Question

Consider a set up that corresponds between alphabetic characters as follows

Assuming the encryption algorithm is C=11m+7 mod 26, where m is the

message and C is the ciphertext.

I. Encrypt the word “CYBERCRIME”

CR,10Marks

II. Derive the decryption algorithm

Answer 1

The given encryption algorithm is C=11m+7 mod 26

A-0, B-1, C-2,D-3,E-4,F-5,G-6,H-7,I-8,J-9,K-10,L-11,M-12,N-13,O-14,P-15,Q-16,R-17,S-18,T-19,U-20,V-21,W-22,X-23,Y-24,Z-25

Now, given word is CYBERCRIME

Now, encrypting the given terms-

c=2

$(11\times 2+7)mod\ 26=29\ mod\ 26$ =3

C=D

Y=24

$(11\times 24+7)\ mod \ 26 = 271 \ mod \ 26 = 11$

y=L

b= 1

$(11\times1+7)\ mod \ 26=18$

B= S

E=4

$(11\times 4+7)\ mod \ 26 = 51 \ mod \ 26 = 25$

E= Z

R=17

$(11\times 17 + 7)\ mod \ 26 = 194 \ mod \ 26 = 12$

R = M,

c=2

$(11\times 2+7)mod\ 26=29\ mod\ 26$ =3

C=D

R=17

$(11\times 17 + 7)\ mod \ 26 = 194 \ mod \ 26 = 12$

R = M,

$I=8$

$(11\times 8 + 7)\ mod \ 26 = 95 \ mod \ 26 = 17$

I=R

M=12

$(11\times 12 + 7)\ mod \ 26 = 139 \ mod \ 26 = 9$

M=J

E=4

$(11\times 4+7)\ mod \ 26 = 51 \ mod \ 26 = 25$

E= Z

Hence, the encrypted word is DLSZMDMRJZ

ii) The decryption algorithm is given below -

$C=(11m+7)\ mod\ 26$

$\Rightarrow 11m=C-7 \ mod \ 26$

$\Rightarrow m= 11^{-1}(C-7 \ mod \ 26)$

Inverse of 11 in mod 26 is 19 Because $11\times 19 \ mod\ 26= 1\ mod \ 26$

Therefore Decryption algorithm will be $m=19(C-7) \ mod \ 26$

$m=19 C - 3\ mod \ 26$