The following table specifies a cryptosystem based around a very simple encrypti

Question

The following table specifies a cryptosystem based around a very simple encryption algorithm with four different plaintexts A, B, C and D (one corresponding to each column) and four different ciphertexts A, B, C and D. The encryption algorithm has five different keys K1, K2, K3, K4, K5 (one corresponding to each row). By writing EK (P)=C to mean that the encryption of plaintext P using encryption key K is C, the entire cryptosystem is defined as follows:

EK1 (A)=B EK2 (A)=B EK3 (A)=D EK4 (A)=A EK5 (A)=C
EK1 (B)=C EK2 (B)=C EK3 (B)=B EK4 (B)=B EK5 (B)=D
EK1 (C)=D EK2 (C)=A EK3 (C)=A EK4 (C)=D EK5 (C)=A
EK1 (D)=A EK2 (D)=D EK3 (D)=C EK4 (D)=C EK5 (D)=B

(1) What is the size of the keyspace?

(2)  If an interceptor sees the ciphertext B then which plaintext can he rule out?

(3)  What is the ciphertext that results from encrypting plaintext B with K3, and is this a problem?

(4) Could we replace the bottom right-hand entry of the table with EK5 (D)=C?

(5)  Suppose that we define a sixth key K6 by the rule EK6 (P)=EK5 (EK1 (P)) for each plaintext P. In other words, for example, EK6 (A)=EK5 (EK1 (A))=EK5 (B)=D. What are the values for EK6 (B), EK6 (C) and EK6 (D)?

(6) Could we use a table such as this to represent a real cryptosystem?

Explanation / Answer

1) In a cryptosystem, the total number of different decryption keys is referred to as the size of the keyspace. In this example, the cryptosystem has 5 different keys, namely, K1, K2, K3, K4, K5. Hence, the size of the keyspace is 5.

2) The ciphertext B can be generated with the encryption of the following combination of plaintext and key.

It is very clear from above data that ciphertext B is obtained from the encryption of either A, B or D. It can never be obtained from plaintext C. Hence, we can rule out C.

3)  The ciphertext that results from encrypting plaintext B with K3 is given as

This is definitely a problem because encrypting plaintext B with K3 gives back B, which is same as plaintext. This means that plaintext B cannot be encrypted using K3 as it gives back the same plaintext after encryption. Hence, we need to modify the cryptosystem.

4) If we replace the bottom right-hand entry of the table with EK5 (D)=C, then the encryption of plaintext using K5 would look like:

If we notice carefully, ciphertext C can come from encryption of either A or D using the same encrytion key, K5. This would result in an ambiguity as now it would be impossible to find out which plaintext from {A or D} was encrypted using K5 to get C. Therefore, we cannot replace the bottom right-hand entry of the table with EK5 (D)=C.

5) EK6 (P)=EK5 (EK1 (P))

6) A real cryptosystem is suspectible to various hackers and crackers. Hence, a real cryptosystem should always be very difficult to crack.

In my opinion, this cryptosystem can be easily cracked using a few brute-force tries. This is beacuse this cryptosystem uses a straight-forward encryption of plaintext using a key and the ciphertext belongs to the set of plaintext too. Anyone with a few attempts could easily identify how to decrypt the ciphertext back as he already knows that ciphertext belongs to the set of plaintext.

Hence, we should not use this table to represent a real cryptosystem.

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