了解树搜索中的PLINQ瓶颈

时间:2013-03-20 08:24:30

标签: c# plinq alpha-beta-pruning

我对PLINQ有一些奇怪的结果,我似乎无法解释。我一直在尝试并行化Alpha Beta树搜索以加快搜索过程,但它实际上正在减慢它的速度。我希望随着我提高并行度,我会每秒线性增加节点数......然后处理其他节点,因为修剪会被推迟到以后。虽然节点数与期望值匹配,但我的时间不会:

非PLINQ, 访问过的节点:61418, 运行时:0:00.67

并行度:1, 访问过的节点:61418, runtime:0:01.48

并行度:2, 访问过的节点:75504, runtime:0:10.08

并行度:4, 访问过的节点:95664, runtime:1:51.98

并行度:8, 访问过的节点:108148, 运行时:1:48.94


有人帮我识别可能的罪魁祸首吗?

相关代码:

    public int AlphaBeta(IPosition position, AlphaBetaCutoff parent, int depthleft)
    {
        if (parent.Cutoff) 
            return parent.Beta;

        if (depthleft == 0) 
            return Quiesce(position, parent);

        var moves = position.Mover.GetMoves().ToList();

        if (!moves.Any(m => true))
            return position.Scorer.Score();

        //Young Brothers Wait Concept...
        var first = ProcessScore(moves.First(), parent, depthleft);
        if(first >= parent.Beta)
        {
            parent.Cutoff = true;
            return parent.BestScore;
        }

        //Now parallelize the rest...
        if (moves.Skip(1)
            .AsParallel()
            .WithDegreeOfParallelism(1)
            .WithMergeOptions(ParallelMergeOptions.NotBuffered)
            .Select(m => ProcessScore(m, parent, depthleft))
            .Any(score => parent.BestScore >= parent.Beta))
        {
            parent.Cutoff = true;
            return parent.BestScore;
        }
        return parent.BestScore;
    }

    private int ProcessScore(IMove move, AlphaBetaCutoff parent, int depthleft)
    {
        var child = ABFactory.Create(parent);
        if (parent.Cutoff)
        {
            return parent.BestScore;
        }
        var score = -AlphaBeta(move.MakeMove(), child, depthleft - 1);
        parent.Alpha = score;
        parent.BestScore = score;
        if (score >= parent.Beta)
        {
            parent.Cutoff = true;
        }
        return score;
    }

然后是跨树级别共享Alpha Beta参数的数据结构...

public class AlphaBetaCutoff
{
    public AlphaBetaCutoff Parent { get; set; }

    private bool _cutoff;
    public bool Cutoff
    {
        get
        {
            return _cutoff || (Parent != null && Parent.Cutoff);
        }
        set
        {
            _cutoff = value;
        }
    }

    private readonly object _alphaLock = new object();
    private int _alpha = -10000;
    public int Alpha
    {
        get
        {
            if (Parent == null) return _alpha;
            return Math.Max(-Parent.Beta, _alpha);
        }
        set
        {
            lock (_alphaLock)
            {
                _alpha = Math.Max(_alpha, value);
            }
        }
    }

    private int _beta = 10000;
    public int Beta
    {
        get
        {
            if (Parent == null) return _beta;
            return -Parent.Alpha;
        }
        set
        {
            _beta = value;
        }
    }

    private readonly object _bestScoreLock = new object();
    private int _bestScore = -10000;
    public int BestScore
    {
        get
        {
            return _bestScore;
        }
        set
        {
            lock (_bestScoreLock)
            {
                _bestScore = Math.Max(_bestScore, value);
            }
        }
    }
}

1 个答案:

答案 0 :(得分:0)

当你只为很少的工作和为所有底层节点设置新线程时,你会在线程上产生巨大的开销。您可能因为Any而处理更多节点,通常处理会停止,但有些节点在找到Any(第一个匹配)之前已经开始处理。当您拥有一组已知的大型底层工作负载时,并行性将更好地发挥作用。如果您只在顶级节点执行并行操作,可以尝试会发生什么。