$${\tilde M}^\dagger_r = (I_2 \otimes (F_m^\dagger
\oplus I_{r/2-m})) R^\dagger (A^\dagger \otimes I_{r/4}) R
Q_r^\dagger.$$ Translating this equation into a quantum circuit, the decomposition of ${\tilde M}_r$ leads to the circuit scheme shown in Figure \[Figur schaltkreis dieder-2pt\].

The operation $A^\dagger$ is decomposed as $$A^\dagger= \sqrt{m} \left( \begin{array}{cccc}\alpha &
\beta & 0 & 0 \\ {\overline \beta} & - \alpha & 0 & 0 \\ 0 & 0 &
\sqrt{1/m} & 0 \\ 0 & 0 & 0 & \sqrt{1/m}
\end{array} \right) 
\sqrt{m} \left( \begin{array}{cccc} \sqrt{1/m} & 0
& 0 &0 \\ 0 & \sqrt{1/m} & 0 & 0 \\ 0 & 0 & \alpha & {\overline \beta}
\\ 0 & 0 & - \beta & \alpha \end{array} \right)$$ corresponding to the second and third gates from the left in Figure \[Figur schaltkreis dieder-2pt\]. We do not have to implement the permutation $R$ explicitly if the controlled one-qubit operations are applied to appropriate qubit pairs. The given circuit can be slightly simplified by merging the first two gates from the left to a single controlled gate with the operation $$\sqrt{m}\left( \begin{array}{cc} \alpha & {\overline \beta} \\ -
\beta & \alpha\end{array} \right) \left( \begin{array}{cc}0&1\\1&0
\end{array} \right) = \sqrt{m} \left(
\begin{array}{cc}{\overline \beta} & 