(-5.12,-5.12)(-4.13,-5.12) (-4.13,-5.12)(-3.42,-5.84) (-3.42,-5.84)(-3.42,-4.84) (-3.42,-5.84)(-2.43,-5.84) (2.4,-5.81)(3.4,-5.79) (3.4,-5.79)(3.39,-4.82) (3.4,-5.79)(4.09,-5.12) (4.09,-5.12)(5.12,-5.12) (5.12,-5.12)(5.12,-4.12) (5.12,-4.12)(5.83,-3.42) (5.83,-3.42)(4.8,-3.42) (5.83,-3.42)(5.82,-2.41)

We can set a cut $\&$ project scheme and window giving rise to the desired point pattern as follows: We set in a physical space $\mathbb{R}^2$ the group $\Gamma $ algebraically generated by four vectors whose coordinates in an orthonormal basis read $$\begin{aligned}
 \begin{tabular}{c c c c} $e_1= (1,0)$ & $e_2 = (\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})$ & $e_3= (0,1)$ & $e_4 = (-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})$
\end{tabular}\end{aligned}$$ These four vectors are algebraically independent, and thus $\Gamma $ is isomorphic with $\mathbb{Z}^4$. Next we set an internal space $\mathbb{R}^2_{int}$ to be a $2$-dimensional real vector space, into which we define a $*-$map through the images of the four above vectors, reading in some orthogonal basis of $\mathbb{R}^2_{int}$ $$\begin{aligned}
 \begin{tabular}{c c c c} $e_1^*= (1,0)\qquad e_2^* = (-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})$ & $e_3^*= (0,-1)$ & $e_4^* = (\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})$
\end{tabular}\end{aligned}$$ The four vectors $\tilde{e_i}:= (e_i,e_i^*)$, $i=1,2,3,4$, are linearly independant in $\mathbb{R}^2_{int}\times \mathbb{R}^2$ and thus form a lattice $\Sigma$, which project into $\mathbb{R}^2_{int}$ into a dense subgroup $\Gamma ^*$. This sets a real cut $\&$ project scheme. We chose the window to be canonical, that is, to be the projection into the internal space of the unit cube of $\mathbb{R}^2_{int}\times \mathbb{R}^2$ 