UGC NET Computer Science July 2018 - II Question 48

Question 48

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Decrypt the message “WTAAD” using the Caesar Cipher with key = 15.

1. (1) LIPPS
2. (2) HELLO
3. (3) OLLEH
4. (4) DAATW

Answer : (2) HELLO

Explanation Question 48

Caesar Cipher Algorithm Mathematical Description:
First we translate all of our characters to numbers, 'a'=0, 'b'=1, 'c'=2, ... , 'z'=25.
We can now represent the Caesar cipher encryption function, e(x), where x is the character we are encrypting, as:
e(x)=(x+k) (mod 26)

Where k is the key (the shift) applied to each letter.
After applying this function the result is a number which must then be translated back into a letter.
The decryption function is :
d(x)=(x-k) (mod 26)

Solution:

Remember alphabet ordering:
Only remember the alphabets that occur at multiplication of 5:
1 st alphabet -> A  (1)
2 nd alphabet -> E (5)
3 rd alphabet -> J  (10)
4 th alphabet -> O (15)
5 th alphabet -> T (20)
6 th alphabet -> Y (25)

Need to remember word EJOTY
for our case number numbering is E- 4, J-9, O-14, T-19 Y-24

Position of our alphabets:
W -> T+3 -> (19+3) = 22
T -> 19
A -> 1
D -> 3

Decryption of given alphabets:
W -> (22 -15) (mod 26) = 7 ->H
T -> (19 -15) (mod 26) = 4  -> E
A -> (1-15) (mod 26) = -14 +26 -> L
A -> (1-15) (mod 26) = -14 + 26 -> L
D -> (3 -15) (mod 26) = -12 + 26 -> O

In the above, the result is in the range 0 to 25; i.e., if x + n or x − n are not in the range 0 to 25, we have to subtract or add 26.)

We can decrypt the the message "WTAAD" using the Caesar Cipher with key=15 to "HELLO".

So, option (2) is correct answer

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