keyspace calculation

[mischief]

in the following code, i am trying to calculate the keyspace for a given charset and password length (no minimum length).

_current.size() returns the size of the password, _charset.size() returns the size of the charset.

however, by using a 5 char charset and a 7char password length i did not get a correct result. can anyone give me some pointers?

Code: Select all

int mischief::brute::calculate_keyspace()
{
   int keyspace = 0;
   for(mischief::brutesize i = 0; i != _current.size(); ++i)
   {
      keyspace += mischief::math::pow(_charset.size(), i);
   }
   return keyspace;
}

int mischief::math::pow(const int number, int exponent)
{
   if(exponent == 0) return 1; // n^0 = 1
   int result = number;
   while (--exponent)
   {
      result *= number;
   }
   return result;
}

the formula for calculating keyspace size is, as per the Cain & Abel help file, as follows:

The key space of all possible combination of passwords to try is calculated using the following formula:

KS = L^(m) + L^(m+1) + L^(m+2) + ........ + L^(M)

where
L = character set length
m = min length of the key
M = max length of the key

For example, when you want to crack an half of a LanManager passwords (LM) using the character set "ABCDEFGHIJKLMNOPQRSTUVWXYZ" of 26 letters, the brute-force cracker have to try KS = 26^1 + 26^2 + 26^3 + ...... + 26^7 = 8353082582 different keys.

[BhaaL]

Give you include the empty password, that example should yield 97656 as keysize.
I wrote that pow, so it can't be wrong
The other stuff, doesn't seem to be wrong either.

You might want to cache the result of _charset.size() for performance, unless your compiler is smart enough to figure that out on his own

[mischief]

my actual result is 19531, BhaaL. using the exact same code.