I corrected one serious error on this page. Someone should carefully check this and fix the others. —Preceding unsigned comment added by 71.179.164.101 (talk • contribs) 15:08, 9 March 2008
- If there are 2n+1 codewords, then k = n+1. Therefore, it was correct as it was. Oli Filth(talk) 17:25, 9 March 2008 (UTC)
- I believe the anonymous user was pointing out that unless the Hadamard matrix of order 2n used to produce the codewords is equivalent to the matrix obtained from Sylvester's construction, the resulting code will not be linear. For example, there are millions of Hadamard matrices of order 32, any of which can be used to construct a (32,64,16) code, but only when Sylvester's matrix is used will it also be a [32,6,16] code.
- As far as I am aware, some authors use “Hadamard code” to refer to this more general construction, which doesn't even require that the Hadamard matrix have power-of-two order.Will Orrick (talk) 04:04, 15 March 2008 (UTC)
- If the code is constructed as the rows of , then there are 2N+1 codes. Therefore, they can be described by an (N+1)-bit index. Therefore the code will be (2N, N+1). How do you get (2N, 2N+1)? In fact, that doesn't even make sense; in an (n,k) code, n cannot be less than k, otherwise you would be achieving perfect compression!
- I'm no expert on Hadamard matrices; by "there are millions of Hadamard matrices of order 32", are you implying that there are order-32 Hadamard matrices that cannot be derived from Sylvester's construction by simple row/column permutation? If so, we do indeed need to clarify that the code is based on matrices from Sylvester's construction. Oli Filth(talk) 15:41, 15 March 2008 (UTC)