通信工程文献翻译——一种新的加密方法

在前一章中，我们注意到对称密钥密码系统受到分发密钥问题的限制。20世纪70年代，一种激进的新方法被开发出来，被称为公共密钥加密技术。基本思想是，典型的用户（Bob）有两个密钥，一个公钥和一个私钥。Alice和其他人使用公钥加密他们希望发送给Bob的消息，Bob使用私钥来解密这些消息。系统的安全性取决于确保Bob的私钥不容易找到，即使每个人都知道他的公钥。Rivest，Shamir和Adleman于1977年发现了一种实现这一想法的实用方法，被称为RSA密码系统。现在还有很多其他的公共密钥加密系统，但RSA仍然被广泛应用，值得研究，因为它简单而优雅地说明了基本原理。在RSA系统中，明文是整数mod n的序列，其中将n确定为具有至少300位十进制数字的数字。可以使用任何简单的规则将“原始”消息转换为纯文本。例如，如果消息写成27符号字母A，我们可以从每个符号开始，以5位二进制数表示： 

L1→00000, A →00001, B →00010, ..., Z →11010.

然后我们可以形成（例如）180个符号的块，使得900个二进制数字的字符串代表具有大约300个十进制数字的数字。这些数字可以被看为整数mod n，只要它们足够大。在实践中这将使用基于ASCII或Unicode标准的系统。关键是可以构建明文，并使用公开的信息将其转换回其“原始”形式。

我们已经在多次观察到，加法和乘法运算的整数mod n是一个环。我们用Zn表示它。通常，不是Zn的每个非零元素都具有乘法反相。我们假设读者熟悉的事实是，当且仅当gcd（x，n）= 1时，非零元素x∈Zn具有乘法反x-1∈Zn。在第13.3节中，我们将描述一个好的方法计算x-1。

定义13.1（φ函数）：

对于任何正整数n，整数x在1≤x≤n范围内，使得gcd（x，n）=1,用φ(n).表示。从上述结果可以看出，φ(n)也是Zn的可逆元素的数量。

例如，取n=14，使得gcd（x，14 ）=1为1,3,5,9,11,13，因此φ(14)=6。在这种情况下，反转容易地被检查：因为3×5=1（mod 14）所以3-1=5，等等。

如果n是质数，说n=p，则p-1数x=1,2，...，p-1都满足gcd（x，p）=1，因此φ(p)=p-1。我们需要以下简单的这个......

A new approach to cryptography

In the previous chapter we noted that symmetric key cryptosystems are limited by the problem of distributing keys. A radical new approach was developed in the 1970s, known as public key cryptography.

The fundamental idea is that a typical user (Bob) has two keys, a public key and a private key. The public key is used by Alice and others to encrypt messages that they wish to send to Bob, and the private key is used by Bob to decrypt these messages. The security of the system depends on ensuring that Bob’s private key cannot be found easily, even though everyone knows his public key.

A practical method of implementing this idea was discovered by Rivest,Shamir, and Adleman in 1977 and is known as the RSA cryptosystem. There are now many other systems of public key cryptography, but RSA is still widely used, and it is worth studying because it illustrates the basic principles simply and elegantly.

In the RSA system the plaintexts are sequences of integers mod n, where it is reasonable to think of n as a number with at least 300 decimal digits.

Any simple rule can be used to convert the ‘raw’ message to plaintext. For instance, if the message is written in the 27-symbol alphabet A, we could begin by representing each symbol as a 5-bit binary number:

→ 00000, A  → 00001, B  → 00010, ..., Z → 11010.

Then we could form blocks of (for example) 180 symbols, making strings of 900 binary digits, which represent numbers with about 300 decimal digits. These numbers can then be treated as integers mod n, provided n is large enough. In practice, a system based on the ASCII or Unicode standards would be used:the point is that the plaintext can be constructed, and converted back into its ‘raw’ form, using publicly available......