1. The Index of Coincidence for English language is approximately
a) 0.068
b) 0.038
c) 0.065
d) 0.048
Explanation: The IC for the English language is approximately 0.065.
2. If all letters have the same chance of being chosen, the IC is approximately
a) 0.065
b) 0.035
c) 0.048
d) 0.038
Explanation: If all letters have the same chance of being chosen, the IC is approximately 0.038, about half of the IC for the English language.
3. Consider the cipher text message with relative frequencies:
4 0 10 25 5 32 24 15 6 11 5 5 1 2 6 6 15 19 10 0 6 28 8 2 3 2
The Index of Coincidence is
a) 0.065
b) 0.048
c) 0.067
d) 0.042
Explanation: Number of letters = 250. From this, IC=0.0676627. This is very strong evidence that the message came from a Monoalphabetic ciphering scheme.
4. Consider the cipher text message:
YJIHX RVHKK KSKHK IQQEV IFLRK QUZVA EVFYZ RVFBX UKGBP KYVVB QTAJK TGBQO ISGHU CWIKX QUXIH DUGIU LMWKG CHXJV WEKIH HEHGR EXXSF DMIIL UPSLW UPSLW AJKTR WTOWP IVXBW NPTGW EKBYU SBQWS
Relative Frequencies –
3 7 2 2 5 5 7 9 11 4 14 4 2 1 3 4 6 5 6 5 7 10 9 8 4 2
The Index of Coincidence is –
a) 0.065
b) 0.048
c) 0.067
d) 0.044
Explanation: Number of letters = 145.From this, IC=0.0438697 .This is very strong evidence that the message came from a polyalphabetic ciphering scheme.
5. A symmetric cipher system has an IC of 0.041. What is the length of the key ‘m’?
a) 1
b) 3
c) 2
d) 5
Explanation: Using the formula for calculating ‘m’ we get m=5, where
m≈0.027n/(I_c (n-1)-0.038n+0.065).
6. 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)
Explanation: 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.
7. 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))
Explanation: 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.
8. If the block size is ‘s’, how many affine transformations are possible ?
a) 2s (2s-1)(2s-1)(2s-12)………(2s-1(s-1))
b) 2s (2s-1)(2s-2)(2s-22)………(2s-2(s-2))
c) 2ss (2s-1)(2s-2)(2s-22)………(2s-2(s-1))
d) 2s (2s-1)(2s-2)(2s-22)………(2s-2(s-3))
Explanation: 2s (2s-1)(2s-2)(2s-22)………(2s-2(s-1)) is the maximum number of affine transformations possible for a block size ‘s’ matrix.
9. What is the number of possible 3 x 3 affine cipher transformations ?
a) 168
b) 840
c) 1024
d) 1344
Explanation: 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 of@ possibilities for A)}=k=7×6×4=168
(Number of affine@ block cipher transformation)}=k×t=8×168 =1344
10. Super-Encipherment using two affine transformations results in another affine transformation.
a) True
b) False
Explanation: 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.