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It might just be that I don't fully understand parity bits, but would it make sense to do an even parity for even values and an odd parity for odd values to add an extra level of redundancy without adding extra bits? Or would that negate the parity.

user3727843
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  • Whether or not a value is odd/even is determined by its least-significant bit; the other bits play no role in it. Beyond this, it's really hard to understand what you're asking. Perhaps you could provide an example to illustrate? – NPE Sep 08 '14 at 19:53
  • The [Wikipedia entry](http://en.wikipedia.org/wiki/Parity_bit). Talking about having even/odd parity with extra bits seem pointless. – Alcanzar Sep 08 '14 at 20:02
  • How would you add a parity bit without adding an extra bit for the parity? – Elliott Frisch Sep 08 '14 at 20:03

1 Answers1

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Figured it out

Code is 1101 > 1101-1 with parity Read as 1100-1 considered incorrect > Code 1100 should be 1100-0. Assuming all bits are present then we know the code should be 1101.

Code is 1100 > 1100-0 with parity Read as 1101-0, considered correct.

So I guess it does negate the parity.

user3727843
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