suppose , codeword is in format :- c0d0c1d1c2
now , if d0 = d1 = 0 then codeword will be 00000
and if d0 = 1 , d1 = 0 then codeword will be 11001
so , minimum distance will be 3 b/w these codewords.
since , for "d" bit error detection , minimum distance >= d+1 ...so d = 2
and for "d" bit error correction , minimum distance >= 2d+1 ...so d = 1
now , we can take any codeword format and put any combination of d0 and d1 like 00,01,10,11 and find the 4 possible codeword for each codeword format..minimum distance will be 3...
because when we put d0 and d1 as 00 then codeword = 00000 for each codeword format. now we can't get any codeword in the same format when hamming distance will be 2 because when we put 01 as data bits then we get check bits as 011 . so we always get one extra 1 due to exor. like this we can analyze how we are not getting minimum distance as 2 for any codeword format..