Cost functions with negative terms: E = (...)^2 - (...)^2 + ....

[Toni Mateos]

Hi, I need to solve a problem with negative terms in the cost function. By that, I mean that the cost function is of the style:

E(x,y,z) = [ c1 Exp(-c2 x^2) ]^2   -  [ c3 Exp(-c4 x^2) ]^2 +  ...

where c1,...,c4 are constants. I can still prove that E > 0 for all x,y,z given the constants of my problem, i.e. the problem is well-posed. But there is now way to write E as a sum without negative signs.

Is there a way I can run the problem with those minus signs? I tried using a LossFunction with trivial -1 scaling  rho(s) = -s, but there are complaints appearing at run-time in various points of the evaluation.

Help is much welcome!!

Toni

[Sameer Agarwal]

Hi Toni,

Generally speaking problems of this form cannot be solved using Ceres.

In your case if you are certain that your objective function is positive, then you could construct a costfunction which evaluates the square root of the objective function. 

I am not sure how many terms you have in your objective, so this may or may not be an effective way of solving the problem.

Sameer

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[Toni Mateos]

Hi Sameer,

I managed to hack in a simple way, and the program is running great now, producing the expected results! 

You might like to consider including this option in future releases, as it really helps (at least, I often encounter problems like these), and it's just a couple of lines.

In the file corrector.xx ,I replaced:

   CHECK_GT(rho[1], 0.0);

   sqrt_rho1_ = sqrt(rho[1]);

   .

   .

   residual_scaling_ =  sqrt_rho1_;

By:

  bool isScalingNegative = (rho[1] < 0) ? true : false;

  if (isScalingNegative) {sqrt_rho1_ = sqrt(-rho[1]);}

  else {sqrt_rho1_ = sqrt(rho[1]);}

  .

  .

  residual_scaling_ = (isScalingNegative) ? -sqrt_rho1_ : sqrt_rho1_;

Note that in my case, where rho is just a sign change, I just have rho[1] = -1, rho[2] = 0.

Thanks!

Toni

[Sameer Agarwal]

Hi Toni,

I am surprised that your hack works, because the underlying math for how the loss functions work does not work anymore.

Sameer

[Toni Mateos]

I know. Actually, the global cost function is perfectly fine, always positive whatever the value of the unknowns. It's just that sometimes it's hard (or impossible) to write it as a sum of positive terms. A simple example is:

E = (x-1)^2  - 1/2  (x-1/2)^2

Which is obviously equivalent to minimizing E = a (x-b) ^2 + c,  where a = 1/2, and b,c are whatever turns out in this case. Clearly, it's perfectly behaved...

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