87 lines
3.0 KiB
Julia
87 lines
3.0 KiB
Julia
#=
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Created on 17 Oct 2021
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@author: David Doblas Jiménez
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@email: daviddoji@pm.me
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Solution for Problem 59 of Project Euler
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https://projecteuler.net/problem=59
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=#
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using BenchmarkTools
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using Combinatorics
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using DelimitedFiles
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function Problem59()
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#=
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Each character on a computer is assigned a unique code and the preferred
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standard is ASCII (American Standard Code for Information Interchange).
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For example, uppercase A = 65, asterisk (*) = 42, and lowercase k = 107.
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A modern encryption method is to take a text file, convert the bytes to
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ASCII, then XOR each byte with a given value, taken from a secret key.
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The advantage with the XOR function is that using the same encryption key
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on the cipher text, restores the plain text; for example, 65 XOR 42 = 107,
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then 107 XOR 42 = 65.
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For unbreakable encryption, the key is the same length as the plain text
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message, and the key is made up of random bytes. The user would keep the
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encrypted message and the encryption key in different locations, and
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without both "halves", it is impossible to decrypt the message.
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Unfortunately, this method is impractical for most users, so the modified
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method is to use a password as a key. If the password is shorter than the
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message, which is likely, the key is repeated cyclically throughout the
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message. The balance for this method is using a sufficiently long password
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key for security, but short enough to be memorable.
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Your task has been made easy, as the encryption key consists of three
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lower case characters. Using p059_cipher.txt, a file containing the
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encrypted ASCII codes, and the knowledge that the plain text must contain
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common English words, decrypt the message and find the sum of the ASCII
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values in the original text.
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=#
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file = "../files/Problem59.txt"
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data = readline(file)
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data = parse.(Int, split(data, ","))
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# encrypted = readdlm(file, ',', Int)
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# print(encrypted)
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# with open('../files/Problem59.txt', 'r') as f:
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# # encrypted = list(map(int, f.read().split(',')))
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# encrypted = [int(char) for char in f.read().split(',')]
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# # print(encrypted)
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# # print(test)
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# plain_text = (length(encrypted) ÷ 3)
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# println(plain_text)
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# ascii_lowercase = join('a':'z')
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# for key in combinations(ascii_lowercase, 3)
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# decrypted = ""
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# for (k, i) in zip(repeat(key, plain_text), encrypted)
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# decrypted += Char(round(BigInt(Int(k) ^ i)))
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# end
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# # assuming Euler will be in the text
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# if "Euler" in decrypted
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# return sum([Int(c) for c in decrypted])
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# end
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# end
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ascisum = 0
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for shift=0:2, c='a':'z'
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decrypted = data[1+shift:3:end] .⊻ Int(c)
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if minimum(decrypted) == Int(' ') && maximum(decrypted) <= Int('z')
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ascisum += sum(decrypted)
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end
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end
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return ascisum
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end
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println("Time to evaluate Problem $(lpad(59, 3, "0")):")
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@btime Problem59()
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println("")
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println("Result for Problem $(lpad(59, 3, "0")): ", Problem59())
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