Algorithm1 (Unique solution of the PIEPB $\left(ϵ,c,s\right)$ ).

1: Input: $c,s,{\left\{{\lambda }_j^{\left(1\right)},{\lambda }_j^{\left(j\right)}\right\}}_{j=1}^n$ . where $n=cs+1$ .

2: $a_1={\lambda }_1^{\left(1\right)}$ .

3: For $j=\left(i-1\right)s+2,i=1,2,\cdots ,c$ do

4: if ${\phi }_{j-1}\left({\lambda }_j^{\left(1\right)}\right)\det \left(B_j^{{\lambda }_j^{\left(j\right)}}\right)-{\phi }_{j-1}\left({\lambda }_j^{\left(j\right)}\right)\det \left(B_j^{{\lambda }_j^{\left(1\right)}}\right)=0$ then

5: ending the algorithm 1.

6: end if

7: if ${\phi }_{j-1}\left({\lambda }_j^{\left(1\right)}\right)\det \left(B_j^{{\lambda }_j^{\left(j\right)}}\right)-{\phi }_{j-1}\left({\lambda }_j^{\left(j\right)}\right)\det \left(B_j^{{\lambda }_j^{\left(1\right)}}\right)\ne 0$ then

8: computing $a_j$ , $b_{1,j}$ and ${ϵ}_{1,j}$ by (7) and (8).

9: end if

10: end for

11: For $j=\left(i-1\right)s+3,i=1,2,\cdots ,c$ do

12: if ${\phi }_{j-1}\left({\lambda }_j^{\left(1\right)}\right){\phi }_{j-2}\left({\lambda }_j^{\left(j\right)}\right)-{\phi }_{j-1}\left({\lambda }_j^{\left(j\right)}\right){\phi }_{j-2}\left({\lambda }_j^{\left(1\right)}\right)=0$ then

13: ending the algorithm 1.

14: end if

15: if ${\phi }_{j-1}\left({\lambda }_j^{\left(1\right)}\right){\phi }_{j-2}\left({\lambda }_j^{\left(j\right)}\right)-{\phi }_{j-1}\left({\lambda }_j^{\left(j\right)}\right){\phi }_{j-2}\left({\lambda }_j^{\left(1\right)}\right)\ne 0$ then

16: computing $a_j$ , $b_{\left(j-1\right),j}$ and ${ϵ}_{\left(j-1\right),j}$ by (9) and (10).

17: end if

18: end for

19: For $j=\left(i-1\right)s+4,\cdots ,is+1,i=1,2,\cdots ,c$ do

20: if ${\phi }_{j-1}\left({\lambda }_j^{\left(1\right)}\right){\phi }_{\left(i-1\right)s+2}\left({\lambda }_j^{\left(j\right)}\right){\displaystyle \prod _{k=\left(i-1\right)s+4}^{j-1}\left({\lambda }_j^{\left(j\right)}-a_k\right)}-{\phi }_{j-1}\left({\lambda }_j^{\left(j\right)}\right){\phi }_{\left(i-1\right)s+2}\left({\lambda }_j^{\left(1\right)}\right){\displaystyle \prod _{k=\left(i-1\right)s+4}^{j-1}\left({\lambda }_j^{\left(1\right)}-a_k\right)}=0$ then

21: ending the algorithm 1.

22: end if

23: if ${\phi }_{j-1}\left({\lambda }_j^{\left(1\right)}\right){\phi }_{\left(i-1\right)s+2}\left({\lambda }_j^{\left(j\right)}\right){\displaystyle \prod _{k=\left(i-1\right)s+4}^{j-1}\left({\lambda }_j^{\left(j\right)}-a_k\right)}-{\phi }_{j-1}\left({\lambda }_j^{\left(j\right)}\right){\phi }_{\left(i-1\right)s+2}\left({\lambda }_j^{\left(1\right)}\right){\displaystyle \prod _{k=\left(i-1\right)s+4}^{j-1}\left({\lambda }_j^{\left(1\right)}-a_k\right)}\ne 0$ then

24: computing $a_j$ , $b_{\left(i-1\right)s+3,j}$ and ${ϵ}_{\left(i-1\right)s+3,j}$ by (11) and (12).

25: end if

26: end for