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 个评论
Bruno Luong
2018-10-19
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
采纳的回答
Matt J
2018-10-19
Is the matrix positive definite? If so, maybe CHOL?
5 个评论
Bruno Luong
2018-10-20
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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