Source code for NiaPy.benchmarks.pinter

# encoding=utf8
# pylint: disable=anomalous-backslash-in-string
import math

__all__ = ['Pinter']


[docs]class Pinter(object): r"""Implementation of Pintér function. Date: 2018 Author: Lucija Brezočnik License: MIT Function: **Pintér function** :math:`f(\mathbf{x}) = \sum_{i=1}^D ix_i^2 + \sum_{i=1}^D 20i \sin^2 A + \sum_{i=1}^D i \log_{10} (1 + iB^2);` :math:`A = (x_{i-1}\sin(x_i)+\sin(x_{i+1}))\quad \text{and} \quad` :math:`B = (x_{i-1}^2 - 2x_i + 3x_{i+1} - \cos(x_i) + 1)` **Input domain:** The function can be defined on any input domain but it is usually evaluated on the hypercube :math:`x_i ∈ [-10, 10]`, for all :math:`i = 1, 2,..., D`. **Global minimum:** :math:`f(x^*) = 0`, at :math:`x^* = (0,...,0)` LaTeX formats: Inline: $f(\mathbf{x}) = \sum_{i=1}^D ix_i^2 + \sum_{i=1}^D 20i \sin^2 A + \sum_{i=1}^D i \log_{10} (1 + iB^2); A = (x_{i-1}\sin(x_i)+\sin(x_{i+1}))\quad \text{and} \quad B = (x_{i-1}^2 - 2x_i + 3x_{i+1} - \cos(x_i) + 1)$ Equation: \begin{equation} f(\mathbf{x}) = \sum_{i=1}^D ix_i^2 + \sum_{i=1}^D 20i \sin^2 A + \sum_{i=1}^D i \log_{10} (1 + iB^2); A = (x_{i-1}\sin(x_i)+\sin(x_{i+1}))\quad \text{and} \quad B = (x_{i-1}^2 - 2x_i + 3x_{i+1} - \cos(x_i) + 1) \end{equation} Domain: $-10 \leq x_i \leq 10$ 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. """ def __init__(self, Lower=-10.0, Upper=10.0): self.Lower = Lower self.Upper = Upper
[docs] @classmethod def function(cls): def evaluate(D, sol): val1 = 0.0 val2 = 0.0 val3 = 0.0 for i in range(D): if i == 0: sub = sol[D - 1] add = sol[i + 1] elif i == D - 1: sub = sol[i - 1] add = sol[0] else: sub = sol[i - 1] add = sol[i + 1] A = (sub * math.sin(sol[i]) + math.sin(add)) B = (math.pow(sub, 2) - 2.0 * sol[i] + 3.0 * add - math.cos(sol[i]) + 1.0) val1 += (i + 1.0) * math.pow(sol[i], 2) val2 += 20.0 * (i + 1.0) * math.pow(math.sin(A), 2) val3 += (i + 1.0) * math.log10(1.0 + (i + 1.0) * math.pow(B, 2)) return val1 + val2 + val3
return evaluate