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Re: Need XOR solution
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Re: Need XOR solution


  • Subject: Re: Need XOR solution
  • From: "Marc K. Myers" <email@hidden>
  • Date: Wed, 07 Aug 2002 16:13:17 -0400
  • Organization: [very little]

> >> Date: Tue, 06 Aug 2002 16:33:17 -0400
> >> Subject: Re: Need XOR solution
> >> From: "Arthur J. Knapp" <email@hidden>
> >> To: <email@hidden>
> >>
> >> > Date: Mon, 05 Aug 2002 18:45:28 -0400
> >> > From: "Marc K. Myers" <email@hidden>
> >> > Subject: Need XOR solution
> >>
> >> > Does anyone know of a reasonably fast way to XOR blocks of data
> >together?
> >
> >> > ... I would appreciate it
> >> > if someone could point me in the right direction.
> >>
> >> This is unlikely to be the right direction, but you may find
> >> my "manual" XORing to be interesting:
> >
> >[...snip!...]
>
> >I find it more than interesting; I find it mind-boggling. I've printed
> >it out and will attempt to understand it gradually at my leisure. I
> >really like the "Bitify" routine which is much simpler than the one I
> >was playing with to get bytes as bit patterns.
>
> You could use the brilliant "Fast ASCII" coverter from Arthur's script to
> get the number value of each byte, and XOR them together with this:
>
> on Xorify(n1, n2)
> set xor to 0
> repeat 8 times
> if n1 mod 2 is not equal to n2 mod 2 then set xor to xor + 256
> set n1 to n1 div 2
> set n2 to n2 div 2
> set xor to xor div 2
> end repeat
> -- return xor
> end Xorify

That works beautifully, but even after "walking" through the eight
iterations with pencil and paper I still didn't understand *how* it
works. Honestly, I was amazed that I got the right answer after the
eighth pass through the loop! Could you explain it, briefly, for those
of us who are less mathematically inclined?

Marc K. Myers <email@hidden>
http://AppleScriptsToGo.com
4020 W.220th St.
Fairview Park, OH 44126
(440) 331-1074

[8/7/02 4:06:27 PM]
--
Those who lack nonvolatile memory are condemned to iterate.
-- John M. Ford
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