AI:图搜索和带曼哈顿启发式的A*实现

Question

弗洛伦萨大学AI课程的This is my project。我必须解决一个经典的游戏:滑动拼图与8和15个细胞。这是我的通用图搜索算法的实现：

public abstract class GraphSearch implements SearchAlgorithm {

protected Queue<Node> fringe;
protected HashSet<Node> closedList;

public GraphSearch() {
    fringe = createFringe();
    closedList = new HashSet<Node>(100);
}

protected abstract Queue<Node> createFringe();

public int getNodeExpanded() {
    return closedList.size();
}

@Override
public Solution search(Puzzle puzzle) {
    fringe.add(new Node(puzzle.getInitialState(), null, null));
    while (!fringe.isEmpty()) {
        Node selectedNode = fringe.poll();
        if (puzzle.getGoalTest().isGoalState(selectedNode.getState())) {
            return new Solution(selectedNode, getNodeExpanded());
        }
        closedList.add(selectedNode);
        LinkedList<Node> expansion = selectedNode.expandNode();
        for (Node n : expansion) {
            if (!closedList.contains(n) && !fringe.contains(n))
                fringe.add(n);
        }
    }
    return new Solution(null, getNodeExpanded());
}

}

这是我的A*代码：

public class AStar extends GraphSearch implements InformedSearch {

private Heuristic heuristic;

public AStar(Heuristic heuristic) {
    setHeuristic(heuristic);
}

public Heuristic getHeuristic() {
    return heuristic;
}

@Override
public void setHeuristic(Heuristic heuristic) {
    this.heuristic = heuristic;
}

@Override
protected Queue<Node> createFringe() {
    return new PriorityQueue<Node>(1000, new Comparator<Node>() {

        @Override
        public int compare(Node o1, Node o2) {
            o1.setH(heuristic.h(o1));
            o2.setH(heuristic.h(o2));
            o1.setF(o1.getG() + o1.getH());
            o2.setF(o2.getG() + o2.getH());
            if (o1.getF() < o2.getF())
                return -1;
            if (o1.getF() > o2.getF())
                return 1;
            return 0;
        }
    });
}

}

这是我在曼哈顿的启发式代码：

    @Override
public int h(Node n) {
    int distance = 0;
    ArrayList<Integer> board = n.getState().getBoard();
    int[][] multiBoard = new int[N][N];
    for (int i = 0; i < N; i++) {
        for (int j = 0; j < N; j++) {
            multiBoard[i][j] = board.get(i * N + j);
        }
    }
    for (int i = 0; i < N; i++) {
        for (int j = 0; j < N; j++) {
            int value = multiBoard[i][j];
            if (multiBoard[i][j] != 0) {
                int targetX = (value - 1) / N;
                int targetY = (value - 1) % N;
                distance += Math.abs(i - targetX) + Math.abs(j - targetY);
            }
        }
    }
    return distance;
}

现在，代码工作并找到了难题的解决方案(难题状态是N*N值的数组，GoalState是1, 2, 3, 4, 5, 6, 7, 8, 9 ,0，空单元格= 0)，但是与其他程序(相同的算法和相同的启发式)相比，我的程序扩展了不同的节点数。我认为在一般的图形search...any想法有一个问题？:D谢谢！

Answer 1

你的启发式计算是错误的。假设9位于棋盘的索引4处。您将其rowRight值计算为3而不是2。这将导致您的算法具有超最佳的性能。您的行右侧计算应为：

int rowRight = (valueFound - 1) / N;