Question

3. Bob is arguing that if you use output feedback (OFB) mode twice in a row to encrypt a long message, M, using the same key each time, it will be more secure. Explain why Bob is wrong, no matter what encryption algorithm he is using for block encryption (15 points).

Answer 1

Bob was wrong in his argument because from the cipher text C₂ only the initialization vector is encrypted two times and XOR with plain text M.

Also it will not provide a better security than the first because it's encryption was done only on initialization vector and not the plain text.

Step-by-step explanation:

Solution

From the given question, in the output feedback mode (OFB) the following Encryption Algorithm is stated as follows:

Where,

V₀ = vector initialization

V₁ = Ek (Vi-₁)

Ek = The encryption algorithm

C₁ =V₁⊕ M₁

Here,

C₁ = this is the cipher text

M₁ = Plain text for it^th session

Thus

The Decryption Algorithm is shown below:

M₁=V₁⊕ C₁

In this case, when bob encrypt the message M for the first time in the OFB method, the cipher text is

C₁ =V₁⊕ M₁

So,

C₁ =Ek (V₀) ⊕ M

Hence the Cipher text C₁ the initialization vector only is encrypted

Thus,

When Bob does a second encryption, the Cipher text will be

C₂ =Ek (V₁) ⊕ C₁

C₂ = Ek (Ek (V₀)) ⊕ Ek (V₀) ⊕ M

Finally, from the cipher text C₂ only the initialization vector encrypted twice and XOR with plain text M. so it will not provide a better security than the first because it's encryption was carried out on initialization vector without the plain text.