Problem Set 5 - Problem 3

[tiagomestreteixeira]

Problem 3 - CiphertextMessage

For this problem, the graders will use our implementation of the Messageand PlaintextMessageclasses, so don't worry if you did not get the previous parts correct.

Given an encrypted message, if you know the shift used to encode the message, decoding it is trivial. If message is the encrypted message, and s is the shift used to encrypt the message, then apply_shift(message, 26-s) gives you the original plaintext message. Do you see why?

The problem, of course, is that you don’t know the shift. But our encryption method only has 26 distinct possible values for the shift! We know English is the main language of these emails, so if we can write a program that tries each shift and maximizes the number of English words in the decoded message, we can decrypt their cipher! A simple indication of whether or not the correct shift has been found is if most of the words obtained after a shift are valid words. Note that this only means that most of the words obtained are actual words. It is possible to have a message that can be decoded by two separate shifts into different sets of words. While there are various strategies for deciding between ambiguous decryptions, for this problem we are only looking for a simple solution.

Fill in the methods in the class CiphertextMessage acording to the specifications in ps6.py. The methods you should fill in are:

• __init__(self, text): Use the parent class constructor to make your code more concise.
• decrypt_message(self): You may find the helper function is_word(wordlist, word) and the string method split() useful. Note that is_word will ignore punctuation and other special characters when considering whether a word is valid.

Hints
Using string.split
You may find the function string.split useful for dividing the text up into words.

>>> 'Hello world!'.split('o')
['Hell', ' w', 'rld!']
>>> '6.00.1x is pretty fun'.split(' ')
['6.00.1x', 'is', 'pretty', 'fun']

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Task of #39