Bottom-up Computing and Discrete Mathematics by P. P. Martin

By P. P. Martin

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Aq (n, d + 1) ≤ Aq (n, d). Proof: Let C be an optimal (n, M, d + 1)-code, so M = Aq (n, d + 1). Choose x, y ∈ C with d(x, y) = d + 1. Assume x, y differ in k-th digit. Remove x from C and replace it with x′ : x′ = πkyk (x) New code C ′ contains x′ and y and d(x′ , y) = d by construction, so d(C ′ ) ≤ d. Let z, w ∈ C ′ . If neither is x′ then z, w ∈ C so d(z, w) ≥ d + 1 > d. If z = x′ (say) then d + 1 ≤ d(x, w) ≤ d(x, x′ ) + d(x′ , w) = 1 + d(z, w) so d(z, w) ≥ d. Thus d(C ′ ) ≥ d, so C ′ ∈ (n, M, d)-cod, so Aq (n, d) ≥ M.

59. [Lagrange] (a) The cosets of a linear code C ⊂ Fqn partition Fqn . (b) Each coset has size q k . Proof: (Idea) Think of C as a subspace (such as a plane in R3 through the origin). We can think of a as shifting this subspace parallel-ly away from the plane; in other words to a new plane not including the origin. We dont have R3 , but the same idea works. 60. Let C be 2-ary generated by G= 1011 0101 That is C = {0000, 1011, 0101, 1110}. Cosets: 0000 + C = C 1000 + C = {1000, 0011, 1101, 0110} = 0011 + C and so on.

The Hamming distance is a metric. Prove it! 9. The minimum distance of a code C ⊂ S n is d(C) = min{d(x, y)|x, y ∈ C, x = y} Examples: C1 00 01 10 11 00 1 1 2 2 1 01 10 1 11 so that d(C) = 1 in case (1). Similarly in case (2) above the min distance is 2; and in case (3) it is 3. 10. (a) If d(C) ≥ t + 1 then C can detect up to t errors; (b) If d(C) ≥ 2t + 1 then C can correct up to t errors by the ‘pick the closest’ strategy. Proof: Exercise, or see below. 11. For any x ∈ S n and r ∈ N the ball of radius r (or r-ball) centred on x is Br (x) := {y ∈ S n |d(x, y) ≤ r} That is, the set of sequences that differ from x in no more than r places.

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