collatz.rb

# Take a positive integer, like 9, and apply the following rule. If it’s odd, multiply it by 3 and add 1; if it’s even, divide it in half. 9 -> 28 -> 14 -> 7 -> 22 -> 11... The Collatz conjecture says that all such chains will terminate in the 1 -> 4 -> 2 -> 1 -> 4... loop. Calling “1” the end of a chain, for what integer less than a million is the Collatz chain the longest?

require 'active_support'
extend ActiveSupport::Memoizable

def apply_collatz_rule(num)
  if num % 2 == 0
    num / 2
  else
    num * 3 + 1
  end
end

def compute_collatz_chain_length(num)
  return 0 if num == 1
  1 + compute_collatz_chain_length(apply_collatz_rule(num))
end
memoize :compute_collatz_chain_length

def find_longest_chain(upper_bound)
  longest_chain = 0
  num_with_longest_chain = 0
  
  1.upto(upper_bound) do |num|
    new_candidate = compute_collatz_chain_length(num)
    
    if new_candidate > longest_chain
      longest_chain = new_candidate
      num_with_longest_chain = num
    end
  end
  
  "#{num_with_longest_chain}: #{longest_chain}"
end

including memoizable into kernel? Just do (assuming this is in a class and not on kernel)

def compute_collatz_chain_length(num)
  if num == 1
    0
  else
    @memo ||= {}
    @memo[num] ||= 1 + compute_collatz_chain_length(apply_collatz_rule(num))
  end
end

"but what if @memo[num] == false"