Linear iterative solver

GraphLab linear solver library is used for solving the linear system Ax = b.

example.

The Jacobi algorithm is one of the simplest methods, it works by applying repeatedly the update rule:

x = (b-(A-diag(diag(A))*x) ./ diag(A)

Input

The input folder is given using the command line –matrix=folder_name. Inside this folder should have a sparse matrix A file with the format, in each line.

row col val

Additional input vector b is given using the command –input_vector=filename, this file should be found inside the folder given by –matrix.

Output

The output of the computation is a solution vector x. File name is x.out_1_of_1.

example.

Assume we have the linear system

A=[  1.8147    0.9134    0.2785
     0.9058    1.6324    0.5469
     0.1270    0.0975    1.9575 ];
b= [ 0.9649    0.1576    0.9706 ]';

The solution will be

x = A \ b = [ 0.6803   -0.4396    0.4736 ]';

To run it in GraphLab we preper a folder named jacobi_testA, inside it we have the input file for A named A:

1 1 1.8147
1 2 0.9134
1 3 0.2785
2 1 0.9058
2 2 1.6324
2 3 0.5469
3 1 0.127
3 2 0.0975
3 3 1.9575

And the input file for b (called vecB)

0.9649
0.1576
0.9706

Note: both A and vecB files are found under jacobi_testA/ folder.

Now we run:

./jacobi --matrix=jacobi_testA/ --initial_vec=vecB --rows=3 --cols=3 --debug=1 --max_iter=10

Running Jacobi
(C) Code by Danny Bickson, CMU 
Please send bug reports to [email protected]
Solution converged to residual: 0.00507232
----------------------------------------------------------
Final Runtime (seconds):   0.649031
Updates executed: 33
Update Rate (updates/second): 50.845

We examine the output:

$ cat x.out_1_of_1
1 0.67806771486642969
2 -0.43984914015767995
3 0.47337333903573475