=begin
The number, 197, is called a circular prime because all rotations of the digits: 197, 971, and 719, are themselves prime.

There are thirteen such primes below 100: 2, 3, 5, 7, 11, 13, 17, 31, 37, 71, 73, 79, and 97.

How many circular primes are there below one million?


detect primes less than one million
check cirucularity
  rotate through all combination of digits in prime
  two digit:
    [0,1].join
    [1,0]
  three digits:
    [0,1,2]
    [2,0,1]
    [1,2,0]

  put string.last >> string[0], continue u

  check if each combination is also prime
add to array of ciruclar primes
calculate size of array
=end


class BaseProblemSolver

  def initialize(limit)
    @limit = limit
  end

  def solve
    circular_numbers = []
    primes_less_than(@limit).each do |prime|
      if circulars_of(prime).all? {|x| is_prime?(x)}
        circular_numbers << prime
      end
    end
    circular_numbers.size
  end

  def primes_less_than(num)
    range = [*2..num-1]
    range.select { |n| is_prime?(n)}
  end

  def is_prime?(num)
    array = [*2..num-1]
    array.all? do |n|
      num % n != 0
    end
  end

  def circulars_of(prime)
    results = []
    counter = 0
    prime_s = prime.to_s
    until counter == prime_s.size - 1
      prime_s = prime_s[1..prime_s.size] + prime_s[0]
      counter += 1
      results << prime_s.to_i
    end
    return results
  end
end

class FasterProblemSolver < BaseProblemSolver
  def initialize(limit)
    super(limit)
    @known_primes = []
    range = [*2..@limit - 1]
    range.delete_if do |n|


    # do the sieve
  end

  def is_prime?(num)
    known_primes.include?(num)
  end
end


CurrentSolver = BaseProblemSolver

p CurrentSolver.new(100).solve