OpenEuphoria

Re: Math Problem - Probability of Password Duplication

• Posted by euphoria at carlw.legend.uk.com Oct 22, 2001
• 453 views

Brian Keene wrote:

> Hi All:
> Can someone who knows math & probability theory better than me (this isn't
hard to
> do :)!) tell me how to figure out how many possible passwords I can
generate by
> having the following parameters?
>
> Password Length = 11 to 14 characters
> Possible Password Content = each character can be any 1 of 89 different
> characters.
> Character Set = a-zA-Z1-9{}[]()^_`~:;<=>? at !#$%&'+,-./
> Excluded Characters = 0|\"
>
> My understanding is that it would be:
>
> char14 = 89*89*89*89*89*89*89*89*89*89*89*89*89*89 -- (1.95641099e+027)
> char11 = 89*89*89*89*89*89*89*89*89*89*89          -- (2.77517307e+021)
>
> result = char14 - char11

result = char14 + char13 + char12 + char11
       = power(89,14)+power(89,13)+power(89,12)+power(89,11)

...since all groups are mutually exclusive.

Here's what makes 'char14 - char11' seem logical (both of these are also the
right answer):

result = (char15 - char11)/88
       = (power(89,15)-power(89,11))/(89-1)

As for the function:

constant validChars="abc..xyzABC..XYZ..etc"

function password()
   sequence p
   integer len
   len = length(validChars)
   p = {}
   for i = 1 to 10+rand(4) -- 11..14
       p &= validChars[rand(len)]
   end for
   return p
end function

Carl