Let $v$ be an element of $Z_2^{4m}$. If $v\in S_0\cup S_1$ then $$|C(v,S_0\cup S_1)|=2^{4m-2}+(-1)^m2^{2m-1}-2.$$

We divide two cases to discuss.

[**Case 1.**]{} $v\in S_0$.\
In this case, assume that $v=4k$ where $1\le k\le m$. For $(x,y)\in C(v,S_0\cup S_1)$, we have $v=x+y$ and thus $x,y\in S_0$ or $x,y\in S_1$. Suppose that $x$ has $i$’s coordinates of $1$ coinciding with that of $v$, that is, $i=v^Tx$.

If $x,y\in S_0$ then $|x|=4t$ and $|y|=4s$ for some $0\le t,s\le m$ ($x,y\ne \mathbf{0}$). Note that $y=v+x$ due to $v=x+y$. By our assumption, $4s=|y|=|v+x|=(4k-i)+(4t-i)$. It follows that $i$ is even, say $i=2j$ where $0\le j\le r_{k,t}=\min\{2k,2t\}$. Therefore, $x$ has exactly ${4k\choose 2j}{4m-4k\choose 4t-2j}$ possible choices and $y$ is uniquely determined whenever $x$ is chosen. Note that $x,y\ne\mathbf{0}$ and so $(v,\mathbf{0})$ and $(\mathbf{0},v)$ should be excluded. Thus, there are exactly $\sum_{t=0}^m\sum_{j=0}^{r_{k,t}}{4k\choose 2j}{4m-4k\choose 4t-2j}-2=\sum_{t=0}^m\sum_{j=0}^{2k}{4k\choose 2j}{4m-4k\choose 4t-2j}-2$ pairs of such $(x,y)$ as ${4k\choose 2j}{4m-4k\choose 4t-2j}=0$ if $j>r_{k,t}$.

If $x,y\in S_1$ then $|x|=4t+1$ and $y=4s+1$ for some $0\le t,s\le m-1$. Note that $y=v+x$ due to $v=x+y$. By our assumption, $4s+1=|y|=|v+x|=(4k-i)+(4t+1-i)$. It follows that $i$ is also even, say $i=2j$ where $0\le j\le r_{k,t}=\min\{2k,2t\}$. Therefore, $x$ has exactly ${4k\choose 2j}{4m-4k\choose 4t+1-2j}$ possible choices and $y$ is uniquely determined whenever $x$ is 