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Power decoding Reed-Solomon codes up to the Johnson radius

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  • Power decoding, or "decoding using virtual interleaving" is a technique for decoding Reed-Solomon codes up to the Sudan radius. Since the method's inception, it has been an open question if it is possible to use this approach to decode up to the Johnson radius - the decoding radius of the Guruswami-Sudan algorithm. In this paper we show that this can be done by incorporating a notion of multiplicities. As the original Power decoding, the proposed algorithm is a one-pass algorithm: decoding follows immediately from solving a shift-register type equation, which we show can be done in quasi-linear time. It is a "partial bounded-distance decoding algorithm" since it will fail to return a codeword for a few error patterns within its decoding radius; we investigate its failure behaviour theoretically as well as give simulation results.

    Mathematics Subject Classification: Primary: 94B35, 11T71; Secondary: 41A28, 41A21, 94A55.


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  • Table 1.  Simulation results. $P_f(\tau)$ denotes the observed probability of decoding failure (no result or wrong result) with random errors of weight exactly $\tau$. $\tau_{\rm bnd}$ indicates the number of errors ${\epsilon}$ for which Proposition 5 yields a bound $< 1$ (where applicable); in parentheses is if the probability estimate of (7) is used instead.

    $[n,k]_q$ $(s,\ell)$ $\tau_{\text{Pow}}$ ${{P}_{f}}(\left\lfloor {{\tau }_{\text{Pow}}} \right\rfloor -1)$ $P_f(\left\lfloor{{\tau_{\text{Pow}}}} \right\rfloor)$ $P_f(\left\lfloor{{\tau_{\text{Pow}}}} \right\rfloor + 1)$ $\tau_{\rm bnd}$
    $[21, 3]_{23}$ $(6,19)$ $14\,{}^{1}\!\!\diagup\!\!{}_{20}\;$ $ 7.43 \times 10^{-3} $ $ 1.97 \times 10^{-1} $ $ 1 $
    $[24, 7]_{25}$ $(2,3)$ $10\,{}^{1}\!\!\diagup\!\!{}_{4}\;$ $0$ $ 2.27 \times 10^{-3} $ $ 1 $ 8 (9)
    $[32, 10]_{37}$ $(2,4)$ $13$ $0$ $ 2.78 \times 10^{-2} $ $ 1 $
    $[64, 27]_{64}$ $(2,3)$ $20\,{}^{1}\!\!\diagup\!\!{}_{4}\;$ $0$ $ 3.10 \times 10^{-4} $ $ 1 $ 19 (19)
    $[68, 31]_{71}$ $(3,4)$ $20$ $0$ $0$ $ 1$
    $[125, 51]_{125}$ $(4,6)$ $42$ $0$ $0$ $ 1$
    $[256, 63]_{256}$ $(2,4)$ $116\,{}^{2}\!\!\diagup\!\!{}_{5}\;$ $0$ $0$ $ 1- 3.00 \times 10^{-4} $
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