Beginner Programming Challenge #10

[schauerlich]

And again in ruby.

DEPTH = 30
MAX_LYCHREL = 10000

class Integer
  def is_lychrel?(d=DEPTH)
    def reverse
      return self.to_s.reverse.to_i
    end

    if d == 0
      return true
    elsif (d != DEPTH) and (self == self.reverse)
      return false
    end

    return (self + self.reverse).is_lychrel?(d-1)
  end
end

puts (1..MAX_LYCHREL).select { |x| x.is_lychrel? }

[JackDarkStone]

I wouldn't consider this a beginner challenge.

[cprofitt]

I wouldn't consider this a beginner challenge.

Jack:

Perhaps, but it is a fairly simple program once you understand the problem.

[Spae]

That was fun and challenging

def rev_N(N):							#reverse a number
    return int(str(N)[::-1])

def main():
    for N in range(11,10001,1):					#Testing numbers 11 to 10000
        attempt_N = 0
        original_number = N
        number = original_number
        reverse = rev_N(number)

        while attempt_N < 30 and number != reverse:		#try 30 times or until a palindrome has been found
            attempt_N = attempt_N + 1 				#counting tries	

            if number != reverse and attempt_N == 30:		#After attempt_N attempts a palindrome has not been formed
                print(original_number,'is a lychrel candidate') #until the 30th iteration
        
            elif number != reverse:				#if the numbers are not equal yet:
                number = number + reverse			#add them together to make new number
                reverse = rev_N(number)				#and new reverse number
main()

#I realise it tests some numbers twice like 59 and 95, and then 95 and 59. Not sure if it's easy to fix that.

[Bachstelze]

Thread closed in 3, 2, 1...

[matehussilvapb]

You can't prove that a number is lychrel number ... You can take n iterations to get to a palindrome and prove it's a non-lychrel number, but if you want to determine it's a lychrel number you'll need to take all the possible iterations, and that's an infinite number of iterations. Sorry man great try !