Exercise # 2) Decrypt the ciphertext message LFDPH LVDZL FRQTX HUHG, which has been encrypted using...

Question

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 transformation C = 15P + 14 (mod 26).
Exercise # 6) Decrypt the message RTOLK TOIK, which was encrypted using the affine transformation C = 3P + 24 (mod 26).
Exercise # 8) The message KYVMR CLVFW KYVBV PZJJV MVEKV VE was encrypted using a shift transformation C = P + k (mod 26). Use frequencies of letters to determine the value of k. What is the plaintext message?
Exercise # 10) If the two most common letters in a long ciphertext, encrypted by an affine transformation C = aP + b (mod 26), are X and Q, respectively, then what are the most likely values for a and b?
Exercise # 16) Find the product cipher obtained by using the transformation C = aP + b (mod 26) followed by the transformation C = cP + d (mod 26), where (a, 26) = (c, 26) = 1.

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INDIVIDUAL ASSIGNMENT, WEEK # 5, BY CHRIS IVY SECTION 8.1 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 transformation C = 15P + 14 (mod 26). Exercise # 6) Decrypt the message RTOLK TOIK, which was encrypted using the affine transformation C = 3P + 24 (mod 26). Exercise # 8) The message KYVMR CLVFW KYVBV PZJJV MVEKV VE was encrypted using a shift transformation C = P + k (mod 26). Use frequencies of letters to determine the value of k. What is the plaintext message? Exercise # 10) If the two most common letters in a long ciphertext, encrypted by an affine transformation C = aP + b (mod 26), are X and Q, respectively, then what are the most likely values for a and b? Exercise # 16) Find the product cipher obtained by using the transformation C = aP + b (mod 26) followed by the transformation C = cP + d (mod 26), where (a, 26) = (c, 26) = 1. SECTION 8.2 Exercise # 2) Decrypt the following message, which was enciphered using the Vigen`ere cipher with encrypting key SECRET: WBRCS LAZGJ MGKMF V. Exercise # 8) Cryptanalyze the given ciphertext, which was encrypted using a Vigen`ere cipher. S I I WZ F D I B N H U D E U WQ J H P J K R N K R L A C T WX B I M MHM P J O F U F P WV E O G P Q P E L V P Z Y D A X I A G P I TMA X F S S S GWP BW I WO F O T FWV F J S X P L B J O T P S U D I J J X F N R F P A F G R P S X I WX J O R P P X S Q I Exercise # 12) Show how we find that the correct key in Example 8.8 (see above) is USA once we know the key has length three. Exercise # 16) Decrypt the ciphertext message UW DM NK QB EK, which was encrypted using the digraphic cipher that sends the plaintext block P1 P2into the ciphertext block...

005_japo-xpsxqwph.docx

Robert · Accepted Answer

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 number
 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 b
a b
 
 

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 b
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 occurrences of  which are 15 positions apart. Again 
 find two occurrences 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 letter
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 ,

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...

Answer To: Exercise # 2) Decrypt the ciphertext message LFDPH LVDZL FRQTX HUHG, which has been encrypted using...

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