By N.J.A. Sloane

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**Additional info for A Short Course on Error Correcting Codes**

**Example text**

U. I i =1 . S1 + bJ. wt(B1 s1 ) ~ 3, = BT ST • , ••• ,e 1 ~ = 1, so for some j, 1 ~j ~ 12, Thus the decoding can be done by computing the weights of the 26 vectors (1 ~ j ~12), B1 S 1 , B1 S1 + bj (1 ~ j ~ 12). Example: S=11100 ... 0, wt(S) = 3, socaseI applies. S1 , S = S1 +bj u 1 + u2 + u 3 , so e = 11100 ••• 0. (3. 6. 2) Decoding Metbad II (Goethals [ 6 7 ] ) This is a clever threshold decoding scheme, using the properties of the designs associated with the code. 4 ), any set of codewords can be used as parity checks for the code.

Use Gleason's theorem to obtain the weight distribution of the code. 4) Example 3 Let ~ be a self-dual code of length n = 8j in which all weights are divisible by 4. What is its greatest possible minimum distance? By Th. 1) (ii), with m = [n/24], w= E · -3k k m e2 , ~c;2 k=O which contains m + 1 parameters ~. 5) § § 4 The Theorems of MacWilliams and Gleason 54 which corresponds to a hypothetical code with minimum distance at least 4m + 4. Of course if it should turn out that A 4m+ 4 =0, the minimum distance would be greater than 4m + 4.

1) Example 1 A self-dual code oflength 12 contains no codewords ofweight 2. What is its weight enumerator W? By Th. 2) Example 2 Is there a self-dual code oflength 32 with"minimum distance 10? By Th. 1) (i), its weight enumerator W has the form Equating coefficients we find that a1 , ... ,a 5 are uniquely determined and that W = x32 + 4960x22 yto - 34 72x'1D y12 + Since a weight enumerator cannot have a negative coefficient, no such code exists. 3) Exercise Take all the codewords in the [24,12,8] extended Golay code which begins either with 00...