Question

Consider the following substitution block cipher:

Plain-text

0. 000 110

1. 001 100

10. 010 111

11. 011 001

100. 100 101

101. 101 000

110. 110 010

111. 111 011

Cipher-text

[10]

                 

Compute the cipher-text belonging to plaintext 001 110 000 101 110 (using a block size of 3 bits) for the Electronic Code Book (ECB) mode and Cipher Block Chaining (CBC) mode taking IV = 111. Show the intermediate steps.

Answer 1

Let's Consider the given substitution block cipher:

Plain Text	Cipher-Text
000	110
001	100
010	111
011	001
100	101
101	000
110	010
111	011

To Compute the cipher-text belonging to plaintext 001 110 000 101 110

Electronic Code Book (ECB): In Electronic Code Book Mode we encrypt each block independently. So We use the Given Substitution cipher and encrypt the message as follow:

Plain Text(Block of 3)	001	110	000	101	110
Cipher-Text(Block of 3)	100	010	110	000	010

The Cipher Text (Computed using ECB) is: 100 010 110 000 010

Cipher Block Chaining (CBC) mode taking IV = 111. Show the intermediate steps.

Solution:

In CBC we use the following Formula for Encrypting the Given Message:

C_i=F_k(P_i⊕C_i-1)

For i=1:

C₁=Fk (P₁⊕IV) For the first Block we use the Initialization Vector.

=F(001⊕ 111)=F(110)=101

C₂=Fk (P2⊕C1)=Fk (110⊕101)=Fk(011)=001

C₃=Fk (P3⊕C2)=Fk (000⊕001)=Fk(001)=100

C₄=Fk (P4⊕C3)=Fk (101⊕100)=Fk(001)=100

C₅=Fk (P5⊕C4)=Fk (110⊕100)=Fk(010)=111

The Computation is Summarized in the following Table:

S.NO	Pi	Ci-1	Pi⊕Ci-1	Cipher Text (ci)
1	001	111	110	101
2	110	101	011	001
3	000	001	001	100
4	101	100	001	100
5	110	100	010	111

The Encrypted Message is: 101 001 100 100 111