Algorithm 2: Collaborative group selection

Input. $\mathcal{T},\left\{r_j|t_j\in \mathcal{T}\right\},{\mathrm{F}}_k,c^{\mathcal{S}},I^S$

Output: a group of users $G_k$

Step 1: Initialization.

1: $G_k-\varnothing ,Q^{\ast }-\varnothing$

Step 2.

2: $G_k-G_k\cup {\mathrm{s}}_k$

3: for $t_j\in \mathcal{T}\cap Q_k$ do

4:      $r_j-r_j-1$

5:      if $r_j=0$, then

6:          $Q^{\ast }-Q^{\ast }\cup t_j$

7:          $\mathcal{T}-\mathcal{T}\backslash \left\{t_j\right\}$

8:      end if

9: end for

10:while $\mathcal{T}\ne \varnothing$; and ${\mathrm{F}}_k\ne \varnothing$; do

11:    for ${\mathrm{s}}_{\mathrm{i}}\in {\mathrm{F}}_k$ do

12:        if $Q_i{}^{\mathrm{s}}\cap \mathcal{T}=\varnothing$, then

13:             ${\mathrm{F}}_k-{\mathrm{F}}_k\backslash \left\{{\mathrm{s}}_i\right\}$

14:        else

15:            calculate ${\Lambda }_i^S$

16:        end if

17:    end for

18:    ${\mathrm{s}}_{\mathrm{i}}\in {\mathrm{F}}_k$18: For all ${\Lambda }_i^S$, sort in a non-decreasing way.

19:    ${\mathrm{s}}_{i^{\ast }}$— has the least number of users ${\Lambda }_i^S$

20:    $G_k-G_k\cup {\mathrm{s}}_{i^{\ast }}$

21:    ${\mathrm{F}}_k-{\mathrm{F}}_k\backslash {\mathrm{s}}_{i^{\ast }}$

22:    for $t_j\in \mathcal{T}\cap Q_{i^{\ast }}$ do

23:        $r_j-r_j-1$

24:        if $r_j=0$, then

25:             $Q^{\ast }-Q^{\ast }\cup t_j$

26:            $\mathcal{T}-\mathcal{T}\backslash \left\{t_j\right\}$

27:        end if

28:    end

29: end while

30: return the user group: $G_k$