A Note on Hamming Code
The Hamming code is a powerful error correcting code. It enables us to detect errors and to
recover the original binary word if one digit goes wrong.
Let s= (s1, s2, s3, s4) be a 4-long binary word (i.e., every sjis either 0 or 1). The Hamming
code,H(s), of sis a 7-long word defined as follows:
H(s)1=H(s)3+H(s)5+H(s)7(mod 2) = s1+s2+s4(mod 2)
H(s)2=H(s)3+H(s)6+H(s)7(mod 2) = s1+s3+s4(mod 2)
H(s)3=s1
H(s)4=H(s)5+H(s)6+H(s)7(mod 2) = s2+s3+s4(mod 2)
H(s)5=s2
H(s)6=s3
H(s)7=s4
For example, if s= (1,0,0,0), then H(s) = (1,1,1,0,0,0,0). Indeed,
H(s)1=H(s)3+H(s)5+H(s)7(mod 2) = s1+s2+s4(mod 2) = 1
H(s)2=H(s)3+H(s)6+H(s)7(mod 2) = s1+s3+s4(mod 2) = 1
H(s)4=H(s)5+H(s)6+H(s)7(mod 2) = s2+s3+s4(mod 2) = 0.
In general, we will refer to the bits coming from the original binary word sas data bits, and the
rest as parity bits. In the above example, the parity bits are the first two occurrences of 1 and the
first occurrence of 0.
Let tbe a 7-long binary word such that
•there is no 4-long binary word sfor which t=H(s),
•if we change a certain bit in t, then it becomes the Hamming code for some 4-long binary
word u.
We claim that ucan be recovered from t.
By assumption there is precisely one incorrect bit in t, but we do not know which one. Consider
the following algorithm. Let the sequence vconsist of the data bits of tand its Hamming code be
H(v). There are two cases.
Case 1: one data bit tiis incorrect. Then some of the parity bits will be different in tand in
H(v). Looking at the definition of the parity bits we can figure out which data bit tiis incorrect.
Note also that there are more than one parity bits in tand H(v) which disagree.
Case 2: one parity bit tiis incorrect. Then all the other parity bits are correct. Thus changing
tiin tyields a binary word such that it is the Hamming code of v.
Finally note that cases 1 and 2 can be distinguished by the number of parity bits that differ
in tand in H(v). Thus we know which one of the cases apply.
As an example let us look at the binary word t= (1,1,1,0,0,1,0). Then v= (1,0,1,0)
and H(v) = (1,0,1,1,0,1,0). Hence the disagreeing parity bits are t2= 1 6=0=H(v)2and
t4= 0 6=1=H(v)4. Thus case 1 above applies and we conclude that we should change t2+4 =t6.
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