java - 使用 Java 内置库的简单 RSA 加密

Question

要求是:

1. 质数 p 和 q 应至少为 1024 位

2. 两个素数之差应该大于2^512(安全起见)

问题:我想知道的是我指定了p和q的位长，并且还有 SecureRandom 实例来随机生成它们，但我被告知差异可能不会大于 2^512。那么我如何指定差异，使其大于 2^512？然后我想我将不再能够随机生成 p 和 q ？

Here是 BigInteger 类构造函数的文档，它表明如果我想手动指定它，我必须使用 bye 数组 byte[]，如果我这样做，则无法随机生成它。

任何提示都会很棒。

谢谢。

这是我遇到问题的方法:

public RSA(int bits) {
    bitlen = bits;
    SecureRandom random = new SecureRandom();
    BigInteger p = new BigInteger(bitlen, 100, random);
    BigInteger q = new BigInteger(bitlen, 100, random);
    n = p.multiply(q);
    BigInteger m = (p.subtract(BigInteger.ONE)).multiply(q
            .subtract(BigInteger.ONE));
    e = new BigInteger(Integer.toString(eValue));
    while (m.gcd(e).intValue() > 1) {
        e = e.add(new BigInteger("2"));
    }
    d = e.modInverse(m);
}

完整的源代码如下:

import java.math.BigInteger;
import java.security.SecureRandom;

public class RSA {

    public static double runningTime;
    private BigInteger n, d, e;
    private int bitlen = 1024;
    static int eValue = 65537;

    /** Create an instance that can encrypt using someone provided public key. */
    public RSA(BigInteger newn, BigInteger newe) {
        n = newn;
        e = newe;
    }

    /** Create an instance that can both encrypt and decrypt. */
    public RSA(int bits) {
        bitlen = bits;
        SecureRandom random = new SecureRandom();
        BigInteger p = new BigInteger(bitlen, 100, random);
        BigInteger q = new BigInteger(bitlen, 100, random);
        n = p.multiply(q);
        BigInteger m = (p.subtract(BigInteger.ONE)).multiply(q
                .subtract(BigInteger.ONE));
        e = new BigInteger(Integer.toString(eValue));
        while (m.gcd(e).intValue() > 1) {
            e = e.add(new BigInteger("2"));
        }
        d = e.modInverse(m);
    }

    /** Encrypt the given plain-text message. */
    public String encrypt(String message) {
        return (new BigInteger(message.getBytes())).modPow(e, n).toString();
    }

    /** Encrypt the given plain-text message. */
    public BigInteger encrypt(BigInteger message) {
        return message.modPow(e, n);
    }

    /** Decrypt the given cipher-text message. */
    public String decrypt(String message) {
        return new String((new BigInteger(message)).modPow(d, n).toByteArray());
    }

    /** Decrypt the given cipher-text message. */
    public BigInteger decrypt(BigInteger message) {
        return message.modPow(d, n);
    }

    /** Generate a new public and private key set. */
    public void generateKeys() {
        SecureRandom random = new SecureRandom();
        BigInteger p = new BigInteger(bitlen, 100, random);
        BigInteger q = new BigInteger(bitlen, 100, random);
        n = p.multiply(q);
        BigInteger m = (p.subtract(BigInteger.ONE)).multiply(q
                .subtract(BigInteger.ONE));
        e = new BigInteger(Integer.toString(eValue));
        while (m.gcd(e).intValue() > 1) {
            e = e.add(new BigInteger("2"));
        }
        d = e.modInverse(m);
    }

    /** Return the modulus. */
    public BigInteger getN() {
        return n;
    }

    /** Return the public key. */
    public BigInteger getE() {
        return e;
    }

    /** Test program. */
    public static void main(String[] args) {
        runningTime = System.nanoTime();
        RSA rsa = new RSA(1024);

        String text1 = "RSA-Encryption Practice";
        System.out.println("Plaintext: " + text1);
        BigInteger plaintext = new BigInteger(text1.getBytes());

        BigInteger ciphertext = rsa.encrypt(plaintext);
        System.out.println("cipher-text: " + ciphertext);
        plaintext = rsa.decrypt(ciphertext);

        String text2 = new String(plaintext.toByteArray());
        System.out.println("Plaintext: " + text2);
        System.out.println("RunningTime: "
                + (runningTime = System.nanoTime() - runningTime) / 1000000
                + " ms");
    }
}

Answer 1

很难随机生成两个相距小于 2⁵¹² 的 1024 位素数。概率是<2^-509(这是不可能的)。所以一般你不应该有问题。如果这种情况仍然发生，您可以在 checkDiff 在 while 循环中返回 false 时重试。这使它成为一种非常快速收敛的拉斯维加斯算法。

final static BigInteger targetDiff = new BigInteger("2").pow(512);

public boolean checkDiff(BigInteger p, BigInteger q){
    BigInteger pDiff = p.subtract(q);
    pDiff = pDiff.abs();

    BigInteger diff = pDiff.subtract(targetDiff);

    return diff.compareTo(BigInteger.ZERO) == 1;
}