问题描述
我工作的一个电梯的项目只是为了好玩，它实际上是硬件。但我认为这更多的是一个软件问题。我不需要有这个功能，其实这完全是多余的，但我很好奇，所以我将它无论如何，所以我可以学到：P
I'm working on an elevator project just for fun, it's actually hardware. But I think this is more of a software question. I don't need to have this feature, in fact it is completely redundant, but I was curious so I'm adding it anyway so I can learn :P
我有一个8位地址，8位数据总线和一个8位的加密code。我有一个主站和许多从设备。主知道从站的地址，并知道加密code。奴隶们知道他们的地址和加密code为好。
I have an 8 bit address, 8 bit data bus, and an 8 bit encryption code.I have a master and many slave devices.The master knows the address of the slaves and knows the encryption code.The slaves know their address and the encryption code as well.
我想一个非常简单的算法，即：
I want a really simple algorithm such that:
主机发送Y，其中，Y =功能（数据加密code）从机接收Y，可以提取由数据=函数2（Y，加密code）
The master sends "y" where, y = function(data,encryption code)The slave receives "y" and can extract data by data = function2(y,encryption code)
我试着用玩弄AND，XOR，OR，等等以及它们的组合，但不能看着办吧。
I tried playing around with AND, XOR, OR, etc... and combinations of them, but couldn't figure it out.
再次声明，我在寻找简单的算法。如果你不介意的话，你可以帮我一个更大的忙，并解释如何，我可以得出这样一个解决方案/函数的一些理论。
Again I'm looking for simple algorithms. If you don't mind you could do me a bigger favour and explain some theory on how I can come to such a solution/functions.
太感谢了！
推荐答案
您可以使用 XOR chiper 这是非常简单的：
You can use XOR chiper it's very simple:
E(x,key)=> y= x XOR key
D(y,key)=> x= y XOR key
很简单！
您可以升级加密，并使其为密码块链接这意味着，例如，你有你需要将其划分为块，让说大小B.块的一个数据D第一块你：
You can upgrade the encryption and to make it to cipher-block chaining that means for example you have a data D you need to divide it to blocks, let say block of size B. for the first block you do:
E(b0,key)=> y0= b0 XOR key
结果它的将是关键的下一个数据块加密：
the result its going to be the key for the next block encryption:
E(b1,y0)=> y1= b0 XOR y0 .... E(bn,yn-1)=> yn= bn XOR yn-1
原始数据是 D = {B0，B1 ..... BN} 和加密数据的现在 E = {Y0， Y1 .... YN} 对加密数据进行解密，你需要做相反的方式！就这样！
The original data was D={b0,b1.....bn} and the encrypted data its now E={y0,y1....yn}to decrypt the encrypted data you need to do the opposite way! that's all!
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