1^b \circ 1 \circ 0^c 1^d)).    \footnote{Note that ${\psi}$ on the right hand side is applied to inputs of length not necessarily $n$.
    However, for the sake of readability, we do not indicate the input length when applying ${\psi}$.
    In other words, ${\psi}$ is a shorthand for a family of functions $\{ {\psi}_n \}_{n \in {{\mathbb N}}}$.}$$ Therefore, it is enough to bound from above the right hand side by $4$. At this point, it is fairly easy to be convinced that the right hand side is bounded by *some* constant. Proving that the constant is $4$ is done by a somewhat tedious case analysis, according to the relations between $a,b,c$ and $d$. We defer the proof of the following claim to Section \[sec:missing claims\].

\[claim:case analysis1\] For every $a,b,c,d \in {{\mathbb N}}$, we have $${{\sf distance}}({\psi}(0^a 1^b \circ 0 \circ 0^c 1^d),{\psi}(0^a 1^b \circ 1 \circ 0^c 1^d)) \leq 4.$$

This completes the proof for ${{\sf maxStretch}}({\psi}) \le 4$.

We now prove that ${{\sf maxStretch}}({\psi}^{-1}) \le 5$, where we use the description of ${\psi}^{-1}$ given in Equation . In order to bound ${{\sf maxStretch}}({\psi}^{-1})$, let us fix an edge in ${\mathcal{B}_{n}}$, that is, take $z \in {\mathcal{B}_{n}}$ and $i 