#!/usr/bin/env python
# Created by "Thieu" at 17:31, 30/07/2022 ----------%
# Email: nguyenthieu2102@gmail.com %
# Github: https://github.com/thieu1995 %
# --------------------------------------------------%
import numpy as np
from opfunu.benchmark import Benchmark
[docs]class OddSquare(Benchmark):
"""
.. [1] Gavana, A. Global Optimization Benchmarks and AMPGO retrieved 2015
.. math::
f_{\text{OddSquare}}(x) = -e^{-\frac{d}{2\pi}} \cos(\pi d) \left( 1 + \frac{0.02h}{d + 0.01} \right )
Where, in this exercise:
.. math::
\begin{cases}
d = n \cdot \smash{\displaystyle\max_{1 \leq i \leq n}}
\left[ (x_i - b_i)^2 \right ] \\
h = \sum_{i=1}^{n} (x_i - b_i)^2
\end{cases}
And :math:`b = [1, 1.3, 0.8, -0.4, -1.3, 1.6, -0.2, -0.6, 0.5, 1.4, 1, 1.3,
0.8, -0.4, -1.3, 1.6, -0.2, -0.6, 0.5, 1.4]`
Here :math:`x_i \in [-5 \pi, 5 \pi]` for :math:`i = 1, ..., n`. `n \leq 20`.
*Global optimum*: :math:`f(x) = -1.00846728102`for :math:`x \approx b`
"""
name = "Odd Square Function"
latex_formula = r'f_{\text{OddSquare}}(x) = -e^{-\frac{d}{2\pi}} \cos(\pi d) \left( 1 + \frac{0.02h}{d + 0.01} \right )'
latex_formula_dimension = r'd = 20'
latex_formula_bounds = r'x_i \in [-5 \pi, 5 \pi]'
latex_formula_global_optimum = r'f(b) = -1.00846728102'
continuous = False
linear = False
convex = True
unimodal = False
separable = False
differentiable = False
scalable = False
randomized_term = False
parametric = False
modality = False # Number of ambiguous peaks, unknown # peaks
def __init__(self, ndim=None, bounds=None):
super().__init__()
self.dim_changeable = True
self.dim_default = 2
self.check_ndim_and_bounds(ndim, bounds, np.array([[-5*np.pi, 5.*np.pi] for _ in range(self.dim_default)]))
self.b = np.array([1, 1.3, 0.8, -0.4, -1.3, 1.6, -0.2, -0.6, 0.5, 1.4, 1, 1.3, 0.8, -0.4, -1.3, 1.6, -0.2, -0.6, 0.5, 1.4])
self.f_global = -1.00846728102
self.x_global = self.b[:self.ndim]
[docs] def evaluate(self, x, *args):
self.check_solution(x)
self.n_fe += 1
b = self.b[0: self.ndim]
d = self.ndim * max((x - b) ** 2.0)
h = np.sum((x - b) ** 2.0)
return (-np.exp(-d / (2.0 * np.pi)) * np.cos(np.pi * d) * (1.0 + 0.02 * h / (d + 0.01)))