Algorithm 1: Block coordinate descent algorithm for X2 and X3

Input: Initial solution $X_2^{\left(k\right)},X_3^{\left(k\right)}$.

Output: Optimal solution $X_2^{\left(k+1\right)},X_3^{\left(k+1\right)}$.

Step 1: m = 1, i = 0, $X_{2, (0)}=X_2^{\left(k\right)}$, $X_{3, (0)}=X_3^{\left(k\right)}$.

Step 2: Update the m-th row vector ${\alpha }_{(i)}^m$ of $X_{2, (i)}$ and ${\beta }_{(i)}^m$ of $X_{3, (i)}$. Input it to Algorithm 2 whose output is the optimal row vector ${\alpha }_{\ast }^m$ and ${\beta }_{\ast }^m$.

Step 3:$\{\begin{array}{l}{X}_{2, (i+1)}=\left[{\alpha }_{(i)}^1,{\alpha }_{(i)}^2,\dots ,{\alpha }_{(i)}^{m-1},{\alpha }_{\ast }^m,{\alpha }_{(i)}^{m+1},\dots ,{\alpha }_{(i)}^M\right]\\ X_{3, (i+1)}=\left[{\beta }_{(i)}^1,{\beta }_{(i)}^2,\dots ,{\beta }_{(i)}^{m-1},{\beta }_{\ast }^m,{\beta }_{(i)}^{m+1},\dots ,{\beta }_{(i)}^M\right]\end{array}$

Step 4: i = i + 1. If $m>M$ is true, m = 1, otherwise m = m + 1.

Step 5: Calculate the objective function using $X_{2, (i)}$ and $X_{3, (i)}$. If the value of the objective function is not decreasing, $X_2^{\left(k+1\right)}=X_{2, (i)},X_3^{\left(k+1\right)}=X_{3, (i)}$ and the optimal matrix is obtained, otherwise jump to Step 2 and continue.