Data structures using C

DATA STRUCTURES
USING C
Chapter 1 – Basic Concepts
1

Overview
 Data: data is a simple value or set of values.
 Entity: Entity is some thing that has certain
attributes or properties which may assigned
values.
 Field: It is a single elementary unit of
information representing an attribute of an
entity.
 Record: A record is a collection of field values
of a given entity.
 File: A file is a collection of records of the
entities in a given entity set.
2

3
Why Data Structures?
 Problem solving is more related to
understanding the problem, designing a
solution and Implementing the solution, then
 What exactly is a solution?
 In a very crisp way, it can be demonstrate as a
solution which is equal to a program and it is
also approximately equal to algorithm.

4
Algorithm
 An algorithm is a sequence of steps that take us
from the input to the output.
 An algorithm must be Correct. It should provide a
correct solution according to the specifications.
 Finite. It should terminate and general.
 It should work for every instance of a problem is
efficient.
 It should use few resources (such as time or
memory).

5
What is Information?
 It is an undefined term that cannot be
explained in terms of more elementary
concepts.
 In computer science we can measure
quantities of information.
 The basic unit of information is bit.
 It is a concentration of the term binary digit.

Definition !
6
 Data may be organized in many different
ways ; the logical / mathematical model of
a particular, organization of data is called
Data Structure.
 Data structure can be also defined as, it is
the mathematical model which helps to
store and retrieve the data efficiently from
primary memory.

System Life Cycle
8
 Large-Scale program contains many complex
interacting parts.
 Programs undergo a development process
called the system life cycle.
 Solid foundation of the methodologies in
1. data abstraction
2. algorithm specification
3. performance analysis and measurement

9
5 phases of system life cycle:
1. Requirements: Begins with purpose of the project.
What inputs (needs), outputs (results)
2. Analysis: Break the problem down into manageable
pieces. bottom-up vs. top-down, divide and conquer
3. Design: data objects and operations creation of
abstract data types, second the specification of
algorithms and algorithm design strategies.
4. Refinement & coding: representations for our data
objects and write algorithms for each operation on
them.
 we should write those algorithms that are
independent of the data objects first.

10
5. Verification: consists of developing correctness proofs for
the program, testing the program with a variety of input data
and removing errors.
 Correctness proofs: proofs are very time-consuming and
difficult to develop fro large projects. Scheduling constraints
prevent the development of a complete set of proofs for a
large system.
 Testing: before and during the coding phase. testing is used
at key checkpoints in the overall process to determine
whether objectives are being met.
 Error Removal: system tests will indicate erroneous code.
Errors can be removed depending on the design & coding
decisions made earlier.

Pointers
11
 For any type T in C there is a corresponding
type pointer-to-T. The actual value of a pointer
type is an address of memory.
 & the address operator.
 * the dereferencing ( or indirection ) operator.
 Declaration
• int i , *pi;
then i is an integer variable and pi is a pointer to an
integer.
• pi = &i;
then &i returns the address of i and assigns it as the
value of pi.

12
• i = 10 or
• *pi = 10;
 In both cases the integer 10 is stored as the value
of i.
 In second case, the * in front of the pointer pi
causes it to be dereferenced, by which we mean
that instead of storing 10 into the pointer, 10 is
stored into the location pointed at by the pointer
pi.
 The size of a pointer can be different on different
computers.
 The size of a pointer to a char can be longer than
a pointer to a float.
 Test for the null pointer in C if (pi == NULL) or

Dynamic Memory Allocation
13
 When you will write program you may not know how
much space you will need. C provides a mechanism
called a heap, for allocating storage at run-time.
 The function malloc is used to allocate a new area of
memory. If memory is available, a pointer to the start
of an area of memory of the required size is returned
otherwise NULL is returned.
 When memory is no longer needed you may free it by
calling free function.
 The call to malloc determines size of storage required
to hold int or the float.
 The notations (int *) and (float *) are type cast
expressions.

Example…!
14
int i,*pi;
float f,*pf;
pi = (int *) malloc (sizeof(int));
pf = (float *) malloc(sizeof(float));
*pi=1024;
*pf=3.124;
printf(“an integer = %d, a float = %fn”,*pi,*pf);
free(pi);
free(pf);

15
 When programming in C it is a wise practice to set all
pointers to NULL when they are not pointing to an
object.
 Use explicit type casts when converting between
pointer types
pi = malloc(sizeof(int));
/* assign to pi a pointer to int */
pf = (float *) pi;
/* casts an int pointer to float pointer */
 In many systems, pointers have the same size as type
int.
 int is the default type specifier.

Algorithm Specification
16
 An algorithm is a finite set of instructions that, if
followed, accomplishes a particular task.
 Algorithms must satisfy following criteria,
 Input: there are zero or more quantities that are
externally supplied.
 Output: at least one quantity is produced.
 Definiteness: Each instruction is clear and
unambiguous.
 Finiteness: for all cases, the algorithm terminates after
a finite number of steps.
 Effectiveness: Every instruction must be basic enough
to be carried out, it must also be feasible.

17
 Difference between an algorithm and a
program : the program does not have to satisfy
the fourth condition (Finiteness).
 Describing an algorithm: natural language
such as English will do. However, natural
language is wordy to make a statement
definite. That is where the Code of a program
language fit in.
 Flowchart: work well only for algorithm, small
and simple.

Example:
Translating a Problem into an
Algorithm18
 Problem
 Devise a program that sorts a set of n integers, where
n>= 1, from smallest to largest.
 Solution I: looks good, but it is not an algorithm
 From those integers that are currently unsorted, find
the smallest and place it next in the sorted list.
 Solution II: an algorithm, written in partially C and English
 for (i= 0; i< n; i++){
Examine list[i] to list[n-1] and suppose that the
smallest integer is list[min];
Interchange list[i] and list[min];
}

Example: Program
19
#define swap(x,y,t) ((t)= (x), (x)= (y), (y)= (t))
void sort(int list[], int n)
{
int i, j, min, temp;
for (i= 0; i< n-1; i++){
min= i;
for (j= i+1; j< n; j++){
if (list[j]< list[min]) min= j;
}
swap(list[i], list[min], temp);
}
void swap(int *x, int *y)
{
int temp= *x;
*x= *y;
*y= temp;
}

Recursive Algorithms
20
 Direct recursion
 Functions call themselves
 Indirect recursion
 Functions call other functions that invoke the
calling function again
 Any function that we can write using
assignment, if-else, and while statements can
be written recursively.

21
int binsearch(int list[], int searchnum, int left, int right)
{// search list[0]<= list[1]<=...<=list[n-1] for searchnum
int middle;
while (left<= right)
{
middle= (left+ right)/2;
switch(compare(list[middle], searchnum))
{
case -1: left= middle+ 1;
break;
case 0: return middle;
case 1: right= middle- 1;
}
}
return -1;
}
int compare(int x, int y)
{
if (x< y) return -1;
else if (x== y) return 0;
else return 1;
}

Recursive Implementation of Binary Search
22
int binsearch(int list[], int searchnum, int left, int right)
{
int middle;
if(left<=right)
{
middle = (left+right)/2;
switch(COMPARE(list[middle],searchnum))
{
case -1: return
binsearch(list,searchnum,middle+1,right);
case 0: return middle;
case 1: return
binsearch(list,searchnum,left,middle-1);
}
}
return -1;
}

24
void Hanoi(int n, char x, char y, char z)
{
if (n > 1)
{
Hanoi(n-1,x,z,y);
printf("Move disk %d from %c to %c.n",n,x,z);
Hanoi(n-1,y,x,z);
}
else
{
printf("Move disk %d from %c to %c.n",n,x,z);
}
}

Data Abstraction
25
 Data Type
A data type is a collection of objects and a
set of operations that act on those objects.
 if program is dealing with predefined or
user-defined data types two aspects i,e
objects & operations must be considered.
 Example of "int"
 Objects: 0, +1, -1, ..., Int_Max, Int_Min
 Operations: arithmetic(+, -, *, /, and %), testing
(equality/inequality), assigns, functions

Data Abstraction &
Encapsulation26
 Abstraction: is extraction of essential
information for a particular purpose and
ingnoring the remainder of the information
 Encapsulation: is the process of binding
together of Data and Code. It can also be
defined as the concept that an object totally
separates its interface from its implementation.
 It has been observed by many software
designers that hiding the representation of
objects of a data type from its users is a good
design strategy.

Abstract Data Type
27
 Abstract Data Type
An abstract data type (ADT) is a data type that is organized in such
a way that the specification of the objects and the operations on the
objects is separated from the representation of the objects and the
implementation of the operations.
 ADT Operations — only the operation name and its parameters
are visible to the user — through interface
 Why abstract data type ?
 implementation-independent
 Abstraction is …
 Generalization of operations with unspecified implementation

Classifying the Functions of a ADT
29
 Creator/constructor
 Create a new instance of the designated type
 Transformers
 Also create an instance of the designated type by
using one or more other instances
 Observers/reporters
 Provide information about an instance of the type,
but they do not change the instance

Abstract data type
NaturalNumber30
 ADT NaturalNumber is
structure NaturalNumber is (denoted by Nat_No)
objects: an ordered subrange of the integers starting at zero
and ending at the maximum integer (INT_MAX) on the computer
functions:
for all x, y  Nat_Number; TRUE, FALSE  Boolean
and where +, -, <, and == are the usual integer operations.
Nat_No Zero ( ) ::= 0
Boolean IsZero(x) ::= if (x) return FALSE
else return TRUE
Nat_No Add(x, y) ::= if ((x+y) <= INT_MAX) return x+y
else return INT_MAX
Boolean Equal(x,y) ::= if (x== y) return TRUE
else return FALSE
Nat_No Successor(x) ::= if (x == INT_MAX) return x
else return x+1
Nat_No Subtract(x,y) ::= if (x<y) return 0
else return x-y
end NaturalNumber
Creator
Observer
Transformer

Cont’d..
31
 ADT definition begins with name of the ADT.
 Two main sections in the definition the objects
and the functions.
 Objects are defined in terms of integers but no
explicit reference to their representation.
 Function denote two elements of the data type
NaturalNumber, while TRUE & FALSE are
elements of the data type Boolean.
 The symbols “::=“ should be read as “is
defined as.”

Cont’d…
32
 First function Zero has no arguments & returns
the natural number zero (constructor function).
 The function Successor(x) returns the next
natural nor in sequence (ex of transformer
function).
 Other transformer functions are Add &
Subtract.

Performance Analysis
33
 Criteria upon which we can judge a program
 Does the program meet the original specifications of
the task ?
 Does it work correctly ?
 Does the program contain documentation that shows
how to use it and how it works ?
 Does the program effectively use functions to create
logical units ?
 Is the program’s code readable ?
Space & Time
 Does the program efficiently use primary and
secondary storage ?
 Is the program’s running time acceptable for the task
?

Cont’d…
34
 Performance evaluation is loosely divided into
two fields:
 First field focuses on obtaining estimates of
time and space that are machine independent.
(Performance Analysis)
 Second field, (Performance measurement),
obtains machine-dependent running times,
these are used to identify inefficient code
segments.

Cont’d…
35
 Evaluate a program
Meet specifications, Work correctly,
Good user-interface, Well-documentation,
Readable, Effectively use functions,
Running time acceptable, Efficiently use space/storage
 How to achieve them?
 Good programming style, experience, and practice
 Discuss and think
 The space complexity of a program is the amount
of memory that it needs to run to completion.
 The time complexity of a program is the amount of
computer time that it needs to run to completion

Space Complexity
36
 The space needed is the sum of
 Fixed space and Variable space
 Fixed space: C
 Includes the instructions, variables, and constants
 Independent of the number and size of I/O
 Variable space: Sp(I)
 Includes dynamic allocation, functions' recursion
 Total space of any program
 S(P)= C+ Sp(Instance)
where C is constant & Sp is instance characteristics

Example1 to find space
complexity37
 Algorithm P1(P,Q,R)
1. Start
2. Total = (P+Q+R*P+Q*R+P)/(P*Q);
3. End
In the above algorithm there is no instance
characteristics and space needed by P,Q,R and
Total is independent of instance characteristics
therefore
S(P1) = 4 + 0
where one space is required for each of P,Q,R &
Total

Ex 2
38
 Algorithm SUM()
1. [Sum the values in an array A]
Sum=0
Repeat for I=1,2,…..N
Sum = sum + A[i]
2. [Finished]
Exit.
In the above algorithm there is an instance characteristic
N, since A must be large enough to hold the N elements
to be summed & space needed by sum, I, and N is the
independent of instance characteristics, we can write
S(Sum) = 3 + N.
N for A[] and 3 for N,I and sum

Program 1.11: Recursive function for
summing a list of numbers
39
 float rsum(float list[ ], int n)
{
if (n) return rsum(list, n-1) + list[n-1];
return 0;
}
 Figure 1.1: Space needed for one recursive call of
Program 1.11
Ssum(I)=Ssum(n)=6n
Type Name Number of bytes
parameter: float
parameter: integer
return address:(used internally)
list [ ]
n
4
4
4(unless a far address)
TOTAL per recursive call 12

Time Complexity
40
 The time T(P) taken by a program, P is the sum
of its compile time and its run (execution) time.
 It depends on several factors
 The input of the program.
 Time required to generate the object code by the
compiler.
 The speed to CPU used to execute the program.
 If we must know the running time, the best
approach is to use the system clock to time the
program
 Total time

Cont’d…
41
 How to evaluate?
 + - * / …
 Use the system clock
 Number of steps performed
 machine-independent
 Instance
 Definition of a program step
 A program step is a syntactically or semantically
meaningful program segment whose execution time is
independent of the instance characteristics.
 We cannot express the running time in
standard time units such as hours, minutes &
seconds rather we can write as “the running
time of such and such an algorithm is
proportional to n”.

Cont’d..
42
 Time complexity for a given algorithm can be
calculated using two steps
1. Separate the particular operations such as
ACTIVE operations that is central to the algorithm
that is executed essentially often as any other.
2. Other operations such as assignments, the
manipulation of index I and accessing values in an
array, are called BOOK KEEPING operations and
are not generally counted.
 After the active operation is isolated, the number
of times that it is executed is counted.

Ex 1
43
 Consider the algorithm to sum the values in a
given array A that contains N values. I is an
integer variable used as index of A
 ALGORITHM SUM()
1. [Sum the values in an array A]
sum = 0
repeat for I=1,2…N
sum=sum+A[I]
2. [Finished]
exit.

Explaination
44
 In the above ex, operation to isolate is the
addition that occurs when another array value
is added to the partial sum
 After active operation is isolated, the nor of
time that it is executed is counted.
 The number of addition of the values in the alg
is N.
 Execution time will increases in proportional to
the number of times the active operation is
executed.
 Thus the above algorithm has execution time
proportional to N.

Tabular Method
45
Step count table for Program Iterative function to sum a list of numbers
Statement s/e Frequency Total steps
float sum(float list[ ], int n)
{
float tempsum = 0;
int i;
for(i=0; i <n; i++)
tempsum += list[i];
return tempsum;
}
0 0 0
0 0 0
1 1 1
0 0 0
1 n+1 n+1
1 n n
1 1 1
0 0 0
Total 2n+3
steps/execution

Recursive Function to sum of a list of numbers
46
Statement s/e Frequency Total steps
float rsum(float list[ ], int n)
{
if (n)
return rsum(list, n-1)+list[n-1];
return list[0];
}
0 0 0
0 0 0
1 n+1 n+1
1 n n
1 1 1
0 0 0
Total 2n+2
Step count table for recursive summing function

Asymptotic Notation (O, ,  )
47
 Exact step count
 Compare the time complexity of two programs that
computing the same function
 Difficult task for most of the programs
 Asymptotic notation
 Big “oh”
 upper bound(current trend)
 Omega
 lower bound
 Theta
 upper and lower bound

Cont’d…
48
 Asymptotic notations are the terminology that
enables meaningful statements about time &
space complexity
 The time required by the given algorithm falls
under three types
1. Best-case time or the minimum time required in
executing the program.
2. Average case time or the average time required
in executing program.
3. Worst-case time or the maximum time required
in executing program.

Break Even Point
49
 If we have two programs with a complexity of c1n2+c2n
& c3n respectively.
 We know one with complexity c3n will be faster that one
with complexity c1n2+c2n for large values on n.
 For small values of n, either would be faster.
 If c1=1,c2=2 & c3=100 then c1n2+c2n for n<=98
 If c1=1,c2=2 & c3=1000 then c1n2+c2n for n<=998
 No matter what the values of c1,c2 &c3 there will be an
n beyond which the program with complexity c3n will be
faster.
 This value of n is called break even point.
 the break-even point (BEP) is the point at which cost
or expenses and revenue are equal: there is no net

Asymptotic Notation BIG OH(O)
50
 Definition
 f(n)= O(g(n)) iff there exist two positive constants c and
n0 such that f(n)<= cg(n) for all n, n>= n0
 Examples
 3n+ 2= O(n) as 3n+ 2<= 4n for all n>= 2
 10n2+ 4n+ 2= O(n2) as 10n2+ 4n+ 2<= 11n2 for n>= 5
 3n+2<> O(1), 10n2+ 4n+ 2<> O(n)
 Remarks
 g(n) is upper bound, the least?
 n=O(n2)=O(n2.5)= O(n3)= O(2n)
 O(1): constant, O(n): linear, O(n2): quadratic, O(n3):
cubic, and O(2n): exponential

Asymptotic Notation BIG
OMEGA()51
 Definition
 f(n)= (g(n)) iff there exist two positive constants c and
n0 such that f(n)>= cg(n) for all n, n>= n0
 Examples
 3n+ 2= (n) as 3n+ 2>= 3n for n>= 1
 10n2+ 4n+ 2= (n2) as 10n2+4n+ 2>= n2 for n>= 1
 6*2n+ n2= (2n) as 6*2n+ n2 >= 2n for n>= 1
 Remarks
 lower bound, the largest ? (used for problem)
 3n+3= (1), 10n2+4n+2= (n); 6*2n+ n2= (n100)

Asymptotic Notation BIG
THETA()52
 Definition
 f(n)= (g(n)) iff there exist positive constants c1, c2, and
n0 such that c1g(n)<= f(n) <= c2g(n) for all n, n>= n0
 Examples
 3n+2=(n) as 3n+2>=3n for n>1 and 3n+2<=4n for all
n>= 2
 10n2+ 4n+ 2=  (n2); 6*2n+n2= (2n)
 Remarks
 Both an upper and lower bound
 3n+2!=(1); 10n2+4n+ 2!= (n)

Example of Time Complexity
Analysis53
Statement Asymptotic complexity
void add(int a[][Max.......) 0
{ 0
int i, j; 0
for(i= 0; i< rows; i++) (rows)
for(j=0; j< cols; j++) (rows*cols)
c[i][j]= a[i][j]+ b[i][j]; (rows*cols)
} 0
Total (rows*cols)

Function values
54
log n
• 0
• 1
• 2
• 3
• 4
• 5
n
• 1
• 2
• 4
• 5
• 16
• 32
n log n
• 0
• 2
• 8
• 24
• 64
• 160
n2
• 1
• 2
• 16
• 64
• 256
• 1024
n3
• 1
• 8
• 64
• 512
• 4096
• 32768
2n
• 2
• 4
• 16
• 256
• 65536
• 4,….

Plot of function values
55
n
logn
nlogn

Performance Measurement
56
 Clocking
 Functions we need to time events are part of C’s library &
accessed through the statement #include<time.h>
 There are actually two different methods for timing events
in C.
 Figure 1.10 shows the major differences between these
two methods.
 Method 1 uses clock to time events. This function gives the
amount of processor time that has elapsed since the
program began running.
 Method 2 uses time to time events. This function returns
the time, measured in seconds, as the built-in type time t.
 The exact syntax of the timing functions varies from
computer to computer and also depends on the operating
system and compiler in use.

57
Method 1 Method 2
Start timing start=clock(); start=time(NULL);
Stop timing stop=clock(); stop=time(NULL);
Type returned clock_t time_t
Result in seconds duration=((double)(stop-
start)/CLK_TCK;
duration=(double)difftime(stop,
start);

Generating Test Data
58
 It is necessary to use a computer program to
generate the worst-case data.
 In these cases, another approach to estimating
worst-case performance is taken.
 For each set of values of the instance
characteristics of interest.
 We generate a suitably large number of random
test data.
 The run times for each of these test data are
obtained.
 The max of these times is used as an estimate of
the worst-case time for a set of values.

Polynomials
59
 Arrays are not only data structures in their own
right.
 Most commonly found data structures: the
ordered or linear list.
 Ex: Days of the week: (Sun,Mon,Tue....)
 Values in a deck of cards: (Ace,2,3,4,5,6,7)
 Years Switzerland fought in WorldWarII: ()
 Above ex is an empty list we denote as ().
 The other lists all contain items that are written in the
form (item0,item1,.....item n-1)

60
 Operations on lists including,
 Finding the length, n , of a list
 Reading the items in a list from left to right or
either.
 A polynomial is a sum of terms where each term
has a form axe , where x is the variable, a is the
coefficient and e is the exponent.
 Ex: A(x) = 3x20+ 2x5+4
 The largest exponent of a polynomial is called its
degree.
 The term with exponent equal to zero does not
show the variable since x raised to a power of zero
is 1.

61
 Standard Mathematical definitions for the sum
& product of polynomials.
 Assume that we have two polynomial...
 Similarly we can define subtraction & division
on polynomials as well other operations

62
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
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
Polynomials A(X)=3X20+2X5+4, B(X)=X4+10X3+3X2+1

63
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

Polynomial representation
64
 Very reasonable first decision requires unique
exponents arranged in decreasing order.
 This algorithm works by comparing terms from
the two polynomials until one of both of the
polynomials become empty.
 The switch statement performs the comparison
and adds the proper term to the new
polynomial d.
 One way to represent polynomials in C is to
typedef to create the type polynomial..

65
#define MAX_DEGREE 101
typedef struct {
int degree;
float coef[MAX_DEGREE];
} polynomial;
 /* 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;
 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

Array representation of two polynomials
66
 A(X)=2X1000+1
 B(X)=X4+10X3+3X2+1
2 1 1 10 3 1
1000 0 4 3 2 0
starta finisha startb finishb avail
coef
exp
0 1 2 3 4 5 6
specification representation
poly <start, finish>
A <0,1>
B <2,5>

Polynomial Addition
Add two polynomials: D = A + B
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;

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,
starta++;
}

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

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;
}
Problem: Compaction is required
when polynomials that are no longer needed.
(data movement takes time.)
*Program 2.6:Function to add anew term (p.65)

Sparse Matrices
71
 Some of the problems require lot of zeros to be
stored as a part of solution.
 Hence storing more number of zeros is a waste of
memory.
 Matrices with relatively high proportion of zero
entries are called sparse matrices.
 Matrix that contains more number of zero
elements are called sparse matrices.
 Sparse matrix is used in almost all areas of the
natural sciences

CHAPTER 272
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
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0002800
0000091
000000
006000
0003110
150220015
col0 col1 col2 col3 col4 col5
row0
row1
row2
row3
row4
row5
8/36
6*6
5*3
15/15
Sparse Matrix
sparse matrix
data structure?

SPARSE MATRIX ABSTRACT
DATA TYPE73
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

74
 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.

Sparse Matrix Representation
75
 Consider a normal matrix containing more
number of zero entries.
 Few steps are…as the first step first row of the
sparse matrix stores the following information
1. Location of[0,0] stores the row size of the original
matrix
2. Location of [0,1] stores the column size of the
original matrix
3. Location of [0,2] stores the number non zero
entries of original matrix.

76
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

77
Create Operation
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]
# of rows (columns)
# of nonzero terms

Transposing a matrix
78
(1) for each row i
take element <i, j, value> and store it
in element <j, i, value> of the transpose.
difficulty: where to put <j, i, value>
(0, 0, 15) ====> (0, 0, 15)
(0, 3, 22) ====> (3, 0, 22)
(0, 5, -15) ====> (5, 0, -15)
(1, 1, 11) ====> (1, 1, 11)
Move elements down very often.
(2) For all elements in column j,
place element <i, j, value> in element <j, i, value>

CHAPTER 2
79
void transpose (term a[], term b[])
/* b is set to the transpose of a */
{
int n, i, j, currentb;
n = a[0].value; /* total number of elements */
b[0].row = a[0].col; /* rows in b = columns in a */
b[0].col = a[0].row; /*columns in b = rows in a */
b[0].value = n;
if (n > 0) { /*non zero matrix */
currentb = 1;
for (i = 0; i < a[0].col; i++)
/* transpose by columns in a */
for( j = 1; j <= n; j++)
/* find elements from the current column */
if (a[j].col == i) {
/* element is in current column, add it to b */

80
b[currentb].row = a[j].col;
b[currentb].col = a[j].row;
b[currentb].value = a[j].value;
currentb++
}
}
}
elements
columns

81
void fast_transpose(term a[ ], term b[ ])
{
/* the transpose of a is placed in b */
int row_terms[MAX_COL], starting_pos[MAX_COL];
int i, j, num_cols = a[0].col, num_terms = a[0].value;
b[0].row = num_cols; b[0].col = a[0].row;
b[0].value = num_terms;
if (num_terms > 0){ /*nonzero matrix*/
for (i = 0; i < num_cols; i++)
row_terms[i] = 0;
for (i = 1; i <= num_terms; i++)
row_term [a[i].col]++
starting_pos[0] = 1;
for (i =1; i < num_cols; i++)
starting_pos[i]=starting_pos[i-1] +row_terms [i-1];
columns
elements
columns

82
for (i=1; i <= num_terms, i++) {
j = starting_pos[a[i].col]++;
b[j].row = a[i].col;
b[j].col = a[i].row;
b[j].value = a[i].value;
}
}
}
Compared with 2-D array representation
O(columns+elements) vs. O(columns*rows)
elements --> columns * rows
O(columns+elements) --> O(columns*rows)
Cost: Additional row_terms and starting_pos arrays are required.
Let the two arrays row_terms and starting_pos be shared.

Sparse Matrix Multiplication
83
Definition: [D]m*p=[A]m*n* [B]n*p
Procedure: Fix a row of A and find all elements in column j
of B for j=0, 1, …, p-1.
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111
111
111
000
000
111
001
001
001

84
void mmult (term a[ ], term b[ ], term d[ ] )
/* multiply two sparse matrices */
{
int i, j, column, totalb = b[].value, totald = 0;
int rows_a = a[0].row, cols_a = a[0].col,
totala = a[0].value; int cols_b = b[0].col,
int row_begin = 1, row = a[1].row, sum =0;
int new_b[MAX_TERMS][3];
if (cols_a != b[0].row){
fprintf (stderr, “Incompatible matricesn”);
exit (1);
}

85
fast_transpose(b, new_b);
/* set boundary condition */
a[totala+1].row = rows_a;
new_b[totalb+1].row = cols_b;
new_b[totalb+1].col = 0;
for (i = 1; i <= totala; ) {
column = new_b[1].row;
for (j = 1; j <= totalb+1;) {
/* mutiply row of a by column of b */
if (a[i].row != row) {
storesum(d, &totald, row, column, &sum);
i = row_begin;
for (; new_b[j].row == column; j++)
;
column =new_b[j].row
}

86
else switch (COMPARE (a[i].col, new_b[j].col)) {
case -1: /* go to next term in a */
i++; break;
case 0: /* add terms, go to next term in a and b */
sum += (a[i++].value * new_b[j++].value);
break;
case 1: /* advance to next term in b*/
j++
}
} /* end of for j <= totalb+1 */
for (; a[i].row == row; i++)
;
row_begin = i; row = a[i].row;
} /* end of for i <=totala */
d[0].row = rows_a;
d[0].col = cols_b; d[0].value = totald;
}

87
 We store the matrices A,B & D in the arrays a,b & d
respectively.
 To place a triple in d to reset sum to 0, mmult uses
storeSum.
 In addition mmult uses several local variables that we
will describe….
 The variable row is the row of A that we are currently
multiplying with the columns in B.
 The variable rowbegin is the position in a of the first
element of the current row, and the variable column in
the column of B that we are currently multiplying with a
row in A.

Compared with matrix multiplication using
array
88
for (i =0; i < rows_a; i++)
for (j=0; j < cols_b; j++) {
sum =0;
for (k=0; k < cols_a; k++)
sum += (a[i][k] *b[k][j]);
d[i][j] =sum;
}

Data structures using C

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