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- Newsgroups: comp.sys.amiga.hardware
- Path: sparky!uunet!ukma!wupost!usc!elroy.jpl.nasa.gov!bret!ceg
- From: ceg@bret.jpl.nasa.gov (Chuck Goodhart)
- Subject: Re: ECC
- Message-ID: <1992Nov17.215225.16065@elroy.jpl.nasa.gov>
- Sender: news@elroy.jpl.nasa.gov (Usenet)
- Nntp-Posting-Host: hyper-bret.jpl.nasa.gov
- Organization: Jet Propulsion Laboratory; Pasadena, CA
- References: <1992Nov11.001328.10434@elroy.jpl.nasa.gov> <1992Nov17.060933.18436@sserve.cc.adfa.oz.au>
- Date: Tue, 17 Nov 1992 21:52:25 GMT
- Lines: 21
-
- In article <1992Nov17.060933.18436@sserve.cc.adfa.oz.au>
- ajft@ajft_sun.cs.adfa.oz.au (Adrian Tritschler) writes:
- >In article <1992Nov11.001328.10434@elroy.jpl.nasa.gov>,
- >ceg@bret.jpl.nasa.gov (Chuck Goodhart) writes:
- >|> By my calculations, 12 bits are required for 1 bit correct, 2 bit detect
- >|> on 8 data bits. 10 bits will only get you 6 data bits.
- >
- >Close, try 11, (8 + 3). The number of detection/correction bits is log (base 2)
- >#databits. I don't believe it can be done in 10.
-
- Check your references, 12 is the right answer.
-
- If "n" is the number of data bits, and "k" is the number of check bits,
- then, in order to be able to correct single bit errors, you must have:
- k >= log2(n+k+1)
-
- See my recent "Re: Parity" message for the essence of the proof of this.
- --
- Chuck Goodhart, ceg@bret.jpl.nasa.gov
- --
- Chuck Goodhart, ceg@bret.jpl.nasa.gov
-