A* Algorithm 8-puzzle

Question

I have read many pseudo code for the A* algorithm, but neither of them actually explain how to output the solution. I believe I understand the concept of using a priority queue for the not yet visited and a has table for the explored, but when I go through the algorithm I don't know at what point to print out the result. Does any one have a pseudo code that actually show how to output the path?

Id really appreciate it. I have been trying to use the algorithm to implement the 8-puzzle problem

Here's my code:

public class Board{

private int[][] squares;
private int f;
private int g;
private int h;

private int size;
private Board parent;

public Board(Board current, Board parent)
{
    this(current);
    g = current.getG();
    h = current.getH();
    f = current.getF();
    this.parent = parent;

}

public void solveH1()
{

    while(!frontier.isEmpty())
    {
        board = frontier.poll();

        ArrayList<Board> successors = new ArrayList<Board>();
        Board b1 = new Board(board.moveDown(),board);
        Board b2 = new Board(board.moveUp(),board);
        Board b3 = new Board(board.moveLeft(),board);
        Board b4 = new Board(board.moveRight(),board);
        if(!b1.equals(board))
            successors.add(b1);
        if(!b2.equals(board))
            successors.add(b2);
        if(!b3.equals(board))
            successors.add(b3);
        if(!b4.equals(board))
            successors.add(b4);
        for(int i=0; i<successors.size(); i++)
        {
            if(successors.get(i).isGoal())
            {
                break;
            }
            int g = board.getG()+1;
            int h = successors.get(i).getH1Cost();
            successors.get(i).setG(g);
            successors.get(i).setH(h);
            successors.get(i).setF(g+h);

            if(frontier.contains(successors.get(i)))
            {
                Iterator<Board> iterator = frontier.iterator();
                Board b = null;
                while(iterator.hasNext())
                {
                    b = iterator.next();
                    if(b.equals(successors.get(i)))
                    {
                        break;
                    }
                }
                if(b.getG() < successors.get(i).getG())
                {
                    break;
                }
            }
            if(exploredSet.contains(successors.get(i)))
            {
                int index = exploredSet.indexOf(successors.get(i));
                if(exploredSet.get(index).getG() < successors.get(i).getG())
                    break;
            }
            else
            {
                frontier.add(successors.get(i));
            }
        }
        exploredSet.add(board);
    }
    printPath();
}
public void printPath()
{

    ArrayList<Board> path = new ArrayList<Board>();
    cursor = board;
    while(cursor.getParent()!=null)
    {
        path.add(cursor);
        cursor = cursor.getParent();
    }
    for(int i=0; i<path.size(); i++)
        System.out.println(path.get(i));
}

for some reason, this just prints one node, and it's bot even the goal. Can anyone tell me what I am missing?

nullpotent · Accepted Answer · 2014-10-26T20:59:38.280

1

When you push the node onto queue you also save (push) the parent of it, that is, the node you've visited it from.

class NodeWrapper {
 public float cost;
 public Node node;
 public Node parentNode;
 public NodeWrapper(Node node, Node parentNode) {
   this.node = node;
   this.parentNode = parentNode;
 }
}

And then

openQueue.push(new NodeWrapper(neihgbouringNode, currentNode));

And when you reach the end node, you then just trace your way back from it.

List<Node> out = new ArrayList<Node>();
while (currentNode != null) {
   out.add(currentNode.node);
   currentNode = currentNode.parentNode;
}
return out;

Here's a demo of a A* pathfinder.

edited Oct 26 '14 at 20:59

answered Oct 25 '14 at 10:27

nullpotent

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So, does that mean I create like a stack or something that holds the parent? I'm still a little confused – Dave Smith Oct 25 '14 at 10:50
No, you don't need to create a separate stack. Use the 'open' priority queue. Just create a plain java Object that holds both the node you're currently visiting and its parent node. – nullpotent Oct 25 '14 at 10:58
You basically create a tree with only parent-node pointers, no child pointers. So there's no way to iterate from the root (start board) to the leaf(goal board), but it's trivial to iterate from the last leaf (goal board) back to the root (starting board). – Martin Oct 25 '14 at 12:25
So, this is the part that I'm getting stuck at. if endNode is of type NodeWrapper then I can not set endNode = endNode.parentNode. I tried creating a node that just have a parent node so I can say node = node.getParent() howver it just prints one node. – Dave Smith Oct 25 '14 at 19:09
Here's my implementation of a a* pathfinder: http://pastebin.com/2MU1CXKz Check the Node wrapper class, I hope it helps. – nullpotent Oct 26 '14 at 20:50
And here's the whole thing: http://pastebin.com/Ei1t7wJD As I said, My Java skills are below par; therefore, please - don't copy/paste anything :) – nullpotent Oct 26 '14 at 20:53

A* Algorithm 8-puzzle

1 Answers1