As we all know, in RS code, when provide erasure (the position of error symbol), the decoding capacity of RS code is stronger. Specifically, $2e+v \leq (n-k)$, where $e$ is the number of errors, $v$ the number of erasures and $(n-k)$ the number of RS code characters (called nsym in the code) [1].

It has been seen in many materials, but there is no analysis about the relationship between the accuracy of erasure and the probability of decoding in RS code. Which means that the provided erasure may be wrong.

Currently, I have split the possibility into three kinds of condition (it may be merged later):

- (1) $E > n-k$;
- (2) $\frac{n-k}2 < E \leq n-k$;
- (3) $E \leq \frac{n-k}2$.

$E$ is the total number of actual errors.

For the condition (1), the probability of successfully decoding is 0.

For the condition (2), because $2e+v = (n-k) = T$, so $e = \frac{T-v}2$, the actual number of error $E$ can be represented as $e+v= \frac{T-v}2 + v = \frac{T+v}2$. Which means that, when provide one erasure, the capacity of block decoding is increased by 1/2, and it is limited by the $2e+v \leq (n-k)$. So, when the correct erasure provided, the number of detected error $e$ is decreased. Because the number of actual errors $E$ is larger than the $\frac{(n-k)}2$, so the probability of successfully decoding block is totally relied on the accuracy of erasure. So, the the probability is easy to get.

**For the condition (3), and this is where I am blocked now.** In this condition, actually, without the provided erasure the RS can decoded successfully 100%. But, as for the erasure provider, it can not know the actual number of errors $E$, thus it always provides the erasure. But I can not analysis the accuracy of provided erasure to the probability of successfully decoding block in this condition.

I have tried some examples that for the RS(n=20,k=12). The $E=4 \leq \frac{(n-k)}2$,

- when I provide 4 erasure (2 correct, 2 wrong) or seven erasure (3 correct , 4 wrong) or eight erasure (4 correct, 4 wrong), RS can still successfully decode the block. But when the any above wrong erasure increases, the RS can not successfully decode the block. And I want to formally analysis this condition like condition (2).

Could you help me to formally analysis this problem? Thanks very much!

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