Graph Traversal: solving the 8-puzzle with basic A.I.

I’ve been working through Peter Norvig and Stuart Russel’s Artificial Intelligence, A Modern Approach (thanks to the Square engineering library) and one of the most helpful chapters involved methodically demonstrating basic graph traversal algorithms for problem solving. If that sounds heady, it’s not — I think you’ll enjoy it.

First,binary tree on Wikipedia let’s talk about graphs. A graph is any set of points (nodes) and the lines (edges) between those points. A simple kind of graph is a tree structure. This is where there’s a single ‘root’ which has 1 or more branches that then have their own branches, etc. Many things in computer science can be expressed using some kind of tree structure graph.

Here’s how this relates to AI: the root node (the one at the top of the tree) is the state the world is in right now. Each other node represents a different state of the world that’s reachable across an edge by the node immediately above it. The line between them is some kind of action. Moving a car’s wheels, turning on a servo, whatever. The A.I. just needs to know that if you start at node A and do action X you’ll get to node B. Your software will appear to be intelligent if it can start at the root node and find its way to a better state of the world through a series of actions.

A world where you try to get a big number

Let’s reuse the above image as an example. Imagine it’s a complete map of all possible states of the world. The only actions that are available to you are “go left” and “go right” (‘cause you’re always starting at the root (top) node and heading downward). Imagine, additionally, that the value of the number of a node is how much we like that node. From a glance you can easily tell that one of the nodes has a value of 11 and is therefore the “best” node in the graph; it’s the “best” state of the world that we could possibly get to. If our software is intelligent it should start at the root node and find the 11 node as it’s destination.

Since we can easily see where the 11 node is the right answer is “go left, then right, then right again.” But how do we teach our code to find the best node?

Here’s the simplest way we can find the right node on the given tree.

best = nil
def walk(node)
  best = node if best.nil? || node > best
  node.children.each {|child| walk child }
end
walk root

Here we “walk” down each branch of the tree all the way to the end using recursion. This is known as recursive depth first search and is a great tool when you think that any path might have a good node really far down so you just want to get really deep really fast. It’s also the least code possible to find a solution to our problem. Unfortunately, the simple implementation of depth-first involves recursion which means we’re limited to only traversing graphs that have a total depth less than our runtime stack frame limit. If you’ve ever seen a “stack overflow error” it’s because there was so much recursion in your program that the computer assumed you were caught in an infinite loop and gave up. To demonstrate this try running this simple program in irb:

def go(n)    # On my machine the last thing printed was 8247 and then I saw:
  puts n     #   SystemStackError: stack level too deep
  go n + 1   # which means Ruby let me use 8,247 stack frames
end          # before giving up. If I were to try going 9,000 nodes deep
go 0         # in a graph I'd get this error.

Now, if you wanted to do a depth-first search without recursion you can but it’s no longer the simplest code so I’ll skip it here. Suffice to say that it would mean you’ll have to manually keep track of which nodes to visit in which order rather than letting your programming language do it for you.

Breadth-first search in a loop

Another simple way to traverse the graph is to look at the nodes from left to right on each level rather than going all the way down one branch and then all the way down another branch. This is called breadth-first search and is better if you think there’s a good node close to the root.

This is also a very simple bit of code and it frees us from having to rely on our runtime’s call stack. Rather than recursing through methods and letting Ruby keep track of our work (as in the above example) we’ll just run in a loop and store our work in an array. The advantage here is you can put more than 9,000 items in an array — it’s only bounded by how much memory is on your machine.

queue = [root]
best = -1
begin
  current = queue.shift
  best = [current, best].max
  current.children.each {|child| queue.push child }
end until queue.empty?

The name “breadth-first” comes from the fact that it’ll look at all the nodes at each level from side to side before proceeding down to the next level. Notice that the array variable is named queue. This is because in a depth-first implementation you’re always going to have a list that you put newly-discovered nodes onto the end of and pull nodes to explore off of the front.

Breadth-first search is easy to reason about, you won’t run out of stack space like when you used recursion (although it’s possible you’ll run out of memory), and if your environment supports concurrency primitives you might be able to run it in parallel quite easily. C# has a Consumer Queue that can help with this and Clojure has multiple ways to iterate through a list in parallel. Ruby requires you to do more work synchronizing when threads get to append their newly-discovered nodes to the end of the queue but otherwise you get the parallelism cheaply.

Solving the sliding-block puzzle

We15 puzzle walk along the edges of a graph from the starting node to a better node. With me so far? If this graph represents a set of world states that are reachable from each other via actions then searching the graph is the same thing as figuring out what to do.

Let’s use this technique to try to solve a problem that has a clear starting state and a clear ending state with many (possibly very many) intermediate states. The sliding-block puzzle (often called an 8-puzzle or, in it’s larger variant, a 15-puzzle) is a great case for us to tackle.

In an 8-puzzle you’ve got a bunch of tiles in the wrong places and just one empty space that you can move around until the tiles are in the right order. How do you know which move to make first? How do you know when you’re on the right track? How do you know if you’ve going in loops?

If we model each possible action as edges in a graph and each potential puzzle state as a node then we just start at the beginning and begin exploring the graph. We’ll stop once we’ve found the solution (or, if we built our code poorly, we’ll stop when we run out of memory or time).

Now, there are two ways we can set up our data for this problem. One is to generate all possible states (nodes) that the puzzle can have and to then connect the adjacent states. We would then have a complete graph we could traverse and we’d even be able to mark the solution ahead of time and know where it was located in the graph. Unfortunately, the number of states is 9-factorial or about 360,000. Generating that many puzzle slide orientations and then iterating through each one would take, at best, (9!)(9!-1)/2 node-comparison operations (the formula for how many edges can exist between nodes in a graph is (n * n-1)/2)

So let’s not do that. Rather, let’s start at the root node (the starting state) and then create branches from each node as we go. We’ll stop when we discover our solution — hopefully long before we examine 360,000 states.

Defining a puzzle class

We’re going to need a few tools. First, let’s put together a way to represent a puzzle board with tiles in a particular position:

class Puzzle
  Solution = [0, 1, 2,              # Let's put all the tiles in ascending order
              3, 4, 5,              # the same way you'd see them on a phone keypad.
              6, 7, 8]              # We use a '0' for the blank cell because
                                    # `nil` doesn't play well with others.
  attr_reader :cells
  def initialize cells
    @cells = cells
  end

  def solution?
    Solution == @cells
  end
end

What did we just do there? That is a Puzzle class where each instance knows whether it’s a solution. The cells/tiles of the puzzle are kept in a list.

Now let’s construct a way to represent a state (a node on the solution graph). A state isn’t just a representation of puzzle tile position but also the history of how that puzzle arrangement was reached from the starting point. This is key: if we don’t keep track of how we got to a solution node on the graph then we’ll never be able to report how to solve the puzzle. So we need to keep a list as we go of which actions we’ve taken to arrive at the current node.

class Puzzle                                          # We're extending the Puzzle class.
  def zero_position                                   # `zero_position` tells us which cell has
    @cells.index(0)                # the '0'.
  end

  def swap swap_index                                 # `swap` tells the puzzle:
    new_cells = @cells.clone                          # "give me you, but with the '0' cell
    new_cells[zero_position] = new_cells[swap_index]  # replaced by the cell at some other
    new_cells[swap_index] = 0                         # location of my choice."
    Puzzle.new new_cells                              # This is how we'll simulate moving a tile.
  end
end

class State                                           # Each `State` instance represents a node
  Directions = [:left, :right, :up, :down]            # in our solution graph. It keeps track of
                                                      # both a puzzle and the list of actions
                                                      # required to arrive at there from the
  attr_reader :puzzle, :path                          # starting node.
  def initialize puzzle, path = []                     
    @puzzle, @path = puzzle, path                     # The `path` is a list of actions
  end                                                 # like 'up', 'down', 'right', right'

  def solution?                                       # Each node in our graph should
    puzzle.solution?                                  # know if it's a solution.
  end

  def branches                                        # Returns all adjacent possible states
    Directions.map do |dir|                           # including steps needed to get there.
      branch_toward dir                               # Most nodes will have 2-4 branches based
    end.compact.shuffle                               # on how many directions the blank tile
  end                                                 # can try to go.

  private

  def branch_toward direction
    blank_position = puzzle.zero_position
    blankx = blank_position % 3
    blanky = (blank_position / 3).to_i
    cell = case direction                      # The only reason this method is so long
    when :left                                 # is because sometimes the blank tile is already
      blank_position - 1 unless 0 == blankx    # at a wall and that direction isn't possible
    when :right
      blank_position + 1 unless 2 == blankx
    when :up
      blank_position - 3 unless 0 == blanky
    when :down
      blank_position + 3 unless 2 == blanky
    end
    State.new puzzle.swap(cell), @path + [direction] if cell
  end
end

This State class knows about one particular arrangement of the puzzle and is able to determine next steps. When we call State#branches we get a list of adjacent puzzle arrangements (anything reachable by moving the empty space over by 1 square) and each of these new states include the full list of steps necessary to reach them.

That’s the setup. Now that we have some problem-specific helpers we can use our breadth-first algorithm from up above to start tackling this.

def search state
  state.branches.reject do |branch|        # Important: don't revisit puzzles
    @visited.include? branch.puzzle.cells  # you've already seen!
  end.each do |branch|                     # The list of places we need to search
    @frontier << branch                    # is known as the 'frontier'
  end
end

require 'set'                           # We'll remember what we've seen in a set, it has
def solve puzzle                        # way better lookup times than an array.
  @visited = Set.new
  @frontier = []
  state = State.new puzzle
  loop {
    @visited << state.puzzle.cells
    break if state.solution?            # This is the `base` or end condition
    search state
    state = @frontier.shift             # Pull another off the list, keep chugging along
  }
  state
end

If we feed in a solveable puzzle we can see that this code works. Let’s try one where the empty tile was moved right and then down. The solution should be to move it up and then left:

p solve(Puzzle.new [1, 4, 2,         # `solve` is going to return a State
                    3, 0, 5,         # instance. We care about it's #path
                    6, 7, 8]).path
## => [:up, :left]

So… it works, but it’s just kinda wandering around until it finds a solution. We gave it a problem that was only 2 steps from a solution so if we gave it something harder would it ever finish? And how long would the solution path be?

Here’s our code running with a puzzle who’s optimal solution is 20 steps away:

p solve(Puzzle.new [7, 6, 2,       # I generated this by creating a solution state and
                    5, 3, 1,       # running `state = state.branches.sample` in a big loop.
                    0, 4, 8]).path
## => [:up, :up, :right, :down, :right, :down, :left, :left,
##     :up, :up, :right, :down, :down, :right, :up, :left,
##     :down, :left, :up, :up]
## Time: 27 seconds

It works! Eventually. But 27 seconds is a bit slow. What if we were tackling the 15-puzzle instead? Rather than the 9! (360K) options we would be searching through 16! (20 trillion) options. That would take almost literally forever.

As we walk the graph we’re keeping a frontier — a list of states we’re hoping to explore in the future. Since we always add to the back of the list and take (shift) from the front it’s technically a FIFO queue rather than just a list.

What if, rather than picking the next element from the queue to explore we tried to pick the best one? Then we wouldn’t have to explore quite so many trillions of nodes in our state graph.

Uniform-cost search entails keeping track of the how far any given node is from the root node and using that as its cost. For our puzzle example that means that however many steps n it takes to get to state s then s has a cost of n. In code: s.cost = steps_to_reach_from_start(s). A variant of this is called Dijkstra’s Algorithm.

There’s one missing piece here though: we don’t want to examine every item in the entire frontier queue every time we want to pick the next lowest-cost element. What we need is a priority queue that automatically sorts its members by some value so looking up an element by cost is cheap and doesn’t slow down the rest of what we’re trying to do.

class PriorityQueue            # This is a terrible implementation of a
  def initialize &comparator   # priority queue. The `#sort!` method iterates
    @comparator = comparator   # through every item every time.
    @elements = []             # What you want is a priority queue backed by a heap data structure.
  end                          # In Ruby you should use the `PriorityQueue` gem
                               # and on the JVM there's a good Java implementation.
  def << element
    @elements << element
    sort!
  end

  def pop                      # `pop` is the typical queue-polling nomenclature.
    @elements.shift            # Your implementation may call it something else.
  end

  private

  def sort!
    @elements = @elements.sort_by &@comparator  # This line is why this implementation 
                                                # sucks. Don't use this IRL.
  end
end

class State
  def cost                           # The cost is pretty simple to calculate here.
    path.size                        # The path contains all the steps, in order that we
  end                                # used to arrive at this state. So the cost
end                                  # is just the number of steps.

require 'set'
def solve puzzle
  @visited = Set.new
  @frontier = PriorityQueue.new {|s| s.cost }
  state = State.new puzzle
  loop {
    @visited << state.puzzle.cells
    break if state.solution?
    search state
    state = @frontier.pop
  }
  state
end

Sidebar: you may be wondering why this gains us any advantage? Sure, we’re now picking the the best node from the queue rather than whichever one was added first but we still have to explore all of them, right? Actually, no. Because we’re sorting by the ‘cost’ of the nodes we can be guaranteed that whenever we find a solution it’s the best one. There may be other paths to solutions in our graph but they are all guaranteed to be of higher cost. So this Uniform-cost search lets us leave a vast section of the queue unexplored.

What does that do to our performance? Well, if we re-run our above 20-step puzzle the time will drop considerably from 27 seconds to 10 seconds (on my machine).

This is a big speedup and, for larger problems, can shave days off the calculation time. But there’s much more we can do.

The uniform-cost search picks the best next state from the frontier. Let’s enhance the code’s understanding of what makes something “best” by calculating not only the distance from the start to where we are but the distance from where we are to the goal.

Old cost function: steps_to_get_to(s)

New cost function: steps_to_get_to(s) + steps_to_goal_from(s)

But, uh, how do we know how far we are from the solution? If we knew how far away the solution was we’d probably already have found it, right? Right. So rather than being exact, let’s just pick a healthy estimate of how far we are from a solution. One approximation would be “how many tiles are out of place?” That would at least differentiate almost-solution nodes from not-even-close ones. But we’d like to be a bit more precise.

So let’s say that the distance cost between a given node and the solution node is the number of tile-movements that would be required if tiles could move through each other and go straight to their goal positions. So a near-solution node might have a distance cost of 3 and a not-even-close node might have a distance cost of 26. That should give us decent precision while also being fair. It’s important that our cost-to-get-to-goal function doesn’t accidentaly deprioritize good near-solution states.

To help us we’ll calculate the Manhattan Distance between each tile and where it’s supposed to be. Manhattan Distance is the distance between two places if you have to travel along city blocks. Essentially, you’re adding up the short sides of a right triangle rather than shortcutting across the hypotenuse. The formula is pretty simple:

class Puzzle
  def distance_to_goal                               # Here we `zip` the current puzzle with the solution
    @cells.zip(Solution).inject(0) do |sum, (a,b)|    # and total up the distances between each cell
      sum += manhattan_distance a % 3, (a / 3).to_i, # and where that cell should be.
                                b % 3, (b / 3).to_i  # This % and / stuff is just turning an integer
    end                                              # into puzzle x,y coordinates
  end
  private

  def manhattan_distance x1, y1, x2, y2  # The manhattan distance of something is just
    (x1 - x2).abs + (y1 - y2).abs        # the distance between x coordinates
  end                                    # plus the distance between y coordinates
end

class State
  def cost
    steps_from_start + steps_to_goal    # Now we have a more informed cost method
  end                                   # so our priority queue should be giving
                                        # us better results.
  def steps_from_start
    path.size
  end

  def steps_to_goal
    puzzle.distance_to_goal
  end
end

require 'set'
def solve puzzle
  @visited = Set.new
  @frontier = PriorityQueue.new {|s| s.cost }
  state = State.new puzzle
  loop {
    break if state.solution?
    search state
    state = @frontier.pop
  }
  state
end

If you’re following along at home (and using a real priority queue) you might think the code is broken because it exited so fast. With a proper priority queue implementation this latest search took 0.07 seconds.

This A* search is able to quickly pick the best candidate to explore in any situation where the distance from the current state to the goal state is knowable. In real-world pathfinding, e.g., you can use the geospatial distance between two points. It doesn’t work at all, however, in situations where you know the goal when you see it but can’t determine how close you are. A robot trying to find a door in unexplored territory would not be able to use this, it would have to just keep bumbling around.

The full reference code for this is on GitHub including a full implentation in Clojure.

Huge thanks to my reviewer Ashish Dixit without whom this post would have been a typo-filled mess of half-conveyed ideas.

A quick recap of the relative time and memory costs for these search algorithms:

uninformed depth-first:               {stack overflow error}
breadth-first w/o tracking `visited`: {out of memory error}
uninformed breadth-first:             27 seconds, 47,892 explored states
uniform-cost (Dijkstra's):            10 seconds, 51,963 explored states
A* search:                            0.07 seconds, 736 explored states