# Symmetric Ciphers Questions and Answers – The AES Algorithm – II

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This set of Cryptography Puzzles focuses on “The AES Algorithm – II”.

1. Conversion of the Plaintext MANIPALINSTITUTE to a state matrix leads to
a)

 M A N I P A L I N S T I T U T E

b)

 M P N T A A S U N L T T I I I E

c)

 M A I L N P I T A N I U S T T E
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d)

 E U T L T I I L T N P A S A N M
Explanation:

 M A N I P A L I N S T I T U T E

The State matrix is Arranged –

 M P N T A A S U N L T T I I I E

2. On encrypting MANIPALINSTITUTE with key ADVANCEDENCRYPTI we get the state matrix –
a)

 FC 1D 1B 0D 15 02 1D 05 10 0F 17 00 20 0D 1B FC

b)

 FC 1D 1B 0D 15 02 1D 05 10 0F 17 00 20 0D 1B 0C

c)

 OC FE 0B 0D D5 02 1D 05 18 09 17 00 08 0D 1B FC

d)

 OC 1E 0B 0D 05 02 1D 05 18 09 17 00 08 0D 1B 0C
Explanation: First convert the state matrix to their equivalent ASCII values and then perform XOR operation.

3. The multiplicative inverse of 0x95 in AES where m(x)=x8+x4+x3+x+1 is
a) 0x8F
b) 0xF8
c) 0x8A
d) 0xA8

Explanation: The multiplicative inverse is 0x8A.

4. 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

Explanation: Ci is the ith bit of byte c with value 0x63 i.e,
c = 01100011.

5. The S-box value for byte stored in cell (6,D)
a) 0x3C
b) 0x7F
c) 0xFD
d) 0x4A

Explanation: We first find the multiplicative inverse of 0x6D. The multiplicative inverse of 0x6D is 0x93. On performing the transformation on 0x93 we get 0x3C.

6. The S-box value for byte stored in cell (B,3)
a) 0x3C
b) 0xB3
c) 0x4F
d) 0x90

Explanation: We first find the multiplicative inverse of 0xB3. The multiplicative inverse of 0xB3 is 0xEF. On performing the transformation on 0xEF we get 0x63.

7. The S-box value for byte stored in cell (3,3)
a) 0xC3
b) 0x3C
c) 0x44
d) 0x9B

Explanation: We first find the multiplicative inverse of 0x33. The multiplicative inverse of 0x33 is 0x6C. On performing the transformation on 0x6C we get 0xC3.

8. 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

Explanation: The value of ‘d’ is 0x05.

9. The Inverse S-box value for byte stored in cell (3,3)
a) 0xC3
b) 0x66
c) 0x1F
d) 0x9B

Explanation: We first find the multiplicative inverse of 0x33. And then perform the matrix transformation to get 0x66.

10. The Inverse S-box value for byte stored in cell (6,3)
a) 0x00
b) 0x11
c) 0x01
d) 0x04

Explanation: We first find the multiplicative inverse of 0x63. And then perform the matrix transformation to get 0x00.

11. The Inverse S-box value for byte stored in cell (D,2)
a) 0x5F
b) 0x2D
c) 0x7F
d) 0x5D

Explanation: We first find the multiplicative inverse of 0xD2. And then perform the matrix transformation to get 0x7F.

12. What is the Shifted Row transformation for the matrix bellow?

 FE 72 2B D7 6B 77 A4 6B AD 01 F0 63 30 D7 AF FE

a)

 FE 72 2B D7 6B 77 A4 6B AD 01 F0 63 30 D7 AF FE

b)

 72 2B D7 FE A4 6B 6B 77 63 AD 01 F0 30 D7 AF FE

c)

 FE 72 2B D7 77 A4 6B 6B F0 63 AD 01 FE 30 D7 AF

d)

 D7 FE 72 2B A4 6B 6B 77 01 AD 63 F0 30 D7 AF FE
Explanation: The Shift Rows transformation consists of:
-Not shifting the first row of the state array at all.
-Circularly shifting the second row by one byte to the left.
-Circularly shifting the third row by two bytes to the left, and
-Circularly shifting the last row by three bytes to the left.

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