FAST- National University of
Computer and Emerging Sciences
CFD Campus
Data Structures and Algorithm
Application of Stack
Asma Sattar
Asma.sattar@nu.edu.pk
Lecture No. 7
Today’s Lecture
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Algebraic Expression
Infix to Prefix
Prefix Evaluation
Infix to Postfix
Postfix Evaluation
Algebraic Expression
• An algebraic expression is a legal combination of operands and the
operators.
– Operand is the quantity on which a mathematical operation is performed.
– Operator is a symbol which signifies a mathematical or logical operation.
Infix, Postfix and Prefix
Expressions
• INFIX: expressions in which operands surround the
operator.
• POSTFIX: operator comes after the operands, also Known
as Reverse Polish Notation (RPN).
• PREFIX: operator comes before the operands, also Known
as Polish notation.
• Example
– Infix: A+B-C Postfix: AB+C- Prefix: +A-BC
Why do we need
PREFIX/POSTFIX?
• Appearance may be misleading, INFIX notations are not as
simple as they seem
• To evaluate an infix expression we need to consider
– Operators’ Priority
– Associative property
– Delimiters
Why do we need
PREFIX/POSTFIX?
• Infix Expression Is Hard To Parse and
difficult to evaluate.
• Postfix and prefix do not rely on operator
priority and are easier to parse.
Why do we need
PREFIX/POSTFIX?
• An expression in infix form is thus converted into
prefix or postfix form and then evaluated without
considering the operators priority and delimiters.
Examples of infix to prefix and post
fix
Infix
PostFix
Prefix
A+B
AB+
+AB
(A+B) * (C + D)
AB+CD+*
*+AB+CD
Prefix example:
2+3*4-5/6^(2-1*3)
Prefix= -+2*34/5^6-2*13 (solve on board)
Prefix Task:
2*5^6/3
Prefix=/*2^563
Postfix example:
2+3*4-5/6^(2-1*3)
Postfix= 234*+56231*-^/-
Postfix Task:
f^(g-h)+(j-k)
Postfix= F g h - ^ j k -+
Algorithm for Infix to postfix
conversion
Algorithm: Q is the given infix expression & we want P.
1.
Push “(“ on STACK and add “)” at the end of Q.
2.
Scan Q from left to right and repeat steps 3 & 6 for each element of Q until
the STACK is empty.
3.
If an operand is encountered, add it to P
4.
If a left parenthesis is encountered, push it onto STACK.
5.
If an operator X is encountered, then:
a.
b.
6.
If a right parenthesis is encountered, then:
a.
b.
7.
Repeatedly pop from STACK and add to P each operator which has same or higher
precedence than X
Push X on STACK
Repeatedly pop from STACK and add to P each operator until a left parenthesis is
encountered
Remove the left parenthesis. [Do not add it to P]
Exit
Infix to Postfix
((A+B)*(C-E))/(F+G)
stack: <empty>
output: []
Infix to Postfix
((A+B)*(C-E))/(F+G))
stack: (
output: []
Infix to Postfix
(A+B)*(C-E))/(F+G))
stack: ( (
output: []
Infix to Postfix
A+B)*(C-E))/(F+G))
stack: ( ( (
output: []
Infix to Postfix
+B)*(C-E))/(F+G))
stack: ( ( (
output: [A]
Infix to Postfix
B)*(C-E))/(F+G))
stack: ( ( ( +
output: [A]
Infix to Postfix
)*(C-E))/(F+G))
stack: ( ( ( +
output: [A B]
Infix to Postfix
*(C-E))/(F+G))
stack: ( (
output: [A B + ]
Infix to Postfix
(C-E))/(F+G))
stack: ( ( *
output: [A B + ]
Infix to Postfix
C-E))/(F+G))
stack: ( ( * (
output: [A B + ]
Infix to Postfix
-E))/(F+G))
stack: ( ( * (
output: [A B + C ]
Infix to Postfix
E))/(F+G))
stack: ( ( * ( output: [A B + C ]
Infix to Postfix
))/(F+G))
stack: ( ( * ( output: [A B + C E ]
Infix to Postfix
)/(F+G))
stack: ( ( *
output: [A B + C E - ]
Infix to Postfix
/(F+G))
stack: (
output: [A B + C E - * ]
Infix to Postfix
(F+G))
stack: ( /
output: [A B + C E - * ]
Infix to Postfix
F+G))
stack: ( / (
output: [A B + C E - * ]
Infix to Postfix
+G))
stack: ( / (
output: [A B + C E - * F ]
Infix to Postfix
G))
stack: ( / ( +
output: [A B + C E - * F ]
Infix to Postfix
))
stack: ( / ( +
output: [A B + C E - * F G ]
Infix to Postfix
)
stack: ( /
output: [A B + C E - * F G + ]
Infix to Postfix
stack: <empty>
output: [A B + C E - * F G + / ]
Task
infix
(a+b-c)*d–(e+f)
postfixt
Evaluating a postfix expression
Algorithm: P is the given postfix expression.
1. Add a right parenthesis “)” at the end of P. [acts as a sentinel]
2. Scan P from left to right and repeat steps 3 & 4 for each
element of P until “)” is encountered.
3. If an operand is encountered, push it on STACK
4. If an operator is encountered, then:
a.
b.
c.
5.
6.
Pop two operands from STACK: A & B
Evaluate: B operator A
Push result on STACK
Set value equal to the top element on STACK
Exit
From Postfix to Answer
• The reason to convert infix to postfix
expression is that we can compute the
answer of postfix expression easier by
using a stack.
From Postfix to Answer
Ex: 10 2 8+* 3 • First, push(10) into the
stack
10
From Postfix to Answer
Ex: 10 + 3 28*
• Then, push(2) into the
stack
2
10
From Postfix to Answer
Ex: 10 2 8 * + 3 • Push(8) into the stack
8
2
10
From Postfix to Answer
Ex: 10 + 3 28*
• Now we see an operator *,
that means we can get an
new number by calculation
8
2
10
From Postfix to Answer
Ex: 10 2 8 * + 3 • Now we see an operator *,
that means we can get an
new number by calculation
8
• Pop the first two numbers
2
10
2
*
8
=
16
From Postfix to Answer
Ex: 10 2 8 * + 3 • Now we see an operator *,
that means we can get an
new number by calculation
• Push the new number back
16
10
2
*
8
=
16
From Postfix to Answer
Ex: 10 2 8 * + 3 • Then we see the next
operator + and perform
the calculation
16
10
10 ++ 16 =
26
From Postfix to Answer
Ex: 10 2 8 * + 3 • Then we see the next
operator + and perform
the calculation
• Push the new number back
26
10 + 16 =
26
From Postfix to Answer
Ex: 10 + 3 28*
• We see the next number 3
• Push (3) into the stack
3
26
Compute the Answer
Ex: 10 2 8 * + 3 • The last operation
26 -
3
=
23
From Postfix to Answer
Ex: 10 2 8 * + 3 • The last operation
23
26 -
3
=
23
answe
r!
Infix to Prefix Conversion
Move each operator to the left of its
operands & remove the parentheses:
( ( A + B) * ( C + D ) )
Infix to Prefix Conversion
Move each operator to the left of its
operands & remove the parentheses:
(+A B *(C+D))
Infix to Prefix Conversion
Move each operator to the left of its
operands & remove the parentheses:
*+A B (C+D)
Infix to Prefix Conversion
Move each operator to the left of its
operands & remove the parentheses:
*+A B +C D
Order of operands does not change!
Algorithm for Infix to prefix
conversion
Algorithm: Q is the given infix expression & we want P.
1.
Push “)“ on STACK and add “(” at the start of Q.
2.
Scan Q from right to left and repeat steps 3 & 6 for each element of Q until the STACK
is empty.
3.
If an operand is encountered, add it to P
4.
If a right parenthesis is encountered, push it onto STACK.
5.
If an operator X is encountered, then:
a.
Repeatedly pop from STACK and add to P each operator which has same or higher
precedence than X
b.
Push X on STACK
6.
If a left parenthesis is encountered, then:
a.
Repeatedly pop from STACK and add to P each operator until a left parenthesis is
encountered
b.
Remove the left parenthesis. [Do not add it to P]
7.
Exit
Example f^(g-h)+(j-k)
Q = (f^ (g-h) +(j-k)
Stack: )
P:
Scan right to left
Q = (f^ (g-h) +(j-k
Stack: ) )
P:
Scan right to left
Q = (f^ (g-h) +(jStack: ))
P:k
Scan right to left
Q = (f^ (g-h) +(j
Stack: ))P:k
Scan right to left
Q = (f^ (g-h) +(
Stack: ))P:jk
Scan right to left
Q = (f^ (g-h) +(
Stack: ))P:-jk
Scan right to left
Q = (f^ (g-h)
Stack: )+
P:-jk
Scan right to left
Q = (f^ (g-h
Stack: )+)
P:-jk
Scan right to left
Q = (f^ (gStack: )+)
P:h-jk
Scan right to left
Q = (f^ (g
Stack: )+)P:h-jk
Scan right to left
Q = (f^ (
Stack: )+)P:gh-jk
Scan right to left
Q = (f^ (
Stack: )+)P:-gh-jk
Scan right to left
Q = (f
Stack: )+^
P:-gh-jk
Scan right to left
Q=(
Stack: )+^
P:f-gh-jk
Scan right to left
Q=(
Stack: )+^
P:+^f-gh-jk
Evaluating a pretfix expression
Algorithm: P is the given prefix expression.
1.
Add parenthesis “(” at the start of P. [acts as a sentinel]
2.
Scan P from right to left and repeat steps 3 & 4 for
each element of P until “)” is encountered.
3.
If an operand is encountered, push it on STACK
4.
If an operator is encountered, then:
5.
6.
a.
b.
c.
Pop two operands from STACK: A & B
Evaluate: A operator B
Push result on STACK
Set value equal to the top element on STACK
Exit
Reading Material
• Nell Dale – Chapter#4
• Schaum’s Outlines – Chapter#6
• D. S. Malik – Chapter#7
• http://www.cs.man.ac.uk/~pjj/cs2121/fix.html
• Examples
–
–
–
animations\in2post\infixtopostfix.exe
animations\evaluation\evaluation.exe
animations\tower\tower.exe
CS 201 - Fall 2013
Algorithm Infix To Postfix
• Assumptions:
1. Space delimited list of tokens represents a
infix expression
2. Operands are single characters.
3. Operators +,-,*,/