polish notations

8/17/2019 Polish Notations

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Linear Data Structures

Stack


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Topics Covered in This Lecture

Applications of stack

– Reversing the string

–Balancing symbols

– Postfix expression evaluation

– Translating infix expression to postfix expression


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Applications of Stack

Reversing the string

– push each character on to a stack as it is read.

– When the line is finished, we then pop characters off

the stack, and they will come off in the reverse order.


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Balancing SymbolsCompilers use a program that checks hether every symbol !like braces" parenthesis"brackets" etc# in a program is balanced$

The simple algorithm for this purpose is%

1. Make an empty stack.

2. Read the characters until end of file.

3. If the character is an open any thing, push it onto the stack.

. If it is a close any thing, then

if the stack is empty report an error.!ther"ise #op the Stack.

If the popped sym$ol is not the corresponding opening sym$ol, thenreport an error.

%. &t the end of the file, if the stack is not empty report an error.


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#olish 'otation

a – b / c + d * e

Precedence?

1. b/c

2. d*e3. a – a1 /a1 = b/c /

4. t2 + a2 / t2 = a – b/c / /a2 = d*e/


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Infi(, Suffi(, #refi(

Infix = a * b + c

((a*b) +c)

Priority:

1. a * b

2. a1 + c / a1 = a * b /

Prefix =

* a b , +a1 c

+*abc

Sffix =

ab* , a1c+

ab*c+


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infix = a- b * c / d + e / f

!ffix =a – bc* / d + e / f

a – bc*d/ + e / f

a – bc*d/ + ef/

abc*d/" + ef/

abc*d/"ef/+

#refix =a " *bc / d + e / f

a " /*bcd + e / f

a " /*bcd + /ef

"a/*bcd + /ef

+"a/*bcd/ef


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$nfix:

a+b*c–d/f+e

Sffix:

abc*+df/"e+

Prefix:

+"+a*bc/dfe


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Postfix &xpression &valuation

– 'hen a number is seen" it is pushed onto the stack

– 'hen an operator is seen" then pop to elements fromstack and push the result onto the stack$

(o e evaluate the folloing postfix expression$

) * + , - . / - , - /0$ The first four are placed on the stack$ The resulting stack is

3

2

5

6

stack


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evaluating the folloing postfix expression$

) * + , - . / - , - /3

2

5

6

stack +$ (ext a - is read" so , and + are popped from the stack and their sum * ispushed$ 5

5

6

stack


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) * + , - . / - , - /

5

5

6

stack

,$ (ext . is read and pushed$ 85

5

6

stack


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) * + , - . / - , - /

1$ (ext a / is seen so . and * are popped as . / * 2 13 is pushed40

5

6

stack

8

5

5

6

stack


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) * + , - . / - , - /

*$ (ext a - is read so 13 and * are popped and 13 - * 2 1* is pushed$

40

5

6

stack

45

6

stack


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) * + , - . / - , - /

)$ (o , is pushed

45

6

stack

3

45

6

stack


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) * + , - . / - , - /

4$ (ext symbol is - so pops , and 1* and pushes 1* - , 2 1." so push 1. in stack$

345

6

stack

48

6

stack


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Applications of StackTranslating infix expressions to postfix expression

– 'hen an operand is read" it is immediately placed onto the output$

– 'hen an operator or left parenthesis comes then save it in the stack

initially stack is empty$

– 6f e see a right parenthesis" then e pop the stack" riting symbols until e encounter a !corresponding# left parenthesis" hich is popped but notoutput$

–

6f e see any other symbol !7-8" 7/8" 7!7" etc# then e pop entries form thestack until e find an entry of loer priority$ 9ne exception is that enever remove a 7!7 from the stack except hen processing a 7#8$

– 'hen the popping is done" e push the operand onto the stack$

– 5inally" if e read the end of input" e pop the stack until it is empty" riting symbols onto the output$


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Translating infix expressions to postfix expression

Convert the folloing infix expression to postfix expression$

a-b/c-!d/e-f#/g

0$ 5irst the symbol a is read" so it is passed through to the output aoutput+$ Then - is read and pushed onto the stack.

stack

4. !e"t a # is read. $he top entry on the operator stack has lower

precedence than #, so nothin% is output and # is put on the .

,$ (ext b is read and passed through to the output$ a&

output #

stack


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Converting the folloing infix expression to postfix expression$

a-b/c-!d/e-f#/g

*$ (ext" c is read and output$

)$ The next symbol is a -$ Checking the stack" e find that priority of stack top

symbol / is higher than - $ So e pop a / and place it on the output" Pop the

other -" hich is not of loer but e:ual priority" and then push -$

stack

a&c#output

#

stack

a&c

output


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a-b/c-!d/e-f#/g

4$ The next symbol read is an 7!7" hich" being of highest precedence" is placed on

the stack$

'

stack

.$ Then d is read and output$ a&c#d

output


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a-b/c-!d/e-f#/g

;$ 'e continue by reading a /$ Since open parenthesis do not get removed except

hen a closed parenthesis is being processed" there is no output and e push /

in stack

#

'

stack

03$ (ext" e is read and output$a&c#de

output


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a-b/c-!d/e-f#/g

0,$ (o e read a 7#8" so the stack is emptied back to the 7!7" e output a -$

01$ 'e read a / next< it is pushed onto the stack$

a&c#de#f

output

stack

#

stack

0*$ (o" g is read and output$ a&c#de#f%

output


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a-b/c-!d/e-f#/g

0)$ The input is no empty" so pop output symbols from the stack until it is empty$

a&c#de#f%#

outputstack

#

stack

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