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)=x8+x4+x3+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.
To practice puzzles on all areas of Cryptography, here is complete set of 1000+ Multiple Choice Questions and Answers.
- Apply for Computer Science Internship
- Check Cryptography and Network Security Books
- Practice Cyber Security MCQ
- Check Computer Science Books
- Practice Computer Science MCQs