# 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

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:
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)

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.

: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)