c# – 优化Poker-Monte-Carlo-Simulation的手动评估算法

我为Hold’em Poker写了一个平衡器,作为一个爱好项目.它工作正常,但仍然有一件事我不满意：
在模拟手的整个过程中,评估手的过程大约占35％的时间.对于我而言,与其他必须完成的工作相似,如迭代和克隆大型数组和内容.

任何关于如何让这种性能更高的想法都会很棒.

这是代码：

private static int getHandvalue(List<Card> sCards)
    {

        // --- Auf Straight / Straight Flush prüfen ---
        if ((sCards[0].Value - 1 == sCards[1].Value) && (sCards[1].Value - 1 == sCards[2].Value) && (sCards[2].Value - 1 == sCards[3].Value) && (sCards[3].Value - 1 == sCards[4].Value))
        {

            if ((sCards[0].Color == sCards[1].Color) && (sCards[1].Color == sCards[2].Color) && (sCards[2].Color == sCards[3].Color) && (sCards[3].Color == sCards[4].Color))
                return (8 << 20) + (byte)sCards[0].Value; // H?chste Karte Kicker

            else
                return (4 << 20) + (byte)sCards[0].Value; // H?chste Karte Kicker (Straight)

        }

        // --- Auf Wheel / Wheel Flush prüfen ---
        if ((sCards[4].Value == Card.CardValue.Two) && (sCards[3].Value == Card.CardValue.Three) && (sCards[2].Value == Card.CardValue.Four) && (sCards[1].Value == Card.CardValue.Five) && (sCards[0].Value == Card.CardValue.Ace))
        {

            if ((sCards[0].Color == sCards[1].Color) && (sCards[1].Color == sCards[2].Color) && (sCards[2].Color == sCards[3].Color) && (sCards[3].Color == sCards[4].Color))
                return(8 << 20) + (byte)sCards[1].Value; // Zweith?chste Karte Kicker

            else
                return(4 << 20) + (byte)sCards[1].Value; // Zweith?chste Karte Kicker (Straight)

        }


        // --- Auf Flush prüfen ---
        if ((sCards[0].Color == sCards[1].Color) && (sCards[1].Color == sCards[2].Color) && (sCards[2].Color == sCards[3].Color) && (sCards[3].Color == sCards[4].Color))
            return (5 << 20) + ((byte)sCards[0].Value << 16) + ((byte)sCards[1].Value << 12) + ((byte)sCards[2].Value << 8) + ((byte)sCards[3].Value << 4) + (byte)sCards[4].Value;


        // --- Auf Vierling prüfen ---
        if (((sCards[0].Value == sCards[1].Value) && (sCards[1].Value == sCards[2].Value) && (sCards[2].Value == sCards[3].Value)) || ((sCards[1].Value == sCards[2].Value) && (sCards[2].Value == sCards[3].Value) && (sCards[3].Value == sCards[4].Value)))
            return (7 << 20) + (byte)sCards[1].Value; // Wert des Vierlings (keine Kicker,da nicht mehrere Spieler den selben Vierling haben k?nnen)


        // --- Auf Full-House / Drilling prüfen ---
        // Drilling vorne
        if ((sCards[0].Value == sCards[1].Value) && (sCards[1].Value == sCards[2].Value))
        {

            // Full House
            if (sCards[3].Value == sCards[4].Value)
                return (6 << 20) + ((byte)sCards[0].Value << 4) + (byte)sCards[3].Value; // Drilling (h?her bewerten)

            // Drilling
            return (3 << 20) + ((byte)sCards[0].Value << 8) + ((byte)sCards[3].Value << 4) + (byte)sCards[4].Value; // Drilling + Kicker 1 + Kicker 2

        }

        // Drilling hinten
        if ((sCards[2].Value == sCards[3].Value) && (sCards[3].Value == sCards[4].Value))
        {

            // Full House
            if (sCards[0].Value == sCards[1].Value)
                return (6 << 20) + ((byte)sCards[2].Value << 4) + (byte)sCards[0].Value; // Drilling (h?her bewerten)

            // Drilling
            return (3 << 20) + ((byte)sCards[2].Value << 8) + ((byte)sCards[0].Value << 4) + (byte)sCards[1].Value; // Drilling + Kicker 1 + Kicker 2

        }

        // Drilling mitte
        if ((sCards[1].Value == sCards[2].Value) && (sCards[2].Value == sCards[3].Value))
            return (3 << 20) + ((byte)sCards[1].Value << 8) + ((byte)sCards[0].Value << 4) + (byte)sCards[4].Value; // Drilling + Kicker 1 + Kicker 2                            


        // --- Auf Zwei Paare prüfen ---
        // Erstes Paar vorne,zweites Paar mitte
        if ((sCards[0].Value == sCards[1].Value) && (sCards[2].Value == sCards[3].Value))
            return (2 << 20) + ((byte)sCards[0].Value << 8) + ((byte)sCards[2].Value << 4) + (byte)sCards[4].Value; // Erstes Paar + Zweites Paar + Kicker

        // Erstes Paar vorne,zweites Paar hinten
        if ((sCards[0].Value == sCards[1].Value) && (sCards[3].Value == sCards[4].Value))
            return (2 << 20) + ((byte)sCards[0].Value << 8) + ((byte)sCards[3].Value << 4) + (byte)sCards[2].Value; // Erstes Paar + Zweites Paar + Kicker

        // Erstes Paar mitte,zweites Paar hinten
        if ((sCards[1].Value == sCards[2].Value) && (sCards[3].Value == sCards[4].Value))
            return (2 << 20) + ((byte)sCards[1].Value << 8) + ((byte)sCards[3].Value << 4) + (byte)sCards[0].Value; // Erstes Paar + Zweites Paar + Kicker


        // --- Auf Paar prüfen ---
        // Paar vorne
        if (sCards[0].Value == sCards[1].Value)
            return (1 << 20) + ((byte)sCards[0].Value << 12) + ((byte)sCards[2].Value << 8) + ((byte)sCards[3].Value << 4) + (byte)sCards[4].Value; // Paar + Kicker 1 + Kicker 2 + Kicker 3

        // Paar mitte-vorne
        if (sCards[1].Value == sCards[2].Value)
            return (1 << 20) + ((byte)sCards[1].Value << 12) + ((byte)sCards[0].Value << 8) + ((byte)sCards[3].Value << 4) + (byte)sCards[4].Value; // Paar + Kicker 1 + Kicker 2 + Kicker 3

        // Paar mitte-hinten
        if (sCards[2].Value == sCards[3].Value)
            return (1 << 20) + ((byte)sCards[2].Value << 12) + ((byte)sCards[0].Value << 8) + ((byte)sCards[1].Value << 4) + (byte)sCards[4].Value; // Paar + Kicker 1 + Kicker 2 + Kicker 3

        // Paar hinten
        if (sCards[3].Value == sCards[4].Value)
            return (1 << 20) + ((byte)sCards[3].Value << 12) + ((byte)sCards[0].Value << 8) + ((byte)sCards[1].Value << 4) + (byte)sCards[2].Value; // Paar + Kicker 1 + Kicker 2 + Kicker 3


        // --- High Card bleibt übrig ---
        return ((byte)sCards[0].Value << 16) + ((byte)sCards[1].Value << 12) + ((byte)sCards[2].Value << 8) + ((byte)sCards[3].Value << 4) + (byte)sCards[4].Value; // High Card + Kicker 1 + Kicker 2 + Kicker 3 + Kicker 4

    }

此方法返回扑克中每个已排序的5卡组合的精确值.它被另一个方法调用：

private static int getHandvalueList(List<Card> sCards)
    {

        int count = sCards.Count;
        if (count == 5) return getHandvalue(sCards);

        int HighestValue = 0;
        Card missingOne;
        int tempValue;

        for (int i = 0; i < count - 1; i++)
        {

            missingOne = sCards[i];
            sCards.RemoveAt(i);

            tempValue = getHandvalueList(sCards);
            if (tempValue > HighestValue) HighestValue = tempValue;

            sCards.Insert(i,missingOne);

        }

        missingOne = sCards[count - 1];
        sCards.RemoveAt(count - 1);

        tempValue = getHandvalueList(sCards);
        if (tempValue > HighestValue) HighestValue = tempValue;

        sCards.Add(missingOne);
        return HighestValue;

    }

此递归方法返回所有可能的5张卡组合的最高值.这一个被最终的公共方法调用：

public static int GetHandvalue(List<Card> sCards)
    {

        if (sCards.Count < 5) return 0;
        sCards.Sort(new ICardComparer());
        return getHandvalueList(sCards);

    }

它最多可以发送7张卡.

更新

到目前为止：每次使用7张牌调用公共功能(大多数情况下就是这种情况),它必须将手动评估方法调用21次(每次可能的5张卡组合一次).

我想在哈希表中为每个可能的5到7张卡片缓存值,然后查看它.
但如果我没有错,它将需要存储超过133.784.560 32位整数值,大约500MB.

什么可能是好的哈希函数将每个可能的组合分配给一个arrayindex？

创建了一个新问题：Hashfunction to map combinations of 5 to 7 cards

更新：
有关接受的答案的进一步改进,请参考：
Efficient way to randomly select set bit