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 |

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)=x^{8}+x^{4}+x^{3}+x+1 is

a) 0x8F

b) 0xF8

c) 0x8A

d) 0xA8

View Answer

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

View Answer

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

View Answer

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

View Answer

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

View Answer

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

View Answer

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

View Answer

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

View Answer

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

View Answer

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.

**Sanfoundry Global Education & Learning Series – Cryptography and Network Security.**

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