The Hadamard multiple-error correcting codes
The (32, 6) 7-error correcting Hadamard code
The Hadamard (32, 6) code was used by NASA for the Mariner
spacecraft in 1969 as it sent pictures
back to the Earth from Mars. The generator matrix is a 32 32
matrix H with entries in {1, 1}.Numbering the rows and columns from
0 to 31, represent i and j in binary as i = a4a3a2a1a0 and
j = b4b3b2b1b0. Then the (i, j) entry Hij in H is given by
Hij = (1)a4b4 + a3b3 + a2b2 + a1b1 + a0b0 .
Thirty-two of the codewords are the rows of H, and the other 32
are the rows of H. A receivedmessage m (a 32-long string of 1s and
1s) can be decoded as follows. Take the dot product of mwith each
row of H. If the message is correct, it will have dot product 0
with all rows but one, and
with that row it will have dot product 32 or 32. If the dot
product is 32 (respectively, 32), thecodeword is that row of H
(respectively, of H). If the message has one error, all dot
products arebetween 2 and 2, except for one, which will be 30 or
30, and that row is the correct codewordor its negative.
This continues as follows:
Number of errors Range for all dot products but one Range for
the right row
2 4 to 4 28 to 32 or 28 to 323 6 to 6 26 to 32 or 26 to 324 8 to
8 24 to 32 or 24 to 325 10 to 10 22 to 32 or 22 to 326 12 to 12 20
to 32 or 20 to 327 14 to 14 18 to 32 or 18 to 32
With eight or more errors, the ranges overlap, and so correction
is not possible. However, this
code has minimum Hamming distance 16, so that one can detect up
to 15 errors.
The (16, 5) 3-error correcting Hadamard code
In this section, we describe the encryption and decryption
procedures for the (16, 5) Hadamard
code. The message words are the 32 five-bit strings that are the
binary representations of the
integers from 0 through 31. There is an encoding (or generating)
4 16 matrix G whose entriesare in {0, 1}, and a parity-check (i.e.
decoding) matrix P whose entries are in {1,1}.
The generating matrix has four rows and 16 columns, and the
columns are numbered 0, 1, . . . , 14, 15.
For 0 j 15, write j = 8j3 + 4j2 + 2j1 + j0; the jth column is
the transpose of the row vector(j3, j2, j1, j0). Thus,
G =
0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 10 0 0 0 1 1 1 1 0 0 0 0 1 1 1 10
0 1 1 0 0 1 1 0 0 1 1 0 0 1 10 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1
Let x4x3x2x1x0 be the five-bit message word. To encode x, do the
following:
1. Remove the high-order bit, and form the vector x = (x3, x2,
x1, x0).
2. Form the 16-long vector y = x G = (y1, . . . , y16) mod 2
3. For i = 1, . . . , 16, set zi = (1)yi .
4. Set z = (z1, . . . , z16) or z = (z1, . . . ,z16) according
as x4 = 0 or x4 = 1.
5. z is the encoding of x.
P is the 16 16 matrix whose jth row is the encoding of j 1, for
j = 1, . . . , 16; it looks like this:
P =
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11
1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 1
1 1 1 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1
1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1
1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1
1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1
11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1
1 1 1 1 1 1 1 1 1 1 1 1 1 1
Sixteen of the codewords are the rows of P ; the other sixteen
are the rows of P , correspondingto the encoding of the numbers 16,
. . . , 31. To decode a 16-long string w of 1s and 1s, computethe
product w P . As it was for the (64, 6) Hadamard code, the
following table tells the tale:
Number of errors Range for all dot products but one Range for
the right row
0 0 16 or 161 2 to 2 14 to 16 or 14 to 162 4 to 4 12 to 16 or 12
to 163 6 to 6 10 to 16 or 10 to 16
As before, the right row is the one that contains either the
codeword (corresponding to a trans-
mitted number from 0 to 15) or its negative (corresponding to a
transmitted number from 16 to
31). With four or more errors, the ranges overlap, and so
correction is not possible. However, this
code has minimum Hamming distance 8, so that one can detect up
to 7 errors.
As an example, the 5-bit string 01011, corresponding to the
integer 11, encodes as the 16-long
(1, 1) vector e = (1,1,1, 1, 1,1,1, 1,1, 1, 1,1,1, 1, 1,1). Let
f = e with the second 1changed to 1, let g = e with the first two
1s changed to 1s, and let h = 3 with the first three1s changed to
1s. Then
e P = (0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 16, 0, 0, 0, 0),f P =
(2, 2, 2,2,2, 2, 2,2,2, 2, 2, 14,2, 2, 2,2),g P = (4, 0, 0,4,4, 0,
0,4,4, 0, 0, 12,4, 0, 0,4), andh P = (6,2,2,6,2, 2, 2,2,6,2,2,
10,2, 2, 2,2).
In each case (with 0, 1, 2 or 3 errors), the outlier is the 12th
entry, corresponding to the number
11. However, if q = e with the first four 1s to 1s, we see
that
q P = (8, 0, 0,8, 0, 0, 0, 0,8, 0, 0, 8, 0, 0, 0, 0),
and we have no outliers. That is because q is also the vector
consisting of all 1s with four 1schanged to 1s, and so q is at a
Hamming distance of four from at least two different codewords.
If A is a matrix, denote its transpose by At. The matrix P has
the property that P P t = 16I,where I denotes the identity matrix
of order 16. Such matrices are called Hadamard matrices, and
have connections with many other areas of mathematics including
cryptography.
