infix to postifix

7/27/2019 Infix to Postifix

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Prefix, Postfix, Infix

NotationUniversity College of

Engineering, Kanchipuram


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Infix Notation

To add A, B, we writeA+B

To multiply A, B, we writeA*B

The operators ('+' and '*') go inbetween the operands ('A' and 'B')This is "Infix"notation.


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Prefix Notation

Instead of saying "A plus B", wecould say "add A,B " and write

+ A B"Multiply A,B" would be written

* A BThis is Prefixnotation.


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Postfix Notation

Another alternative is to put theoperators after the operands as in

A B +and

A B *This is Postfixnotation.


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The terms infix, prefix, and postfixtell us whether the operators gobetween, before, or after theoperands.


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Parentheses

Evaluate 2+3*5.+ First:

(2+3)*5 = 5*5 = 25* First:

2+(3*5) = 2+15 = 17Infix notation requires Parentheses.


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What about Prefix Notation?

+ 2 * 3 5 == + 2 * 3 5= + 2 15 = 17

* + 2 3 5 =

= * + 2 3 5= * 5 5 = 25

No parentheses needed!


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Postfix Notation

2 3 5 * + == 2 3 5 * += 2 15 + = 17

2 3 + 5 * =

= 2 3 + 5 *= 5 5 * = 25

No parentheses needed here either!


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Conclusion:

Infix is the only notation thatrequires parentheses in order tochange the order in which theoperations are done.


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Infix to Prefix Conversion

Move each operator to the left of itsoperands & remove the parentheses:

( ( A + B) * ( C + D ) )


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( + A B * ( C + D ) )


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* + A B ( C + D )


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* + A B + C D

Order of operands does not change!


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Infix to Postfix

( ( ( A + B ) * C ) - ( ( D + E ) / F ) )

A B + C * D E + F / -

Operand order does not change!Operators are in order of evaluation!


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Computer AlgorithmFPE Infix To Postfix

Assumptions:1. Space delimited list of tokens

represents a FPE infix expression2. Operands are single characters.

3. Operators +,-,*,/


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FPE Infix To Postfix

Initialize a Stack for operators,

output listSplit the input into a list of tokens.for each token (left to right):

if it is operand: append to outputif it is '(': push onto Stackif it is ')': pop & append till '('


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FPE Infix to Postfix

( ( ( A + B ) * ( C - E ) ) / ( F + G ) )

stack: output: []


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( ( A + B ) * ( C - E ) ) / ( F + G ) )

stack: (output: []


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( A + B ) * ( C - E ) ) / ( F + G ) )

stack: ( (output: []


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A + B ) * ( C - E ) ) / ( F + G ) )

stack: ( ( (output: []


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+ B ) * ( C - E ) ) / ( F + G ) )

stack: ( ( (output: [A]


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B ) * ( C - E ) ) / ( F + G ) )

stack: ( ( ( +output: [A]


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) * ( C - E ) ) / ( F + G ) )

stack: ( ( ( +output: [A B]


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* ( C - E ) ) / ( F + G ) )

stack: ( (output: [A B + ]


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( C - E ) ) / ( F + G ) )

stack: ( ( *output: [A B + ]


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C - E ) ) / ( F + G ) )

stack: ( ( * (output: [A B + ]


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- E ) ) / ( F + G ) )

stack: ( ( * (output: [A B + C ]


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E ) ) / ( F + G ) )

stack: ( ( * ( -output: [A B + C ]


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) ) / ( F + G ) )

stack: ( ( * ( -output: [A B + C E ]


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) / ( F + G ) )

stack: ( ( *output: [A B + C E - ]


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/ ( F + G ) )

stack: (output: [A B + C E - * ]


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( F + G ) )

stack: ( /output: [A B + C E - * ]


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F + G ) )

stack: ( / (output: [A B + C E - * ]


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+ G ) )

stack: ( / (output: [A B + C E - * F ]


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G ) )

stack: ( / ( +output: [A B + C E - * F ]


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) )

stack: ( / ( +output: [A B + C E - * F G ]


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)

stack: ( /output: [A B + C E - * F G + ]


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stack: output: [A B + C E - * F G + / ]


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Problem with FPE

Too many parentheses.Establish precedence rules:

My Dear Aunt SallyWe can alter the previous program to

use the precedence rules.


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Infix to Postfix

Initialize a Stack for operators, outputlistSplit the input into a list of tokens.for each token (left to right):

if it is operand: append to outputif it is '(': push onto Stackif it is ')': pop & append till '('if it in '+-*/':

while peek has precedence it:pop & append

push onto Stackpop and append the rest of the Stack.

infix to postifix

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