Compartilhar via


Path Finding Using A* in C# 3.0, Part Four

Finally we are ready to translate our pseudocode into C# 3.0. 

What do we need to make the algorithm run? We need a start node, a destination node, a function which tells us the exact distance between two neighbours, and a function which tells us the estimated distance between the last node on a proposed path and the destination node.

We also need the ability to take any node and get a list of its neighbours. We'll represent that by an interface:

interface IHasNeighbours<N>
{
IEnumerable<N> Neighbours { get; }
}

Recall that the pseudocode looked like this:

closed = {}
q = emptyqueue;
q.enqueue(0.0, makepath(start))
while q is not empty
p = q.dequeueCheapest
if closed contains p.last then continue;
if p.last == destination then return p
closed.add(p.last)
foreach n in p.last.neighbours
newpath = p.continuepath(n)
q.enqueue(newpath.TotalCost + estimateCost(n, destination), newpath)
return null

And now we're all set. The translation from pseudocode to real code is quite straightforward:

static public Path<Node> FindPath<Node>(
Node start,
Node destination,
Func<Node, Node, double> distance,
Func<Node, double> estimate)
where Node : IHasNeighbours<Node>
{
var closed = new HashSet<Node>();
var queue = new PriorityQueue<double, Path<Node>>();
queue.Enqueue(0, new Path<Node>(start));
while (!queue.IsEmpty)
{
var path = queue.Dequeue();
if (closed.Contains(path.LastStep))
continue;
if (path.LastStep.Equals(destination))
return path;
closed.Add(path.LastStep);
foreach(Node n in path.LastStep.Neighbours)
{
double d = distance(path.LastStep, n);
var newPath = path.AddStep(n, d);
queue.Enqueue(newPath.TotalCost + estimate(n), newPath);
}
}
return null;
}

UPDATE: I’ve been asked to clarify why the estimation function takes only one parameter. You could certainly declare the parameter as “Func<Node, Node, double> estimate”, and then call estimate(n, destination) if you’d rather. I prefer to use partial application when possible. If I have an estimation function that takes two nodes, then in the caller I would simply pass n=>estimate(n, destination) as the estimation function. Either way works; it’s a stylistic thing.

You might wonder why we check to see if the current-best-path ends in a "closed" node after the dequeue, rather than testing that before we do the enqueue. Think about it this way: suppose we take the best path off the queue, note that its end node is "closed" so that we don't come back to it, and put all of its extensions on the queue. There might be at present a number of other paths which end in that node on the queue. All of them must be, by definition, worse than the path we just took off the priority queue. We should therefore not even consider them as possbilities anymore. If we happen to put more paths with that property on the queue, that's unfortunate. But when we eventually get to them we'll discard them automatically so its not a big deal.

A nice property of the A* algorithm is that it finds the optimal path in a reasonable amount of time provided that:

  • the estimating function never overestimates the distance to the goal. (Think about what happens if the estimating function sometimes overestimates the distance; if the optimal path is one of the ones overestimated then it will possibly not make it to the front of the priority queue before a nonoptimal solution is found.)
  • calling the estimating function does not take very long.

One way to ensure that the estimating function never overestimates the distance is to always estimate zero. If you do so then this becomes Dijkstra's algorithm. However, the better your estimating function can get without going over (this really should have been called the "The Price Is Right" algorithm...) the faster this will identify the truly optimal path.

Unfortunately, the space complexity of this algorithm can be really quite high on complicated graphs. There are more complex versions of this algorithm which can deal with the space complexity, but I'm not going to go there.

I have a nice example of using A* to find the shortest path around a lake (but not the longest shortest path!) but I'm not going to post it here until Orcas ships. I'll just put up the whole project file at that time and you can check it out all at once.

Next time: at long last, Eric talks about variance in C#.

Comments

  • Anonymous
    October 10, 2007
    PingBack from http://www.artofbam.com/wordpress/?p=6818

  • Anonymous
    October 10, 2007
    This has been a great series of posts Eric.  I've been tinkering with programming SLAM for an iRobot Create with a CharmedLabs Qwerk controller, and this A* will come in quite handy.

  • Anonymous
    October 10, 2007
    Shouldn't that be "IHaveNeighbours<N>"? :)

  • Anonymous
    October 10, 2007
    > Next time: at long last, Eric talks about variance in C#. I'm really looking forward to it! Lack of generic variance support has been and continues to be my major gripe with C#.

  • Anonymous
    October 11, 2007
    I felt like I was in school again :-)

  • Anonymous
    October 14, 2007
    Welcome to the thirty-third edition of Community Convergence. This week we have a new video called Programming

  • Anonymous
    November 26, 2007
    I realize that VS2008 was JUST released, but I was wondering if, when you had a moment, we could see the solution and demo?  I really enjoyed reading this.  Thanks!

  • Anonymous
    June 25, 2008
    The first thing I want to be able to do in MicroQuest is move my &quot;unit&quot; around the &quot;game

  • Anonymous
    June 08, 2009
    Suppose you have to build a road to connect two cities on different sides of a lake. How would you plan

  • Anonymous
    June 20, 2009
    Hi Eric, Standing on the shoulders of giants just like you, this is my try to put a running sample of the code you presented in this magnific series. http://www.leniel.net/2009/06/astar-pathfinding-search-in-csharp.html Thanks, Leniel Macaferi http://leniel.net

  • Anonymous
    December 13, 2009
    Implemented your code in our XNA game "Climate Conflict". http://klimakonflikt.codeplex.com/ It's elegant, it's fast and it was a cinch to implement! :) Thanx!

  • Jakob