您正在寻找的是函数形式的伪随机排列，例如 f ，该函数映射从1开始的数字以伪随机双射的方式从N到N到1到N的数字上.然后，要生成伪随机排列的 n 个数字，只需返回 f(n)

What you are looking for is a pseudo-random permutation in the form of a function, say f, that maps numbers from 1 to N onto numbers from 1 to N in a pseudo-random bijective way. Then, to generate the nth number in a pseudo-random permutation, you just return f(n)

这与加密本质上是相同的问题.带密钥的分组密码是伪随机双射函数.如果您以某种顺序一次将所有可能的纯文本块送入，它将以不同的伪随机顺序一次一次返回所有可能的密文块.

This is essentially the same problem as encryption. A block cipher with a key is a pseudo-random bijective function. If you feed it all possible plaintext blocks exactly once in some order, it will return all possible ciphertext blocks exactly once in a different, pseudo-random order.

因此，要解决像您这样的问题，您实际上要做的是创建一个密码，该密码适用于从1到N的数字，而不是256位块或其他任何数字.您可以使用密码学中的工具来做到这一点.

So to solve a problem like yours, what you essentially do is create a cipher that works on numbers from 1 to N instead of 256-bit blocks or whatever. You can use the tools from cryptography to do this.

例如，您可以使用Feistel结构构造置换函数( https://en.wikipedia. org/wiki/Feistel_cipher )，就像这样:

For example, you can construct your permutation function with a Feistel structure (https://en.wikipedia.org/wiki/Feistel_cipher) like this:

1. 让W为floor(sqrt(N))，让该函数的输入为x
2. 如果x< W ^ 2，然后将x分为2个字段，从0到W-1:h = floor(x/W)和l = x％W.哈希h以产生从0到W-1的值，并设置l =(l + hash)％W.然后交换字段-让x = l * W + h
3. x =(x +(N-W ^ 2))％N
4. 重复步骤(2)和(3)几次.您做得越多，结果看起来就越随机.步骤(3)确保x ＜0. W ^ 2在许多回合中都是正确的.

由于此函数由多个步骤组成，每个步骤都将以双射的方式将0到N-1的数字映射到0到N-1的数字，因此整个函数也将具有此属性.如果您输入0到N-1之间的数字，则会以伪随机顺序将它们取回来.

Since this function consists of a number of steps, each of which will map the numbers from 0 to N-1 onto the numbers from 0 to N-1 in a bijective way, the entire function will also have this property. If you feed it the numbers from 0 to N-1, you'll get them back out in a pseudo-random order.