Exercise # 2) Decrypt the ciphertext message LFDPH LVDZL FRQTX HUHG, which has been encrypted using the Caesar cipher. Exercise # 4) Encrypt the message THE RIGHT CHOICE using the affine...

SECTION 8.1
Sol: (2)
By using ROT23 transformation, L is equal to I . After repeating the same procedure,
we get ICAME ISAWI CONQU ERED.

Sol: (4)
On converting the each letter of the given plaintext into the number, we get the numbe
19, 7, 4, 17, 8, 6, 7, 19, 2, 7, 14, 8, 2, 4. Now on applying the affine transformation
15 14 (mod 26) to aC P  ll the numbers. Taking the first number, 15 19 + 14 13
(mod 26), we get 13. After repeating the same procedure to each numbers, we get 15,
22, 9, 4, 0, 15, 13, 18, 15, 16, 4, 18, 22 which ie equal to
 
NPWJE APNSP QSW.
Sol: (6)
The inverse of 3 · 9 = 27 1 (mod 26) is 9. Thus,
24 3 (mod 26) and
P 9( 24) 9(C + 2) 9C + 18 (mod 26).
By using the above equation we g
C P
C

 
   
et the message Phone Home . 
Sol: (8)
From the given message, V is occuring eight times , and K is occuring three times.
So that E is mapped to V and T is mapped to K . That is, (4 21) and (19 10).
By putting these in equation (mod 2C P k
 
  6) gives a value of 17 for . Using this
we decrypt the ciphertext to: THE VALUE OF THE KEY IS SEVENTEEN.
k
Sol: (10)
Let two most common letters in a long ciphertext, encrypted by an affine transformation
are E and T, so 4, 19, 16, 23.
So 23 4 (mod 26),
E T Q and X
a b and
   
 
16 19 (mod 26),
From these two equations, we get the solution 3; 11.
a
a
 
 
Sol: (16)
The product cipher obtained by using the transformation (mod 26)
followed by the transformation (mod 26) is given by,
( ) (mod 26)
C aP
C cP d
C cP d c aP b d acP bc d
 
 
       
SECTION 8.2
Sol: (2)
We know that
(mod 26), .............. (1),
Now the numerical equivalents of the letters,
SECRET 18 4
i i ip c k 
 2 17 4 19
and the numerical equivalents of the letters of the ciphertext,
WBRCS LAZGJ MGKMF V 22 1 17 2 18 11 0 25 6 9 12 6 10 12 5 21.
From the equation (1), Subtracting the key gives 4 23

15 11 14 18 8 21 4 18 8 13 18
8 3 4, So that the letter is EXPLOS IVESIN SIDE.
Sol: (8)
From the given ciphertext, two occu
ences of which are 15 positions apart. Again
find two occu
ences of which are 42 positions apart, the period of these two
occurance is (15, 42) = 3. Now o
GP
WV
n taking the indexes of coincidence for each of the 3
groups of letters which are 0.057, 0.053, 0.095, respectively, and the period of
these groups is 3. Now we count the frequencies of the lette
and
s in positions 1, 4, 7,. . . ,
we find 7 ' , and 4 each of , . Now we compare the differences of these
letters are , , , , , we see that 6 6, so that
and , which
W s A J and P
E T N R I and O P J and T N P T
J N
    
 1
2
implies ...