1. In brute force attack, on average half of all possible keys must be tried to achieve success.
a) True
b) False
Answer
Answer: a [Reason:] In brute force attack the attacker tries every possible key on a piece of cipher-text until an intelligible translation into plaintext is obtained.
2. If the sender and receiver use different keys, the system is referred to as conventional cipher system.
a) True
b) False
Answer
Answer: b [Reason:] Such a system is called asymmetric, two-key, or public-key cipher system.
3. Divide (HAPPY)26 by (SAD)26. We get quotient –
a) KD
b) LD
c) JC
d) MC
Answer
Answer: a [Reason:] Dividing (HAPPY)26 by (SAD)26 gives us KD with a remainder MLP.
4. Dividing (11001001) by (100111) gives remainder –
a) 11
b) 111
c) 101
d) 110
Answer
Answer: d [Reason:] Dividing (11001001) by (100111) gives us (110).
5. pi in terms of base 26 is
a) C.DRS
b) D.SQR
c) D.DRS
d) D.DSS
Answer
Answer: c [Reason:] On converting using base conversions we get 3.1415926 as D.DRS.
6. The time required to convert a k-bit integer to its representation in the base 10 in terms of big-O notation is
a) O(log2 n)
b) O(log n)
c) O(log2 2n)
d) O(2log n)
Answer
Answer: a [Reason:] Let n be a k-bit integer in binary. The conversion algorithm is as follows. Divide 10 = (1010) into n. The remainder – which will be one of the integers 0, 1, 10, 11, 100, 101, 110, 11 1, 1000, or 1001 – will be the ones digit d0. Now replace n by the quotient and repeat the process, dividing that quotient by (1010), using the remainder as d1 and the quotient as the next number into which to divide (1010). This process must be repeated a number of times equal to the number of decimal digits in n, which is [log n/log 10] +1 = O(k).
We have O(k) divisions, each requiring O(4k) operations (dividing a number with at most k bits by the 4 bit number (1010)). But O(4k) is the same as O(k) (constant factors don’t matter in the big-0 notation, so we conclude that the total number of bit operations is O(k). O(k) = 0(k2). If we want to express this in terms of n rather than k, then since k = O(1og n), we can write
Time(convert n to decimal) = 0(log2n).
7. In base 26, multiplication of YES by NO gives –
a) THWOE
b) MPAHT
c) MPJNS
d) THWAE
Answer
Answer: c [Reason:] Convert the alphabets into their respective values in base 26 and proceed with base 26 multiplications.
8. Division of (131B6C3) base 16 by (lA2F) base 16 yeilds –
a) 1AD
b) DAD
c) BAD
d) 9AD
Answer
Answer: d [Reason:] Base 16 division to be followed where A-F stand for 10-15.
9. An encryption scheme is unconditionally secure if the ciphertext generated does not contain enough information to determine uniquely the corresponding plaintext, no matter how much cipher text is available.
a) True
b) False
Answer
Answer: a [Reason:] The above statement is the definition for unconditionally secure cipher systems.
10. The estimated computations required to crack a password of 6 characters from the 26 letter alphabet is-
a) 308915776
b) 11881376
c) 456976
d) 8031810176
Answer
Answer: a [Reason:] The required answer is 26^6 = 308915776.
Reduce the following big-O natations:
11. O[ ax7 + 3 x3 + sin(x)] =
a) O[ax7].
b) O[sin(x)].
c) O[x7].
d) O[x7 + x3].
Answer
Answer: c [Reason:] O[ ax7 + 3 x3 + sin(x)] = O(ax7) = O(x7)
12. O[ en + an10] =
a) O[ an10 ].
b) O[ n10 ].
c) O[ en ].
d) O[ en + n10 ].
Answer
Answer: c [Reason:] O[ en + an10] = O[ en ].
13. O [ n! + n50 ] =
a) O [ n! + n50 ].
b) O [ n! ].
c) O [ n50].
d) None of the Mentioned
Answer
Answer: b [Reason:] O [ n! + n50 ] = O [ n! ].
