Algorithm 1

procedure EBRM($\mathbf{V}$, $\mathbf{T}$) 2: for $k\leftarrow 1,K$ do $w_m^{(1)}$ and ${\widehat{\mathbf{T}}}_m^{(1)}$    for $i\leftarrow 1,I$ do 4:    for $b\leftarrow 1,B$ do $V_{i,b}^{*}=(v_1^{*},v_2^{*},...,v_N^{*})$ and $T_{j,b}^{*}=(t_1^{*},t_2^{*},...,t_N^{*})$,       ${\overline{V}}_{i,b}^{*}=\frac{1}{N}{\sum }_{n=1}^N{v}_n^{*}$ and ${\overline{T}}_{j,b}^{*}=\frac{1}{N}{\sum }_{n=1}^N{t}_n^{*}$, 6:     ${\mathbf{V}}_i^{*}=({\overline{V}}_{i,1}^{*},{\overline{V}}_{i,2}^{*},...,{\overline{V}}_{i,B}^{*})$ and ${\mathbf{T}}_j^{*}=({\overline{t}}_{j,1}^{*},{\overline{t}}_{j,2}^{*},...,{\overline{t}}_{j,B}^{*})$       ${\tilde{\mathbf{V}}}^{*}=({\mathbf{V}}^{*}|w_m^{(k)})$ 8:     ${\mathbf{U}}^{*}=\left[{\tilde{\mathbf{V}}}^{*}{\widehat{\mathbf{T}}}_m^{(k)}\right]$      end for 10:  end for    call learning: $\mathbf{back}-\mathbf{propagation}$ $\left\{{\widehat{f}}^{*k}({\mathbf{U}}_m^{*},{\mathbf{T}}_m^{*})\right\}$, 12:  output: ${\widehat{\mathbf{T}}}_m^{*}$, $\forall m=1$ to M    ${\epsilon }_{max}=\underset{m=1,...,M}{max}{[{\widehat{\mathbf{T}}}_m^{*}-{\mathbf{T}}_m^{*}]}^2$ 14:  ${\epsilon }_m=\frac{{[{\widehat{\mathbf{T}}}_m-{\mathbf{T}}_m^{*}]}^2}{{\epsilon }_{max}}$    $\overline{\epsilon }={\sum }_{m=1}^M{\epsilon }_m{w}_m^{(k)}$ 16:  ${\beta }_k=\frac{\overline{\epsilon }}{1-\overline{\epsilon }}$    $w_m^{(k+1)}=w_m^{(k)}{\beta }_k^{(1-{\epsilon }_m)}$ 18:  $w_m^{(k+1)}=\frac{w_m^{(k+1)}}{{\sum }_m{w}_m^{(k+1)}}$   end for 20: end procedure