Computer Science Test

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Computer Science Test - 3

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Question 1

2 / -0.66

In a J-K flip flop, when J = 1 and K = 1 then it will be considered as:

set condition

reset condition

no change

toggle condition

Solution

Concept:

JK flip flop:

The truth table of JK flipflop:

T flip-flop is formed by combining both J and K inputs of the JK-flipflop

In the above truth table when J = K = 1, its output is toggled.

Characteristic Table of JK flip flop

J	K	Q_n	Q_n+1
0	0	0	0
0	0	1	1
0	1	0	0
0	1	1	0
1	0	0	1
1	0	1	1
1	1	0	1
1	1	1	0

Question 2
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Consider 4 matrices Q, R, S and T with dimensions 13 × 12, 12 × 30, 30 × 15 and 15 × 18 respectively. What is the least number of scalar multiplications needed to find the product QRST using the basic matrix multiplication method?

A
11150

B
11250

C
11350

D
11450

Solution

Given matrices Q, R, S, and T with dimensions 13 × 12, 12 × 30, 30 × 15, and 15 × 18 respectively. From the given matrices sequence of dimensions can be seen as:

<P₀, P₁, P₂, P₃, P₄> = <13, 12, 30, 15, 18>

Let c[i, j] be the minimum number of scalar multiplications needed to compute the product QRST.

Using Dynamic Programming recursive definition of c[i, j] is given as-

Hence the minimum number of scalar multiplications needed to compute the product QRST is 11250 using (Q(RS))T).
Question 3
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Directions For Questions

Consider a Turing machine M given below that accepts the language.

L = {bⁿaⁿ⁺¹ |n ≥ 1}

B in ‘M’ represents Blank symbol.

...view full instructions

What will be the missing part/transition for P and Q respectively in the given Turing machine?

A
(a, x, L) and (a, y, R)

B
(a, a, R) and (a, y, L)

C
(a, y, L) and (a, a, R)

D
(a, y, R) and (a, x, L)
Solution
The above given Turing machine is accepting L={bnan+1} as for example

"bbaaa" is accepted by the Turing machine.
- (q0, b) → (q1, x, R)
- (q1, b) → (q1, b, R)
- (q1, a) → (q2, y, L)
- (q2, b) → (q2, b, L)
- (q2, x) → (q1, x, R)
- (q1, y) → (q1, y, R)
- (q1, a) → (q2, y, L)
- (q2, y) → (q2, y, L)
- (q2, x) → (q0, x, R)
- (q0, y) → (q3, y, R)
- (q3, y) → (q3, y, R)
- (q3, a) → (q4, a, R)
- (q4, B) → (q5, B, L)
Question 4
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[A] is a square matrix which is neither symmetric nor skew-symmetric and [A]^T is its transpose. The sum and difference of these matrices are defined as [S] = [A] + [A]^T and [D] = [A] - [A]^T, respectively. Which of the following statements is true?

A
Both [S] and [D] are symmetric

B
Both[S] and [D] are skew-symmetric

C
[S] is skew-symmetric and [D] is symmetric

D
[S] is symmetric and [D] is skew-symmetric

Solution
Question 5
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To derive the string length 4, How many minimum productions are required for Chomsky normal form ?

A
4

B
5

C
6

D
7

Solution

The correct answer is 7.

Concept:

Chomsky's normal form:

A Chomsky normal form follows the,

V→VV (Exactly 2 non-terminals)

V is Non-terminals

V→T (Exactly one terminal)

T is terminal

Explanation:

If length n=1, Number of productions = 1

S→a

If length n=2, Number of productions = 3

S→AB

A→a

B→b

If length n=3, Number of productions = 5

S→AX

X→BC

A→a

B→b

C→c

According to this, If length n, the number of productions = (2n-1) is required.

n=4,

then,

(2(4)-1)= 7 productions

Hence the correct answer is 7.

Question 6

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A router uses the following routing table:

Destination	Mask	Interface
144.16.0.0	255.255.0.0	eth0
144.16.64.0	255.255.224.0	eth1
144.16.68.0	255.255.255.0	eth2
144.16.68.64	255.255.255.224	eth3

A packet bearing an estimation address 144.16.68.117 arrives at the router. On which interface will it be forwarded?

eth0

eth1

eth2

eth3

Solution

The correct answer is "option 3".

CALCULATION:

To calculate the interface, perform the following steps:

1.Bitwise AND between IP address 144.16.68.117 & subnet mask.

2.Match the output network address with the destination address given corresponding to the subnet mask.

DESTINATION	ADDRESS (GIVEN)	MASK	BITWISE AND (ADDRESS & MASK)
144.16.0.0	144.16.68.117	255.255.0.0	144.16.0.0
144.16.64.0	144.16.68.117	255.255.224.0	144.16.64.0
144.16.68.0	144.16.68.117	255.255.255.0	144.16.68.0
144.16.68.64	144.16.68.117	255.255.255.224	144.16.68.96

Option 3:

	1^st Octet	2^nd Octet	3^rd Octet	4^th Octet
144.16.68.117	10010000	00010000	01000100	1110101
255.255.255.0 (&)	11111111	11111111	11111111	00000000
Result	10010000	00010000	01000100	00000000
	144	16	68	0

Hence, the packet with IP address 144.16.68.117 will be forwarded to the interface eth2.

Question 7
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Box A contains 2 black and 3 red balls while box B contains 3 black and 4 red balls. Out of these two boxes one is selected at random and the probability of choosing box A is double that of box B. If a red ball is drawn from the selected box, then the probability that it has come from box B is

A
21/31

B
12/31

C
13/41

D
10/31

Solution

Concept:

Let A be the event of choosing box A, and B be the event of choosing box B, and R is the event of drawing a red ball.
Question 8
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For the given IP address: 150.32.4.1/22, which of the following cannot be valid host IDs on the LAN interface attached to the router?

A
150.32.4.1

B
150.32.3.28

C
150.32.6.254

D
150.32.7.250

Solution

Concept:

150.32.2.1/22

Therefore 22 bits are in NID and only 10 bits for Host id.

Explanation:

Last two-byte of IP address is represented in binary

Ethernet: 150.32.4.1/22

Range of IP address: 150.32.0000 0100.0000000 to 150.32.0000111.11111111

Range of IP address: 150.32.4.0 to 150.32.7.255

First and last IP address is reserved

Valid Host IP address: 150.32.4.1 to 150.32.7.254

Option 2:150.32.3.28

It does not fall in 150.32.4.0 to 150.32.7.255
Question 9
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A
log_e 2

B
1

C

D

Solution
Question 10
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Find the decimal equivalent of the bit pattern 10111111110100000000000000000000 as per the IEEE 754 standard. In IEEE 754 a single bit precision binary floating point format is stored in 32 bits.

S ( 1 bit )

E ( 8 bits )

M ( 23 bits )

Exponent is biased with 127.

A
1.625

B
-2.854

C
-1.625

D
0.854

Solution

1

01111111

10100000000000000000000

S E M

Sign bit: 1 (negative)

Actual exponent : Biased exponent – Bias = 127 – 127 = 0

Mantissa = 1010000000000000000000

In IEEE-754 format, 32-bit (single precision)

(-1)s × 1.M × 2^{E – 127}

-1)s × 1.101 × 2^{127 – 127}

Data: -( 2⁰ × 1 + .2^-1× 1 + 2^-3 × 1)

Data: - (1 + 0.5 + 0.125

Data: - 1.625