Understand sharp ridge function

[yedidyakfir]

Hi, I am using the Coco dataset for my research. I am trying to reverse-engineer some of the functions. I went over the c code for the sharp ridge function. It seems the function equation is sqrt(sum(x_i ** 2)) + sum(x_i ** 2), but I cant find out the rotation and other manipulation you do for each instance of the function. Can you help me with that?

Thanks

[FMGS666]

Hello,

the transformations are done in the f_sharp_ridge_bbob_problem_allocate function. If you want to find where these transformations are done you need to look at the f_sharp_ridge_bbob_problem_allocate function in the f_sharp_ridge.c file. You can see that all the transformations of the instance of the function are done in lines 82-106.

In general, the code of each test function follows pretty much the following structure:

f_<function-name>_raw which Implements the raw and unaltered function ($f:\Omega \rightarrow \mathbb{R}$ WITHOUT any random transformation) without connections to any COCO structures. It takes as input the decision vector $x\in\Omega$ and simply returns the scalar value $f(x)$

f_<function-name>_evaluate Which evaluates the solution on a given instance of coco_problem_t (passed as first input to the function) by wrapping around the f_<function-name>_raw function

f_<function-name>_bbob_problem_allocate which handles the actual allocation of the coco_problem_t instance and also computes all the random transformations.

I hope this could help

[yedidyakfir]

the instance of the function are done

Can you explain a bit more? I see there is a calculation of 2 rotation matrices there, but where do we apply them to the data? Or make any use of them? In other words, how do this lines affect the evaluate method?

[brockho]

FYI, the mathematical definitions of the bbob functions can be found here: https://numbbo.github.io/gforge/downloads/download16.00/bbobdocfunctions.pdf

[FMGS666]

Let us only consider the transformation that shifts vertically the objective function.

For a given function $f: \mathbb{R}^d \rightarrow \mathbb{R}$ and a solution $x \in \mathbb{R}^d$, this transformation is simply given by $f(x) + f{opt}$ where $f{opt} \in \mathbb{R}$ is determined in a pseudo random fashion

Let's ignore now how to determine the value of $f_{opt} $ and let's imagine that we have computed it and we have it stored in a double variable named fopt (You can see that it is computed by using the bbob2009_compute_fopt function)

Each transformation is defined as a wrapper around the coco_problem_t instance that implements the target transformation by nesting coco_problem_t inside each other (using coco_problem_t -> data and coco_problem_transformed_data_t -> inner_problem) and wrapping around the coco_problem_t->evaluate_function member function to apply the desired transformation at each nesting level.

In the f_sharp_ridge_bbob_problem_allocate function, still at the beginning, we read the declaration of a NULL pointer to a coco_problem_t:

  coco_problem_t *problem = NULL;

Now the problem is initialized by means of the f_sharp_ridge_allocate that allocates the coco_problem_t for the sharp ridge function with the following line

  problem = f_sharp_ridge_allocate(dimension);

The next line reads:

  problem = transform_obj_shift(problem, fopt);

This function returns the wrapped problem that applies the transformation as I mentioned before.

You can think of this new problem as a wrapper around the initial inner problem (i.e.: the one that we have allocated with the f_sharp_ridge_allocate function), whose evaluate_function member function calls the evaluate_function of the inner problem (in our case f_sharp_ridge_evaluate) and adds fopt to its result.

after this line the new problem variable will be the transformed version of the problem, while the original problem will end up in problem->data->inner_problem, if I am not wrong.

When we call the problem->evaluate_function member function, we call it on the new "external" problem and what will actually happen (in a stylized synthesis) is that we will be first calling problem->data->inner_problem and then applying the transformation, so that we will be returning the wanted (transformed) value.

I hope this was clearer.

[olafmersmann]

Can this issue be closed, or are there open questions left about the function definition?