[no subject]

What I was talking about was the development branch (pre 2.1).

Can you try it?

Benoit

2009/6/8 Benoit Jacob <jacob.benoit.1@xxxxxxxxx>:
> There's been discussion recently over this value and indeed we should
> reduce it a bit, perhaps down to 1e-13.
>
> There's no way that it can make sense to make it lower than the
> machine epsilon for double which is roughly 1e-16. See the source code
> of ei_isMuchSmallerThan, and ei_isApprox : basically, this would turn
> these fuzzy comparisons into exact comparisons !!
>
> 2009/6/8 Scott Stephens <stephens.js@xxxxxxxxx>:
>> I was having a problem with a Cholesky decomposition failing with the
>> indication that the matrix was not positive definite. =A0I was surprised
>> because I was able to successfully perform the Cholesky decomposition
>> in MATLAB. =A0I tracked this down to the value returned by
>> precision<double>() as defined in src/Core/MathFunctions.h, the square
>> root of which is being used as the cut-off for numerically equivalent
>> to zero in the LLT code. =A0The return value is hard-coded to 1e-11; I
>> was able to match MATLAB by setting this to 1e-18 instead.
>
> If you really had to use such a small value, smaller than machine
> epsilon, that probably means that your matrix isn't significantly
> distinguishable from an indefinite one and then you should use LDLT
> instead of LLT. Or else you found a bug in our code. It would be
> interesting if you could run a SVD on this matrix and find out its
> condition number. if it's greater than 1e15 then your matrix isn't
> significantly definite.
>
>>
>> Is there a particular reason 1e-11 was chosen for this value? =A0Would
>> it make sense to go with a smaller value by default, or at least offer
>> the ability to configure this number as part of the LLT interface?
>
> The current value is determined by a formula that Keir found in a
> paper by Higham (see source code) and that we later rederived
> experimentally (see thread "LU precision tuning"). So we can be very
> confident that this is the right formula and there's hardly a chance
> that the user might know better... can you check the condition number
> of your matrix, check whether Matlab uses plain LLT or LDLT (that
> would explain it), check the accuracy of the result it gives...
>
>> =A0Or is there a better way for me to solve this problem? (Maybe some ki=
nd
>> of scaling before the decomposition, then rescaling when I'm done?)
>
> Scaling won't change the condition number :)
> Then I'm no big expert in numerical questions. All I can say is try
> LDLT... maybe someone will know better.
>
> Benoit
>