Faster approach of LU decomposition for a symmetric sparse matrix than lu in Matlab?
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Dear All,
I have a symmetric sparse matrix A. I want to obtain its LU decomposition. But I found lu(A) took me about 1 second. I am wondering if there is a faster approach than lu.
Thanks a lot in advance.
Bei
1 Comment
Bruno Luong
on 19 Oct 2018
Make sure using LU with 4 or 5 output arguments, because you get filled LU without row/column permutations and this is strongly not recommended for sparse matrix
Accepted Answer
Matt J
on 19 Oct 2018
Is the matrix positive definite? If so, maybe CHOL?
5 Comments
Bruno Luong
on 20 Oct 2018
It takes sparse, but you have to call with 4 arguments, sparse matrix needs permutation for remains sparse, the exact same comment I put or LU. If you don't want to bother with permutation, work with FULL matrix, no point to make a fixation on SPARSE.
>> A=sprand(10,10,0.3)
A =
(7,1) 0.2575
(10,1) 0.4733
(7,2) 0.8407
(9,2) 0.1966
(10,2) 0.3517
(2,3) 0.5060
(7,3) 0.2543
(4,4) 0.9593
(6,4) 0.1493
(10,4) 0.8308
(2,5) 0.6991
(5,5) 0.5472
(8,5) 0.2435
(7,6) 0.8143
(1,7) 0.7513
(3,7) 0.8909
(9,7) 0.2511
(10,7) 0.5853
(5,8) 0.1386
(8,8) 0.9293
(9,8) 0.6160
(10,8) 0.5497
(1,9) 0.2551
(8,10) 0.3500
(10,10) 0.9172
>> [L,D,P,S] = ldl(A)
L =
(1,1) 1.0000
(2,1) 0.3415
(2,2) 1.0000
(7,2) 0.2903
(9,2) 1.1320
(3,3) 1.0000
(6,3) 0.2265
(7,3) 0.0493
(8,3) 1.0000
(9,3) 0.0735
(10,3) 0.5116
(4,4) 1.0000
(5,5) 1.0000
(7,5) 0.3815
(8,5) 1.0000
(6,6) 1.0000
(8,6) -0.3815
(9,6) -0.2903
(10,6) 0.5141
(7,7) 1.0000
(8,8) 1.0000
(9,9) 1.0000
(10,9) -0.8834
(10,10) 1.0000
D =
(1,1) 1.0000
(2,2) 0.8834
(4,3) 1.0000
(3,4) 1.0000
(5,5) 1.0000
(7,6) 1.0000
(6,7) 1.0000
(7,7) -0.0345
(8,8) -1.0000
(9,9) -1.1320
(10,10) 0.8834
P =
(5,1) 1
(8,2) 1
(3,3) 1
(7,4) 1
(4,5) 1
(1,6) 1
(10,7) 1
(6,8) 1
(9,9) 1
(2,10) 1
S =
(1,1) 4.6979
(2,2) 3.2506
(3,3) 21.0084
(4,4) 1.0210
(5,5) 1.3518
(6,6) 6.5604
(7,7) 0.1872
(8,8) 1.0374
(9,9) 1.5648
(10,10) 0.4498
>>
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