Basic Cryptography questions and answers on “Overview”.
1. 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
Answer: a
Clarification: ci is the ith bit of byte c with value 0x63 i.e,
c = 01100011
2. 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
Answer: b
Clarification: The value of ‘d’ is 0x05.
3. What is the block size in the Simplified AES algorithm?
a) 8 bits
b) 40 bits
c) 16 bits
d) 36 bits
Answer: b
Clarification: The block size for the AES algorithm is 16 bits.
4. Which function can be used in AES multiplication –
a) m(x)=x7+x4+x3
b) m(x)=x8+x4+x3+x+1
c) m(x)=x8+x3+x2+x+1
d) m(x)=x8+x5+x3+x
Answer: b
Clarification: m(x)=x8+x4+x3+x+1 stands for 100011011 which is an irreducible polynomial. Others are not irreducible polynomials.
5. In the DES algorithm the Round Input is 32 bits, which is expanded to 48 bits via ______________
a) Scaling of the existing bits
b) Duplication of the existing bits
c) Addition of zeros
d) Addition of ones
Answer: a
Clarification: The round key is 48 bits. The input is 32 bits. This input is first expanded to 48 bits (permutation plus an expansion), that involves duplication of 16 of the bits.
6. Using Differential Crypt-analysis, the minimum computations required to decipher the DES algorithm is-
a) 256
b) 243
c) 255
d) 247
Answer: d
Clarification: Differential Crypt-analysis requires only 247 computations to decipher the DES algorithm.
7. Using SDES, the Plaintext for the Ciphertext 00001111, given that the key is 1111111111 is,
a) 01100111
b) 00001010
c) 11111111
d) 01101101
Answer: c
Clarification: Perform the SDES Decryption algorithm and compute the cipher text.
8. 12. For p = 11 and q = 17 and choose e=7. Apply RSA algorithm where Cipher message=11 and thus find the plain text.
a) 88
b) 122
c) 143
d) 111
Answer: a
Clarification: n = pq = 11 × 19 = 187.
C=M^e mod n ; C=11^23 mod 187 ; C = 88 mod 187.
9. Reduce the following Big-O notation. O[ ax7 + 3 x3 + sin(x)] =
a) O[ax7].
b) O[sin(x)].
c) O[x7].
d) O[x7 + x3].
Answer: c
Clarification: O[ ax7 + 3 x3 + sin(x)] = O(ax7) = O(x7).
10. Reduce the following Big-O notation. O[ en + an10] =
a) O[ an10].
b) O[ n10 ].
c) O[ en ].
d) O[ en+ n10 ].
Answer: c
Clarification: O[ en + an10 ] = O[ en ].