RSA算法的實現不是在java中

Question

理論上我知道，如果n=33，e(public key)=3和d(private key)=7我可以用BigInteger類modPow(e, n)加密plaintext，並與modPow(d,n)解密正常工作，但解密後plaintext是不一樣的第一。

這裏是我的代碼：

public class KeyTest { 
private BigInteger n = new BigInteger("33"); 
private BigInteger e = new BigInteger("3"); 
private BigInteger d = new BigInteger("7"); 

public static void main(String[] args) { 
    KeyTest test = new KeyTest(); 

    BigInteger plaintext = new BigInteger("55"); 
    System.out.println("Plain text: " + plaintext); 

    BigInteger ciphertext = test.encrypt(plaintext); 
    System.out.println("Ciphertext: " + ciphertext); 

    BigInteger decrypted = test.decrypt(ciphertext); 
    System.out.println("Plain text after decryption: " + decrypted); 
} 

public BigInteger encrypt(BigInteger plaintext) { 

    return plaintext.modPow(e, n); 
} 

public BigInteger decrypt(BigInteger ciphertext) { 

    return ciphertext.modPow(d, n); 
} 
} 

輸出是：

Plain text: 55 Ciphertext: 22 Plain text after decryption: 22 

Answer 1

你的明文（55）比模量（33）大，所以你不能真正加密信息。考慮下面的稍微不同的實施例：

• p = 11
• q = 17
• n = 187
• phi(n) = 160
• 選擇e = 3
• 如果d = 107然後e * d = 321 = 1 mod phi(n)

所以你的代碼更改爲：

private BigInteger n = new BigInteger("187"); 
    private BigInteger e = new BigInteger("3"); 
    private BigInteger d = new BigInteger("107"); 

    public static void main(String[] args) { 
    KeyTest test = new KeyTest(); 

    BigInteger plaintext = new BigInteger("55"); 
    System.out.println("Plain text: " + plaintext); 

    BigInteger ciphertext = test.encrypt(plaintext); 
    System.out.println("Ciphertext: " + ciphertext); 

    BigInteger decrypted = test.decrypt(ciphertext); 
    System.out.println("Plain text after decryption: " + decrypted); 
    } 

    public BigInteger encrypt(BigInteger plaintext) { 

    return plaintext.modPow(e, n); 
    } 

    public BigInteger decrypt(BigInteger ciphertext) { 

    return ciphertext.modPow(d, n); 
    } 
} 

輸出：

Plain text: 55 
Ciphertext: 132 
Plain text after decryption: 55