Question

A one-time pad was used to generate the ciphertext “AOS”. This one-time pad used the PRNG (5xi + 3) mod 19, starting from 7 to generate the shift amounts. That is, 7 was the first shift amount. Decrypt the ciphertext to generate the plaintext.

Answer 1

One time pad encryption uses a sequence of random number to encrypt plain text.

And this sequence of random number is shared with the receiver so that he can decrypt it.

In this we generate random number by Pseudo random number generation process with

Xi+1=(5Xi +3)mod 19

and it is given that seed(X0)=7

Therefore X1=(5X0+ 3)mod 19=((5X7)+3)mod 19=(35+3)mod19=38mod19=0 (mod means remainder)

Similarly X2=(5X1+ 3)mod 19=((5X0)+3)mod 19=(0+3)mod19=3mod19=3

Now, as we know that in one time pad cipher length of plain text., key and cipher text are same.

And as we are given that Cipher text="AOS" with length 3

So, key(sequence of random numbers) will be of length 3 also as X0X1X2

Now, Decryption algorithm for one time pad cipher is as follows:

Let cipher text is as C0C1 ......Cn and key is X0X1 ......Xn

then plain text will be P0P1 ......Pn

where Pi=(Ci - Xi) mod 26

our cipher text is C="AOS"

(A=0, B=1,C=2,........,Z=25)

C0="A"=0  

C1 ="O"=14

C2 ="S"=18

Therefore P0=(C0-X0)mod 26=(0-7)mod 26=19="T"

P1=(C1-X1)mod 26=(14-0)mod 26=14="O"

P2=(C2-X2)mod 26=(18-3)mod 26=15="P"

Therefore plain text is "TOP".

Hence cipher text is decrypted to generate plain text.