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- Xref: sparky sci.crypt:2764 comp.security.misc:855
- Newsgroups: sci.crypt,comp.security.misc
- Path: sparky!uunet!gatech!destroyer!ubc-cs!newsserver.sfu.ca!erick
- From: erick@fraser.sfu.ca (Erick Bryce Wong)
- Subject: Re: choosing passwords (was: Re: Crypt should be based on MD5)
- Message-ID: <1992Jul28.200252.5570@sfu.ca>
- Sender: news@sfu.ca
- Organization: Simon Fraser University, Burnaby, B.C., Canada
- References: <1992Jul26.202154.15664@sfu.ca> <7memoB5w165w@elrond.toppoint.de>
- Date: Tue, 28 Jul 1992 20:02:52 GMT
- Lines: 19
-
- freitag@elrond.toppoint.de (Claus Schoenleber) writes:
- >erick@fraser.sfu.ca (Erick Bryce Wong) writes:
- >
- >I don't think so. Here you have to write (11 - i) if i=1 is the leftmost
- >position, because the first digit has to be multiplied with 10, the following
- >with decrement 1, etc. Then the sum has to be subtracted from the next greater
- >number dividable by 11. Your formula stops at the same point as my first
- >(and wrong) one.
-
- But, don't the two operations of subtracting i from 11 and of subtracting the
- total from a multiple of 11 cancel out? (Two wrongs make a right? :-)
-
- >There are several ways to write it down, of course. It was meant as an
-
- Quite true. I just figured that simplifying it would make it look, well,
- um, er, simpler... ;)
-
- --
- -- Erick, the perfect square :-)
-