data structures it

7/25/2019 Data Structures IT

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dvanced Data Structures

Chapter II:

Data Structures


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Terminologies

Data Types:Refer to the different kinds of data that a variable may hold in aprogramming languages

Examples:

integer

real

floatchar

string


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Terminologies

Data Objects:Refer to the set of elements

Example: data object integer refers to

D = {0, 1, 2, 3, }


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Terminologies

Data Structures:

Describe the set of objects and how they are related.

Describe the set of operations that can be applied on theelements of a data object.

Typically this describe the more complex data types such asarrays, stacks, queues, trees, graphs and operations of how

to manipulate these.


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Records

A structure that can have a number of heterogeneouselements.Declaration: Recor dType

t ypedef st r uct

{

f i el d1;

f i el d2;

f i el d3; f i el dN;

}Recor dType;

f i el d1

f i el d2

f i el d3

.

.

f i el dN


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Records

Field1, field2, field3 fieldN can be of any data type (i.e. integer,float, real, record)

To define a variable for the record:

Recor dType A;Assigning a value to a record field:

A. f i el d1 = ;

A. f i el d2 = ;To retrieve a value from a record field:

pr i nt f ( %d, A. f i el d1) ;

pr i nt f ( %c, A. f i el d2) ;

pr i nt f ( %d: 4: 2, A. f i el d3) ;


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Arrays

Consecutive set of memory locations is a set of pairs index and a value finite, ordered set of homogeneouselements.

Forms:

- one-dimensional array

- n-dimensional array


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Arrays

2 data types Base type of component type

Index type

Declaration i nt A[ 10] ; char B[ 45] ;

2 basic operations Extraction

Storing


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Arrays

If an array is declared to be A[n], then:

n = number of elements

If an array is declared to be A[n][m], thenn*m = number of elements

if given n-dimensional array declarationA[b][c][d][n] = b,c, n


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Arrays

To determine the ith element of a Single-dimensionarray:

A[ i ] = + ( i ) * esi ze

wher e:

- base or st ar t i ng addr ess

i el ementesi ze el ement si ze i n byt es

Exampl e: Det er mi ne t he addr ess of 5t h el ement of an

i nt eger ar r ay A wi t h a st ar t i ng addr ess of 2000


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Arrays

To determine the ith element of a Two-dimension array:

A[ i ] [ j ] = + [ ( i ) *( UB2) +( j ) ] * esi ze

wher e:

UB2 upper bound of t he 2nd di mensi on

- base or st ar t i ng addr ess

esi ze el ement si ze i n byt es


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Arrays

To determine the ith element of a Three-dimension array:

A[ i ] [ j ] [ k] = +[ ( i ) *( UB2) *( UB3) +( j ) *( UB3) +( k) ] *esi ze

wher e:

UB3 upper bound of t he 3r d di mensi on

UB2 upper bound of t he 2nd di mensi on - base or st ar t i ng addr ess

esi ze el ement si ze i n byt es


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Arrays

Exercises:

1. Given A[10][3][3][6], =2000, esize=4 bytes:

a. f i nd t he f or mul a t o r epr esent an el ement i n a 4-di mensi onal ar r ay.

b. f i nd t he t ot al number of el ement s

c. f i nd t he addr ess of A[ 2] [ 2] [ 0] [ 4]

2. Given X[8][3][6][2][3], =3000, esize=3 bytes:

a. f i nd t he t ot al number of el ement s

b. f i nd t he addr ess of X[ 0] [ 2] [ 5] [ 1] [ 2]


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Arrays

3. Consider the following declaration:

t ypedef st r uct {

i nt A;

char B[ 10] ;

f l oat C;

char D;

}r ect ype;

t ypedef r ect ype mat r i x[ 121] [ 4] [ 5] ;

mat r i x A1;

a. comput e t he addr ess of el ement A1[ 120] [ 3] [ 3] gi ven t hebase addr ess at 2000.

b. Assume t hat we do not know t he si ze of RECTYPE i nnumber of byt es but we know t hat t he addr ess of

A1[ 20] [ 2] [ 3] i s 2160. Gi ve t he si ze of RECTYPE i nbyt es. Assume t he base addr ess i s at 2000.


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Stacks

An ordered list in which all insertions and deletions aremade at one end called the TOP.

LIFO (Last In First Out)A B C D E F

TOP

G

A B C D E F G

TOP


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Stacks

Operations: Create(top) Create an empty stack

Push(Stack, top, item) Inserts an element item into the stack

Pop(Stack, top, item) Removes the top element of the stack

and stores the value in item S_top(Stack, top) Returns the top element of the stack

Empty(top) Determines whether the stack is empty or not.

Stack full: Top = n-1 Stack empty: Top = -1


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Stacks

Representation: One-dimensional array

Singly linked-list

TOP

A B C D E F

A B

TOP

C D


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Stacks

Declaration

#def i ne n

t ypedef el ement t ype;

t ypedef el ement t ype St ack[ n] ;

Example:

Processing of procedure calls and their terminations


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Stacks

Procedures for Stack Operations:

voi d cr eat e( i nt *t op)

{*t op = - 1; }

voi d push( st ack S; i nt *t op; el ement t ype i t em){

i f ( t op==n- 1)

st ackf ul l ;

el se {*t op++;

S[ *t op] = i t em;

}

}


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Stacks

voi d pop( st ack S; i nt *t op; el ement t ype *i t em)

{ i f ( t op==1) st ackempt y;

el se{ *i t em = S[ t op] ;

* top- - ;}

}el ement t ype s_t op( st ack S; i nt t op)

{ i f ( t op==- 1) er r or _rout i ne;

el se r et ur n S[ t op] ;

}

i nt empt y( i nt t op)

{ i f ( t op == 1) r et urn 1;

el se r et ur n 0;

}


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Stacks

voi d mai n( )

{ cr eat e( &t op1) ; cr eat e( &t op2) ;s_empt y = empt y( t op1) ;

pr i nt f ( %d, s_empt y) ;

push( s1, &t op1, 16) ;

s_empt y = empt y( t op1) ;

pr i nt f ( %d, s_empt y) ;

push( s1, &t op1, 10) ; push( s1, &t op1, 9) ;

push( s1, &t op1, 8) ; push( s1, &t op1, 7) ;j =s_t op( s1, t op1) ;

pr i nt f ( Top i s %d, j ) ;

pr i nt f ( %d\ n, t op1) ;


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Stacks

s_empt y = empt y( t op2) ;

pr i nt f ( %d, s_empt y) ;push( s2, &t op2, 10) ; push( s2, &t op2, 9) ;



pop( s1, &t op1, &i t em) ;

pop( s2, &t op2, &i t em) ;

j =s_t op( s2, t op2) ; pr i nt f ( Top i s %d, j ) ;

pr i nt f ( %d\ n, t op2) ;}


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Evaluation of Expressions

Expression is made up of operands, operators anddelimiters.

Operands can be any legal variable names or constantsin programming languages.

Operations are described by operators:Basic arithmetic operators: + - * /

Unary operators: - +

Relational operators: ==, >, =,


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Our concern: how the expressions are evaluated

The compiler accepts expressions and produce correct result byreworking the expressions into postfix form. Other forms include

infix and prefix.

Prefix :

Postfix : Infix :


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Expression: A + B

Prefix : +AB

Postfix: AB+

Expression: (A+B*C)/D

Prefix : /+A*BCD

Postfix: ABC*+D/


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Conversion from Infix to Postfix

using Stacks

IN-Stack Priority (ISP) The priority of the operator asan element of the stack.

IN-Coming Priority (ICP) The priority of the operatoras current token.

SYMBOL ISP ICP

) -- --^ 3 4

*, / 2 2

+,- 1 1

( 0 4


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using Stacks

Rule:

Operators are taken out of the stack (POP) as long astheir ISP is greater than or equal to the ICP of the newoperator.

Note: ISP and ICP of # sign = -1


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using Stacks

Algorithm:voi d POSTFI X( expr essi on E)

t oken x, y;

st ack[ 0] = # ; t op = 0;x = next t oken( E) ; / / t akes t he f i r st t oken and r emove i t

f r om t he or i gi nal expr essi on.

whi l e x ! = #

{ i f x i s an oper and pr i nt f ( x) ;el se i f x == ) {/ / unst ack unt i l (

whi l e st ack[ t op] ! = (

{ pop( y) ; pr i nt f ( y) ; }

pop( y) ; / / del et e (

}


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using Stacks

el se{ whi l e i sp[ st ack[ t op] ] >= i cp[ x]

{ pop( y) ; pr i nt f ( y) ; }push( x) ;

}

x = next t oken( E) ;

}

i f ( ! empt y( st ack) )

{ whi l e st ack[ t op] ! = #

{ pop( y) ;pr i nt f ( y) ;

}

}


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Queues

An ordered list which all insertions take place at oneend, called the REAR, while all deletions take place atthe other end, called the FRONT.

FIFO (First-In-First-Out)

Elements are processed in the same order as they werereceived. The first element inserted in the queue will bethe first one to be removed.


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Queues

Conventions for FRONT and REAR: Front is always 1 less than the actual front of the queue.

Rear always points to the last element in the queue.

Initial value: Front = Rear = -1

Operations: Createq(Front, Rear) creates an empty queue

Insert(Queue, Rear, Item) inserts the element item to the rear

of the queue. Delete(Queue, Front, Rear, Item) removes the front element

from the queue and assigns is to variable item.

Qfront(Queue, Front, Rear) returns the front element of thequeue


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Queues

Qempty(Front, Rear) determines if the queue is empty or not

Returns 1 if true

Otherwise return 0

Front = Rear means queue is empty.


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Queues

Representation

One-dimensional Array

Singly linked-list

Rear

A B C D E F G

A B C D

Front

Rear Front


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Queues

Declaration:

#def i ne n

t ypedef el ement t ype;

t ypedef el ement t ype Queue[ n] ;t ypedef i nt FR

FR f r ont , r ear ;

Queue Q;


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Queues

Example:Processing of customers transactions (i.e. cashiers, bank-related)

Procedures for Queue Operations:

voi d cr eat eq( FR *f r ont , FR *r ear )

{ *f r ont = *r ear = - 1 }

voi d i nser t ( queue Q, FR *r ear , el ement t ype i t em){ i f ( r ear == n- 1) queuef ul l ;

el se { ( *r ear ) ++;

Q[ *r ear ] = i t em; }

}


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Queues

voi d del et e( queue Q, FR *f r ont , FR *r ear , el ement t ype

*i t em){ i f ( *f r ont == r ear ) queueempt y;

el se { ( *f r ont ) ++;

i t em = Q[ f r ont ] ; }

}

el ement t ype qf r ont ( queue Q, FR f r ont , FR r ear )

{ i f ( f r ont == r ear ) er r or r out i ne;

el se r et ur n( Q[ f r ont + 1] )}

i nt quempt y( )

{ i f ( f r ont == r ear ) r et ur n 1;

el se r et ur n 0;

}


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Queues

Notes:

QUEUEFULL signal does not necessary imply thatthere are N elements in the queue.

To solve this, move the entire queue to the left so thatthe first element is again at Q[0] and front = -1.


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Circular Queues

n- 1

0

1

23

. .

. .

4


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Circular Queues

voi d i nser t ( queue *Q, FR f r ont , FR *r ear , el ement t ype

i t em){ i f ( f r ont == ( r ear +1) %n) queuef ul l ;

el se { *r ear = ( *r ear +1) %n;

Q[ *r ear ] = i t em; }

}

voi d del et e( queue Q, FR *f r ont , FR r ear , el ement t ype

*i t em){ i f ( f r ont == r ear ) t hen cqueueempt y;

el se { *f r ont = ( *f r ont +1) %n;

*i t em = Q[ f r ont ] ; }

}


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Circular Queues

el ement t ype cqf r ont ( queue Q, FR f r ont , FR r ear )

{ i f ( f r ont == rear ) er r or r out i ng;

el se r et ur n( Q[ f r ont +1] %n) ;

}

i nt cquempt y( )

{ i f ( f r ont == r ear ) r et ur n 1;

el se r et ur n 0;

}


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Exercises

Convert the following infix expressions to postfix and

prefix:

1. A+B*C/ D

2. A/ B C+D*E- A*C

3. ( A+B) *D+E/ ( F+A*D) +C

4. A+B*D E F G/ H J *K- L5. ! ( A&&! ( BD) ) | | ( C


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Exercises

Convert the following prefix expressions to infix:

6. - +- ABC*D EFG

7. +- ABC+D- EF

8. | | | | &&ABC!


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Exercises

Convert the following infix expressions to postfix

using stack:12. B+C ( E+F*G H) / ( K- L/ M) +N

13. B+C/ D ( E+F*G H) +Z

14. A+B- C*D*( E+F- G*H I J ) +L/ M+( N*O/ P+

( Q R S T) ) +U

data structures it

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