Cryptography MCQ set number 00522

1. In affine block cipher systems if f(m)=Am + t, what is f(m1+m2) ?
a) f(m1) + f(m2) + t
b) f(m1) + f(m2) + 2t
c) f(m1) + t
d) f(m1) + f(m2)

Answer

Answer: a [Reason:] In general f(∑(i=1 to n) m_i = ∑(i=1 to n) f(m_i) + tδ_n) where δ_n=0 if n is odd and 1 if n is even.

2. In affine block cipher systems if f(m)=Am + t, what is f(m1+m2+m3) ?
a) f(m1) + f(m2) + f(m3) + t
b) f(m1) + f(m2) + f(m3) +2t
c) f(m1) + f(m2) + f(m3)
d) 2(f(m1) + f(m2) + f(m3))

Answer

Answer: c [Reason:] In general f(∑(i=1 to n) m_i =∑(i=1 to n) f(m_i ) + tδ_n) where δ_n=0 if n is odd and 1 if n is even.

3. If the block size is ‘s’, how many affine transformations are possible ?
a) 2^s (2^s-1)(2^s-1)(2^s-1²)………(2^s-1^(s-1))
b) 2^s (2^s-1)(2^s-2)(2^s-2²)………(2^s-2^(s-2))
c) 2^ss (2^s-1)(2^s-2)(2^s-2²)………(2^s-2^(s-1))
d) 2^s (2^s-1)(2^s-2)(2^s-2²)………(2^s-2^(s-3))

Answer

Answer: c [Reason:] 2^s (2^s-1)(2^s-2)(2^s-2²)………(2^s-2^(s-1)) is the maximum number of affine transformations possible for a block size ‘s’ matrix.

4. What is the number of possible 3 x 3 affine cipher transformations ?
a) 168
b) 840
c) 1024
d) 1344

Answer

Answer: d [Reason:] Since ‘A’ cannot have columns of ‘0’s. so there are ‘7’ choices i.e. 001/010/011/100/101/110/111. ‘a1’ is chosen for first column of ‘A.
We have ‘6’ choices for second column, let ‘a2’ be chosen for second column.
The final column can be any 3-tuple except 0, a1, a2, a1+a2. That means any one of the remaining ‘4’ 3-tuples may be chosen for the final column.
(Total number possibilities for A)}=k=7×6×4=168
(Number of block cipher transformation)}=k×t=8×168 =1344

5. Super-Encipherment using two affine transformations results in another affine transformation.
a) True
b) False

Answer

Answer: a [Reason:] f(g(m))=A_1 g(m)+c_1
f(g(m))=A_1 (A_2 m+c_2)+c_1 f(g(m))=A_1 A_2 m+A_1 c_2+c_1 f(x)=A_3 m+c_3
where
A_3=A_1 A_2
c_3=A_1 c_2+c_1
This results in another affine transformation, and does not improve the security.

6. If the key is 110100001, the output of the SP network for the plaintext: 101110001 is
a) 110100011
b) 110101110
c) 010110111
d) 011111010

Answer

Answer: b [Reason:]

7. If the key is 110100001 where,
If ki=0, then S_i (x)=((1 1 0 | 0 1 1 | 1 0 0 ))x+((1 1 1))
and If ki=1, then S_i (x)=((0 1 1 | 1 0 1 | 1 0 0))x+((0 1 1))
then the output of the SP network for the plaintext: 101110001 is
a) 010110011
b) 111000011
c) 110110111
d) 010110110

Answer

Answer: a [Reason:]

8. Confusion hides the relationship between the ciphertext and the plaintext.
a) True
b) False

Answer

Answer: b [Reason:] Confusion hides the relationship between the ciphertext and the key.

9. The S-Box is used to provide confusion, as it is dependent on the unknown key.
a) True
b) False

Answer

Answer: a [Reason:] The S-Box is used to provide confusion, as it is dependent on the unknown key.
The P-Box is fixed, and there is no confusion due to it, but it provides diffusion.

10.
This is an example of
a) SP Networks
b) Feistel Cipher
c) Hash Algorithm
d) Hill Cipher

Answer

Answer: b [Reason:] The figure is the Feistel Cipher Structure.

11. Which of the following slows the cryptographic algorithm –
1) Increase in Number of rounds
2) Decrease in Block size
3) Decrease in Key Size
4) Increase in Sub key Generation

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

Answer

Answer: b [Reason:] Increase in any of the above 4 leads to slowing of the cipher algorithm i.e. more computational time will be required.