The dining philosophers problem in Ruby, solved using the resource hierarchy solution

rons.rb

require 'thread'
 
class Ron
  def initialize(name, left_fork, right_fork)
    @name = name
    @left_fork = left_fork
    @right_fork = right_fork
    while true
      think
      dine
    end    
  end
 
  def think
    puts "Ron #@name is thinking..."
    sleep(rand())
    puts "Ron #@name is hungry..."
  end
 
  def dine
    while true
      @left_fork.lock
      puts "Ron #@name has one fork..."
      if @right_fork.try_lock
        break
      else
        puts "Ron #@name cannot pickup second fork"
        @left_fork.unlock
      end
    end
    puts "Ron #@name has the second fork!"
 
    puts "Ron #@name eats..."
    sleep(rand())
    puts "Ron #@name belches"
 
    @left_fork.unlock
    @right_fork.unlock
  end
end
 
n = 5
 
forks = []

(1..n).each do |i|
  forks << Mutex.new
end

threads = []

(1..n).each do |i|
  threads << Thread.new do
    if i < n
      left_fork = forks[i]
      right_fork = forks[i+1]
    else
      # special case for philosopher #5 because he gets forks #5 and #1
      # and the left fork is always the lower id because that's the one we try first.
      left_fork = forks[0]
      right_fork = forks[n]
    end
    Ron.new(i, left_fork, right_fork)
  end
end

threads.each {|thread| thread.join}