How to determine which algorithm mldivide (A\b) uses

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Jonathan
Jonathan 2015 年 10 月 27 日
回答済み: Grzegorz Knor 2017 年 4 月 19 日
I have a sparse, banded matrix, A, for which I am solving the linear system Ax=b using A\b in Matlab.
The documentation for mldivide shows a flow chart that explains which solver it will use. However, I would like to automate the process of determining the best algorithm. Therefore, I would really just like to know which algorithm mldivide chooses for different inputs. Is there a way to get this information out?

回答 (3 件)

Grzegorz Knor
Grzegorz Knor 2017 年 4 月 19 日
For sparse matrices check function spparms :
spparms('spumoni',2)
sparse(rand(3))\rand(3,1)
On my computer it produces:
sp\: bandwidth = 2+1+2.
sp\: is A diagonal? no.
sp\: is band density (1) > bandden (0.5) to try banded solver? yes.
sp\: is LAPACK's banded solver successful? yes.

the cyclist
the cyclist 2015 年 10 月 27 日
Look at the Algorithms section of this documentation page.
  1 件のコメント
Jonathan
Jonathan 2015 年 10 月 27 日
Yes, thanks, as I stated in my question, I already looked at that. However, what I what to know is if I can give mldivide various inputs and have it automatically tell me which algorithm it choses.
In other words, I want to automate this process.

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Jan
Jan 2015 年 10 月 27 日
編集済み: Jan 2015 年 10 月 27 日
I'm not sure if I understand your question. You explain:
The documentation for mldivide shows a flow chart that explains which solver it will use.
Is this not a valid answer of your question already? Or are you looking for the name of the library function instead of the name of the solver? Matlab does not offer the source code for mldivide.
  3 件のコメント
Steven Lord
Steven Lord 2015 年 10 月 30 日
Even if it DID tell you, for what purpose would you use that information?
Perhaps you would be better served by the LINSOLVE function, which allows you to specify various properties of the coefficient matrix for the system you're trying to solve. This can avoid some of the testing for those matrix properties (which for large matrices could be expensive.)

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