I want to transmit a 32bit message in eight groups of 5bit each. This leaves me with 8bits to use for error checking.

Overall, a group is likely transmitted without error, but when there is an error transmitting a group, there are probably multiple bits wrong.

If I use one parity bit per group, I have a 50% chance to detect a wrong group. But I don't need to know which group of a message is wrong, I want to check the entire message.

I want 100% chance of detecting if one group of the message is incorrect, regardless of how many bits are flipped in that group. If possible, I also want to be able to check wheather two neighboring groups have been switched.

Which algorithm should I use for error checking / How should I code the data?

  • $\begingroup$ Can you make your error model more precise? What guarantee should your code have? $\endgroup$ – Yuval Filmus Sep 5 '19 at 11:51
  • $\begingroup$ @YuvalFilmus I added a bit more detail, although I'm not sure how much error checking I can expect $\endgroup$ – Cephalopod Sep 5 '19 at 20:16
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    $\begingroup$ I'd look for a code, not an algorithm. But for two neighboring groups have been switched, this looks like the standard case for a block code. Two groups switched can be viewed as a burst error of twice the length of a group (eight or ten bits, depending how you look at it) - $n$-bit burst error-correcting code with $n$ check bits doesn't look promising. $\endgroup$ – greybeard Sep 5 '19 at 21:03
  • $\begingroup$ Given the errors you're interested in, it might be better to think of your message as 8 symbols from an alphabet of size 32. It is a pity that 5 doesn't divide 32 – otherwise you could have used a linear code over a field of size 32. $\endgroup$ – Yuval Filmus Sep 5 '19 at 21:07
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    $\begingroup$ The fact that the numbers just agree leads me to wonder where this question came from. $\endgroup$ – Yuval Filmus Sep 5 '19 at 21:12

Think of your message as seven 5-bit numbers $x_2,\ldots,x_8 \in \{0,\ldots,30\}$. This gives you $7\log_2 31 \approx 34.68$ message bits. Calculate $x_1 = -\sum_{i=2}^8 ix_i \bmod{31}$, so that $$ \sum_{i=1}^8 ix_i \equiv 0 \pmod{31}. $$

It is easy to check that every single symbol error is detected, since $31$ is prime.

Now suppose that $x_i$ and $x_j$ are switched, but the equation above still holds. Then $ix_i + jx_j \equiv ix_j + jx_i$, and so $(i-j)(x_i - x_j) \equiv 0$, implying that either $i = j$ or $x_i = x_j$.

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