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Majority logic decoding

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In error detection and correction, majority logic decoding is a method to decode repetition codes, based on the assumption that the largest number of occurrences of a symbol was the transmitted symbol.

Theory

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In a binary alphabet made of 0 , 1 {\displaystyle 0,1} {\displaystyle 0,1}, if a ( n , 1 ) {\displaystyle (n,1)} {\displaystyle (n,1)} repetition code is used, then each input bit is mapped to the code word as a string of n {\displaystyle n} {\displaystyle n}-replicated input bits. Generally n = 2 t + 1 {\displaystyle n=2t+1} {\displaystyle n=2t+1}, an odd number.

The repetition codes can detect up to [ n / 2 ] {\displaystyle [n/2]} {\displaystyle [n/2]} transmission errors. Decoding errors occur when more than these transmission errors occur. Thus, assuming bit-transmission errors are independent, the probability of error for a repetition code is given by P e = k = n + 1 2 n ( n k ) ϵ k ( 1 ϵ ) ( n k ) {\displaystyle P_{e}=\sum _{k={\frac {n+1}{2}}}^{n}{n \choose k}\epsilon ^{k}(1-\epsilon )^{(n-k)}} {\displaystyle P_{e}=\sum _{k={\frac {n+1}{2}}}^{n}{n \choose k}\epsilon ^{k}(1-\epsilon )^{(n-k)}}, where ϵ {\displaystyle \epsilon } {\displaystyle \epsilon } is the error over the transmission channel.

Algorithm

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Assumption: the code word is ( n , 1 ) {\displaystyle (n,1)} {\displaystyle (n,1)}, where n = 2 t + 1 {\displaystyle n=2t+1} {\displaystyle n=2t+1}, an odd number.

  • Calculate the d H {\displaystyle d_{H}} {\displaystyle d_{H}} Hamming weight of the repetition code.
  • if d H t {\displaystyle d_{H}\leq t} {\displaystyle d_{H}\leq t}, decode code word to be all 0's
  • if d H t + 1 {\displaystyle d_{H}\geq t+1} {\displaystyle d_{H}\geq t+1}, decode code word to be all 1's

This algorithm is a boolean function in its own right, the majority function.

Example

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In a ( n , 1 ) {\displaystyle (n,1)} {\displaystyle (n,1)} code, if R=[1 0 1 1 0], then it would be decoded as,

  • n = 5 , t = 2 {\displaystyle n=5,t=2} {\displaystyle n=5,t=2}, d H = 3 {\displaystyle d_{H}=3} {\displaystyle d_{H}=3}, so R'=[1 1 1 1 1]
  • Hence the transmitted message bit was 1.

References

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  1. Rice University, https://web.archive.org/web/20051205194451/http://cnx.rice.edu/content/m0071/latest/

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