Figure 2.4:Sparse matrix and its transpose stored as triples (p.69)
(1) Represented by a two-dimensional array.
Sparse matrix wastes space.
(2) Each element is characterized by <row, col, value>.
row, column in ascending order
# of rows (columns)
# of nonzero terms
transpose

CHAPTER 2 78
Sparse_matrix Create(max_row, max_col) ::=
#define MAX_TERMS 101 /* maximum number of terms +1*/
typedef struct {
int col;
int row;
int value;
} term;
term a[MAX_TERMS]
* (P.69)
# of rows (columns)
# of nonzero terms

Basics of Data structure using C describing basics concepts

More Related Content

Similar to Basics of Data structure using C describing basics concepts (20)

Recently uploaded (20)

Basics of Data structure using C describing basics concepts