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By John C. Bowman

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Exercise: Show that the ternary Golay [11, 6] code generated by the first 11 columns of the generator matrix   1 0 0 0 0 0 0 1 1 1 1 1 0 1 0 0 0 0 1 0 1 2 2 1   0 0 1 0 0 0 1 1 0 1 2 2  G12 =  0 0 0 1 0 0 1 2 1 0 1 2   0 0 0 0 1 0 1 2 2 1 0 1 0 0 0 0 0 1 1 1 2 2 1 0 has minimum distance 5. 2 (Nonexistence of (90, 278 , 5) codes) There exist no (90, 278 , 5) codes. Proof: Suppose that a binary (90, 278 , 5) code C exists. 2, without loss of generality we may assume that 0 ∈ C. Let Y be the set of vectors in F290 of weight 3 that begin with two ones.

This is important, since to make computing the secret key d (from knowledge of n and e alone) difficult, d must be chosen to be about as large as n. Instead of computing m = cd directly, we first compute a = cd (mod p) and b = cd (mod q). This is very easy since Fermat’s Little Theorem says that cp−1 = 1 (mod p), so these definitions reduce to a = cd mod(p−1) (mod p), b = cd mod(q−1) (mod q). The Chinese Remainder Theorem then guarantees that the system of linear congruences m = a (mod p), m = b (mod q) has exactly one solution in {0, 1, .

Remark: The fact that all elements in a field Fq can be expressed as powers of a primitive element can be exploited whenever we wish to multiply two elements together. We can compute the product αi αj simply by determining which element can be expressed as α raised to the power (i + j) mod(q − 1), in exactly the same manner as one uses a table of logarithms to perform multiplication. Remark: The primitive element of a finite field Fq need not be unique. 4 that the number of such elements is ϕ(q − 1).

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