Chapter 2. Parity checks Modular arithmetic Weighted codes. Simple Block Code Parity Checks. Idea: Break a string of bits into blocks of size ( n 1) and append (or prepend) an additional bit for even or odd parity (even # of 1’s, odd # of 1’s), to obtain n bit blocks. parity bit.
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Idea: Break a string of bits into blocks of size (n 1) and append (or prepend) an additional bit for even or odd parity (even # of 1’s, odd # of 1’s), to obtain n bit blocks.
Other codes that admit parity checks: 2-out-of-5, 3-out-of-7 (van Duuren code)
Assume independent error probability p for each bit.
The probability of no error = (1 – p)n.
The probability of one error = np(1 – p)n-1.
Assume noise comes in “bursts” of length ≤ L, and is otherwise independently distributed.
Instead of computing a parity check over contiguous sequences of bit positions, use a “checksum” over words of length L.
Mod 2 addition
= logical AND
= logical XOR
The base can be any number (usually a prime) and the rules are similar to those for base 2.
In mod (modulo) 2 arithmetic, 2 is the base (modulus) and there are no numbers other than 0 and 1. Any higher number mod 2 is obtained by dividing it by 2 and taking the remainder. For instance, 3 ≡ 1 mod 2 and 4 ≡ 0 mod 2.
Mod 5 addition
Mod 5 multiplication
Fact: If a≡ma′ and b ≡mb′, then a + b ≡ma′ + b′ and a ∙ b ≡ma′ ∙ b′.
Proof: a≡ma′ and b ≡mb′ means that a = a′ + im and b = b′ + jm, so a + b = a′ + b′ + (i + j)m and a ∙ b = a′ ∙ b′ + a′jm + b′im + ijm2. QED
Multiplicative inverses may not exist for some numbers.
Example: 2 × 5 ≡ 0 mod 10. Does 2 have a multiplicative inverse? Suppose it does, then 2 × 2−1 ≡ 1 mod 10. However, multiplying both sides by 5 yields 0 ≡ 5 mod 10, which is false.
Note: If the modulus is a prime, p, then numbers not congruent to p can’t multiply together to be congruent to p, so you don’t have the problem in the above example. Here is a proof that multiplicative inverses (for nonzero) numbers always exist in a prime modulus, p: Let q , 0 < q < p, then GCD(p, q) = 1. (They are relatively prime). By the Euclidean Algorithm, we can always find k, l such that kp + lq = 1, lq≡ 1 mod p l = q−1.
Deals with transcription errors (i.e. human non-binary error):
Example: Assign numerical values 1 … 37 to the symbols consisting of letters, digits, and a space. Take a message of length less than 37 and weight each symbol therein according to its position in the message, appending a final check symbol s1 chosen so the weighted sum is congruent to zero:
For n < 37,
sn… … s1
transposing adjacent digits [e.g. 67 → 76]: ksk + (k+1)sk+1 becomes (k+1)sk + ksk+1 whose difference is sk+1− sk ≢ 0 (mod 37)
repeating an adjacent digit [e.g. 667 → 677]: difference is
ksk − ksk+1 = k(sk − sk+1) ≢ 0 provided 0 < k < 37 and sk ≠ sk+1
10 digit weighted code (mod 11) where X stands for ten in the check (last) digit