It was written in the presentation that the volume bound inequality
assuming minimum distance between 2 codewords id d=3, is:
(2^k)*(n+1) <= 2^n
I thought about the case that d>3, and i came up with this inequality:
(2^k)*((n^A)+1) <= 2^n
A = floor((n-1)/2)
Do you agree with that claim?