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# Table 2 Low-complexity suboptimal two-step algorithm

Two-step algorithm
1. Initialization: S = , K = , L =
2.      Clustering Procedure:
3.          For each resource pattern with only one RB rb {1,2,…,K rb }
4.             Find the CUE sub-cluster and DUE sub-cluster $$\left\{\overline{k},\overline{l}\right\}= \arg \underset{\left\{k,l\right\}}{ \max }{t}_{k,l,rb}$$
5.               $$K=K\cup}\kern0.5em \overline{k}$$
6.               $$L=L\;\cup}\kern0.5em \overline{l}$$
7.           End
8. Searching Space Reduction Procedure:
9.      For each $$\widehat{k}\in K;\widehat{l}\in L$$ sets
10. $${s}_0=\left\{i\Big|\left(\widehat{k}-1\right)*{N}_{all,d}{N}_{all,rb}+\left(\widehat{l}-1\right)*{N}_{all,rb}+1\le i\le \left(\widehat{k}-1\right)*{N}_{all,d}{N}_{all,rb}+\widehat{l}*{N}_{all,rb}\right\}$$
12.    S = S s 0
13.      End
14. Joint Solution Procedure:
15.             Obtain the reduced-dimension optimization problem
16.                   $${x}_s=\underset{{\mathbf{x}}_S}{ \min}\left\{-{\mathbf{t}}_{\boldsymbol{S}}^{\boldsymbol{T}}{\mathbf{x}}_S\right\}$$, s.t. $${\mathbf{R}}_{:,S}{\mathbf{x}}_S={\mathbf{1}}_{\left({N}_c{N}_d{N}_{rb}+{K}_c+{K}_d\right)\times 1}$$
17.             Find its solution based on the BBS algorithm
18. Return $${x}_s$$
19. End 