My simple "Escape from Zurg" solver

Escape from Zurg in Ruby.rb

require "pp"

DUDES = [ { name: "Alice",   time: 5 },
          { name: "Bob",     time: 10 },
          { name: "Candace", time: 20 },
          { name: "Dave",    time: 25 } ]

INITIAL_STATE = { left: [ ], right: (0...DUDES.length).to_a, time_left: 60 }

def backward(state, path)
  time_left, left, right = state[:time_left], state[:left], state[:right]
  return [path] if right.empty?
  left_candidates = left.select { |x| DUDES[x][:time] <= time_left }  
  left_candidates.inject([ ]) do |acc, p|
    time_for_p = DUDES[p][:time]
    new_left, new_right = left - [p], right + [p]
    acc + forward({ time_left: time_left - time_for_p,
                    left: new_left, right: new_right },
      path + ["#{DUDES[p][:name]} goes backward"])
  end
end

def forward(state, path)
  time_left, left, right = state[:time_left], state[:left], state[:right]
  right_candidates = right.select { |x| DUDES[x][:time] <= time_left }
  right_candidates.inject([ ]) do |acc, p1|
    time_for_p1 = DUDES[p1][:time]
    right_without_p1 = (right - [p1]).select do |x|
      DUDES[x][:time] <= time_for_p1
    end
    acc + right_without_p1.inject([ ]) do |acc, p2|
      new_left, new_right = left + [p1, p2], right - [p1, p2]
      acc + backward({ time_left: time_left - time_for_p1,
                       left: new_left, right: new_right },
        path + ["#{DUDES[p1][:name]} and #{DUDES[p2][:name]} go forward"])
    end
  end
end

pp forward(INITIAL_STATE, [ ])