## 3 Proposed MIDE Algorithm
### Modified IDE-Based Detection Algorithm with Self-Update Damping  
The performance of the IDE-based detection algorithm is influenced by the choice of the damping factor. In early studies, this parameter was fixed, e.g., 0.05 in [25]. However, a fixed damping factor is not applicable for all cases, and an optimal damping factor is not easy to obtain. We propose an MIDE herein to decide a proper damping factor \(\alpha\), by which the \(\mathbf{\hat{x}}_{d}\)-update step is analyzed as follows. \[\mathbf{\hat{x}}_{d}^{1} =\left(1-\alpha^{0}\right)\mathbf{\hat{x}}_{d}^{0}+\alpha^{0} \mathbf{\hat{x}}^{1},\] \[\mathbf{\hat{x}}_{d}^{2} =\left(1-\alpha^{1}\right)\mathbf{\hat{x}}_{d}^{1}+\alpha^{1} \mathbf{\hat{x}}^{2}\] \[=\left(1-\alpha^{1}\right)\left(\left(1-\alpha^{0}\right)\mathbf{ \hat{x}}_{d}^{0}+\alpha^{0}\mathbf{\hat{x}}^{1}\right)+\alpha^{1}\mathbf{\hat {x}}^{2}\] \[\vdots\] \[\mathbf{\hat{x}}_{d}^{k+1} =\left(1-\alpha^{k}\right)\mathbf{\hat{x}}_{d}^{k}+\alpha^{k} \mathbf{\hat{x}}^{k+1}\] \[=\left(1-\alpha^{k-1}\right)\cdots\left(1-\alpha^{0}\right) \mathbf{\hat{x}}_{d}^{0}+\left(1-\alpha^{k-1}\right)\cdots\left(1-\alpha^{1 }\right)\alpha^{0}\mathbf{\hat{x}}^{1}+\cdots+\alpha^{k-1}\mathbf{\hat{x}}^{k +1}\]  
Since there is no prior information of the final result \(\mathbf{\hat{x}}_{d}\), the initial value of \(\mathbf{\hat{x}}_{d}^{0}\) can be set as a zero vector. Hence, the expression for \(\mathbf{\hat{x}}_{d}\) in Equation (16) is composed of the solution of \(\mathbf{\hat{x}}\) with different values of the damping factor. Based on the expression in Equation (16), the convergence of the iterations can be measured by the difference between \(\mathbf{\hat{x}}_{d}^{k}\) and \(\mathbf{\hat{x}}^{k+1}\). Specifically, the Euclidean distance, which is one of the widely-used approaches for measuring the distance between two vectors [30], is defined as:  
\[d^{k}\left(\mathbf{\hat{x}}_{d}^{k}\mathbf{\hat{x}}^{k+1}\right)=\sqrt{\sum_{i= 1}^{N_{t}}\left(\mathbf{\hat{x}}_{d}^{k}\left(i,1\right)-\mathbf{\hat{x}}^{k+1 }\left(i,1\right)\right)^{2}} \tag{17}\]We can notice from Equation (17) that the smaller the distance \(d\), the closer the two vectors, implying the convergence of the IDE iterations. Based on this discussion, the following heuristic damping factor at the \(k^{\text{th}}\) iteration is defined as:  
\[\alpha^{k}=\frac{d^{k}}{d^{k}+q} \tag{18}\]  
where \(q\) is a positive constant. Obviously, a higher value of \(d\) leads to a higher \(\alpha\), and vice versa. In other words, when \(d^{k}\to 0\), \(\alpha^{k}\to 0\), and when \(d^{k}\to 1\), \(\alpha^{k}\to 1\). We utilize the first iteration result \(d^{1}\) to obtain the value of \(p\). According to Equation (16), \(d^{1}\) can be calculated as:  
\[ \begin{split} d^{1}&=\sqrt{\sum_{i=1}^{N_{t}}\left( \mathfrak{K}_{d}^{0}\left(i,1\right)-\mathfrak{K}^{1}\left(i,1\right)\right)^ {2}}\\ &=\sqrt{\sum_{i=1}^{N_{t}}\left(\mathfrak{K}^{1}\left(i,1\right) \right)^{2}}\\ &=\sqrt{\sum_{i=1}^{N_{t}}\left(\Re\left(\mathfrak{K}^{1}\left( i,1\right)\right)\right)^{2}+\left(\Im\left(\mathfrak{K}^{1}\left(i,1\right) \right)\right)^{2}}\end{split}  \tag{19}\]  
It is noticed that the value of \(d^{1}\) is varying and decided based on several factors such as the modulation method, the noise in the system, etc. For the ease of calculation, the expectation of \(d^{1}\) is computed instead of the direct calculation of \(d^{1}\). Since the real part and the imaginary part of the vector \(\mathfrak{K}^{1}\) have the same uniform distribution, the expectation for \(d^{1}\) is obtained as:  
\[E\left(d^{1}\right)=2E\left(\sqrt{\sum_{i=1}^{N_{t}}\left(\Re\left(\mathfrak{K }^{1}\left(i,1\right)\right)\right)^{2}}\right) \tag{20}\]  
The expectation of \(d^{1}\) is based on the constellation points of the modulation scheme. With 16-QAM, for example, \(\Re\left(\mathfrak{K}^{1}\left(i,1\right)\right)\) can be \(\left\{-3,-1,+1,+3\right\}\), and the probability \(p\) of each possible value of the point is the same without the prior information, i.e., \(p=0.25\) in this case.