project euler #44 Pentagonal Numbers in Ruby. This is probably the most inefficient, naive way to solve it, but it solved the problem. blehhhhh

44a_pentagonal_numbers.rb

=begin

Pentagonal numbers are generated by the formula, Pn=n(3n−1)/2. The first ten pentagonal numbers are:

1, 5, 12, 22, 35, 51, 70, 92, 117, 145, ...

It can be seen that P^4 + P^7 = 22 + 70 = 92 = P^8.
However, their difference, 70 − 22 = 48, is not pentagonal.

Find a pair of pentagonal numbers P^j and P^k,
for which their sum and difference are pentagonal
and D = |P^k - P^j| is minimised (the smallest). What is the value of D?

=end

require 'set'
# require 'benchmark' 
# (I don't know how to use benchmark properly yet)

beginning = Time.now

=begin
no longer using this
class Array
  def contains? element
    self.any? { |x| x == element }
  end
end
=end

class Integer
  def pentafy
    self*(self*3 - 1)/2
  end

  #@@first_million = (1..1e6).to_a.map{ |x| x.pentafy }
  @@first_set = Set.new (1..1e5).to_a.map{ |x| x.pentafy }

  def pentagonal?
    #@@first_million.contains? self
    @@first_set.include? self
  end

end

puts "Hi, I just made the class: "
p Time.now - beginning
beginning = Time.now

#boolean: sum AND difference are pentagonal?
def sad?(pent1, pent2)
  (pent1+pent2).pentagonal? and (pent1-pent2).abs.pentagonal?
end

puts "...and I just defined sad?"
p Time.now - beginning
beginning = Time.now


first_some = (1..1e5).to_a.map{ |x| x.pentafy }
#first_ten = [1, 5, 12, 22, 35, 51, 70, 92, 117, 145]
#first_five = first_ten.take(5)

first_set = first_some.take(10000)

puts "Finally, I defined first_set"
p Time.now - beginning
beginning = Time.now

# [1,5] [1,12] [1,22]
pairs = []
first_set.each do |n|
  first_set.each do |m|
    if n < m
      pairs << [n,m]
    end
  end
end
puts "Oh, I am so tired.  I just defined pairs"
p Time.now - beginning
beginning = Time.now

pentagonal_pairs = pairs.select { |p| sad?(p[0],p[1]) }

puts "I work so hard for you.  I just defined pentagonal_pairs"

p pentagonal_pairs
p Time.now - beginning
beginning = Time.now

# and, inspired by @Spooner
puts "\n But don't ask me anymore, I can't be bothered to do any more work after all of this!"



# for any n, m
# only ONE of these three conditions can be true
# n < m
# n == m
# n > m

(last five lines from Nicky)