In a code with distance $d$, the maximum number of correctable errors is $\lfloor \frac{1}{2}(d-1) \rfloor$.
But apparently, the maximum number of correctable erasures is $d-1$. How come? Isn't an erasure essentially also just an error?
By constructing some examples, I can see that this is true. But is there a more formal way, like a proof-sketch or even proof to show it is true in general?
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I was thinking about the problem a bit more and came up with an answer to explain that "d-1 erasures are correctable" is generally true. As it might help others with the same question, I would like to share my thoughts here.
Suppose the receiver receives a codeword. If it has errors then the location of the errors are unknown. Hence, the location of uncorrupted/correct bits are also unknown, and thus cannot be directly used to recover the correct codeword. Instead, using minimum distance decoding, the codeword with the closest distance to the errorneous codeword is chosen. Hence errors up to $\lfloor \frac{1}{2}(d-1) \rfloor$ are correctable.
If the codeword contains erasures however, the location of erasures are known. Assuming the rest of the bits are correct, the receiver knows which bits are uncorrupted and can be directly used to recover the correct codeword. Now, given this - within a codeword with distance d, each two codewords differ in at least d bits, i.e. there are at least d bits in each codeword that distinguishes it from all other codewords from the code. Hence, if the received codewords has d-1 erasures, i.e. d-1 unknown bits, there is still at least 1 bit in that word to distinguish it from all codewords in the code except for the one, which it is supposed to represent. Hence, any codeword with d-1 erasures can be recovered to the correct codeword, and thus, up to d-1 erasures are correctable.
The decoder needs to fill in $d-1$ (known) coordinates in order to recover the transmitted codeword. Suppose that there are two different possible ways of filling in these $d-1$ coordinates and these give us two different codewords $c$ and $\hat{c}$. Can you show that the Hamming distance between $c$ and $\hat{c}$ is $d-1$ or smaller, which contradicts the assumption that any two distinct codewords must differ in at least $d$ places? Note that this argument cannot be applied to the correction of errors since the assumption is that the locations of the errors are unknown.