Question

In Hill cipher, the ciphertext is computed by: C = PK (mod 26), where P is the plaintext matrix and K is the key. All computations are done in (mod 26) for English alphabet. (a) Explain how Eve can break the system if enough plaintext-ciphertext pairs are provided. (b) What is the type of this attack?

Answer 1

a) According to the question it says C=PK(mod26) where P is plaintext matrix and K is the Key and C is Ciphertext

now let us analyze the following Encryption code

TR:DOG and CR: FQI

TR:GOD and CR:IQF

You can easily guess that algo is caesar cipher and key is +2!!!! By looking at the difference between D and F and then between G and I

It is evident only in those cases where weak algorithms and keys are used.In case of known plaintext attack (KPA) both TR and CR are available with Evans and he can easily break it.

But for Industry standard a strong algo and strong keys should be implemented.A CryptoSystem should pass through COA(cipher text only),CPA(chosen -plaintext),ACPA(Adaptive chosen plaintext) and RKA( Related key attacks) to avoid any sort of security bleach

b) This type of attack is known as KPA( known Plaintext Attack)