1. Math Problem - Probability of Password Duplication
- Posted by brian_keene at yahoo.com Oct 22, 2001
- 363 views
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 Also, could anyone give me a small function that would generate a random password using this character set? Thanks In Advance Regards Brian Keene ===== Regards Brian Keene
2. Re: Math Problem - Probability of Password Duplication
- Posted by "C. K. Lester" <cklester at yahoo.com> Oct 22, 2001
- 356 views
> 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. > > 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 My first instinct is that you should probably add these. Your 11-character passwords are unique from the 14-character passwords. You would also add the 12- and 13-character length password combinations. > Also, could anyone give me a small function that would generate a random password > using this character set? chars = { "...your valid character set... " } for t=1 to passwordLength do password &= chars[rand(length(chars))] end for Something like that... ck
3. Re: Math Problem - Probability of Password Duplication
- Posted by "Thomas Parslow (PatRat)" <patrat at rat-software.com> Oct 22, 2001
- 364 views
> 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 As far as I can see it would be: result = power(89,11)+power(89,12)+power(89,13)+power(89,14) > Also, could anyone give me a small function that would generate a random > password > using this character set? constant PASSWORD_CHARS = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ123456789{}[]()^_`~:;<=>? at !#$%&'+,-./" constant MIN_LENGTH = 11 constant MAX_LENGTH = 14 function GenPass() sequence pass pass = {} for i = 1 to rand(MAX_LENGTH - MIN_LENGTH + 1) + MIN_LENGTH - 1 do pass &= PASSWORD_CHARS[rand(length(PASSWORD_CHARS))] end for return pass end function puts(1,GenPass()) > Thanks In Advance > Regards > Brian Keene > ===== > Regards > Brian Keene Thomas Parslow (PatRat) ICQ #:26359483 Rat Software http://www.rat-software.com/ Please leave quoted text in place when replying
4. Re: Math Problem - Probability of Password Duplication
- Posted by brian_keene at yahoo.com Oct 22, 2001
- 350 views
CK: Sorry. What I meant was that the passwords needed to be a minimum of 11 chars. & a maximum of 14 chars. NOT that they would need to be 11 OR 14 chars long. So your function looks perfectly fine to me. Thanks! The only question my boss wants to know is how many possible unique passwords can this generate? Is my formula correct or is my math wrong? Which is quite possible! :) Regards Brian Keene --- "C. K. Lester" <cklester at yahoo.com> wrote: > > > 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. > > > > 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 > > My first instinct is that you should probably add these. Your 11-character > passwords are unique from the 14-character passwords. You would also add the > 12- and 13-character length password combinations. > > > Also, could anyone give me a small function that would generate a random > password > > using this character set? > > chars = { "...your valid character set... " } > for t=1 to passwordLength do > password &= chars[rand(length(chars))] > end for > > Something like that... > ck > > > ===== Regards Brian Keene
5. Re: Math Problem - Probability of Password Duplication
- Posted by Evan Marshall <evan at net-link.net> Oct 22, 2001
- 351 views
6. Re: Math Problem - Probability of Password Duplication
- Posted by euphoria at carlw.legend.uk.com Oct 22, 2001
- 356 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
7. Re: Math Problem - Probability of Password Duplication
- Posted by "C. K. Lester" <cklester at yahoo.com> Oct 22, 2001
- 363 views
> The only question my boss wants to know is how many possible unique passwords can > this generate? Thomas answered this one thusly, and I think it's correct: > As far as I can see it would be: > result = power(89,11)+power(89,12)+power(89,13)+power(89,14) That comes to 1,978,642,898,134,390,000,000,000,000, according to MS Excel.
8. Re: Math Problem - Probability of Password Duplication
- Posted by euphoria at carlw.legend.uk.com Oct 22, 2001
- 374 views
CK Lester wrote: > > As far as I can see it would be: > > result = power(89,11)+power(89,12)+power(89,13)+power(89,14) > > That comes to 1,978,642,898,134,390,000,000,000,000, according to MS Excel. bc under linux says 1,978,642,898,134,388,772,961,847,220, but Euphoria only matches Excel in this regard. Carl
9. Re: Math Problem - Probability of Password Duplication
- Posted by "C. K. Lester" <cklester at yahoo.com> Oct 22, 2001
- 374 views
> > > As far as I can see it would be: > > > result = power(89,11)+power(89,12)+power(89,13)+power(89,14) > > > > That comes to 1,978,642,898,134,390,000,000,000,000, according to MS > Excel. > > bc under linux says 1,978,642,898,134,388,772,961,847,220, but Euphoria only > matches Excel in this regard. I'm not surprised. Windows and Excel are the Behemoth's products... I just did a test using 89^11 with the standard Windows calculator, and it comes out more precise than Excel!! 2,775,173,073,766,990,340,489 vs. 2,775,173,073,766,990,000,000 MS = Massive Stupidity? MS = Mathematical Simpletons