Data structure lecture7

Lecture - 6(Stacks)
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Data structures

Lecture Outline

• What is a Stack?
• Array implementation of stacks
• Operations on a Stack
• Arithmetic expressions
• stacks are used to evaluate postfix expressions
• Infix expressions into postfix Expressions
• Quicksort

Stacks

• What is a Stack?
• DEFINITION:
A stack is a homogeneous collection of elements in which an
element may be inserted or deleted only at one end, called the top of the stack.

• Formally this type of stack is called a Last In, First Out (LIFO) stack.

• Special terminology is used for two basic operations associated with stacks:
a) “Push” is the term used to insert an element into a stack.
b) “Pop” is the term used to delete an element from a stack.

Diagram of stacks
Suppose the following 6 elements are pushed, in order, onto an empty stack:
AAA, BBB, CCC, DDD, EEE, FFF
Frequently designate the stack by writing:
STACK : AAA, BBB, CCC, DDD, EEE, FFF

AAA BBB CCC DDD EEE FFF

1 2 3 4 5 6 7 8 9 10 N-1
N
TOP

Array implementation of stacks

• To implement a stack, items are inserted and removed at the same end
(called the top)
• To use an array to implement a stack, you need both the array itself and an
integer
• The integer tells you either:
– Which location is currently the top of the stack, or
– How many elements are in the stack

Array Representation of Stacks
STACK

XXX YYY ZZZ

1 2 3 4 5 6 7 8

MAXSTK 8
TOP 3

STACK : A linear array
TOP : A pointer variable, Which contains the location of the top element of the
stack.
MAXSTK : Gives the maximum number of elements that can be held by the
stack.
TOP = 0 or NULL will indicate that the stack is empty

Operations on a Stack

Maximum size of n.

Push : This operation adds or pushes another item onto the stack. The
number of items on the stack is less than n.
Pop: This operation removes an item from the stack. The number of items
on the stack must be greater than 0.
Top: This operation returns the value of the item at the top of the stack.
Note: It does not remove that item.
Is Empty: This operation returns true if the stack is empty and false if it is
not.
Is Full: This operation returns true if the stack is full and false if it is not.
These are the basic operations that can be performed on a stack.

Push :
Push is the function used to add data items to the stack.
In order to understand how the Push function operates, we need to look at the
algorithm in more detail.
Procedure 6.1 : PUSH(STACK, TOP, MAXSTK, ITEM)

1. If TOP = MAXSTK, then : Print :”OVERFLOW”, and return.
2. Set : TOP := TOP+1.
3. STACK[TOP]) := ITEM.
4. Return.

In order to understand the algorithm, let's break it apart line by line.

PUSH(STACK, TOP, MAXSTK, ITEM)

First, Push accepts a parameter - item. This parameter is of the same type as the rest
of the stack. Item is the data to be added to the stack.
if top = MAXSTK, then stack is full.
This line performs a check to see whether or not the stack is full.

top := top+1; If the stack is not full, top is increased by a value equal to the size of
another item. This allocates room for the insertion.

Pushing
0 1 2 3 4 5 6 7 8 9
STACK 17 23 97 44

top = 3 N=4

• If ITEM=80 is to be inserted, then TOP=TOP+1=4
• STACK[TOP]:= STACK[4]:= ITEM=80
• N=5
• MAXSTACK=10
0 1 2 3 4 5 6 7 8 9
STACK 17 23 97 44 80

top = 4 N=5

Pop :
Data is removed from a stack by using Pop. From a procedural perspective, pop is
called with the line Pop(item), where item is the variable to store the popped item in.
Once again, we will begin by looking at the algorithm.

Procedure 6.2 POP (STACK, TOP, ITEM);
1. If TOP = 0 then : Print :”UNDERFLOW”, and return.
2. ITEM : =STACK[TOP]) .
3. Set : TOP := TOP-1
4. Return.

If TOP = 0 then, stack is empty, there is no item to pop off the stack. Control
can then be passed to an error handling routine.

ITEM : =STACK[TOP]) Set item to be equal to the data in the top .

TOP := TOP-1; This statement removes the top item from the stack.
Decrement top by 1 so that it now accesses the new top of the stack.

Poping

0 1 2 3 4 5 6 7 8 9
STACK 17 23 97 44 80

top = 4 N=5

• If ITEM=80 is to be deleted, then ITEM:=STACK[TOP]= STACK[4]= 80
• TOP=TOP-1=3
• N=4
• MAXSTACK=10

0 1 2 3 4 5 6 7 8 9
STACK 17 23 97 44

top = 3 N=4

Arithmetic expressions
• Polish Notation (prefix) : Polish notation, also known as prefix notation, is a form of
notation for logic, arithmetic, and algebra. Its distinguishing feature is that it places
operators to the left of their operands.
• Lacking parentheses or other brackets.

• Syntax : operator operand1 operand2
• Example : -*+ABC/D+EF

Infix notation
• Infix notation is the conventional notation for arithmetic expressions. It is called infix
notation because
• each operator is placed between its operands,
• operands (as in the case with binary operators such as addition, subtraction,
multiplication, division,……).
• When parsing expressions written in infix notation, you need parentheses and
precedence rules to remove ambiguity.

• Syntax: operand1 operator operand2
• Example: (A+B)*C-D/(E+F)

Arithmetic expressions

• Postfix notation
• In postfix notation, the operator comes after its operands. Postfix notation is also known
as reverse Polish
• notation (RPN) and is commonly used because it enables easy evaluation of
expressions.

• Syntax : operand1 operand2 operator
• Example : AB+C*DEF+/-

Polish Notation

Q an arithmetic expression involving constants and operations
The binary operation in Q may have different levels of precedence :

Highest : Exponentiation(↑)
Next highest : Multiplication(*) and division(/)
Lowest : Addition (+) and subtraction (-)
Ex :
2 ↑ 3 + 5 *2 ↑ 2 – 12 / 6
Evaluating the Exponentiations :
8 + 5 * 4 -12 / 6
Evaluating the Multiplication and division
8 + 20 -2
Evaluating the Addition and subtraction
26

Polish notation

• Infix expressions into polish notation :
(A+B)*C = [+AB]*C = *+ABC
A+(B*C) = A +[*BC] = +A*BC
(A+B)/(C-D) =[+AB]/[-CD] = /+AB-CD
Fundamental property of polish notation is that the order in which the operation are to
be performed is completely determined by the positions of the operators and operands
in the expression.

Polish notation

An example shows the ease with which a complex statement in prefix notation
can be deciphered through order of operations:
- * / 15 - 7 + 1 1 3 + 2 + 1 1 =
- * / 15 - 7 2 3 + 2 + 1 1 =
- * / 15 5 3 + 2 + 1 1 =
-*33+2+11=
-9+2+11=
-9+22=
-94=
5
An equivalent in-fix is as follows: ((15 / (7 - (1 + 1))) * 3) - (2 + 1 + 1) = 5

Arithmetic expressions(Reverse Polish notation )
Reverse Polish notation provided a straightforward solution for calculator or
computer software mathematics because it treats the instructions (operators) and
the data (operands) as "objects" and processes them in a last-in, first-out (LIFO)
basis.
This is called a "stack method". (Think of a stack of plates. The last plate you put
on the stack will be the first plate taken off the stack.)

The computer evaluates an arithmetic expression written in infix notation in two
steps.
1. It converts the expression to postfix notation
2. Then it evaluates the postfix expression
In each step stack is main tool that is used to accomplish the given task

Stacks are used to evaluate postfix expressions

Evaluation of a postfix expression
Let P is an arithmetic expression written in postfix notation. The following algorithm,
which uses a STACK to hold operands, evaluates P

Algorithm 6.3 : This algorithm finds the VALUE of an arithmetic expression P written
in postfix notation.
1. Add a right parenthesis “)” at the end of P
2. Scan P from left to right and repeat steps 3 and 4 for each element of P until
the sentinel “)” is encountered.
3. If an operand is encountered, put it on STACK.
4. If an operator x is encountered, then :
(a) Remove the two top element of STACK, where A is the top
element and B is the next – to – top element.
(b) Evaluate B x A.
(C) Place the result of (b) back on STACK.
[End of if structure]
[End of step 2 loop]
5. Set VALUE equal to the top element on STACK.
6. Exit.

Example 6.5
Postfix notation P : 5, 6, 2, +, *, 12, 4, /, -)

Symbol Scanned STACK
(1) 5 5
(2) 6 5, 6
(3) 2 5, 6, 2
(4) + 5, 8
(5) * 40
(6) 12 40, 12
(7) 4 40, 12, 4
(8) / 40, 3
(9) - 37
(10) )

Infix expressions into postfix Expressions

Algorithm 6.4 :
Suppose Q is an arithmetic Expressions written in Infix notation. This algorithm finds
the equivalent postfix expression P.
1. Push “(“ onto STACK, and add “)” to the end of Q.
2. Scan Q from left to right and repeat steps 3 to 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) Repeatedly pop from STACK and add to P each operator (on the top of
STACK) which has the same precedence as or higher precedence than X
(b) Add X to STACK
6. If a right parenthesis is encountered, then :
(a) Repeatedly pop from STACK and add to P each operator (on the top of
STACK) until a left parenthesis is encountered
(b) Remove the left parenthesis
[End of step 2 loop]
7. Exit.

Infix expression Q: A+(B*C-(D/E F)*G)*H

Infix into postfix expression

Infix Postfix
2+3 23+

2+3*6 36*2+

(2 + 3) * 6 23+6*

A / (B * C) + D * E - A * C ABC*/DE*+AC*-

Quicksort
Quicksort is a divide-and-conquer method for sorting.
Divide and conquer method:
It works by partitioning an array into parts, then sorting each
part independently.

Divide: Partition into sub arrays (sub-lists),
Select a splitting element (pivot)
Rearrange the array (sequence/list)
v
Conquer: Recursively sort 2 sub arrays
Combine : the sorted S1 (by the time returned from
recursion), followed by v, followed by the sorted S2 (i.e., S
nothing extra needs to be done)

v
S1 S2

Quicksort

– How do we partition the array efficiently?
• choose partition element to be leftmost element
• scan from left for larger element
• scan from right for smaller element
• exchange
• repeat until pointers cross

Example

We are given array of n integers to sort:

40 20 10 80 60 50 7 30 100

Pick Pivot Element
There are a number of ways to pick the pivot element. In this example,
we will use the first element in the array:

40 20 10 80 60 50 7 30 100
[0] [1] [2] [3] [4] [5] [6] [7] [8]

LOC=0 ; LEFT = 0 and RIGHT = 8

1. Set LEFT = BEG, RIGHT = END and LOC = LEFT
2. (a) Repeat A[LOC] <= A[RIGHT] and LOC ≠ RIGHT
RIGHT := RIGHT -1
(b) if LOC =RIGHT then return
(c) If A[LOC] > A[RIGHT]
(i) interchange each other
(ii) LOC =RIGHT
(iii) Goto Step 3
3. (a) Repeat A[LOC] > A[LEFT] and LOC ≠ LEFT
LEFT := LEFT +1
(b) if LOC = LEFT then return
(c) If A[LEFT] > A[LOC]
(ii) LOC =LEFT
(iii) Goto Step 2

LOC = 0 40 20 10 80 60 50 7 30 100

[0] [1] [2] [3] [4] [5] [6] [7] [8]

LEFT = 0 RIGHT = 8


RIGHT := RIGHT -1
(ii) LOC =RIGHT
(iii) Goto Step 3
LEFT := LEFT +1
(ii) LOC =LEFT
(iii) Goto Step 2

LOC = 0 40 20 10 80 60 50 7 30 100

[0] [1] [2] [3] [4] [5] [6] [7] [8]

LEFT = 0 RIGHT = 8


RIGHT := RIGHT -1
(ii) LOC =RIGHT
(iii) Goto Step 3
LEFT := LEFT +1
(ii) LOC =LEFT
(iii) Goto Step 2

LOC = 0 40 20 10 80 60 50 7 30 100

[0] [1] [2] [3] [4] [5] [6] [7] [8]

LEFT = 0 RIGHT = 7


RIGHT := RIGHT -1
(ii) LOC =RIGHT
(iii) Goto Step 3
LEFT := LEFT +1
(ii) LOC =LEFT
(iii) Goto Step 2

LOC = 7 30 20 10 80 60 50 7 40 100

[0] [1] [2] [3] [4] [5] [6] [7] [8]

LEFT = 0 RIGHT = 7


RIGHT := RIGHT -1
(ii) LOC =RIGHT
(iii) Goto Step 3
LEFT := LEFT +1
(ii) LOC =LEFT
(iii) Goto Step 2

LOC = 7 30 20 10 80 60 50 7 40 100

[0] [1] [2] [3] [4] [5] [6] [7] [8]

LEFT = 1 RIGHT = 7


RIGHT := RIGHT -1
(ii) LOC =RIGHT
(iii) Goto Step 3
LEFT := LEFT +1
(ii) LOC =LEFT
(iii) Goto Step 2

LOC = 7 30 20 10 80 60 50 7 40 100

[0] [1] [2] [3] [4] [5] [6] [7] [8]

LEFT = 2 RIGHT = 7


RIGHT := RIGHT -1
(ii) LOC =RIGHT
(iii) Goto Step 3
LEFT := LEFT +1
(ii) LOC =LEFT
(iii) Goto Step 2

LOC = 7 30 20 10 80 60 50 7 40 100

[0] [1] [2] [3] [4] [5] [6] [7] [8]

LEFT = 3 RIGHT = 7


RIGHT := RIGHT -1
(ii) LOC =RIGHT
(iii) Goto Step 3
LEFT := LEFT +1
(ii) LOC =LEFT
(iii) Goto Step 2

LOC = 3 30 20 10 40 60 50 7 80 100

[0] [1] [2] [3] [4] [5] [6] [7] [8]

LEFT = 3 RIGHT = 7


RIGHT := RIGHT -1
(ii) LOC =RIGHT
(iii) Goto Step 3
LEFT := LEFT +1
(ii) LOC =LEFT
(iii) Goto Step 2

LOC = 3 30 20 10 40 60 50 7 80 100

[0] [1] [2] [3] [4] [5] [6] [7] [8]

LEFT = 3 RIGHT = 6


RIGHT := RIGHT -1
(ii) LOC =RIGHT
(iii) Goto Step 3
LEFT := LEFT +1
(ii) LOC =LEFT
(iii) Goto Step 2

LOC = 6 30 20 10 7 60 50 40 80 100

[0] [1] [2] [3] [4] [5] [6] [7] [8]

LEFT = 3 RIGHT = 6


RIGHT := RIGHT -1
(ii) LOC =RIGHT
(iii) Goto Step 3
LEFT := LEFT +1
(ii) LOC =LEFT
(iii) Goto Step 2

LOC = 6 30 20 10 7 60 50 40 80 100

[0] [1] [2] [3] [4] [5] [6] [7] [8]

LEFT = 4 RIGHT = 6


RIGHT := RIGHT -1
(ii) LOC =RIGHT
(iii) Goto Step 3
LEFT := LEFT +1
(ii) LOC =LEFT
(iii) Goto Step 2

LOC = 4 30 20 10 7 40 50 60 80 100

[0] [1] [2] [3] [4] [5] [6] [7] [8]

LEFT = 4 RIGHT = 6


RIGHT := RIGHT -1
(ii) LOC =RIGHT
(iii) Goto Step 3
LEFT := LEFT +1
(ii) LOC =LEFT
(iii) Goto Step 2

LOC = 4 30 20 10 7 40 50 60 80 100

[0] [1] [2] [3] [4] [5] [6] [7] [8]

LEFT = 4 RIGHT = 5


RIGHT := RIGHT -1
(ii) LOC =RIGHT
(iii) Goto Step 3
LEFT := LEFT +1
(ii) LOC =LEFT
(iii) Goto Step 2

LOC = 4 30 20 10 7 40 50 60 80 100

[0] [1] [2] [3] [4] [5] [6] [7] [8]

LEFT = 4 RIGHT = 4

30 20 10 7 40 50 60 80 100

[0] [1] [2] [3] [4] [5] [6] [7] [8]

First sublist second sublist

Apply the above procedure repetitively until each sub list contains one element

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