## 3 Conventional Data Detection Algorithm Based on Manifold Optimization
### Problem Formulation and Solving
For \(\mathrm{Grad}\,F(\mathbf{\tilde{S}})\), since the multiplication \(\mathbf{\tilde{H}}\mathbf{\tilde{S}}\) has already been analyzed and does not need to be repeated, we only need to consider the remaining multiplications. The multiplication \(\mathbf{\tilde{H}}^{T}\mathbf{\tilde{Y}}\) has a complexity of \(\mathrm{O}\Big{(}4K^{3}{N_{r}}^{2}{N_{t}}{N_{f}}^{3}\Big{)}\). The multiplication \(\mathbf{\tilde{H}}^{T}\mathbf{\tilde{H}}\mathbf{\tilde{S}}\) also has a complexity of \(\mathrm{O}\Big{(}4K^{3}{N_{r}}^{2}{N_{t}}{N_{f}}^{3}\Big{)}\). Therefore, the complexity of calculating \(\mathrm{Grad}\,F(\mathbf{\tilde{S}})\) is \(\mathrm{O}\Big{(}4K^{3}{N_{r}}^{2}{N_{t}}{N_{f}}^{3}\Big{)}\). Let \(T_{2}\) denote the number of iterations of the Riemannian conjugate gradient algorithm in Algorithm 2. The overall complexity of calculating these steps in Algorithm 2 is \(\mathrm{O}\Big{(}4K^{3}{N_{r}}^{2}(2{N_{t}}+{N_{r}}){N_{f}}^{3}T_{2}\Big{)}\). Similarly to Algorithm 2, the complexity of calculating \(F(\mathbf{\tilde{S}})\) is \(\mathrm{O}\Big{(}4K^{3}{N_{r}}^{2}({N_{t}}+{N_{r}}){B}^{3}\Big{)}\) and the complexity of calculating \(\mathrm{Grad}\,F(\mathbf{\tilde{S}})\) is \(\mathrm{O}\Big{(}4K^{3}{N_{r}}^{2}{N_{t}}{B}^{3}\Big{)}\) in Algorithm 3. Let \(T_{3}\) denote the number of iterations of the Riemannian conjugate gradient algorithm in Algorithm 3. Since we divide one data frame of \({N_{f}}\) symbols into subframes of \(B\) symbols each for computation, the overall complexity of calculating these steps in Algorithm 3 is \(\mathrm{O}\Big{(}4K^{3}{N_{r}}^{2}(2{N_{t}}+{N_{r}}){B}^{2}{N_{f}}T_{3}\Big{)}\). ```
1:Input \(\bigg{\{}\big{\{}\mathbf{y}_{n,k}\big{\}}_{n=1}^{N_{f}}\bigg{\}}_{k=1}^{K}\)
2:Output: \(\Big{\{}\big{\{}\mathbf{\hat{x}}_{n,k}\big{\}}_{n=1}^{N_{f}}\Big{\}}_{k=1}^{K}\)
3:(1) The same as step (1) in Algorithm 2;
4:(2) The same as step (2) in Algorithm 2;
5:(3) Divide the data frame into \({N_{B}}={N_{f}}/B\) subframes;
6:for\(i=1\) to \({N_{B}}\)do
7: (i) Set \(j=0\);
8: (ii) Similar to Algorithm 2, construct matrices \(\mathbf{\hat{H}}_{i,\mathrm{ini}}\in\mathbb{R}^{2K{N_{B}}\times 2K{N_{B}}}\), \(\mathbf{\tilde{Y}}_{i}\in\mathbb{R}^{2K{N_{B}}\times 2K{N_{B}}}\), and \(\mathbf{\hat{S}}_{i,\mathrm{ini}}^{(j)}\in\mathbb{R}^{2K{N_{B}}\times 2K{N_{B}}}\) according to \(\big{\{}\mathbf{H}_{n,k}\big{\}}_{n=1+(i-1)B}^{iB}\), \(\Big{\{}\mathbf{y}_{n,k}\Big{\}}_{n=1+(i-1)B}^{iB}\), and \(\big{\{}\mathbf{x}_{n,k}\big{\}}_{n=1+(i-1)B}^{iB}\);
9: (iii) Solve (43) to obtain the solution \(\mathbf{\hat{S}}_{i}^{(j)}\), where \(\mathbf{\tilde{H}}\), \(\mathbf{\tilde{Y}}\), and \(\mathbf{\tilde{S}}\) are replaced by \(\mathbf{\hat{H}}_{i,\mathrm{ini}}\), \(\mathbf{\tilde{Y}}_{i}\), and \(\mathbf{\tilde{S}}_{i,\mathrm{ini}}^{(j)}\), respectively;
10: (iv) Similarly to Algorithm 2, obtain the estimated transmission signal \(\Big{\{}\mathbf{\hat{x}}_{n,k}^{(j)}\Big{\}}_{n=1+(i-1)B}^{iB}\), and then obtain the estimated pilot symbols \(\bigg{\{}\mathbf{\hat{x}}_{n_{f}^{\prime},k_{i}^{\prime}}^{B}\bigg{\}}\);
11:if\(\mathbf{\hat{x}}_{n_{f}^{\prime},k_{i}^{\prime}}^{p,(j)}\
eq\mathbf{x}_{n_{f}^{\prime},k_{ i}^{\prime\prime}}^{p}\), \(\forall p\in\{1,\cdots,N_{t}\}\)then
12: (a) Set \(\mathbf{\hat{S}}_{i,\mathrm{ini}}^{(j)}=\mathbf{\hat{S}}_{i}^{(j)}\) and \(j\gets j+1\);
13: (b) Return to (iii). 14:endif
15:endfor
```

**Algorithm 3** The proposed less-complex algorithm. After observing a large number of simulation experiments, we can conclude that \(T_{2}\) is usually around 7 times and \(T_{3}\) is usually around 10 times during the simulation process. In summary, when \({N_{f}}=14\), the computational complexity of Algorithm 2 and the three Algorithms 3 taking different subframe lengths is shown in Table 2. Under the condition of \({N_{f}}=14\) and \(B=1\), the complexity of calculating the above steps is \(\mathrm{O}\left(76832K^{3}N_{r}{}^{2}(2N_{t}+N_{r})\right)\) in Algorithm 2 and \(\mathrm{O}\left(560K^{3}N_{r}{}^{2}(2N_{t}+N_{r})\right)\) in Algorithm 3.