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copied!<p>In the absence of "nice" algebraic structure, I suspect that it's going to be hard to find a concise scheme that gets you all the way to 10**4 codewords, since information-theoretically, there isn't a lot of slack. (The one below can use GF(5) for 5**5 = 3125.) Fortunately, the problem is small enough that you could try Shannon's greedy code-construction method (find a codeword that doesn't conflict with one already chosen, add it to the set).</p> <hr> <p>Encode up to 35 bits as a quartic polynomial f over GF(128). Evaluate the polynomial at eight predetermined points x0,...,x7 and encode as 0f(x0) 1f(x1) 0f(x2) 1f(x3) 0f(x4) 1f(x5) 0f(x6) 1f(x7), where the alternating zeros and ones are stored in the MSB.</p> <p>When decoding, first look at the MSBs. If the MSB doesn't match the index mod 2, then that byte is corrupt and/or it's been shifted left by a deletion. Assume it's good and shift it back to the right (possibly accumulating multiple different possible values at a point). Now we have at least seven evaluations of a quartic polynomial f at known points, of which at most one is corrupt. We can now try all possibilities for the corruption.</p> <p>EDIT: bmm6o has advanced the claim that the second part of my solution is incorrect. I disagree.</p> <p>Let's review the possibilities for the case where the MSBs are 0101101. Suppose X is the array of bytes sent and Y is the array of bytes received. On one hand, Y[0], Y[1], Y[2], Y[3] have correct MSBs and are presumed to be X[0], X[1], X[2], X[3]. On the other hand, Y[4], Y[5], Y[6] have incorrect MSBs and are presumed to be X[5], X[6], X[7].</p> <p>If X[4] is dropped, then we have seven correct evaluations of f.</p> <p>If X[3] is dropped and X[4] is corrupted, then we have an incorrect evaluation at 3, and six correct evaluations.</p> <p>If X[5] is dropped and X[4] is corrupted, then we have an incorrect evaluation at 5, and six correct evaluations.</p> <p>There are more possibilities besides these, but we never have fewer than six correct evaluations, which suffices to recover f.</p>

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