加密单个 int 的方法 [英] Way to encrypt a single int

问题描述

如何以低成本对 32 位 int 进行双向加密，以便每个数字都映射到该空间中的其他 int 并以难以预测的方式返回?

How can you inexpensively two-way encrypt a 32 bit int, such that every number maps to some other int in that space and back in a way that's difficult to predict?

当然不需要在映射表中预先存储 42.9 亿个整数.

And doesn't require pre-storing 4.29 billion ints in a mapping table, of course.

推荐答案

您需要的是 32 位分组密码.不幸的是，由于 短块大小的弱点.如果您可以处理两倍于输入的加密 int，那么您可以只使用 Blowfish、TDES 或其他一些经过良好审查的 64 位分组密码.

What you want is a 32-bit block cipher. Unfortunately, most block ciphers are 64-bits or more due to the weaknesses of a short block size. If you can handle the encrypted int being twice as large as the input, then you can just use Blowfish, TDES, or some other nicely vetted 64-bit block cipher.

如果你真的需要 32 位并且不介意降低安全性，那么它很容易将 Feistel 网络密码(如 Blowfish)修剪到任何块长度，该长度是 2 的倍数且小于起始密码.对于 Blowfish，只需在两个半块之间平均分配您的输入数字，然后修剪F 函数的输出和 P 值下降到目标块大小的 1/2.这一切都可以在像往常一样键入算法后完成.

If you really need 32 bits and don't mind the decreased security then its easy enough to trim a Feistel network cipher like Blowfish down to any block length that's a multiple of 2 and less than the starting cipher. For Blowfish, just split your input number evenly between the two half blocks, and trim the output of the F function and the P-values down to 1/2 your target block size. This can all be done after keying the algorithm as usual.

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