Question

我试图为这一问题找到最佳解决办法。在[0......N]范围内,我有一套明确的非否定性分类。我需要能够代表本套中的每一份数字,并能够将这种str倒数改为原来的数字。因此,这应该是双管齐下的职能。

Additional requirements are:

String representation of a number should obfuscate original number at least to some degree. So primitive solution like f(x) = x.toString() will not work.
String length is important: the less the better.
If one knows the string representation of K, I would like it to be non-trivial (to some degree) to guess the string representation of K+1.

For p.1 & p.2 the obvious solution is to use something like Base64 (or whatever BaseXXX to fit all the values) notation. But can we fit into p.3 with minimal additional effort? Common sense tells me that I additionally need a bijective "String <-> String" function for BaseXXX values. Any suggestions? Or maybe there s something better than BaseXXX to use to fit all 3 requirements?

Answer 1

如果你不需要too安全,那么在基三十编码后,你只能使用一个简单的不对称的电离层。例如,你可以选择关键的分类体格[n1, n2, n3...],然后使用。

环流背后的基本思想是每个特性C的简单编码,即C+K(mod 26),K是关键因素之一。随着你们的到来,我们刚刚从下一个特点的关键中得出下一个数字,一俟你摆脱了关键价值观,就将紧接着。

你在这方面确实有两种选择:你可以首先将一些数字转换为第三十基点的扼杀,然后进行加密,或者你可以把同一想法仅仅作为单一特性对每个数字进行加密。在这种情况下,您希望将其从mod 26 改为mod N + 1。



要考虑这一点,一个更为简单的选择是,从关键和价值来看,只读到xor。 (与使用Vgenere公式相反) 我认为,这同样有助于消除混乱。

Answer 2

This method meets requirements 1-3, but it is perhaps a bit too computationally expensive:

find a prime p > N+2, not too much larger
find a primitive root g modulo p, that is, a number whose multiplicative order modulo p is p-1
for 0 <= k <= N, let enc(k) = min {j > 0 : g^j == (k+2) (mod p)}
f(k) = enc(k).toString()

Answer 3


编制一个长度表<代码>M。 该表应绘制0至M-1的地图,以随机顺序标明短体。 用表格中的插图表示数字。 直截了当的反射。

有了<代码>M=26,你只能用一封信给每位数字。或采用<代码>M=256,每位数使用一个星号。

Not even remotely a good cryptographic approach!

Answer 4

因此,你需要说明,混淆了原来的编号,但允许在知道地(K)时确定地(K+1)?

How about just doing f(x) = (x + a).toString(), where a is secret? Then an outside user can t determine x from f(x), but they can be confident that if they have a string "1234", say, for an unknown x then "1235" maps to x+1.

Answer 5

第1页和第3页略为矛盾,也含糊不清。

我将提议使用X射线表示。

17 => 0x11
123123 => 1E0F3

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