Question

Alice has at her disposal 3 encryption algorithms, they are: a transposition cipher, a monoalphabetic substitution cipher and a polyalphabetic substitution cipher. She encrypts the following plaintext: Hello World For each of the following ciphertexts write down which type of cipher, or ciphers, Alice could have used to generate it. Explain your reasoning. In addition if you believe that more than one cipher could have generated the ciphertext explain which one is the most likely. (a) ollerldWHO (3 marks) (b) LippsAsvph (3 marks) (c) ZUFLDARJBX (3 marks) (d) oller (3 marks)

Answer 1

In transposition cipher, the plaintext is jumbled in a way to generate cipher text.

In monoalphabetic substitution cipher, each letter in plaintext is substituted with some other letter. In this one to one mapping is used. This means if 'a' is substituted by 'B' then in whole plaintext a will be substituted by 'B'.

In polyalphabetic substitution cipher, one letter can be replaced by more than one letter in different occurrences. For example: If at one occurrence 'a' is substituted by 'B' then at other occurrence 'a' may be substituted by 'C' in the ciphertext.

allerld WHO oll In this cipher tent we can observe that all the letters are same as that of plantent but they are gumbled. This is most likely a case when transposition cipher is used. ♡ Lipps Asuph Here we observe that in place of dee, we get PP: Also in place of 6 of World, we get po in cipher Also è of Hello is replaced with's' and same is the case with o' of World: so, this is most likely a case of mono alphabetic substitution cipher. 0 0

© ZUFLDARIB X Here for ll, FL is used. So do get this cipher tent we have to use polyalphabetic substitution cipher @ aller Ciphortent is lesser in size than plaintest None of the 3 tipte encryption algorithms can give this plain ciphertent

6.

a. Number of intrusions = Number of actual positives = 450+50 = 500

b. Number if times alarm is raised = Number of predicted positives = 450 + 40 = 490

c. Accuracy = Rightly classified / Total = (450 + 460) / 1000 = 910/1000 = 0.91 OR 91%.

d. Accuracy of second detector = (400 + 390) / 1000 = 790/1000 = 0.79 OR 79%

So, first detector performed better.

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