Source code for niapy.problems.whitley

# encoding=utf8

"""Whitley function."""

import numpy as np
from niapy.problems.problem import Problem

__all__ = ['Whitley']


[docs]class Whitley(Problem): r"""Implementation of Whitley function. Date: 2018 Authors: Grega Vrbančič and Lucija Brezočnik License: MIT Function: **Whitley function** :math:`f(\mathbf{x}) = \sum_{i=1}^D \sum_{j=1}^D \left(\frac{(100(x_i^2-x_j)^2 + (1-x_j)^2)^2}{4000} - \cos(100(x_i^2-x_j)^2 + (1-x_j)^2)+1\right)` **Input domain:** The function can be defined on any input domain but it is usually evaluated on the hypercube :math:`x_i ∈ [-10.24, 10.24]`, for all :math:`i = 1, 2,..., D`. **Global minimum:** :math:`f(x^*) = 0`, at :math:`x^* = (1,...,1)` LaTeX formats: Inline: $f(\mathbf{x}) = \sum_{i=1}^D \sum_{j=1}^D \left(\frac{(100(x_i^2-x_j)^2 + (1-x_j)^2)^2}{4000} - \cos(100(x_i^2-x_j)^2 + (1-x_j)^2)+1\right)$ Equation: \begin{equation}f(\mathbf{x}) = \sum_{i=1}^D \sum_{j=1}^D \left(\frac{(100(x_i^2-x_j)^2 + (1-x_j)^2)^2}{4000} - \cos(100(x_i^2-x_j)^2 + (1-x_j)^2)+1\right) \end{equation} Domain: $-10.24 \leq x_i \leq 10.24$ Reference paper: Jamil, M., and Yang, X. S. (2013). A literature survey of benchmark functions for global optimisation problems. International Journal of Mathematical Modelling and Numerical Optimisation, 4(2), 150-194. """
[docs] def __init__(self, dimension=4, lower=-10.24, upper=10.24, *args, **kwargs): r"""Initialize Whitley 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"""$f(\mathbf{x}) = \sum_{i=1}^D \sum_{j=1}^D \left(\frac{(100(x_i^2-x_j)^2 + (1-x_j)^2)^2}{4000} - \cos(100(x_i^2-x_j)^2 + (1-x_j)^2)+1\right)$"""
def _evaluate(self, x): xi = np.tile(x, (self.dimension, 1)).T xj = np.tile(x, (self.dimension, 1)) tmp = 100.0 * (xi ** 2 - xj) ** 2 + (1 - xj) ** 2 return np.sum((tmp ** 2) / 4000.0 - np.cos(tmp) + 1.0)