带有二进制搜索树的Alpha Beta修剪

时间:2018-10-14 15:54:44

标签: java algorithm binary-search-tree minimax alpha-beta-pruning

我正在研究带有here的Alpha-Beta修剪示例的Minimax算法。在示例中,他们使用数组来实现搜索树。我遵循了该示例,但也尝试使用二进制搜索树来实现它。这是我在树中使用的值:3、5、6、9、1、2、0,-1。

最后的最佳值应该是5。通过BST实施,我不断得到2。

我认为这是问题所在,但我不知道该如何解决:
如果遇到尝试检查下一个值的叶子节点阻止获取空指针异常的情况,我编写了代码以退出递归。但是,相反,我认为它太早停止了搜索(基于我在调试器中逐步执行代码时所看到的内容)。但是,如果我删除该检查,代码将在空指针上失败。

有人可以指出我正确的方向吗?我在做什么错了?

代码如下:

public class AlphaBetaMiniMax {

    private static BinarySearchTree myTree = new BinarySearchTree();
    static int MAX = 1000;
    static int MIN = -1000;
    static int opt;

    public static void main(String[] args) {
        //Start constructing the game
        AlphaBetaMiniMax demo = new AlphaBetaMiniMax();

        //3, 5, 6, 9, 1, 2, 0, -1
        demo.myTree.insert(3);
        demo.myTree.insert(5);
        demo.myTree.insert(6);
        demo.myTree.insert(9);
        demo.myTree.insert(1);
        demo.myTree.insert(2);
        demo.myTree.insert(0);
        demo.myTree.insert(-1);

        //print the tree
        System.out.println("Game Tree: ");
        demo.myTree.printTree(demo.myTree.root);

        //Print the results of the game
        System.out.println("\nGame Results:");

        //run the  minimax algorithm with the following inputs
        int optimalVal = demo.minimax(0, myTree.root, true, MAX, MIN);
        System.out.println("Optimal Value: " + optimalVal);

    }

    /**
     * @param alpha = 1000
     * @param beta = -1000
     * @param nodeIndex - the current node
     * @param depth - the depth to search
     * @param maximizingPlayer - the current player making a move
     * @return - the best move for the current player
     */
    public int minimax(int depth, MiniMaxNode nodeIndex, boolean maximizingPlayer, double alpha, double beta) {

        //Base Case #1: Reached the bottom of the tree
        if (depth == 2) {
            return nodeIndex.getValue();
        }

        //Base Case #2: if reached a leaf node, return the value of the current node
        if (nodeIndex.getLeft() == null && maximizingPlayer == false) {
            return nodeIndex.getValue();
        } else if (nodeIndex.getRight() == null && maximizingPlayer == true) {
            return nodeIndex.getValue();
        }

        //Mini-Max Algorithm
        if (maximizingPlayer) {
            int best = MIN;

            //Recur for left and right children
            for (int i = 0; i < 2; i++) {

                int val = minimax(depth + 1, nodeIndex.getLeft(), false, alpha, beta);
                best = Math.max(best, val);
                alpha = Math.max(alpha, best);

                //Alpha Beta Pruning
                if (beta <= alpha) {
                    break;
                }
            }
            return best;
        } else {
            int best = MAX;

            //Recur for left and right children
            for (int i = 0; i < 2; i++) {

                int val = minimax(depth + 1, nodeIndex.getRight(), true, alpha, beta);
                best = Math.min(best, val);
                beta = Math.min(beta, best);

                //Alpha Beta Pruning
                if (beta <= alpha) {
                    break;
                }
            }
            return best;
        }
    }
}

输出:

Game Tree: 
-1 ~ 0 ~ 1 ~ 2 ~ 3 ~ 5 ~ 6 ~ 9 ~ 
Game Results:
Optimal Value: 2

1 个答案:

答案 0 :(得分:1)

您的问题是您的迭代取决于2的循环控制,而不是node == null的nodeIndex.getRight()(最大)getLeft(最小)

记住一棵树有 1个头(第一级)

第二级= 2

第三级= 4

4 8 等等。因此,您的循环算法甚至不会下降3级。

df=pd.read_csv("AREX.csv", sep = ";")
df[~df['Date'].isin(pd.date_range(start='20100101', end='20171231'))]        
print(df)
df.drop(["Open","High","Low","Volume","Open interest"],axis = 1, inplace=True)
print(df)

更改循环以正确控制迭代,您应该轻松找到最大值。