uniformly spaced.

Consider the information symbol alphabet $\Lambda=\{\lambda_1,\ldots,\lambda_8\}$ with $M=8$. For the $\Lambda_\mathsf{par}$, we could pick $\Lambda_\mathsf{par}=\{\lambda_1, \lambda_6, \lambda_7,\lambda_8\}$ with $p_X(\lambda)=\{0.25, 0.25, 0.25, 0.25\}$ and $M_\mathsf{par}=4$.

To find the function $\bm{s}(\cdot )$ that maps the parity symbols onto $\lambda\in\Lambda_\mathsf{par}$, we use a greedy algorithm as described in [Algorithm]{}. It starts with , i.e., $\Lambda_\mathsf{par}=\{\lambda_1, \lambda_M\}$. For each of the remaining symbols $\lambda\in\Lambda\setminus\Lambda_\mathsf{par}$, it calculates the time-scaled of $\lambda\cup\Lambda_\mathsf{par}$, finds the symbol $\lambda$ for which the time-scaled of $\lambda\in\Lambda\setminus\Lambda_\mathsf{par}$ is maximized, and adds it to $\Lambda_\mathsf{par}$. All symbols with a greater or equal imaginary part than $\lambda$ are removed, i.e., the eigenvalues $\{\lambda'\in\Lambda : {\mathfrak{I}\{\lambda'\}}\geq{\mathfrak{I}\{\lambda\}}\}$ are removed. This process is repeated until there are no symbols left. We then choose the set of symbols that gives the highest time-scaled as $\Lambda_\mathsf{par}$. We note that this procedure does not guarantee an optimal solution. However, for $M=\{4,8\}$ an exhaustive search gives the same result as that of [Algorithm]{}.

Constellation $\Lambda$ Constellation $\Lambda_\mathsf{par}$ $\Lambda_\mathrm{placed} = \{\lambda_1, \lambda_M\}$ $\Lambda_\mathsf{par} = \{\lambda_1, \lambda_M\}$

$\Lambda_\mathrm{not\,placed} = \Lambda \setminus \Lambda_\mathsf{par}$ Calculate ${\mathsf{I}(\Lambda_\mathrm{placed} \cup \lambda_i)}$ $\lambda_\mathrm{max} := \arg\max {\mathsf{I}(\cdot)}$ $\Lambda_\mathrm{placed} = \Lambda_\mathrm{placed} \cup \lambda_\mathrm{max}$ $\Lambda_\mathsf{par} = \Lambda_\mathrm{placed}$ $\Lambda_\mathrm{not\,placed} = \Lambda_\mathrm{not\,placed}\setminus\{\lambda:\lambda\in\Lambda_\mathrm{not\,placed}, {\mathfrak{I}\{\lambda\}}\geq{\mathfrak{I}\{\lambda_\mathrm{max}\}}\}$

$\Lambda_\mathsf{par}$

In [Fig.]{}, we show $\Lambda_\mathsf{par}$ for different 