Ruby中简单的凯撒密码分析

Question

我真的很喜欢Ruby，我想做得更好。如有任何意见，我们将不胜感激。

描述

该程序提示用户输入一个字符串，该字符串是凯撒密码给定(英文)明文和范围0，25上的密钥所产生的密文。然后，该程序使用英文字符频率表执行简单的密码分析，以确定前5种最可能的移位密钥，并在这些密钥用于解密密文时显示所产生的明文。

还应该指出的是，唯一需要输入的是字母和空格，但我认为它适用于大多数标点符号。

代码：

class Cryptanalysis
  def initialize
    @alphabet = ('A'..'Z').to_a.join

    @english_frequency = { :A => 0.080, :B => 0.015, :C => 0.030,
                           :D => 0.040, :E => 0.130, :F => 0.020,
                           :G => 0.015, :H => 0.060, :I => 0.065,
                           :J => 0.005, :K => 0.005, :L => 0.035,
                           :M => 0.030, :N => 0.070, :O => 0.080,
                           :P => 0.020, :Q => 0.002, :R => 0.065,
                           :S => 0.060, :T => 0.090, :U => 0.030,
                           :V => 0.010, :W => 0.015, :X => 0.005,
                           :Y => 0.020, :Z => 0.002 }
  end

  def decrypt_caesar(ciphertext, shift)
    i = shift % @alphabet.size
    decrypt = @alphabet
    encrypt = @alphabet[i..-1] + @alphabet[0...i]

    ciphertext.tr(encrypt, decrypt)
  end

  def analyze(ciphertext)
    chars = ciphertext.gsub(/[^A-Z]/i, '').split(//)
    ciphertext_frequency = character_frequency(chars)

    correlation_of_frequency = Hash.new

    alphabet = @alphabet.split(//)
    (0..25).each { |i|
      sum = 0.0
      alphabet.each { |c|
        e = alphabet.index(c)
        sum += ciphertext_frequency[c.to_sym] * @english_frequency[alphabet[(26 + e - i) % 26].to_sym]
      }
      correlation_of_frequency[i] = sum
    }

    correlation_of_frequency.sort_by { |_, v| -v }.take(5).to_h
  end

  def character_frequency(chars)
    frequency = Hash.new
    inc = 1.0 / chars.size

    @alphabet.each_char { |c| frequency[c.to_sym] = 0.0  }
    chars.each          { |c| frequency[c.to_sym] += inc }

    frequency
  end
end

crypt = Cryptanalysis.new

puts 'Enter a string to analyze'
text = gets.chomp.upcase
result = crypt.analyze(text)

puts 'Top 5 possible shifts: '
choice = 1

result.each do |k, v|
  puts '%10s) Shift amount: %2s Correlation: %%%.4f Result: %s' % [choice, k, v, crypt.decrypt_caesar(text, k)]
  choice += 1
end

Answer 1

干得好。您可以通过以下方法获得一些简单的改进：

• 利用ruby的函数方法消除临时变量，使代码更具有声明性
• 酌情使用常量
• 通过在模块上而不是在类上使用静态方法来使代码反映您的风格--您实际上并没有将该类用于原始代码中的任何内容。或者，您可以在类构造函数中传入保存密文，然后执行类似于crypt_instance.analyze的操作。
• 避免像25或26这样的“神奇数字”。

重写：

module Cryptanalysis
  ALPHABET = ('A'..'Z').to_a

  ENGLISH_FREQUENCY = { :A => 0.080, :B => 0.015, :C => 0.030,
                        :D => 0.040, :E => 0.130, :F => 0.020,
                        :G => 0.015, :H => 0.060, :I => 0.065,
                        :J => 0.005, :K => 0.005, :L => 0.035,
                        :M => 0.030, :N => 0.070, :O => 0.080,
                        :P => 0.020, :Q => 0.002, :R => 0.065,
                        :S => 0.060, :T => 0.090, :U => 0.030,
                        :V => 0.010, :W => 0.015, :X => 0.005,
                        :Y => 0.020, :Z => 0.002 }

  def self.decrypt_caesar(ciphertext, shift)
    decrypt = ALPHABET
    encrypt = ALPHABET.rotate(shift)
    ciphertext.tr(encrypt, decrypt)
  end

  def self.analyze(ciphertext)
    freqs = frequencies(clean(ciphertext))

    (0..ALPHABET.size)
    .map{|i| [i, score(freqs, i)]}
    .sort_by { |x| -x[1] }
    .take(5)
    .to_h
  end

  def self.clean(text)
    text.upcase.gsub(/[^A-Z]/i, '')
  end

  def self.score(freqs, shift)
    ENGLISH_FREQUENCY
    .values
    .rotate(shift)
    .zip(freqs.values)
    .map{|x| x.first * x.last}
    .reduce(:+)
  end

  def self.frequencies(text)
    ENGLISH_FREQUENCY
    .keys
    .map {|k| [k.to_sym, text.count(k.to_s) / text.size.to_f]}
    .to_h
  end
end

text = "the quick brown fox jumps over the lazy dog"
puts 'Top 5 possible shifts: ' 
puts Cryptanalysis.analyze(text)