Preeja Unit1 Stack

8/11/2019 Preeja Unit1 Stack

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Preeja

References:

Data Structures and Algorithms:Alfred V. Aho, John E. Hopcroft, Jeffrey D. Ullman

A Practical Approach to Data Structures andAlgorithms: Sanjay Pahuja


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STACK An Abstract Data Type

Special List : insertions & deletions at one

end(top)

Also called LIFO(Last In First Out) list.

8/27/2014 2Data Structures - Module1


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8/27/2014 Data Structures - Module1 3


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Operations performed on stack MAKENULL(S) : Make stack S an empty stack.

TOP(S) : Return the top element of stack S.

POP(S) : Delete the top element from the stack. {check

for stack empty condition}

PUSH(x,S) : Insert the element x at the top of stack S.{ Stack overflow condition should be checked}

EMPTY(S) : Returns true if S is an empty stack.



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Stack Implementation Static implementation(Using arrays)

Dynamic implementation(Using pointers: Linked list)



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Definition of Array-Based Stack typedef struct s

{

intdata[MAX];inttop;

}s;



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Push Operation on StackWe have declared data array in the above declaration.

Whenever we add any element in the data array then

it will be called as Pushing Data on the Stack.



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Initial stack


3

2

1

0

Top = -1


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Push (12, S)


s.top:=s.top+1= -1 + 1 = 0;

s.elements[s.top]:= s.elements[0] = 12;

3

2

1

0 12 Top = 0


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Push (47, S)


s.top:=s.top+1= 1 + 1 = 2;

s.elements[s.top]:= s.elements[2] = 47;

3

2 47

1 34

0 12

Top = 2


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POP Operation on StackWhenever we try to remove element from the stack

then the operation is called as POP Operation on

Stack.



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But pseudocode does not use specific programminglanguage syntax and therefore could be understood byprogrammers who are familiar with differentprogramming languages.

Additionally, transforming an algorithm presented inpseudocode to programming code could be mucheasier than converting an algorithm written in naturallanguage.



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Operation PUSH(Stack, Top, MAX Stack, Item)

1.Stack is already filled?

If top=MAX STK

Print :Over flow and Return

2.Set top=top+1

3.Set stack[top]=item

4.return


Algorithm for push


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POP(Stack,top,Item)

1. Stack has an item to be removed.

If top=0

Print :underflow and return

2. Set item =stack[top]

3. Set top=top-1

Return


Algorithm for POP


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Applications of stack Explicit and implicit recursive calls

Conversion of Infix to postfix

Postfix expression evaluation

Conversion of infix to prefix

Prefix expression evaluation



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Applications of stack Expression : operands together with operator

Infix notation(A+B)

Prefix notation( +AB)

Postfix notation( AB+) Operator precedence

^ (Exponential ) .Highest

*, /(mul, div) Next highest+, - (add,sub) Lowest



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Infix ,Prefix and Postfix Prefix is easy to program for in computers, but postfix

uses less memory.

Infix notation:X + Y Operators are written in-between

their operands. This is the usual way we writeexpressions.

An expression such asA * ( B + C ) / Dis usually takento mean something like: "First add B and C together,

then multiply the result by A, then divide by D to givethe final answer."



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Infix ,Prefix and Postfix Prefix notation : + X Y Operators are written before

their operands.

The expressions given above are equivalent

to / * A + B C D

As for Postfix, operators are evaluated left-to-right andbrackets are superfluous. Operators act on the two

nearest values on the right.



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Infix ,Prefix and Postfix



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Applications of stack

Conversion from infix to postfix

Rules

1. Parenthesize the expression starting from left to right

2. The operands associated with the operator having higherprecedence will be parenthesized first.

3. The sub expression which has been converted to postfix

is to be treated as single operand.

4. Once the expression is converted to postfix form,

remove the parenthesis



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Infix to postfix conversion

algorithm1. In this Algorithm we are reading token from Left to

Right and Postfix expression is generated.

2. So Entered Token maybe -a) Alphabet from A-Z or a-Z

b) Numerical Digit from 0-9

c) Operator

d) Opening And Closing Braces ( , )3. If Entered Character isAlphabetthen Following

Action Should be taken-

a. Print Alphabet as Output



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4. If Entered Character isDigitthen Following

Action Should be taken-1. Print Digit as Output

5. If Entered Character isOpening BracketthenFollowing Action Should be taken-

1. Push ( Onto Stack2. If any Operator Appears before ) then Push it onto

Stack.

3. If Corresponding ) bracket appears then Start

Removing Elements [Pop] from Stack till ( isremoved.



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6. If Entered Character isOperatorthen Following

Action Should be taken-a) Check Whether There is any Operator Already present

in Stack or not.

b) If Stack is Empty then Push Operator Onto Stack.

c) If Present then Check Whether Priority of IncomingOperatoris greater than Priority of Topmost StackOperator.

d) If Priority of Incoming Operator is Greater then Push

Incoming Operator Onto Stack.e) Else Pop Operator From Stack again goto Step 6.



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Infix to Prefix ConversionStep 1. Push ) onto STACK, and add ( to end of the A

Step 2. Scan A from right to left and repeat step 3 to 6 for

each element of A until the STACK is empty

Step 3. If an operand is encountered add it to B

Step 4. If a right parenthesis is encountered push it onto

STACK



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Step 5. If an operator is encountered then:

a. Repeatedly pop from STACK and add to B each

operator (on the top of STACK) which has same orhigher precedence than the operator.

b. Add operator to STACK



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Step 6. If left parenthesis is encountered then

a. Repeatedly pop from the STACK and add to B

(each operator on top of stack until a left parenthesis isencounterd)

b. Remove the left parenthesis

Step 7. Exit



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Evaluating Postfix expression(1) Add a right parenthesis )at the end of P[this acts as a

sentinel].

(2) Scan P from left to right and repeat steps 3 and 4 foreach element of P until the sentinel )is encountered.

(3) If an operand is encountered, put it on STACK.(4) If an operator # is encountered, then

(a) Remove the two top elements of STACK ,where A isthe top element and B is the next-to-top element.

(b) Evaluate B # A.(c) Place the result onto STACK.

5) RESULT equal to the top element on stack.

6) Exit



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Evaluate 45+72-*


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Evaluate 59+2*65*+



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Infix to prefix conversion algorithm

I => infix expressionO => output expression

P => prefix expression

(1) Push )onto stack and add (to the beginning of I.(2) Scan I from right to left and repeat steps 3 to 6 for each

element of I until the stack is empty.

(3) If an operand is encountered, add it to O.

(4) If an operator # is encountered, then

(a) Repeatedly pop from stack and add to O eachoperator (on the top of stack) which has higher

precedence than #.



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(b) Add # to stack.

5. If a right parenthesis is encountered, push it onto thestack.

6. If a left parenthesis is encountered then

(a) Repeatedly pop from stack and add to O, until a

right parenthesis is encountered.

(b) Remove the right parenthesis.

[ do not add the right parenthesis to O].

7. Reverse O to get the prefix expression P.8. Exit

# Symbolizes any operator in I.8/27/2014 36Data Structures - Module1


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Convert (A*B-F*H)^D


Symbol Scanned Operator Stack Output Expression

D ) D^ )^ D

) )^) D

H )^) DH

* )^)* DH

F )^)* DHF

- )^)- DHF*

B )^)- DHF*B

* )^)-* DHF*B

A )^)-* DHF*BA( )^ DHF*BA*-

( (Empty) DHF*BA*-^

Result: ^-*AB*FHD


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Home Work Convert to prefix:

A+ [ B+C - D+E] * F / G

+A/*+-+BCDEFG (ab)/c*(d + e f / g)

*/-abc-+de/fg

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