C++ 暴力破解 XOR 密码

Question

我已经实现了这样的异或加密算法:

string XOR(string data, const char* key)
{
    string xorstring = data; //initialize new variable for our xordata
    for (int i = 0; i < xorstring.length(); i++) { //for loop for scrambling bits in the string
    xorstring[i] = data[i] ^ key[i]; //scrambling the string/descrambling it
    } 
    return xorstring;
}

效果很好，例如:string ciphertext = XOR("test", "1234"); 将返回密文，解密时:string plaintext = XOR(ciphertext, "1234"); 它将返回“测试”。

所以，我想创建一种算法，通过暴力破解来破解异或密码，所以基本上是尝试使用所有可能的 key 组合来解密密文。

它(应该)像这样工作:

• 从字母表的字符数组生成字符串
• 将给定的密文与给定的字符串异或(解密)得到明文
• 然后对生成的明文进行异或(加密)，并在if语句中进行比较，看是否与原始密文匹配
• 如果匹配，则找到用于解密密文的正确 key 。

就这么简单，但我发现自己在算法上苦苦挣扎:

const char Numbers[11] = "0123456789";
const char AlphabetUpper[27] = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
const char AlphabetLower[27] = "abcdefghijklmnopqrstuvwxyz";

string XOR(string data, const char* key)
{
   string xorstring = data; //initialize new variable for our xordata
   for (int i = 0; i < xorstring.length(); i++) { //for loop for scrambling   bits in the string
    xorstring[i] = data[i] ^ key[i]; //scrambling the string/descrambling it
   }
   return xorstring;
 }

 string cipher, plain; //store the ORIGINAL ciphertext and plaintext

 int main()
 {
    plain = "test"; //set the plaintext
    cipher = XOR(plain, "1234"); //encrypt it with the xor function

    cout << plain << endl;  //output
    cout << cipher << endl; //output

    cout << "press enter to start bruteforcing!" << endl;
    getchar();

    while (true) //loop for bruteforcing
    {
       static int stringlength = 1; //the keylength starts from 1 and then
       //expands to 2,3,4,5, etc...
       BruteForce(stringlength, cipher, ""); //call the brute force function
       stringlength++; //increment the keylength
    }
    return 0;
}

void BruteForce(int length, string ciphertext,  string tempKey)
{
    static int count = 0; // for counting how many times key was generated
    string decipher, recipher; //for storing new XORed strings.
    if (length == 0) 
    {
       //decrypt the given ciphertext with the random key
       decipher = XOR(ciphertext, tempKey.c_str());
       //encrypt it again with the same key for comparison
       recipher = XOR(decipher, tempKey.c_str());

       cout << deciphered << endl; //output

       //compare the two ciphertexts
       if (ciphertext == recipher)
       {
         //....
          cout << "Key found! It was: '" << tempKey << "'" << endl;
          cout << "it took " << count << " iterations to find the key!";
          getchar();
       }
      return;
   }
   count++;
   //generate the keys.
   for (int i = 0; i < 26; i++) {
       std::string appended = tempKey + AlphabetLower[i];
       BruteForce(length - 1, ciphertext, appended);
    }
    for (int i = 0; i < 26; i++) {
       std::string appended = tempKey + AlphabetUpper[i];
       BruteForce(length - 1, ciphertext, appended);
     }
    for (int i = 0; i < 10; i++) {
       std::string appended = tempKey + Numbers[i];
       BruteForce(length - 1, ciphertext, appended);
   }
}

由于未知原因，该算法不 起作用。

很奇怪，理论上应该可以。当程序运行时，它表示在每次执行 bruteforce() 函数时都找到了 key 。你自己试试吧。有人可以指出我在这里做错了什么吗？感谢帮助。谢谢。

Answer 1

它不起作用，因为您的任务是不可能完成的。一次一密 (xor) 具有称为 perfect secrecy 的属性这意味着您的密文可以等概率地解密为任何纯文本。所以不可能知道哪个纯文本是原始的。

对于您的方案，(message XOR key) XOR key 始终只是再次返回的消息。