chosen. Thus, there are exactly $\sum_{t=0}^m\sum_{j=0}^{r_{k,t}}{4k\choose 2j}{4m-4k\choose 4t+1-2j}=\sum_{t=0}^m\sum_{j=0}^{2k}{4k\choose 2j}{4m-4k\choose 4t+1-2j}$ pairs of such $(x,y)$ as ${4k\choose 2j}{4m-4k\choose 4t+1-2j}=0$ if $j>r_{k,t}$.

By arguments above, we have $$\begin{array}{lll}|C(v,S_0\cup S_1)|&=&\displaystyle \sum_{t=0}^m\sum_{j=0}^{2k}{4k\choose 2j}{4m-4k\choose 4t-2j}+\sum_{t=0}^m\sum_{j=0}^{2k}{4k\choose 2j}{4m-4k\choose 4t+1-2j}-2\\
&=&\displaystyle\sum_{t=0}^m\sum_{j=0}^{2k}{4k\choose 2j}{4m-4k+1\choose 4t-2j+1}-2.\end{array}$$ By Theorem \[thm-z-2\], we have $$|C(v,S_0\cup S_1)|=2^{4m-2}+(-1)^m2^{2m-1}-2.$$

[**Case 2.** ]{} $v\in S_1$.\
In this case, assume that $|v|=4k+1$ where $0\le k\le m-1$. For $(x,y)\in C(v,S_0\cup S_1)$, we have $v=x+y$ and thus $x\in S_0$, $y\in S_1$ or $x\in S_1$, $y\in S_0$. Suppose that $x$ has $i$’s coordinates of $1$ coinciding with that of $v$, that is, $i=v^Tx$.

If $x\in S_0$ and $y\in S_1$ then $|x|=4t$ and $y=4s+1$ for some $0\le t\le m$ and $0\le s\le m-1$. Note that $y=v+x$ due to $v=x+y$. By our assumption, $4s+1=|y|=|v+x|=(4k+1-i)+(4t-i)$. It follows that $i$ is even, say $i=2j$ where $0\le j\le r_{k,t}=\min\{2k,2t\}$. Therefore, $x$ has exactly ${4k+1\choose 2j}{4m-4k-1\choose 4t-2j}$ possible choices and $y$ is uniquely determined when $x$ is chosen. Note that $x\ne \mathbf{0}$ and so $(\mathbf{0},v)$ should be excluded. Thus, there are exactly $\sum_{t=0}^m\sum_{j=0}^{r_{k,t}}{4k+1\choose 2j}{4m-4k-1\choose 4t-2j}-1=\sum_{t=0}^m\sum_{j=0}^{2k}{4k+1\choose 2j}{4m-4k-1\choose 4t-2j}-1$ pairs of such $(x,y)$ as ${4k+1\choose 2j}{4m-4k-1\choose 4t-2j}=0$ if $j>r_{k,t}$. By the symmetry of $x$ and $y$, there are also $\sum_{t=0}^m\sum_{j=0}^{2k}{4k+1\choose 2j}{4m-4k-1\choose 4t-2j}-1$ pairs of $(x,y)$ such that $v=x+y$, $x\in S_1$ 