C#bruteforce md5哈希字符串

时间:2015-01-29 09:00:10

标签: c# loops hash cryptography md5

我尝试创建一个C#应用程序,通过使用bruteforce恢复MD5哈希密码(仅用于测试,限制为小写a-z)

我遇到了循环问题。它需要在尝试a-z后添加一个新字符并启动一个尝试所有可能组合的新循环。

以下代码有效,但远非最佳。关于如何制作一个聪明的"循环?

public void CompareMD5()
{
    string outputMD5 = "";
    int chars = 1;
    int loop = 0;
    int i = 0;
    int ii = 0;
    while (!outputMD5.Equals(targetMD5))
    {
        if (chars == 1)
        {
            Console.WriteLine("trying " + charSet[i]);
            outputMD5 = GenerateMD5(charSet[i]);
            i++;
            if (i >= charSet.Length)
            {
                Console.WriteLine("\n" + "*** ADDING CHAR ***" + "\n");
                i = 0;
                chars++;
            }
        }

        if (chars == 2)
        {
            Console.WriteLine("trying " + charSet[i] + charSet[ii]);
            outputMD5 = GenerateMD5(charSet[i] + charSet[ii]);
            i++;
            if (i >= charSet.Length)
            {
                i = 0;
                ii++;
            }
            if (ii >= charSet.Length)
            {
                ii = 0;
                i++;
                loop++;
            }
            if (loop == 1)
            {
                Console.WriteLine("\n" + "*** ADDING CHAR ***" + "\n");
                i = 0;
                ii = 0;
                chars++;
            }
        }
    }
}

1 个答案:

答案 0 :(得分:1)

试试这个:

  // 'a', ..., 'z', 'aa', ..., 'zz', 'aaa', ..., 'zzz'
  int maxLength = 3;

  for (int length = 1; length <= maxLength; ++length) {
    // initial combination "a...a" ('a' length times)
    StringBuilder Sb = new StringBuilder(new String('a', length));

    while (true) {
      String value = Sb.ToString();
      //TODO: Test MD5 here
      // if (value.Equals(targetMD5)) {...}

      // Is this the last combination? (all 'z' string)
      if (value.All(item => item == 'z'))
        break;

      // Add one: aaa -> aab -> ... aaz -> aba -> ... -> zzz
      for (int i = length - 1; i >= 0; --i)
        if (Sb[i] != 'z') {
          Sb[i] = (Char) (Sb[i] + 1);

          break;
        }
        else
          Sb[i] = 'a';
    }
  }
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