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发布于 2024-10-05 22:01:48 字数 666 浏览 2 评论 0原文

今天，我团队的一些成员正在讨论密码存储和一般安全问题。不管怎样，讨论简要讨论了 GPU 加速的暴力攻击与传统的仅 CPU 实现相比有多快。

这引起了我的兴趣，所以我决定尝试一些代码。因为我以前从未写过这样的东西，所以我决定写一个简单的（仅 CPU）暴力破解器。我最初的实现涉及固定长度（4 位数字）密码。出于测试目的，我按照 la 的方式实现了它：

for(char a = '0'; a <= '9'; ++a)
{
  for(char b = '0'; b <= '9'; ++b)
  {
    for(char c = '0'; c <= '9'; ++c)
    {
      for(char d = '0'; d <= '9'; ++d)
      {
        candidate[0] = a; candidate[1] = b;
        candidate[2] = c; candidate[3] = d;

        // Test 'candidate'...
      }
    }
  }
}

这效果很好，但显然不灵活。我试图概括上述内容以处理任何密码长度，但未能做到这一点。出于某种原因，我无法理解这些暴力破解者在给定“字母表”的情况下运行 1-n 个字符可能性的逻辑。

是否有一些通用的算法可以帮助您完成此任务？欢迎任何例子。

原文

Some members of my team were discussing password storage and general security issues today. Anyway, the discussion briefly touched upon how quick GPU-accelerated brute-force attacks are compared to the traditional CPU-only implementations.

This got me interested, so I decided to play around with some code. Since I've never written anything like this before, I decided to write a simple (CPU-only) brute-forcer. My initial implementation dealt with a fixed length (4 digit) password. For testing purposes, I implemented it a la:

for(char a = '0'; a <= '9'; ++a)
{
  for(char b = '0'; b <= '9'; ++b)
  {
    for(char c = '0'; c <= '9'; ++c)
    {
      for(char d = '0'; d <= '9'; ++d)
      {
        candidate[0] = a; candidate[1] = b;
        candidate[2] = c; candidate[3] = d;

        // Test 'candidate'...
      }
    }
  }
}

This works well, but is obviously inflexible. I attempted to generalize the above to handle any password length but have failed to do so. For some reason, I can't get my head around the logic that these brute-forcers use to run through 1-n character possibilities given an "alphabet".

Is there some common algorithm that allows you to accomplish this? Any examples welcome.

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苦笑流年记忆 2024-10-12 22:01:48

搜索引擎是你的朋友:)
C++ 暴力破解示例

（向下滚动一点..）

回复收藏 0 原文

违心° 2024-10-12 22:01:48

这是一个迭代版本......以下仅适用于小写字母，但可以轻松修改......

public static String nextLexographicWord(String txt)
{
    char [] letters = txt.toCharArray();
    int l = letters .length - 1;
    while(l >= 0)
    {
        if(letters[l] == 'z')
            letters[l] = 'a';
        else
        {
            letters[l]++;
            break;
        }
        l--;
    }
    if(l < 0) return 'a' + (new String(letters));
    return new String(letters); 
}

Here's a Iterative Version....The following works only for lowercase, but can easily be modified....

public static String nextLexographicWord(String txt)
{
    char [] letters = txt.toCharArray();
    int l = letters .length - 1;
    while(l >= 0)
    {
        if(letters[l] == 'z')
            letters[l] = 'a';
        else
        {
            letters[l]++;
            break;
        }
        l--;
    }
    if(l < 0) return 'a' + (new String(letters));
    return new String(letters); 
}

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~没有更多了~