凯撒密码的蛮力解密

Question

这是我使用经典凯撒密码进行暴力破解的代码。蛮力解密显示出奇怪的结果，因为当我将密钥设置为 3 时，它会多次发布相同的解密字符串。在其他密钥中，例如 7，它甚至不显示正确的解密字符串。

我的蛮力解密方法是根据字母表改变加密消息中的每个字母，因此有一个26次的for循环，以及另一个消息长度的for循环。它使用 StringBuilder 的 setCharAt 方法来改变字符串中的字符。

这是我仅使用暴力破解的代码：

    void decryptbruteforce(String encryptmessage) {
    
    //Get the standard alphabet
    String standalpha = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
    
    //Convert this message to uppercase
    String encryptmessageupper = encryptmessage.toUpperCase();
    
    StringBuilder sbdecrypt = new StringBuilder(encryptmessageupper);
    
    
    int key;
    int i;
    int index;
    char currentchar;
    char newchar;
    
    //Loop through the 26 keys in the alphabet.
    for (key = 1; key < 27; key++) {
        
        //Loop through the encrypted message
        for (i = 0; i < sbdecrypt.length(); i++) {
            
            //Get the encrypted character
            currentchar = sbdecrypt.charAt(i);
            
            //Get the index in the alphabet
            index = standalpha.indexOf(currentchar);
            
            //If the currentchar is in the alphabet
            if (index != -1) {
                
                //Reduce the character by the key in the alphabet
                index = index - key;
                
                //If the character goes below 0, aka 'A', go back to the end of the alphabet
                if (index < 0) {
                    index = index + 26;
                    
                    //Get the new character in the alphabet
                    newchar = standalpha.charAt(index);
                    
                    //Set the character in the stringbuilder
                    sbdecrypt.setCharAt(i, newchar);
                }
                
                else {
                    
                    //Get the new character in the alphabet
                    newchar = standalpha.charAt(index);
                    
                    //Set the character in the stringbuilder
                    sbdecrypt.setCharAt(i, newchar);
                }
            }
        }
        
        //Print the key and the resulting string
        System.out.println("Key: " + key + " Decrypted String: " + sbdecrypt);
    }
}

这是我的输出： Caesar Cipher key of 3 Caesar Cipher key of 7

如果有帮助，这是我的整个代码的链接：https://gist.github.com/cliven-hew/35e9458c24d5f5b1ace97b7146ec429a

Answer 1

您的问题是您总是在编辑同一个StringBuilder sbdecrypt 对象。因此，当您使用键 1 进行移位时，您的下一次尝试不是基于移位原始字符串，而是基于先前移位的字符串。这就是为什么您在跳过 M.... 的同时从 N.... 转到 L.... 并且这种模式一直持续的原因。

您可以通过在 for (key = 1; key < 27; key++) 行后添加 sbdecrypt = new StringBuilder(encryptmessageupper); 来解决此问题。

  public static void decryptbruteforce(String encryptmessage) {

    String standalpha = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
    String encryptmessageupper = encryptmessage.toUpperCase();
    StringBuilder sbdecrypt = new StringBuilder(encryptmessageupper);
    
    int key;
    int i;
    int index;
    char currentchar;
    char newchar;

    for (key = 1; key < 27; key++) {
      //Ensure you do your shift on the same string each time.
      //This makes a new object each time
      sbdecrypt = new StringBuilder(encryptmessageupper);

      for (i = 0; i < sbdecrypt.length(); i++) {
        currentchar = sbdecrypt.charAt(i);
        index = standalpha.indexOf(currentchar);
        if (index != -1) {
          index = index - key;
          if (index < 0) {
            index = index + 26;
          }
          newchar = standalpha.charAt(index);
          sbdecrypt.setCharAt(i, newchar);
        }
      }
      System.out.println("Key: " + key + " Decrypted String: " + sbdecrypt);
    }
  }