Suppose we define an encryption scheme as follows. The key will be four elements

Question

Suppose we define an encryption scheme as follows. The key will be four elements k1,k2,k3 and k4 in Z26 . The message space will be sequences of elements of  of length 6. The ciphertext space will be the same as the message space.

The encryption algorithm is the following. Given a key  and  and a message , the corresponding ciphertext is , where:

b1 = k1a1 + k2a2 (mod 26)

b2 = k3a1 + k4a2 (mod 26)

b3 = k1a3 + k2a4 (mod 26)

b4 = k3a3 + k4a4 (mod 26)

b5 = k1a5 + k2a6 (mod 26)

b6 = k3a5 + k4a6 (mod 26)

Suppose that any four elements in Z26 can be a key.

1. How many keys are there in the Hill Cipher described above?

2. Suppose we regard the Hill Cipher as described in the description as a block cipher. The blocks are elements of  (i.e. sequences of letters of length 6). How many distinct blocks are there?

Explanation / Answer

1. As described in the question any letter from Z26 can be a letter in key. The key will be four elements from Z26.

In the equations, there are K1, K2, K3 and K4 from Z26.

We can create multiple keys from these letters. The total no of four letter keys will be (4*4*4*4 = 256). We can create 256 different keys by all the permutations of K1, K2, K3 and K4

Example - Keys can be K1K1K1K1, K1K2K3K4, K1K3K2K4, K3K3K3K3 ....etc

2. Blocks will be of length 6. we have a1, a2, a3, a4, a5 and a6 a total of 6 letters and we have to define different sequences of 6 letters. So there can be total (6*5*4*3*2*1 = 720) blocks possible of 6 letters.

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