Please assume that no more than two errors have occured during transmission using the [8,4] extended binary Hamming code.
Now, a received word using this code will have $2^{8-4}=16$ possible syndromes:
if no error has occurred, the coset leader corresponding to the syndrome is $00000000$;
if only one error has occurred, the coset leader corresponding to the syndrome is one of the eight possible weight-$1$ error words;
if two errors have occurred, there are $7$ possible syndromes (which will not correspond to any column of the check matrix) and $8\choose2$$=28$ possible weight-$2$ error words; I know that in this case the code will be unable to (uniquely) correct the received word (meaning that each syndrome will not correspond to a unique coset leader), but I'm interested to know how the $28$ weight-$2$ error words are distributed among the $7$ remaining syndromes. Without having to exhaustively construct the table myself, can someone please point me to a syndrome table/dictionary for the [8,4] Hamming code? In particular, I want to know if it is the case that each such syndrome corresponds to four error words, and how many coset leaders correspond to each such syndrome. Many thanks.