Notes to Chapter 4.

n = 6
d = 3 
l = n - 2d + 1 = 6 - 6 + 1

Restricted code: 

Distance 3, block length 5, Theorem says 2d / (2d-n) = 6




Exercise 2.



Let C be a code und fix 0 <= w <= n. For a point x in {0,1}^n consider the Hamming sphere S_w(x) and in particular |C \cap S_w(x)|. We build a bipartite graph:

- Left side is {0,1}^n, right side is C. 

- Point x on the left side is connected to each y in C with d(x,y) = w.

- Each y in C is connected to every x in S_w(y), so the right degree of the graph is {n choose w}

- The neighborhood of each x in C (call it C_x) gives a distance-d-code of constant Hamming weight w if we xor every codeword with x. So the left degree is at most A(n,d,w).



|Right side| * right-degree = |edges in this graph| <= |left side| * (left side upper bound)

|C| * {n \choose w} <= 2^n * A(n,d,w).

|C| <= 2^n / {n \choose w} * A(n,d,w).



Johnson and Elias-Bassalygo bounds


