AugustoPedraza

# Given the word list, count how many words are in the dictionary.
# Note: This isn't the most efficient way to do thise.  If I was going for
#   speed, I would have made each dicitonary word the key of a hash, and
#   had O(n) lookup times.
def process_words(words, dictionary)
  puts "Starting thread #{Thread.current.object_id}\n"

  found = 0
  words.each do |word|
    if dictionary.include? word