complexity even for the case of $N_r=1$, where the complexity lies in the interval $3 \times 2^m \leq \mathcal{C}_{\rm{Rx-SD}} \leq 6 \times 2^m$. We note that the SM–Rx solution requires no pre–computations with respect to the ML–optimum detector. In fact, $\tilde N_r \left( {\ell,s} \right)$ for $\ell \in \left\{ {1,2, \ldots N_t } \right\}$ and $s \in \left\{ {s_1 ,s_2 , \ldots s_M } \right\}$ in are implicitly computed when solving the detection problem in .

SM–Tx
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The computational complexity of SM–Tx can be upper–bounded by,

$$\label{Eq:Txcomp} \mathcal{C}_{{\rm{Tx - SD}}} \leq \mathcal{C}_{\Theta_R }  + 3N_t{\rm{card}}\left\{ {\Theta _R } \right\}$$

where $\rm{card\{\cdot\}}$ denotes the cardinality of a set, and $\mathcal{C}_{\Theta_R}$ is the complexity of finding the points in the subset $\Theta_R$,

$$\mathcal{C}_{\Theta_R} = \mathcal{C}_{\text{Pre-Comp}} + \mathcal{C}_{\text{Interval}}$$

where,

1.  $\mathcal{C}_{\text{Pre-Comp}}$ is the number of operations needed to compute the Cholesky decomposition. Calculating the upper triangular matrix $\mathbf{\bar D}$ using Cholesky decomposition has the complexity [@book:gl9601],

    $$\mathcal{C}_{\rm CH} = 4N_t^3/3$$

    It can be easily shown that the calculation of $\bar{\bf G}$, $\bar \rho$ and $\mathbf{\bar z}$ requires $2N_rN_t(2N_t+1)$, $2N_t(2N_t+2N_r+1)$ and $N_t(2N_t+1)$ real operations respectively, where back–substitution algorithm was used for calculating $\bar \rho$ [@book:gl9601].

    Hence, $$\mathcal{C}_{\text{Pre-Comp}} = \mathcal{C}_{\rm CH} + N_t(4N_rN_t+6N_r+6N_t+3)$$

2.  