Discrete Mathematics Questions and Answers – Trees – Interconversion for Prefix Postfix Infix Notations

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This set of Discrete Mathematics Question Paper focuses on “Trees – Interconversion for Prefix Postfix Infix Notations”.

1. Evaluation of expression a/b+c*d-e in postfix notation.
a) ab+cd/*-e
b) ab/cd*+e-
c) abc/+d*-e
d) abcd/+*-e
View Answer

Answer: b
Explanation: The expression=a/b+c*d-e
={(ab/)+(cd*)}-e
={(ab/)(cd*)+}-e
={(ab/)(cd*)+}e-
So the output is: ab/cd*+e-
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2. Evaluation of 4*5+3/2-9 in prefix notation.
a) *45-/32+9
b) *+453/-29
c) -+*45/329
d) *+/45932
View Answer

Answer: c
Explanation: The expression=4*5+3/2-9
={(4*5)+(3/2)-9}
={(*45)+(/32)-9}
={+(*45)(/32)}-9
=-{+(*45)(/32)9
So the output is; -+*45/329.

3. What is the output of the following if funct1(7)?

Void main()
{
    int n;
    long int func;
    scanf(“%d”,&n);
    func=func1 (n)
    printf(“%ld!=%ld”,n,func);
}
long int func1(int n)
{
    if(n==0)
    {
        Return 1;
    }
    else
    {
        return(n*func1(n-1));
    }
}

a) 128
b) 4320
c) 720
d) 5040
View Answer

Answer: d
Explanation: This is a factorial function of an integer using recursive approach. By running the function on integer 7 we get 5040.
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4. Infix to prefix conversion can be done using __________
a) two queues
b) two stacks
c) one stack and two queues
d) one stack
View Answer

Answer: b
Explanation: In the infix expression, the operator appears between the operands and in infix notation if the operator appears before the operands in the expression. For the conversion between them two stacks are used efficiently. The idea is to use one stack for operators and other to store operands.

5. Conversion from prefix to postfix expression can be done _______________
a) using bubble sort
b) using radix sort
c) using two queues
d) in a direct manner
View Answer

Answer: d
Explanation: In a postfix expression, the operators appear after the operands. Conversion from prefix to postfix is done directly which is better than converting the prefix expression in infix and then infix to postfix expression. It gives better efficiency.

6. What is the postfix expression of 9+3*5/(10-4)?
a) 9 3 + * 5 / 10 4 –
b) 9 3 5 + * / 10 4 –
c) 9 3 + 5 * / 10 4 –
d) 9 3 5 * / + 10 – 4
View Answer

Answer: c
Explanation: The expression, 9+3*5/(10-4)
= 9+3*5/(10 4-)
= 9+35/*(10 4-)
= 935/*+(10 4-)
So the output is:9 3 5 / * + 10 4 -.

7. What is the postfix expression of (A+B)-C*(D/E))+F?
a) A B + C D E / * – F +
b) A B C D E + / * F – +
c) A B C + * D E / F + –
d) A B + C – * D E / F +
View Answer

Answer: a
Explanation: The expression is (A+B)-C*(D/E))+F
= (A+B)-C*(DE/)+F
= (A+B)-C*(DE/)F+
= (A+B)-C(DE/)*F+
= (A+B)C(DE/)*-F+
= (AB+)C(DE/)*-F+
So the output is: AB+CDE/*-F+.
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8. Convert the following expression into prefix notation.

(g-(f^e/d+c)-ba)

a) ^-/gfed+c-ab
b) -ab/+-ec^dgf
c) -ab-+c/d^efg
d) ab/+-^cde-fg
View Answer

Answer: c
Explanation: Convert it first in postfix notation, we can have
(g-(f^e/d+c)-ba)
= (g-(f^e/dc+)-ba)
= (g-(f^ed/c+)-ba)
= (g-(fe^d/c+)-ba)
= (g-(fe^d/c+)ba-)
= (gfe^d/c+-ba-)
By reversing this expression gives the prefix expression, i.e
-ab-+c/d^efg.

9. What is the postfix expression of the given expression, (2*4-(5+7/3^4)-8)10?
a) 2 4 5 * 7 3 4 ^ / + 8 – – 10
b) 2 4 * ^ 5 7 3 4 / + 8 10 – –
c) 2 4 * 5 7 ^ 3 4 / + – 8 10 –
d) 2 4 * 5 7 3 4 ^ / + – 8 – 10
View Answer

Answer: d
Explanation: By solving we can have,
(2*4-(5+7/3^4)-8)10
= (2*4-(5+7/34^)-8)10
= (2*4-(5+734^/)-8)10
= (2*4-(5734^/+)-8)10
= (2*45734^/+–8)10
= 2*45734^/+-8-10
= 24*5734^/+-8-10
So the output is: 2 4 * 5 7 3 4 ^ / + – 8 – 10.

10. Prefix expression can be evaluated _________
a) from right to left
b) from left to right
c) from the exact middle
d) from second right element
View Answer

Answer: b
Explanation: Expressions are usually evaluated from left to right manner. Prefix expressions follow the normal rule i.e from left to right. Postfix expressions can be evaluated from right to left.
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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn