java - 追踪 Minimax 的最佳移动

Question

我知道以前有人问过此类问题，但我无法解决我的疑问。我有一个简单的黑白棋引擎(实际上它玩得很好)，它使用下面的类来获得最佳棋步:

import java.util.*;
import java.util.concurrent.*;

public class MinimaxOthello implements Runnable
{
  private CountDownLatch doneSignal;    
  private int maxDepth;
  private int calls;    
  private OthelloMove bestFound;
  private OthelloBoard board;
  private static float INFINITY = Float.MAX_VALUE/1000;    
  private boolean solve = false;
  private Comparator<OthelloMove> comparator = Collections.reverseOrder(new MoveComparator());

public MinimaxOthello (OthelloBoard board, int maxDepth, CountDownLatch doneSignal, boolean solve)
{
    this.board = board;        
    this.bestFound = new OthelloMove();
    bestFound.setPlayer(board.getCurrentPlayer());
    this.maxDepth = maxDepth; 
    this.doneSignal = doneSignal;                
    this.solve = solve;
}

public OthelloMove getBestFound()
{       
    return this.bestFound;
}
public void run()
{        
    float val = minimax(board, bestFound, -INFINITY, INFINITY, 0);
    System.out.println("calls: " + calls);
    System.out.println("eval: " + val);
    System.out.println();
    doneSignal.countDown();        
}

private float minimax(OthelloBoard board, OthelloMove best, float alpha, float beta, int depth)
{
    calls++;             
    OthelloMove garbage = new OthelloMove();             
    int currentPlayer = board.getCurrentPlayer();

    if (board.checkEnd())
    {                        
        int bd = board.countDiscs(OthelloBoard.BLACK);
        int wd = board.countDiscs(OthelloBoard.WHITE);

        if ((bd > wd) && currentPlayer == OthelloBoard.BLACK)
        {                
            return INFINITY/10;
        }
        else if ((bd < wd) && currentPlayer == OthelloBoard.BLACK)
        {                
            return -INFINITY/10;
        }
        else if ((bd > wd) && currentPlayer == OthelloBoard.WHITE)
        {                
            return -INFINITY/10;
        }
        else if ((bd < wd) && currentPlayer == OthelloBoard.WHITE)
        {                
            return INFINITY/10;
        }
        else 
        {                
            return 0.0f;
        }
    }
    if (!solve)
    {
        if (depth == maxDepth)
            return OthelloHeuristics.eval(currentPlayer, board);
    }

    ArrayList<OthelloMove> moves = board.getAllMoves(currentPlayer);
    if (moves.size() > 1)
    {
        OthelloHeuristics.scoreMoves(moves);        
        Collections.sort(moves, comparator);
    }

    for (OthelloMove mv : moves)
    {                                    
        board.makeMove(mv);            
        float score = - minimax(board, garbage, -beta,  -alpha, depth + 1);           
        board.undoMove(mv);             

        if(score > alpha)
        {  
            alpha = score;                
            best.setFlipSquares(mv.getFlipSquares());
            best.setIdx(mv.getIdx());        
            best.setPlayer(mv.getPlayer());                              
        }

        if (alpha >= beta)
            break;                

    }            
    return alpha;
 }  
}

我有一个bestFound实例变量，我的疑问是，为什么必须调用

OthelloMove garbage = new OthelloMove(); 

并传递它？该代码有效，但对我来说似乎很奇怪!

是否有“更好”的方法来获得最佳着法或主要变体？我真的不是递归专家，这非常难以调试和可视化。谢谢!

**PS:您可以在 https://github.com/fernandotenorio/ 克隆它

Answer 1

看来您可以将 best 参数去掉为 minimax，从而消除对 garbage 的需要，然后替换 best 与 this.bestFound。仅当深度 = 0 时才设置 bestFound 的属性。

您可以通过将 this.bestFound 设置为初始空列表来获取主要变体。在 moves 循环之前，创建一个新的移动。在 if (score > alpha) 部分，将其属性设置为与现在相同。将移动推到循环之后的列表中。主要变化将与列表相反。

如果这很重要，您可以进行一些更改来提高类的多线程能力:

• 不要将 bestFound 列表存储为实例变量，而是将其设为 run 中的局部变量，并将其作为参数添加到 minimax
• Make Board.makeMove 不会修改棋盘，而是返回应用了移动的棋盘的新实例。您可以通过克隆棋盘并将移动代码应用到克隆来实现这一点，而不是改变this。然后，将克隆的板传递给下一次调用 minimax。