Date: Fri, 03 Mar 2000 09:07:19 -0800 From: Eric Rescorla <ekr@rtfm.com> To: Kuzak <kuzak@kuzak.net> Cc: freebsd-stable@FreeBSD.ORG Subject: Re: Password Length Message-ID: <200003031707.JAA37304@romeo.rtfm.com> In-Reply-To: Your message of "Fri, 03 Mar 2000 08:49:16 PST." <200003031700.e23H0Wp41410@alpha.dgweb.com>
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> >Passwords are ASCII, so the total number of 8-byte 8^8=2^64. > > > >It should be obvious by inspection that 8^8 << 73! > >Incidentally, the number of atoms in a glass of water > >is on the order of 10^25 >> 8^8. > I see it more like this... > > 64^8... Assume 64 possible characters to choose from... > > for the first char in the passwd you have 64 choices, > for the second char in the passwd you have 64 choices, > and so on 8 times.. > > So by the multiplication principal you will have a total > of 64*64*64*64*64*64*64*64 or 64^8 possible permutations > which is >> than 8^8 Though much less than 73! Correct. I was totally on crack when I wrote 8^8. What I meant was (2^8)^8 == 256^8 == 2^64. Nevertheless, 256 > 10^3, thus 256^8 < (10^3)^8 Since (10^3^8) = 10^24, 10^25 >> 2^64. I stand by the rest of my message. -Ekr To Unsubscribe: send mail to majordomo@FreeBSD.org with "unsubscribe freebsd-stable" in the body of the message
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