Re: [eigen] Optimization advice for a specific expression

[Christoph Hertzberg <chtz@xxxxxxxxxxxxxxxxxxxxxxxx>]

On 2016-02-04 17:02, Alberto Luaces wrote:
> > Other than that: * What exactly do you mean by homogeneous?
>
> G = [ v3x3 | 0]
>      [  0ᵀ  | 1]
>
> where "v3x3" is a 3x3 matrix, and "0" a vector of 3 zero-valued elements.
>
> > Can you make any assumptions on the determinant of G? (Calculating the
> > det didn't not seem to be the bottleneck, though)
>
> No, I cannot, but as you point, removing the determinant computation for
> testing purposes does not yield any improvement.

Given the above, you have:
  G.determinant() == v3x3.determinant(),
which should be significantly faster than the full determinant.
If v3x3 happens to be a rotation matrix, G.determinant()==1;

> > Does it have any special form (e.g. low rank, or Identity+lowRank)
>
> It is just symmetric:
>
> Eplus << 2 , 1 , 1 , 5,
>           1 , 2 , 1 , 5,
>           1 , 1 , 2 , 5,
>           5 , 5 , 5 , 20;

The upper left is Identity + [1,1,1]^T*[1,1,1]. I'll abbreviate V=v3x3.
If you then define w=V*[1,1,1]^T == V.rowwise().sum(), the resulting 
summand is (without the determinant factor):
  [ V*V^T + w*w^T,  5*w ]
  [ 5*w^T        ,  20  ]

Again, if V happens to be a rotation, then V*V^T=Identity

So with a bit of hand-coding you may gain a lot efficiency here. It may 
be tricky to get it right that vectorization can still be used. E.g., 
instead of V.rowwise().sum() you can calculate Vector4d 
w=G.leftCols<3>().rowwise().sum(), which should be near optimal [again, 
check the generated assembly]
Multiplication by 5 can be done at the end (if I4x4 initially is 0), and 
the lower right element is just 20*{sum of determinants}.

Cheers,
Christoph



--
 Dipl. Inf., Dipl. Math. Christoph Hertzberg

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