如何获得第n个随机的"nextInt".价值? [英] How to obtain the nth random "nextInt" value?

问题描述

在使用类java.util.Random时，如何才能获得通过N次调用nextInt()方法获得的值，但是以一种更为有效的方式(特别是在O(1)中)?

When using class java.util.Random, how can one get the value obtained from calling the method nextInt() N times, but in a much more efficient way (in O(1) specifically)?

例如，如果我构造一个具有特定种子值的Random对象，并且想要获得第100,000个"nextInt()值"(即，在调用方法nextInt()100,000次后获得的值)，一种快速的方法，我可以吗?

For example, if I construct a Random object with a particular seed value, and I want to get the 100,000th "nextInt() value" (that is, the value obtained after calling the method nextInt() 100,000 times) in a fast way, could I do it?

为简单起见，假设JDK版本1.7.06，因为可能需要知道类Random中某些私有字段的确切值.可以这样说，我发现以下字段与随机值的计算有关:

Assume, for simplicity, version 1.7.06 of the JDK, since it may be required to know the exact values of some private fields in class Random. And speaking of, I found the following fields to be relevant in the calculation of a random value:

private static final long multiplier = 0x5DEECE66DL;
private static final long addend = 0xBL;
private static final long mask = (1L << 48) - 1;

在研究了一些随机性之后，我发现使用线性同余生成器获得了随机值.执行该算法的实际方法是方法next(int):

After exploring a bit about randomness, I found that random values are obtained using a Linear congruential generator. The actual method that executes the algorithm is method next(int):

protected int next(int bits) {
    long oldseed, nextseed;
    AtomicLong seed = this.seed;
    do {
        oldseed = seed.get();
        nextseed = (oldseed * multiplier + addend) & mask;
    } while (!seed.compareAndSet(oldseed, nextseed));
    return (int)(nextseed >>> (48 - bits));
}

该算法的相关行是获得下一个种子值的行:

The relevant line for the algorithm is the one that obtains the next seed value:

nextseed = (oldseed * multiplier + addend) & mask;

因此，更具体地说，是否有一种方法可以使该公式通用化以获得第n个nextseed"值?我在这里假设，有了这个之后，我可以通过使变量"bits"为32来简单地获取第n个int值(方法nextInt()只需调用next(32)并返回结果).

So, to be more specific, is there a way that I can generalize this formula to obtain the "nth nextseed" value? I'm assuming here that after having that, I can then simply obtain the nth int value by letting the variable "bits" be 32 (the method nextInt() simply calls next(32) and returns the result).

预先感谢

PS:也许这是一个更适合 mathexchange 的问题?

PS: Perhaps this is a question more suitable for mathexchange?

推荐答案

您可以在 O(log N)时间内完成此操作.从 s(0)开始，如果我们暂时忽略模数(2 ⁴⁸ )，我们可以看到(使用 m 和 a 作为乘数和加数的简写)

You can do it in O(log N) time. Starting with s(0), if we ignore the modulus (2⁴⁸) for the moment, we can see (using m and a as shorthand for multiplier and addend) that

s(1) = s(0) * m + a
s(2) = s(1) * m + a = s(0) * m² + (m + 1) * a
s(3) = s(2) * m + a = s(0) * m³ + (m² + m + 1) * a
...
s(N) = s(0) * m^N + (m^(N-1) + ... + m + 1) * a

现在，通过重复平方，通过模幂运算可以轻松地以 O(log N)步骤计算 m ^ N(mod 2 ^ 48).

Now, m^N (mod 2^48) can easily be computed in O(log N) steps by modular exponentiation by repeated squaring.

另一部分要复杂一些.现在暂时忽略模数，其几何和为

The other part is a bit more complicated. Ignoring the modulus again for the moment, the geometric sum is

(m^N - 1) / (m - 1)

计算此模 2 ^ 48 有点不平凡的原因是 m-1 并非模的互质.但是，由于

What makes computing this modulo 2^48 a bit nontrivial is that m - 1 is not coprime to the modulus. However, since

m = 0x5DEECE66DL

m-1 的最大公约数，模数为4，(m-1)/4 具有模数反函数 inv 以 2 ^ 48 为模.让

the greatest common divisor of m-1 and the modulus is 4, and (m-1)/4 has a modular inverse inv modulo 2^48. Let

c = (m^N - 1) (mod 4*2^48)

然后

(c / 4) * inv ≡ (m^N - 1) / (m - 1) (mod 2^48)

所以

• 计算 M≡m ^ N(mod 2 ^ 50)
• 计算 inv

获得

s(N) ≡ s(0)*M + ((M - 1)/4)*inv*a (mod 2^48)

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