Algebraic Coding Theory

1
Chapter 11

• Algebraic Coding Theory

2
Single Error Detection

• M (1, 1, , 1) is the m ? 1 parity check matrix
  for single error detection. If c (0, 1, 0, ,
  1) is an n-bit codeword, then McT 0 (use XOR
  addition). If c' c e contains an error, then
  the syndrome is
• M c'T M(c e)T McT MeT MeT 1
• This detects errors 0 for none, 1 for an error.

11.4
3
Single Error Correction / Double Error Detection
1 1 1 1 1 1 1 1 0 0 0 0 1 1 1 1 0 0 1 1 0 0 1 1 0
1 0 1 0 1 0 1
for double-error detection (optional)
the 3 x 7 parity check matrix for a single-error
correcting Hamming code
M'
If c is a 7 bit code word, then McT (0 0 0)T
0 and Mc'T MeT the syndrome for the error
(e.g. try 0110011 ? 0100011). In general, we can
choose any parity check matrix for M, provided
the rows are linearly independent (otherwise the
checks themselves are redundant), and McT 0.
But, to correct a double-error, we must have the
property that the syndrome M(e1 e2)T Me1T
Me2T is unique for every pair of columns
(assuming e1 and e2 are single-error vectors).
11.2
4
Polynomials
degree
coefficients
leading coefficient

• ?2x bnxn b0 n 0 bi 0, 1 bn
  1
• Associate with each n-bit vector the
  corresponding polynomial. Like a generating
  function. Arithmetric operations (, ) apply to
  ?2x, provided we do arithmetic on the
  coefficients in ?2 (mod 2).

unless n 0
A polynomial P(x) is prime (or irreducible) if it
cannot be factored (over ?2x) into lower-order
polynomials. Examples x, x 1, x2 x 1, x3
x 1, x3 x2 1 all prime composite x2, x2
1 (x 1)2, x2 x, x3, x3 1, x3 x2 x 1
(x 1)3.
Primitive roots e2pi/n generates all roots of xn
- 1 over ?. If n is prime, any root (except 1)
will work.
11.5, 11.6, 11.7
5
Polynomial arithmetic

• Consider arithmetic in ?2x / p(x) modulo an
  irreducible polynomial p(x) of degree n. Since
  xn p(x) - xn (mod p(x)) there are no
• polynomials of degree n. In fact, ?2x / p(x)
  has 2n elements bn-1xn-1 b0, which form a
  field, and hence there exists a polynomial g(x)
  whose powers generate all the non-zero elements.
• E.g. The powers of g(x) x mod p(x) x3 x 1
  are
• 1, x, x2, x 1, x2 x, x2 x 1, x2 1.
  Writing these as column vectors of a matrix gives
  a rearranged Hamming code
• n 3
• 2n 8 ?2x / p(x)
• 7 8 - 1 ( non-zero)

x2 x1 x0
11.8
6
Polynomial Coding

• A sent codeword should have McT 0, which means
  that c(x), as a polynomial, is evenly divisible
  by p(x), the modulus polynomial. Why?
• Because c(x) c6x6    c0 , and McT c(x)
  by definition of matrix multiplication, and since
  the powers of x that make up the columns of M are
  modulo p(x),
• McT 0 ? c(x)  0 (mod p(x)).
• For a received codeword c', Mc'T MeT
  syndrome, and similarly, the corresponding
  polynomial s(x) has c'(x) - s(x) 0 mod p(x) ?
  c'(x)  s(x) and that the column matched, xi
  s(x) is the error location (assuming e contains
  only one error, i.e. s(x) xi).

11.8 end
7

• Encode Place message in colums 6, 5, 4, 3, and
  compute
• b6x6 b5x5 b4x4 b3x3 modulo p(x). Put
  remainder,
• r(x) b2x2 b1x b0, in the last columns.

message
checks
b6 b5 b4 b3 b2 b1 b0
Decode Divide received polynomial by p(x), and
use the remainder r(x) gi (x) to locate error
position i). no error ? i doesnt exist.
Example 1 0 0 1 ? x6 x3
? /\ /\ 1 0 0 1 1 1 0 ? x2 1 x
1 x2 x
11.8
8
Double error-correcting Code

• An illustration start with a 15 bit Hamming
  code having 4 parity checks.
• Verify that x4 x 1 is prime, and that the
  primitive x generates ?2x / (x4 x 1) \ 0.
• The resulting parity check matrix M1 is single
  error-correcting, but x  x2 x12 x14 so
  double error-correction is impossible.
• Now, pick another primitive generator x3 of the
  same ?2x / (x4  x  1), with resulting matrix
  M2.
• Let s1 xi xj M1 c'T be the first syndrome
  and s2  (x3)i  (x3)j  M2 c'T be the second
  syndrome.
• Then s2 (xi)3 (xj)3 (xi xj)((xi)2
  (xi)(xj) (xj)2) s1xixj s12
• So that xi xj s1 and xixj s12 s2/s1

11.11