Weight of words in Binary Reed-Muller code

Question

Assuming we construct a $(2^m, 2^{m+1}, 2^{m-1})$ Reed-Muller code from the $(4,8,2)$ even-weight code and the $(4,2,4)$ repetition code, how many words will there be of each weight?

Answer 1

We will show that in a Reed-Muller code, any word different from $0$ and $1$ has a weight equal to $2^{m-1}$.

Any word of the $RM(0,m)$ code, different from $0$ and $1$ is of the form $(u,u)$ or $(u,u+1)$ with $u\in RM(1,m-1)$.

If the codeword is of the form $(u,u)$,then $u$ can be neither $0$ or $1$ in $RM(1,m-1)$ since then,the codeword would be $0$ or $1$ in $RM(1,m)$. So,it follows from induction(*) that $u$ has weight equal to $2^{m-2}$ and $(u,u)$ has weight equal to $2(2^{m-2})=2^{m-1}$.

Now,let's assume that the codeword is of the form $(u,u+1)$.
If $u=0$, then $u+1$ is $1$ and it's weight is $2^{m-1}$.
If $u=1$, then $u+1$ is $0$ and it's weight is $2^{m-1}$
For any other $u\in RM(1,m-1)$ half of it's coordinates (it's symbols)are equal to $1$. Thus,in $u+1 \in RM(1,m)$ we likewise have half of its coordinates equal to $1$ and it follows that the weight of $(u,u+1)=2(2^{m-2})=2^{m-1}$

(*) Assuming that $RM(1,m-1)$ is a $[2^{m-1},m,2^{m-2}]$ code,we show that $RM(1,m)$ is a $[2^m,m+1,2^{m+1}]$ code.