Hi,
I'm using the UnconvMat website to analyze eBR and aBR. I am trying to understand the results.
I tried the example from here, so what do the last two columns mean?
Best,
Dongsheng
There are 2 solutions for eBR decomposition.
2
1 1@5 Ag@9e ( 1) : 0; 0;
2 2@5 Au@9e ( 1) : 0; 0;
3 3@5 Bg@9e ( 1) : 0; 0;
4 4@5 Bu@9e ( 1) : 0; 0;
5 1@6 Ag@9d ( 1) : 0; 0;
6 2@6 Au@9d ( 1) : 0; 0;
7 3@6 Bg@9d ( 1) : 0; 0;
8 4@6 Bu@9d ( 1) : 0; 0;
9 1@8 A1g@3b ( 1) : 1; 2;
10 2@8 A1u@3b ( 1) : 0; 0;
11 3@8 A2g@3b ( 1) : 0; 0;
12 4@8 A2u@3b ( 1) : 0; 1;
13 5@8 Eg@3b ( 1) : 0; 0;
14 6@8 Eu@3b ( 1) : 0; 0;
15 1@9 A1g@3a ( 1) : 1; 0;
16 2@9 A1u@3a ( 1) : 0; 0;
17 3@9 A2g@3a ( 1) : 0; 0;
18 4@9 A2u@3a ( 1) : 1; 0;
19 5@9 Eg@3a ( 1) : 0; 0;
20 6@9 Eu@3a ( 1) : 1; 1;
'''
Hi,
I'm using the UnconvMat website to analyze eBR and aBR. I am trying to understand the results.
I tried the example from here, so what do the last two columns mean?
Best,
Dongsheng