lecture 3 stack

8/7/2019 Lecture 3 Stack

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Data Structures & Algorithms

Lecture # 3


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Today

Topics

What are Stacks?

Representation of Stacks

Operations on Stacks Push Pop

Evaluation of Expressions

Infix to Postfix Conversion


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Stack application Function calls

Reversing Data: We can use stacks to reverse data.

(example: files, strings)

Very useful for finding palindromes.Consider the following pseudocode:

1) read (data)

2) loop (data not EOF and stack not full)

1) push (data)

2) read (data)

3) Loop (while stack notEmpty)

1) pop (data)

2) print (data)


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Stack application


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What are Stacks?

In array insertion and deletion can take place atboth end, i.e. start and end,

But if insertion and deletion are restricted from oneend, must used STACKS , QUEUES.

Stack can be defined as:

A stack is a linear data structure which canbe accessed only at one of its ends for

storing and retrieving data. OR In which item may b inserted or deleted only at one

end called top of the stack. There are two ways of implementing a stack:

Array (Static) Link List (Dynamic)


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Example of Stack A common model of a

stack is a plate or coinstacker. Plates are

"pushed onto to the topand "popped" off from thetop.

Stacks form Last-In-First-Out (LIFO) queues.


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Stack terminology

Top: end where insertion and

deletion take place

Push: is the term used to insert an

element into an array

Pop: is the term used to delete an

element from a stack.


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Basic operations of a StackNew Push

Pop Peek

?

Empty?


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Array implementation of Stacks

A stack can be implemented with an arrayand an integer.

A pointer variable TOP which contains thelocation of the top element of the stack;

Variable MAXSTK which gives the maximum

number of elements that can be held by thestack.

Condition TOP=0, or MAXSTK=NULLindicate that the stack is empty.


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Note that all operations need check arraybounds

Pushing onto a full stack: Overflow When TOP=MAXSTK

Popping an empty stack: Underflow

TOP=NULL

Stack implemented in an array


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Operations on Stack A stack is generally implemented with only

two principle operations (apart from aconstructor and destructor methods):

initializeStack: initializes the stack to an emptystate

destroyStack: Removing all the element 4m thestack

Is EmpthyStack: check whether stack is empty or

not, and return true or false. IsFullStack: returns true if full else false. Top: returns the top elements of the stack; Push: adds an item to a stack Pop: extracts the most recently pushed item from

the stack


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How the stack routines work

empty stack; push(a), push(b); pop

empty stacktop = 0

a

push(a)top = 1

b

a

push(b)top = 2

a

pop()top = 1


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Implementation of the operation Initialize Stack: Destroy Stack: EmpthyStack: Full Stack:

Top: returns the top elements of the stack; Push: tow step process, adding and

incrementing the top pointer. Pop: extracts the most recently pushed item

from the stack


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Algorithms of Push OperationPush (STACK , TOP, MAXST, ITEM)

Algorithm for push element into the Stack

1. [stack already filled]IfTOP= MAXSTK, then: print: OVERFLOW,and return.

2. [increases TOP by 1] Set TOP:= TOP + 1

3. [insert ITEM in new TOP position ]

Set STACK[TOP]: =ITEM4. RETRUN


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Algorithms of Pop OperationPop (STACK, TOP, ITEM)

Algorithm for pop element from the Stack

1. [Stack has an item to be removed]

IfTOP= NUL, then

Print: UNDERFLOWand return

2. [assignTOP element to ITEM]

Set ITEM := STACK[TOP]

3. [DecreaseTOP by 1]Set TOP:= TOP -1

4. Return


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Algorithm for Display Stack

elementsDisplay_Stack ()

Algorithm for display stack elements

1. Start2. Repeat step 3 For i = 1 to TOP by 1

3. Print S[i]

4. End


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Algorithm for isempty Stackisempty ()

Algorithm for return the Stack status

1. Start2. ifTOP = 0 then

Return True

else

Return False3. End


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Algorithm for return the top of

stack valuetop ()

Algorithm for return the top of Stackvalue

1. Start

2. Return TOP

3. End


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Evaluation of Expression Order in which the expression is evaluated is:

If the expression has parenthesis, then they areevaluated first

Exponential (^) is given highest priority

Multiplication (*) and division (/) have the nexthighest priority

Addition (+) and subtraction (-) have the lowestpriority


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Evaluation of Expression Steps to evaluate the following expression:

(2^3 + 6) * 2 9 / 3

= (8 + 6) * 2 9 / 3

= 14 * 2 9 / 3

= 28 3

= 25


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Infix, Prefix (Polish) and Postfix

(Reverse Polish) NotationInfix Prefix

(Polish Notation)

Postfix

(Reverse PolishNotation)

A+B +AB AB+

A*B *AB AB*

A/B /AB AB/

A-B -AB AB-


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Evaluation of Expression Computer evaluates an expression given in infix notation by

converting it into postfix notation.

Stack is used to perform this operation

Following steps are take to evaluate a postfix expression:

Expression is scanned from left to right until the end of theexpression

When an operand is encountered, it is pushed into stack

When an operator is encountered, then Top two operands of stack are removed Arithmetic operation is performed Computed result is pushed back to the stack

When end of the expression is reached, the top value from thestack is picked. It is the computed value of the expression


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Infix to Postfix Conversion Stack is used to convert an infix expression to postfix.

Stack is used to store operators and operands and then pass tothe postfix expression according to their precedence.

Infix expression is converted into postfix expression accordingto the following rules:

Infix expression is scanned from left to right until end ofthe expression.

Operands are passed directly to the output.

Operators are first passed to the stacks.

Operand is encountered, it is added to the output.


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Infix to Postfix Conversion Each time an operator is read, the stack is repeatedly

popped and operands are passed to the output, until anoperator is reached that has a lower precedence than themost recently read operator. The most recently readoperator is then pushed onto the stack.

When end of the infix expression is reached, all operatorsremaining in the stack are popped and passed to theoutput in the same sequence.

Parentheses can be used in the infix expression but theseare not used in the postfix expression.

During conversion process, parentheses are treated asoperators that have higher precedence than any otheroperator.


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Infix to Postfix Conversion Left parenthesis is pushed into the stack, when

encountered.

Right parenthesis is never pushed to the stack.

Left parenthesis is popped only when rightparenthesis is encountered.

The parentheses are not passed to the output postfixexpressions. They are discarded.

When end of expression is reached, then all operatorsfrom stack are popped and added to the output.


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Example (Infix to Postfix

Conversion)Convert infix expression

A+B*C+(D*E+F)*G into postfix

A + B * C + ( D * E + F ) * G

1. Scanned from left to right. Firstoperand read is A and passed tooutput

Stack

Output: A


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Conversion)A + B * C + ( D * E + F ) * G

2. Next the + operator is read, atthis stage, stack is empty.

Therefore no operators arepopped and + is pushed into thestack. Thus the stack and outputwill be:

+

Stack

Output: A


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3. Next the B operand is read andpassed to the output. Thus the

stack and output will be:

+

Stack

Output: AB


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4. Next the * operator is read, Thestack has + operator which has

lower precedence than the *operator. Therefore no operatorsare popped and * is pushed intothe stack. Thus the stack andoutput will be:

*

+

Stack

Output: AB


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5. Next the C operand is read andpassed to the output. Thus the


*

+

Stack

Output: ABC


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6. Next the + operator is read, Thestack has * operator which has

higher precedence than the +operator. The Stack is poppedand passed to output. Next stackhas + operator which has sameprecedence than the + operator.The Stack is popped and passedto output. Now stack is empty,therefore no operators arepopped and + is pushed into thestack. Thus the stack and outputwill be:

+

Stack

Output: ABC*+


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Example (Infix to PostfixConversion)

A + B * C + ( D * E + F ) * G

7. Next the left parenthesis ( isread, Since all operators have

lower precedence than the leftparenthesis, it is pushed into thestack. Thus the stack and outputwill be:

(

+

Stack

Output: ABC*+


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A + B * C + ( D * E + F ) * G

8. Next the D operand is read andpassed to the output. Thus the


(

+

Stack

Output: ABC*+D


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A + B * C + ( D * E + F ) * G

9. Next the * operator is read.Now, left parenthesis ( has

higher precedence than *; it cannot be popped from the stackuntil a right parenthesis ) hasbeen read. Thus the stack is notpopped and * is pushed into thestack. Thus the stack and outputwill be:

*

(

+

Stack

Output: ABC*+D


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A + B * C + ( D * E + F ) * G

10. Next the E operand is read andpassed to the output. Thus the


*

(

+

Stack

Output: ABC*+DE


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A + B * C + ( D * E + F ) * G

11. Next the + operator is read, Thestack has * operator which has

higher precedence than the +operator. The Stack is poppedand passed to output. Next stackhas left parenthesis ( which hasnot been popped and + operatoris pushed into the stack. Thusthe stack and output will be:

+

(

+

Stack

Output: AB*+DE*


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A + B * C + ( D * E + F ) * G

12. Next the F operand is read andpassed to the output. Thus the


+

(

+

Stack

Output: ABC*+DE*F


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A + B * C + ( D * E + F ) * G

14. Next the * operator is read, Thestack has + operator which has

lower precedence than the *operator. Therefore no operatorsare popped and * is pushed intothe stack. Thus the stack andoutput will be:

*

+

Stack

Output: ABC*+DE*F+


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A + B * C + ( D * E + F ) * G

15. Next the G operand is read andpassed to the output. Thus the


*

+

Stack

Output: ABC*+DE*F+G


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A + B * C + ( D * E + F ) * G

16. The end of expression isencountered. The operators are

popped from the stacked andpassed to the output in the samesequence in which these arepopped. Thus the stack andoutput will be:

Stack

Output: ABC*+DE*F+G*+


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Example (Evaluation of Expression)

Postfix Expression

A B C * + D E * F + G * +

A=5, B=6, C=9, D=2, E=4, F=8, G=3

1. Scanned from left to right. In first,second and third iteration, the valueof A, B and C are pushed into thestack. Thus the stack will be: 9

6

5

Stack


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Postfix Expression

A B C * + D E * F + G * +

A=5, B=6, C=9, D=2, E=4, F=8, G=3

2. In fourth iteration, the operator * isread. So the two values 9 and 6are popped from the stack andmultiplication is perform. i.e 9*6=54.The computed value pushed back tothe stack. Thus the stack will be: 54

5

Stack


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Postfix Expression

A B C * + D E * F + G * +

A=5, B=6, C=9, D=2, E=4, F=8, G=3

3. In fifth iteration, the operator + isread. So the two values 54 and 5are popped from the stack andaddition is perform. i.e 54+5=59.The computed value pushed back tothe stack. Thus the stack will be:

59

Stack


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Postfix Expression

A B C * + D E * F + G * +

A=5, B=6, C=9, D=2, E=4, F=8, G=3

4. In sixth and seventh iteration, thevalue of D and E are pushed into thestack. Thus the stack will be:

4

2

59

Stack


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Postfix Expression

A B C * + D E * F + G * +

A=5, B=6, C=9, D=2, E=4, F=8, G=3

5. In eighth iteration, the operator * isread. So the two values 4 and 2are popped from the stack andmultiplication is perform. i.e 2*4=8.The computed value pushed back tothe stack. Thus the stack will be:

8

59

Stack


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Postfix Expression

A B C * + D E * F + G * +

A=5, B=6, C=9, D=2, E=4, F=8, G=3

6. In ninth iteration, the value of F ispushed into the stack. Thus the stackwill be:

8

8

59

Stack


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Postfix Expression

A B C * + D E * F + G * +

A=5, B=6, C=9, D=2, E=4, F=8, G=3

7. In tenth iteration, the operator + isread. So the two values 8 and 8are popped from the stack andaddition is perform. i.e 8+8=16. Thecomputed value pushed back to thestack. Thus the stack will be:

16

59

Stack


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Postfix Expression

A B C * + D E * F + G * +

A=5, B=6, C=9, D=2, E=4, F=8, G=3

8. In eleventh iteration, the value of Gis pushed into the stack. Thus thestack will be:

3

16

59

Stack


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Postfix Expression

A B C * + D E * F + G * +

A=5, B=6, C=9, D=2, E=4, F=8, G=3

9. In twelve iteration, the operator * isread. So the two values 3 and 16are popped from the stack andmultiplication is perform. i.e3*16=48. The computed valuepushed back to the stack. Thus the

stack will be:

48

59

Stack


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Postfix Expression

A B C * + D E * F + G * +

A=5, B=6, C=9, D=2, E=4, F=8, G=3

10. In thirteen iteration, the operator +is read. So the two values 48 and59 are popped from the stack andaddition is perform. i.e 48+59=107.The computed value pushed back tothe stack. Thus the stack will be:

107

Stack


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A + B * C + ( D * E + F ) * G

A=5, B=6, C=9, D=2, E=4, F=8, G=3

= 5 + 6 * 9 + (2 * 4 + 8) * 3

= 5 + 54 + (8 + 8) * 3

= 59 + 16 * 3

= 59 + 48

=107


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Next Lecture

What are Queues?

Representation of Queues

Operations on Queues

QInsert

QDelete


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References

Structures: A Pseudocode Apporoach with C. Gilberg, RichardF., Forouzan, Data Behrouz A.. PWS Publishing Company:

1998

lecture 3 stack

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