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For a Reed-Solomon codeword over GF(2 3 ), there will be seven three-bit symbols. For location and correction of one symbol, there must be two redundant symbols P and Q, leaving A-E for data.
The following expressions must be true, where a is the primitive element of x 3 • x • 1 and • is XOR throughout:
(1) |
(2) |
Dividing equation (2) by a :
a 6 A • a 5 B • a 4 C • a 3 D • a 2 E • a P • Q = 0
= A • B • C • D • E • P • Q
Cancelling Q, and collecting terms:
( a 6 • 1)A • ( a 5 • 1)B • ( a 4 • 1)C • ( a 3 1)D • ( a 2 • 1)E
= ( a • 1)P
Using section 3.16 to calculate ( a n + 1), e.g. a 6 + 1 = 101 + 001 = 100 = a 2 :
a 2 A • a 4 B • a 5 C • a D • a 6 E = a 3 P
a 6 A • a B • a 2 C • a 5 D • a 3 E = P
Multiplying equation (1) by a 2 and equating to equation (2):
a 2 A • a 2 B • a 2 C • a 2 D • a 2 E • a 2 P • a 2 Q = 0
= a 7 A • a 6 B • a 5 C • a 4 D • a 3 E • a 2 P • a Q
Cancelling terms a 2 P and collecting terms (remember a 2 • a 2 = 0):
( a 7 • a 2 )A • ( a 6 • a 2 )B • ( a 5 • a 2 )C • ( a 4 • a 2 )D •
( a 3 • a 2 )E = ( a 2 • a )Q
Adding powers according to section 3.16, e.g.
a 7 • a 2 = 001 • 100 = 101 = a 6 :
a 6 A • B • a 3 C • a D • a 5 E = a 4 Q
a 2 A • a 3 B • a 6 C • a 4 D • a E = Q
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