Source code for niapy.problems.katsuura

# encoding=utf8

"""Implementations of Katsuura functions."""

from math import fabs
from niapy.problems.problem import Problem

__all__ = ['Katsuura']


[docs]class Katsuura(Problem): r"""Implementations of Katsuura functions. Date: 2018 Author: Klemen Berkovič License: MIT Function: **Katsuura Function** :math:`f(\textbf{x}) = \frac{10}{D^2} \prod_{i=1}^D \left( 1 + i \sum_{j=1}^{32} \frac{\lvert 2^j x_i - round\left(2^j x_i \right) \rvert}{2^j} \right)^\frac{10}{D^{1.2}} - \frac{10}{D^2}` **Input domain:** The function can be defined on any input domain but it is usually evaluated on the hypercube :math:`x_i ∈ [-100, 100]`, for all :math:`i = 1, 2,..., D`. **Global minimum:** :math:`f(x^*) = 0`, at :math:`x^* = (420.968746,...,420.968746)` LaTeX formats: Inline: $f(\textbf{x}) = \frac{10}{D^2} \prod_{i=1}^D \left( 1 + i \sum_{j=1}^{32} \frac{\lvert 2^j x_i - round\left(2^j x_i \right) \rvert}{2^j} \right)^\frac{10}{D^{1.2}} - \frac{10}{D^2}$ Equation: \begin{equation} f(\textbf{x}) = \frac{10}{D^2} \prod_{i=1}^D \left( 1 + i \sum_{j=1}^{32} \frac{\lvert 2^j x_i - round\left(2^j x_i \right) \rvert}{2^j} \right)^\frac{10}{D^{1.2}} - \frac{10}{D^2} \end{equation} Domain: $-100 \leq x_i \leq 100$ Reference: http://www5.zzu.edu.cn/__local/A/69/BC/D3B5DFE94CD2574B38AD7CD1D12_C802DAFE_BC0C0.pdf """
[docs] def __init__(self, dimension=5, lower=-100.0, upper=100.0, *args, **kwargs): r"""Initialize Katsuura problem.. Args: dimension (Optional[int]): Dimension of the problem. lower (Optional[Union[float, Iterable[float]]]): Lower bounds of the problem. upper (Optional[Union[float, Iterable[float]]]): Upper bounds of the problem. See Also: :func:`niapy.problems.Problem.__init__` """ super().__init__(dimension, lower, upper, *args, **kwargs)
[docs] @staticmethod def latex_code(): r"""Return the latex code of the problem. Returns: str: Latex code. """ return r'''$f(\textbf{x}) = \frac{10}{D^2} \prod_{i=1}^D \left( 1 + i \sum_{j=1}^{32} \frac{| 2^j x_i - round\left(2^j x_i \right) |}{2^j} \right)^\frac{10}{D^{1.2}} - \frac{10}{D^2}$'''
def _evaluate(self, x): val = 1.0 for i in range(self.dimension): val_t = 1.0 for j in range(1, 33): val_t += fabs(2 ** j * x[i] - round(2 ** j * x[i])) / 2 ** j val *= (1 + (i + 1) * val_t) ** (10 / self.dimension ** 1.2) - (10 / self.dimension ** 2) return 10 / self.dimension ** 2 * val