Trying to improve the Klement-solver

[Deltadahl]

Trying to solve #20 and #19.
I tried the approach described in #18

But I have not used the singularity test @YingboMa explained:
"Or F = lu(J), and the singularity check is then abs(F.factors[end, end]) < singular_tol"
Although that solution was faster I was afraid that the LU-decomposition lu(J) could result in an error if J was singular.

Therefore I read about good approaches to checking for a singular matrix and found that

cond(J) > tolerance_value

was a good approach (https://stackoverflow.com/questions/13145948/how-to-find-out-if-a-matrix-is-singular and https://discourse.julialang.org/t/checking-for-singularity-of-matrix/54746/4)

Is this OK, or do you want me to change something?

[codecov]

Codecov Report

Merging #22 (8915554) into main (e681591) will decrease coverage by 1.85%.
The diff coverage is 93.18%.

@@            Coverage Diff             @@
##             main      #22      +/-   ##
==========================================
- Coverage   91.32%   89.47%   -1.86%     
==========================================
  Files           8        8              
  Lines         219      247      +28     
==========================================
+ Hits          200      221      +21     
- Misses         19       26       +7     

Impacted Files	Coverage Δ
src/klement.jl	90.56% <91.89%> (+0.24%)	⬆️
src/broyden.jl	92.59% <100.00%> (ø)
src/utils.jl	100.00% <100.00%> (ø)
src/falsi.jl	84.09% <0.00%> (-11.37%)	⬇️

📣 We’re building smart automated test selection to slash your CI/CD build times. Learn more

[ChrisRackauckas]

Although that solution was faster I was afraid that the LU-decomposition lu(J) could result in an error if J was singular.

Just do lu(A, check = false).

  lu(A, pivot = RowMaximum(); check = true) -> F::LU

  Compute the LU factorization of A.

  When check = true, an error is thrown if the decomposition fails. When check = false, responsibility for checking the decomposition's validity (via issuccess) lies with the user.

[ChrisRackauckas]

Do this to Broyden too

[Deltadahl]

I didn't manage to get the program non-allocating on static-arrays due to @ballocated lu(J, check = false) > 0.
To get the program non-allocating for scalars, I split the program into 2 cases.

Just tell me if I can improve anything,
And thx for all the good feedback, I'm learning a lot from it! :)

[ChrisRackauckas]

This does one more LU factorization than is needed. Instead, move the factorization to the top

[YingboMa]

This allocates unnecessarily.

[Deltadahl]

I didn't get abs(F.factors[end, end]) to work for static arrays (F.factors gives error)

[Deltadahl]

And for non-static arrays abs(F.factors[end, end]) worked fine if J was non-singular.
But it didn't work as I excpected when J was singular, i.e. it gave abs(F.factors[end, end])=3.0 and abs(F.factors[end-1, end-1])=0.0, resulting in the solver ending up with NaN as output.

[ChrisRackauckas]

Let's open up an issue on this. It may need to be specialized more, but for now we can go with this.