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Subject: Internetworkers: http://www.ibiblio.org/internetworkers/
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- From: Greg Cox <glcox AT pobox.com>
- To: "Internetworkers: http://www.ibiblio.org/internetworkers/" <internetworkers AT lists.ibiblio.org>
- Subject: Re: [internetworkers] Math Problem
- Date: Fri, 25 Feb 2005 23:44:12 -0500 (EST)
> So!, assuming use of the 95 non-control characters of a standard English
> keyboard, what would be the formula to calculate the number of possible
> combinations for a password allowed N characters long (remember the
> password *could* be 1-N characters long in actuality). Is it as simple as
> 95^N??
For large values, you could approximate that. The limit tends towards X^N.
> > X characters, strings of max N length, A = answer.
X=95 characters, strings of max length N.
> > A = (X^(N+1) - X) / (X-1)
A = (95^(N+1) - 95) / 94
Note that 95^N+1, minus a tiiiny amount, divided by almost 95,
works out to about 95^N.
For a standard 8 character max:
A = 6.70478095 × 10^15
95^8 (the rough value) = 6.63420431 × 10^15
-
[internetworkers] Math Problem,
Alan MacHett, 02/25/2005
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Re: [internetworkers] Math Problem,
John Broome, 02/25/2005
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Re: [internetworkers] Math Problem,
Alan MacHett, 02/25/2005
- Re: [internetworkers] Math Problem, Ian Meyer, 02/25/2005
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Re: [internetworkers] Math Problem,
Alan MacHett, 02/25/2005
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Re: [internetworkers] Math Problem,
Greg Cox, 02/25/2005
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Re: [internetworkers] Math Problem,
Alan MacHett, 02/25/2005
- Re: [internetworkers] Math Problem, Greg Cox, 02/25/2005
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Re: [internetworkers] Math Problem,
Alan MacHett, 02/25/2005
- <Possible follow-up(s)>
- Re: [internetworkers] Math Problem, DonBartholf, 02/25/2005
- Re: [internetworkers] Math Problem, DonBartholf, 02/26/2005
-
Re: [internetworkers] Math Problem,
John Broome, 02/25/2005
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