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Developing a double-error correcting Code
GF(16)
5
Addition in GF(16):
1
2
3
4
5
6
7
8
9
10
11
12
13
14
0
1
10
4
2
10
3
12
4
5
14
6
7
2
9
8
7
9
13
10
11
12
13
14
3
0
Power Notation Binary Notation
0
1
0010
2
3
4
5
6
1100
7
8
0101
9
10
0111
11
12
13
14
H
1
= [1,
4
2
8
5
10
3
14
9
7
6
13
11
12
]
1
H
H
2
= [1, ,
2
3
4
5
6
7
8
9
10
11
12
13
14
]
2
H
2 12 13 14
3 3 3 2 3 12 3 13 3 14 3
H
2 3 4 5 6 7 8 9 10 11 12 13 14
3 3 6 3 6 9
12
H
3
H
Homework 8
Due April 21, 2009
- Complete the addition in GF(16) table (at least above the main diagonal) on your
class handout.
- Perform the following computations in GF(16):
a) (1001)×(1011) + (0101)÷(1100) b) (1000) ÷ (1101)
c) (1111)
d) (0101)
1/
e) (1000)
1/
f) (1110)
1/
- Using the double-error correcting code defined in class, find the position(s) of any
errors in words whose syndromes are:
a) syn =
0
8
b) syn =
6
3
Info digits:
- Using the parity check matrix H 3 for the double-error correcting BCH code developed
in class, determine the location of any errors in the following received vectors – and
report the corrected word.
a) r 1
= [ 0100 1011 0101 011]
b) r 2
= [ 0110 1111 1001 100]
c) r 3
= [ 1110 0101 1001 100]
d) r 4
= [ 1110 1000 1111 0 10]
e) r 5
= [ 110 0 1101 0101 10 1]
f) r 6
= [ 11 01 1100 1101 000]
- Find a quadratic polynomial with coefficients from GF(16) that is irreducible.
Examples:
- Recall that the Hamming (15,11) code has a 4×15 parity check matrix while the
double-error correcting BCH code has an 8×15 parity check matrix. This tells us that
both codes have length 15.
a) For the BCH code, what is k (the number of information digits)?
b) How many codewords are there in this BCH code?
- Let a x ( ) be a binary polynomial. Prove that
2 2
( ( )) a x a x ( ).
Hint: Think Induction on the degree of the polynomial.
Polynomials of degree zero are 0, and 1.
Polynomials of degree 1 have the form 0
a x ( ) x a
...
