Tutorial 3: Infix, Prefix, Postfix & Tower of Hanoi

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Consider the following operation performed on a stack of size 5.

Push(1);Pop();Push(2);Push(3);Pop();Push(4);Pop();Pop();Push(5);

a) 1b) 2c) 3d) 4

Consider the following operation performed on a stack of size 5.

Push(1);Pop();Push(2);Push(3);Pop();Push(4);Pop();Pop();Push(5);

a) 1b) 2c) 3d) 4

If the SIZE is the size of the array used in the implementation of circular queue, array index start with 0, front point to the first element in the queue, and back point to the last element in the Queue. Which of the following condition specify that circular queue is completely EMPTY?

a) Front=Back=0b) Front=Back=-1 c) Front=Back+1d) Front=(Back+1)% SIZE

If the SIZE is the size of the array used in the implementation of circular queue, array index start with 0, front point to the first element in the queue, and back point to the last element in the Queue. Which of the following condition specify that circular queue is completely EMPTY?

a) Front=Back=0b) Front=Back=-1 c) Front=Back+1d) Front=(Bear+1)% SIZE

Consider the following operations on a Queue. After the codebelow executes, how many elements would remain in the queue?

Queue q = new Queue( );q.enqueue(3); q.enqueue(5);q.enqueue(9);q.dequeue( );

q.enqueue(2); q.enqueue(4);q.dequeue( );q.dequeue( );q.enqueue(1); q.enqueue(8);

a) 0b) 4c) 5d) 6

Consider the following operations on a Queue. After the codebelow executes, how many elements would remain in the queue?

Queue q = new Queue( );q.enqueue(3); q.enqueue(5);q.enqueue(9);q.dequeue( );

q.enqueue(2); q.enqueue(4);q.dequeue( );q.dequeue( );q.enqueue(1); q.enqueue(8);

a) 0b) 4c) 5d) 6

What is the reason for using a "circular queue" instead of a regular one?

a) The running time will improveb) Can traverse all the elements easily c) Re-Use empty spacesd) None of the above

What is the reason for using a "circular queue" instead of a regular one?

a) The running time will be improveb) Can traverse all the elements easily c) Re-Use empty spacesd) None of the above

Tutorial 3: Infix, Prefix, Postfix & Tower of Hanoi

Agenda● Infix, Postfix, Prefix● Postfix & Prefix Calculation using Stack● Tower of Hanoi

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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● Operand may be a variable like x, a, z or a constant

like 3,7,8,1,0…● Operator: +, -, /…

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a2+b2

=c2C=π⋅d=2⋅π⋅r

E=mc2 f (x )=1

σ √2πe−

( x−μ)2

2σ2

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Infix, Postfix, Prefix● Infix: The expressions in which operands surround the

operator. Example: X+Y● Postfix: Reverse Polish Notation(RPN)

✔ Operators are written after their operands. Example: XY+

● Prefix: Polish Notation(PN)✔ Operators are written before their operands. Example: +XY

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Infix Postfix Prefix

A*B+C/D AB*CD/+ +*AB/CD

A*(B+C)/D ABC+*D/ /*A+BCD

A*(B+C/D) ABCD/+* *A+B/CD

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● A+B/C*(D+E)-F● (X+Y/Z*(A+B)-C)● A*B^C+D● (a+b+c)*(e/f)+(g-h/i)● (4+8)*(6-5)/(3-2)*(2+2)

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Infix to Postfix using Stack● Order: ( ) { } [ ], Exponent, Multiplication, Divide, A, S

✔ ^: Priority 3✔ *, /: Priority 2✔ +,-: Priority 1

● No two operators of same priority can stay together● If operator is in between ( and ), that will be popped out

from stack

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Infix: A*B^C+D

Characters Stack Postfix

A Empty A

* * A

B * AB

^ * ^ AB

C * ^ ABC

+ * ^ + ABC^

ABC^*

D + ABC^*D

Postfix: ABC^*D+

Infix to Prefix using Stack● Procedure

✔ Reverse the infix✔ Calculate postfix✔ Then, Reverse again

● Infix: (A*B)+C✔ C+(B*A)✔ Postfix: CBA*+✔ Prefix: +*ABC

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Tower of Hanoi● Tower of Brahma(Lucas Tower)● 1883: Edouard Lucas

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● moveTower(N, Source, Spare, Dest): General Notation● Solution

✔ moveTower(N-1, Source, Dest, Spare)✔ moveTower(1, Source, Spare, Dest)✔ moveTower(N-1, Spare, Source, Dest)

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TOH Solution (N=No. of Disks)

moveTower(N, Source, Spare, Dest)IF n==1move that disk from Source to Dest

ELSEmoveTower(N-1, Source, Dest, Spare)moveTower(1, Source, Spare, Dest)moveTower(N-1, Spare, Source, Dest)

TOH Solution (N=No. of Disks)

moveTower(N, Source, Spare, Dest)IF n==1move that disk from Source to Dest

ELSEmoveTower(N-1, Source, Dest, Spare)moveTower(1, Source, Spare, Dest)moveTower(N-1, Spare, Source, Dest)

O(2n)

public class towerHanoi{ public towerHanoi() { Scanner sc=new Scanner(System.in); System.out.print("Enter Number Of Discs:"); int n=sc.nextInt(); TOH(n,1,2,3); }

public void TOH(int n, int A, int B, int C) { if(n>0){ TOH(n-1,A,C,B); System.out.println("Move Disk "+n +" from Tower "+ A +" to Tower "+C); TOH(n-1,B,A,C); } } public static void main(String args[]) { towerHanoi obj=new towerHanoi(); }}

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● 3 Disks: 23-1=7 Steps● 4 Disks: 24-1= 15 Steps● The no of moves doubles if we add a single disk● Increasing disks from 1 to n, the computation will

double at each stage So O(2n)

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Proof by Induction● Claim: f(n)=2n-1● Basis:

✔ Suppose n=1, f(1)=1 (TRUE)✔ Suppose n=3, f(3)=7 (TRUE)

● Induction Step✔ The claim is true for n. So, It must support for n+1 too.

✔ f(n)=2n-1 for all n PROVED

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Thank You

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