GRS-example - A GRS decoding example Consider the code...

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A GRS decoding example Consider the code GRS 10 , 4 ( α , v ) over F 11 with α = v = (10 , 9 , 8 , 7 , 6 , 5 , 4 , 3 , 2 , 1) , hence (by direct calculation or Problems 5.1.3/5.1.5) we may take u = (1 , 1 , 1 , 1 , 1 , 1 , 1 , 1 , 1 , 1) . Note that here r = 10 - 4 = 6, so we can correct up to r/ 2 = 3 errors. We wish to decode the received word p = (0 , 0 , 8 , 0 , 0 , 2 , 0 , 0 , 7 , 0) . First calculate the syndrome polynomial: S p ( z ) = 10 X j =1 u j · p j 1 - α j z = 1 · 8 1 - 8 z + 1 · 2 1 - 5 z + 1 · 7 1 - 2 z (mod z 6 ) = 8( 1 +8 z +9 z 2 +6 z 3 +4 z 4 +10 z 5 ) +2( 1 +5 z +3 z 2 +4 z 3 +9 z 4 + z 5 ) (mod z 6 ) +7( 1 +2 z +4 z 2 +8 z 3 +5 z 4 +10 z 5 ) = 6 + 0 z + 7 z 2 + 2 z 3 + 8 z 4 + 9 z 5 . We now (partially) calculate gcd( z 6 , 9 z 5 + 8 z 4 + 2 z 3 + 7 z 2 + 6) = 1 over F 11 , the Euclidean Algorithm example discussed in class (and on a handout). In our Euclidean Algorithm example, Step 3, where r 3 ( z ) = 10 z 2 + 5 z + 7, is the first step j for which the degree of r j ( z ) is less than r/ 2 = 3. So, in decoding, we stop at this step. We have t 3 ( z ) = 2 z 3 + 10 z + 3, hence t 3 (0) - 1 = 3 - 1 = 4. We thus set σ ( z ) = 4(2 z 3 +10 z +3) = 8 z 3 +7 z +1 and ω ( z ) = 4(10 z 2 +5 z +7) = 7 z 2 +9 z +6
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