所以我一直在寻找http://codahale.com/how-to-safely-store-a-password/#,并成为好奇哈希有多快不同时,稍微强大的台式计算机上bruteforced和被诱惑,以测试它并行蛮力算法
大部分的算法我见过,虽然是单线程,它让我感到这对于使用c#4.0 Parallel.net/Plinq扩展和并发结构(如ConcurrentBag和IProducerConsumer)是一个非常有趣的挑战。
所以我的任务如下,使用并行化构建一个密码长度为n和长度为charset [x]的最有效/高性能的bruteforce检查器,即生成给定字符集和长度的所有可能字符串,直到找到匹配。假设至少两个核心和RAM的合理量
我要试一试自己,让最好的男人/女人赢得:)
编辑
第一次尝试没有性能比较但和范围有限的和已知的密码长度
char[] chars = new char[] { '0', '1', '2', '3', '4', '5', '6', '7', '8', '9', 'a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z' };
public long NrCombinations(int nrChars, int stringLength)
{
Func<long, int, long> power = null;
power = (i, p) => p == 1 ? i : i * power(i, p - 1);
return power(nrChars, stringLength);
}
public static bool StringArrayEquals(char[] a, char[] b)
{
if (a.Length != b.Length)
return false;
for (int i = 0; i < a.Length; i++)
{
if (!a[i].Equals(b[i]))
return false;
}
return true;
}
public char[] GenerateString(int i, int stringLength)
{
char[] current = new char[stringLength];
for (int i = 0; i < stringLength; i++)
{
double remainder = i % this.chars.Length;
i = i/this.chars.Length;
current[i] = this.chars[(int) remainder];
}
return current;
}
public bool IsMatch(int i, char[] password)
{
return StringArrayEquals(GenerateString(i, password.Length), password);
}
private int GetMatching(string passwordString)
{
char[] password = passwordString.ToArray();
int nrCombinations = (int)NrCombinations(this.chars.Length, password.Length);
return ParallelEnumerable.Range(0, nrCombinations).WithDegreeOfParallelism(10).FirstOrDefault(i => IsMatch(i, password));
}
下一次尝试
使用ParallelEnumerable不是很聪明,因为它的大小限制为int,即使我怀疑这会持续很长一段时间,并且需要较大的密码字符集,但您很快就需要至少很长的时间。猜想你要么必须去BigInt,要么就开始以某种方式分解它。
public long NrCombinations(int nrChars, int stringLength)
{
Func<long, int, long> power = null;
power = (i, p) => p == 1 ? i : i * power(i, p - 1);
return power(nrChars, stringLength);
}
public string GenerateString(long number, int sentenceLength)
{
char[] current = new char[sentenceLength];
for (int i = 0; i < sentenceLength; i++)
{
double remainder = number % this.chars.Length;
number = number/this.chars.Length;
current[i] = this.chars[(int) remainder];
}
return new string(current);
}
public bool IsMatch(string hash, long i, int passwordLength)
{
string generated = GenerateString(i, passwordLength);
string hashed = GetMasterHash(generated, this.site);
return string.Equals(hashed, hash);
}
private string GetMatching(string hash,int passwordLength)
{
string result = string.Empty;
int stringlength = passwordLength;
long nrCombinations = NrCombinations(this.chars.Length, stringlength);
long x = 0;
Parallel.For(0, nrCombinations, (i, loopState) =>
{
if (IsMatch(hash,i, passwordLength))
{
x = i;
loopState.Stop();
return;
}
});
if (x > 0)
{
result = this.GenerateString(x, passwordLength);
}
return result;
}
彩虹桌更凉爽 – thejh 2010-12-16 17:06:35
HAVAL-3-128或MD2? – 2010-12-16 17:17:36
只是生成字符串组合..通过散列运行它们总是可以在最后加上 – Homde 2010-12-16 17:19:12