$1 \le i_1 < i_2 < \cdots < i_t \le n$ are the coordinates in which $y$ contains ${\sqcup}$, then there exists some $0 \le \ell \le t$ such that $x_{i_1} = \cdots = x_{i_\ell} = 0$ and $x_{i_{\ell+1}} = \cdots = x_{i_t} = 1$. That is, $x$ is located at the $(\ell+1)^{\text{st}}$ position of the chain $C_y$, when counting from the top. The function ${\psi}$ outputs the vertex located at the $({{\lfloor\ell/2\rfloor}}+1)^{\text{st}}$ position in the chain, concatenated with 1 or 0, depending on the parity of $\ell$. In other words, we obtain ${\psi}(x)$ by keeping intact all the bits of $x$ in the coordinates other than $i_{{{\lfloor\ell/2\rfloor}}+1}, \ldots, i_\ell$, and by setting ${\psi}(x)_{i_{{{\lfloor\ell/2\rfloor}}+1}} = \cdots = {\psi}(x)_{i_{\ell}} = 1$. Then, we append 1 to the obtained string if $\ell$ is even, and append 0 otherwise.

For example, let us consider the Boolean cube ${\{0,1\}^{4}}$, whose partition is presented in Figure \[fig:the partition\], and write explicitly where each vertex in the chain of length 5 is mapped under ${\psi}$. $$\begin{aligned}
&{\psi}(1111) = 1111 \circ 1,\\
&{\psi}(0111) = 1111 \circ 0,\\
&{\psi}(0011) = 0111 \circ 1,\\
&{\psi}(0001) = 0111 \circ 0,\\
&{\psi}(0000) = 0011 \circ 1,\end{aligned}$$ where we write the concatenation 