This article originally appeared upon on texperts.com

“Escape from Zurg” is a puzzle (featuring characters from the movie “Toy Story”) which has been used to teach students Logic Programming (in, for example, Prolog). Recently I came across a very interesting paper which presents an elegant Haskell solution to the riddle. This got me wondering about whether I could come up with a similarly satisfying solution in Ruby.

After a few false starts (mainly a result of me trying to transliterate the Prolog or Haskell solutions rather than come up with something which worked well in Ruby) I’ve come up with something that I’m pretty happy with. Not only is it (I think) concise and clear, but it’s also efficient. Given that Ruby doesn’t have built-in searching (like Prolog) and and doesn’t support lazy evaluation or list comprehension (like Haskell), the fact that the Ruby implementation is as straightforward as it turns out to be is, I think, a testament to the power of the language.

Here’s the puzzle:

Buzz, Woody, Rex, and Hamm have to escape from Zurg. They merely have to cross one last bridge before they are free. However, the bridge is fragile and can hold at most two of them at the same time. Moreover, to cross the bridge a flashlight is needed to avoid traps and broken parts. The problem is that our friends have only one flashlight with one battery that lasts for only 60 minutes (this is not a typo: sixty). The toys need different times to cross the bridge (in either direction):

Buzz: 5 minutes Woody: 10 minutes Rex: 20 minutes Hamm: 25 minutes Since there can be only two toys on the bridge at the same time, they cannot cross the bridge all at once. Since they need the flashlight to cross the bridge, whenever two have crossed the bridge, somebody has to go back and bring the flashlight to those toys on the other side that still have to cross the bridge.

The problem now is: In which order can the four toys cross the bridge in time (that is, in 60 minutes) to be saved from Zurg?

Like the Haskell solution, the Ruby solution depends on a `SearchProblem`

class which looks after the task of generating candidate solutions:

SearchProblem = Struct.new(:initial) do def each_candidate(&proc) step [], initial, &proc end def step(history, state, &proc) if state.terminal? yield history else state.each_successor do |move, state| step history + [move], state, &proc end end end end

`SearchProblem`

takes an initial `State`

which provides two methods; `terminal?`

returns true if the state is terminal (*i.e.* in our case, all the toys are on the right hand side of the bridge) and `each_successor`

calls a block with each valid move from that state, together with the new state after that move.

`SearchProblem`

provides an `each_candidate`

method which calls the provided block with each candidate solution in turn. Crucially, candidate solutions are generated as needed, so we don’t have to hold the entire search tree in memory (not a big deal for a problem of this size, but definitely an issue for a “real world” problem!). The Haskell solution achieves a similar effect through lazy evaluation (contrast with this Erlang solution which does have to generate the entire search tree).

To define the particular problem at hand, we first define our `Toys`

:

Toy = Struct.new(:name, :time) Toys = [ Toy.new(:buzz, 5), Toy.new(:woody, 10), Toy.new(:rex, 20), Toy.new(:hamm, 25)]

Next, we define `Move`

which we will use to represent transitions between states. In this case, a move consists of a direction (right or left) and a group of toys (two toys for a move to the right, one toy for a move to the left):

Move = Struct.new(:direction, :who) do def cost who.collect(&:time).max end end

`Move`

provides one method, `cost`

which returns the time taken to complete the move (the implementation of which makes use of the `Symbol#to_proc`

trick).

Most of the heavy lifting is performed by the `State`

class:

State = Struct.new(:pos, :group) do def terminal? group.empty? end def each_successor case pos when :left group.each_pair do |pair| yield Move.new(:right, pair), State.new(:right, group - pair) end when :right (Toys - group).each do |toy| yield Move.new(:left, [toy]), State.new(:left, group + [toy]) end end end end

A `State`

comprises two attributes, `pos`

which represents the current flashlight position and `group`

which represents the toys remaining on the left-hand side of the bridge (so the toys on the right-hand side will be `Toys - group`

).

The implementation of `State#terminal?`

is obvious — if `group`

is empty (*i.e.* if there are no toys on the left-hand side) then we’re done.

`State#each_successor`

has to deal with two cases, depending on the position of the flashlight. The implementation utilises a small utility method on `Array`

which allows all unique pairs of elements of the array to be generated:

class Array def each_pair each_with_index do |x, i| self[i + 1 ... length].each {|y| yield [x, y]} end end end

Finally, we create an instance of `SearchProblem`

with our initial state and print out the first candidate solution we find in which the time taken is no more than 60 minutes:

problem = SearchProblem.new State.new(:left, Toys) puts problem.each_candidate {|history| break history if history.inject(0) {|acc, move| acc + move.cost} <= 60 }

As it happens (by sheer fluke) with the `Toys`

array initialized in the natural order, the very first solution generated turns out to be the correct one, so the amount of work performed is the bare minimum. Clearly we can’t always rely on being so lucky in general, however!

To my mind, this solution is at least as easy to understand (and as flexible) as the Haskell equivalent (and both are a significant improvement on the Prolog). I would very much welcome comments and suggestions, however!

The complete source code can be downloaded here.

## Update (2007-09-11)

There’s an interesting discussion about this problem on the ruby-talk mailing list.