Encrypting text using Caeser cipher in CFB mode

Question

I am interested in learning and understanding cryptography and its application in every day life. One thing I can't seem to get my head around is encrypting plaintext using a block mode such as CFB. I want to use a Caesar block cipher in CFB mode to encrypt the following plaintext: lucozade with a key: 3.

So in this case, I get that using the Caeser block cipher the ciphertext for lucozade would be oxfrcdgh as the letters have shifted three places to the right. What I can't do is this encryption using CFB mode. How do I go about doing that? Like would I have to use the ciphertext oxfrcdgh as the IV?

Any help would be great

Answer 1

Caesar cipher

You are confusing Caesar cipher with block ciphers.

In Caesar cipher the ciphertext is calculated as

$$c_i = p_i + 3 \bmod 26$$

There is no security in Caesar Cipher. A ciphertext only attack possible, as worst scenario.

CFB Caesar cipher

$c_i = c_{i-1} + 3 + p_i \bmod 26 $, with $c_0 = IV$

And, the decryption;

$p_i = c_i - 3 -c_{i-1} \bmod 26 $, with $c_0 = IV$

Example : $P=lucozade = \{11,20,2,14,25,0,3,4\}$

Let choose $IV = f = 5$

\begin{align} c_1 &\equiv IV + 3 + p_1 \bmod 26 \equiv 5 + 3 + 11 \bmod 26 \equiv 19 \text { that is } T\\ c_2 &\equiv c_1 + 3 + p_2 \bmod 26 \equiv 19 + 3 + 20 \bmod 26 \equiv 16 \text { that is } Q\\ c_3 &\equiv c_2 + 3 + p_3 \bmod 26 \equiv 16 + 3 + 2 \bmod 26 \equiv 19 \text { that is } T\\ c_4 &\equiv c_3 + 3 + p_4 \bmod 26 \equiv 19 + 3 + 14 \bmod 26 \equiv 10 \text { that is } K\\ \end{align}