The XOR gates are ideal for testing parity because even-parity words produce a _____ output and odd-parity words produce a _____ output

An odd-parity generator produces an odd-parity bit to go along with the data. The parity of the transmitted data is _____ An XOR gate can test each received word for parity rejecting words with _____ parity.

A

odd, even

B

high, low

C

low, high

D

even, odd

E

None of the above

In AES, to make the s-box, we apply the transformation b’_i = b_i XOR b_(i+4) XOR b(i+5) XOR b_(i+6) XOR b_(i+7) XOR c_i What is c_i in this transformation?

A

c_i is the ith bit of byte c with value 0x63

B

c_i is the ith bit of byte c with value 0x25

C

c_i is the ith bit of byte c with value 0x8F

D

c_i is the ith bit of byte c with value 0x8A

In AES, to make the s-box, we apply the transformation – b’i = bi XOR b(i+4) XOR b(i+5) XOR b(i+6) XOR b(i+7) XOR ci What is ci in this transformation?

A

ci is the ith bit of byte c with value 0x63

B

ci is the ith bit of byte c with value 0x25

C

ci is the ith bit of byte c with value 0x8F

D

ci is the ith bit of byte c with value 0x8A

In each question below is given a statement numbered I, II and III. An assumption is something supposed or taken for granted. You have to consider the following assumption and decide which of the assumption is implicit in the statement. Statement: The Amazon.com Inc. founder, Jeff Bezos’s net worth broke $150 billion in New York on Monday morning, according to the Bloomberg Billionaires Index. That’s about $55 billion more than Microsoft Corp. co-founder Bill Gates, the world’s second-richest person. Bezos, 54, has now topped Gates in inflation-adjusted terms. The $100 billion mark that Gates hit briefly in 1999 at the height of the dot-com boom would be worth about $149 billion in today’s dollars. Assumptions: I. Jeff Bezos is the richest person in modern history. II. That makes the Amazon founder richer than anyone else on earth since at least 2000 when Forbes published its inaugural wealth ranking. III. Gates would have had a net worth of more than $150 billion if he’d held onto assets that he’s given away, largely to the Bill & Melinda Gates Foundation.

A

Only I is implicit

B

All except II, I is implicit

C

Neither I, II nor III is implicit

D

Only II is implicit

E

Both I and II are implicit

Which of the following statements are true? i) practical application of XOR linked lists are in environments with limited space requirements, such as embedded devices. ii)xor lists are not suitable because most garbage collectors will fail to work properly with classes or structures that don’t contain literal pointers iii)in order to calculate the address of the next node you need to remember the address of the previous node iv)xor lists are much efficient than single, doubly linked lists and arrays

A

i, ii, iii, iv

B

i, ii, iii

C

i, ii

D

i

The Chi function is defined as ← a XOR ( aXOR 1) AND a) b) a ← a AND ( aXOR 1) AND a) c) a ← a AND ( a XOR a) d) a ← a XOR ( a AND a)

A

a ← a XOR ( aXOR 1) AND a)

B

x

C

x

D

x

Calculate the minimum number of AND gates, OR gates and NOT gates required to realize Exclusive-OR operation if only using AND, OR and NOT gates are allowed.

A

2 AND, 1 OR and 2 NOT gates or 1 AND, 2 OR and 2 NOT gates

B

2 AND, 2 NOT and 2 OR gates

C

3 AND, 1 NOT and 2 OR gates

D

2 AND and 2 OR gates

Below some statements are given. You have to take the given statements to be true even if they appear to be at variance with commonly known facts, and then decide which of conclusions logically follows from the given statements. For each question, mark out an appropriate answer choice that you think is correct. Statements: All gates are libraries. Some gates are poles. No pole is room. Some rooms are halls. Conclusions: I) Some poles are library. II) Some gates are room. III) Some libraries are gates.

A

Conclusion I follows

B

Conclusion II follows

C

Both conclusions I and III follow

D

Conclusion III follows

The inverse s-box permutation follows, b’_i = b_(i+2) XOR b(i+5) XOR b_(i+7) XOR d_i Here d_i is

A

d_i is the ith bit of a byte ‘d’ whose hex value is 0x15

B

d_i is the ith bit of a byte ‘d’ whose hex value is 0x05

C

d_i is the ith bit of a byte ‘d’ whose hex value is 0x25

D

d_i is the ith bit of a byte ‘d’ whose hex value is 0x51

The inverse s-box permutation follows, b’i = b(i+2) XOR b(i+5) XOR b(i+7) XOR di Here di is –

A

di is the ith bit of a byte ‘d’ whose hex value is 0x15

B

di is the ith bit of a byte ‘d’ whose hex value is 0x05

C

di is the ith bit of a byte ‘d’ whose hex value is 0x25

D

di is the ith bit of a byte ‘d’ whose hex value is 0x51

The XOR gates are ideal for testing parity because even-parity words produce a _ output and odd-parity words produce a _ output

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