Rabin Karp字符串匹配算法

Question

我在网站的论坛上看到了这个Rabin Karp字符串匹配算法,我有兴趣尝试实现它,但我想知道如果有人能告诉我为什么变量ulong Q和ulong D分别是100007和256:S ？这些价值观带有什么意义？

static void Main(string[] args)
{
    string A = "String that contains a pattern.";
    string B = "pattern";
    ulong siga = 0;
    ulong sigb = 0;
    ulong Q = 100007;
    ulong D = 256;
    for (int i = 0; i < B.Length; i++)
    {
        siga = (siga * D + (ulong)A[i]) % Q;
        sigb = (sigb * D + (ulong)B[i]) % Q;
    }
    if (siga == sigb)
    {
        Console.WriteLine(string.Format(">>{0}<<{1}", A.Substring(0, B.Length), A.Substring(B.Length)));
        return;
    }
    ulong pow = 1;
    for (int k = 1; k <= B.Length - 1; k++)
        pow = (pow * D) % Q;

    for (int j = 1; j <= A.Length - B.Length; j++)
    {
        siga = (siga + Q - pow * (ulong)A[j - 1] % Q) % Q;
        siga = (siga * D + (ulong)A[j + B.Length - 1]) % Q;
        if (siga == sigb)
        {
            if (A.Substring(j, B.Length) == B)
            {
                Console.WriteLine(string.Format("{0}>>{1}<<{2}", A.Substring(0, j),
                                                                    A.Substring(j, B.Length),
                                                                    A.Substring(j + B.Length)));
                return;
            }
        }
    }
    Console.WriteLine("Not copied!");
}

Answer 1

关于神奇的数字保罗的回答非常明确.

就代码而言,Rabin Karp的主要思想是在字符串的滑动部分和模式之间执行哈希比较.

每次在整个子串上都不能计算哈希值,否则计算复杂度将是二次O(n^2)而不是线性的O(n).

因此,应用滚动散列函数,例如在每次迭代时,仅需要一个字符来更新子串的散列值.

那么,让我们评论你的代码:

for (int i = 0; i < B.Length; i++)
{
    siga = (siga * D + (ulong)A[i]) % Q;
    sigb = (sigb * D + (ulong)B[i]) % Q;
}
if (siga == sigb)
{
    Console.WriteLine(string.Format(">>{0}<<{1}", A.Substring(0, B.Length), A.Substring(B.Length)));
    return;
}

^这篇文章计算了pattern B(sigb)的哈希值,以及A相同长度的初始子字符串的哈希码B.实际上它并不完全正确,因为哈希可以碰撞¹,因此,有必要修改if语句:if (siga == sigb && A.Substring(0, B.Length) == B).

ulong pow = 1;
for (int k = 1; k <= B.Length - 1; k++)
    pow = (pow * D) % Q;

^这是计算pow执行滚动哈希所必需的.

for (int j = 1; j <= A.Length - B.Length; j++)
{
    siga = (siga + Q - pow * (ulong)A[j - 1] % Q) % Q;
    siga = (siga * D + (ulong)A[j + B.Length - 1]) % Q;
    if (siga == sigb)
    {
        if (A.Substring(j, B.Length) == B)
        {
            Console.WriteLine(string.Format("{0}>>{1}<<{2}", A.Substring(0, j),
                                                                A.Substring(j, B.Length),
                                                                A.Substring(j + B.Length)));
            return;
        }
    }
}

^ 最后,扫描剩余的字符串(即从第二个字符到结束),更新A子串的哈希值,并与B的哈希值(在开头计算)进行比较.

如果两个哈希值相等,则比较子字符串和模式¹,如果它们实际上相等则返回一个消息.

¹ 哈希值可能会发生碰撞 ; 因此,如果两个字符串具有不同的散列值,它们肯定是不同的,但如果两个散列相等则它们可以相等或不相同.

Answer 2

该算法使用散列进行快速字符串比较.Q和D是编码器可能通过一些试验和错误得到的神奇数字,并且为这个特定算法提供了良好的散列值分布.

您可以看到用于散列许多地方的这些类型的幻数.下面的示例是.NET 2.0字符串类型的GetHashCode函数的反编译定义:

[ReliabilityContract(Consistency.WillNotCorruptState, Cer.MayFail)]
public override unsafe int GetHashCode()
{
    char* chrPointer = null;
    int num1;
    int num2;
    fixed (string str = (string)this)
    {
        num1 = 352654597;
        num2 = num1;
        int* numPointer = chrPointer;
        for (int i = this.Length; i > 0; i = i - 4)
        {
            num1 = (num1 << 5) + num1 + (num1 >> 27) ^ numPointer;
            if (i <= 2)
            {
                break;
            }
            num2 = (num2 << 5) + num2 + (num2 >> 27) ^ numPointer + (void*)4;
            numPointer = numPointer + (void*)8;
        }
    }
    return num1 + num2 * 1566083941;
}

以下是R#为样本类型生成的GetHashcode覆盖函数的另一个示例:

    public override int GetHashCode()
    {
        unchecked
        {
            int result = (SomeStrId != null ? SomeStrId.GetHashCode() : 0);
            result = (result*397) ^ (Desc != null ? Desc.GetHashCode() : 0);
            result = (result*397) ^ (AnotherId != null ? AnotherId.GetHashCode() : 0);
            return result;
        }
    }