Step by Step solution about How to resolve the algorithm Population count step by step in the Ruby programming language

The provided Ruby code extends the Integer class with two instance methods: popcount and evil?. It also includes some sample usage of these methods. Here's a detailed explanation:

1. Integer#popcount: This method counts the number of 1 bits in the binary representation of an integer. In Ruby 2.4 and later, it uses the digits method to convert the integer to a base-2 array and then counts the occurrences of 1 in that array. In Ruby versions prior to 2.4, it converts the integer to a binary string using to_s(2) and counts the occurrences of "1" in the string.

2. Integer#evil?: This method returns true if the integer is non-negative and the number of 1 bits in its binary representation is even. Otherwise, it returns false. An "evil" number is a non-negative integer with an even number of 1 bits in its binary representation.

3. After defining the instance methods, the code includes some sample usage:

   • It calculates and prints the popcount for powers of 3 from 0 to 29 using the map method and join method.
   • It prints the first 30 evil numbers using the step method, lazy method, and select method to filter out non-evil numbers.
   • It prints the first 30 non-evil numbers using the step method, lazy method, and reject method to filter out evil numbers.

class Integer
 
  def popcount
    digits(2).count(1)     #pre Ruby 2.4: self.to_s(2).count("1")
  end
 
  def evil?
    self >= 0 && popcount.even?
  end

end
 
puts "Powers of 3:",  (0...30).map{|n| (3**n).popcount}.join(' ')
puts "Evil:"  , 0.step.lazy.select(&:evil?).first(30).join(' ')
puts "Odious:", 0.step.lazy.reject(&:evil?).first(30).join(' ')