SUPER_LU_D0:
  Simple 5 by 5 example of SUPER_LU solver.
Use natural column ordering.

CompCol matrix Matrix A:
Stype 0, Dtype 1, Mtype 0
nrow 5, ncol 5, nnz 12
nzval: 19.000000  12.000000  12.000000  21.000000  12.000000  12.000000  21.000000  16.000000  21.000000  5.000000  21.000000  18.000000  
rowind: 0  1  4  1  2  4  0  2  0  3  3  4  
colptr: 0  3  6  8  10  12  

SuperNode matrix Factor L:
Stype 2, Dtype 1, Mtype 1
nrow 5, ncol 5, nnz 15, nsuper 0
nzval:
0	0	1.900000e+01
1	0	6.315789e-01
2	0	0.000000e+00
3	0	0.000000e+00
4	0	6.315789e-01
0	1	0.000000e+00
1	1	2.100000e+01
2	1	5.714286e-01
3	1	0.000000e+00
4	1	5.714286e-01
0	2	2.100000e+01
1	2	-1.326316e+01
2	2	2.357895e+01
3	2	0.000000e+00
4	2	-2.410714e-01
0	3	2.100000e+01
1	3	-1.326316e+01
2	3	7.578947e+00
3	3	5.000000e+00
4	3	-7.714286e-01
0	4	0.000000e+00
1	4	0.000000e+00
2	4	0.000000e+00
3	4	2.100000e+01
4	4	3.420000e+01

nzval_colptr: 0  5  10  15  20  25  
rowind: 0  1  2  3  4  
rowind_colptr: 0  5  5  5  5  5  
col_to_sup: 0  0  0  0  0  
sup_to_col: 0  5  

CompCol matrix Factor U:
Stype 0, Dtype 1, Mtype 4
nrow 5, ncol 5, nnz 15
nzval: 
rowind: 
colptr: 0  0  0  0  0  0  

Dense matrix Solution X:
Stype 5, Dtype 1, Mtype 0
nrow 5, ncol 1, lda 5

nzval: -0.031250  0.065476  0.013393  0.062500  0.032738  

  The exact solution:

0  -0.031250
1  0.065476
2  0.013393
3  0.062500
4  0.032738

perm_r
0	0
1	1
2	2
3	3
4	4

SUPER_LU_D0:
  Normal end of execution.