Source code for NiaPy.benchmarks.zakharov

# encoding=utf8
"""Implementations of Zakharov function."""

from NiaPy.benchmarks.benchmark import Benchmark

__all__ = ['Zakharov']

[docs]class Zakharov(Benchmark): r"""Implementations of Zakharov functions. Date: 2018 Author: Klemen Berkovič License: MIT Function: **Levy Function** :math:`f(\textbf{x}) = \sum_{i = 1}^D x_i^2 + \left( \sum_{i = 1}^D 0.5 i x_i \right)^2 + \left( \sum_{i = 1}^D 0.5 i x_i \right)^4` **Input domain:** The function can be defined on any input domain but it is usually evaluated on the hypercube :math:`x_i ∈ [-5, 10]`, for all :math:`i = 1, 2,..., D`. **Global minimum:** :math:`f(\textbf{x}^*) = 0` at :math:`\textbf{x}^* = (0, \cdots, 0)` LaTeX formats: Inline: $f(\textbf{x}) = \sum_{i = 1}^D x_i^2 + \left( \sum_{i = 1}^D 0.5 i x_i \right)^2 + \left( \sum_{i = 1}^D 0.5 i x_i \right)^4$ Equation: \begin{equation} f(\textbf{x}) = \sum_{i = 1}^D x_i^2 + \left( \sum_{i = 1}^D 0.5 i x_i \right)^2 + \left( \sum_{i = 1}^D 0.5 i x_i \right)^4 \end{equation} Domain: $-5 \leq x_i \leq 10$ Reference: https://www.sfu.ca/~ssurjano/levy.html """ Name = ['Zakharov']
[docs] def __init__(self, Lower=-5.0, Upper=10.0): r"""Initialize of Zakharov benchmark. Args: Lower (Optional[float]): Lower bound of problem. Upper (Optional[float]): Upper bound of problem. See Also: :func:`NiaPy.benchmarks.Benchmark.__init__` """ Benchmark.__init__(self, Lower, Upper)
[docs] @staticmethod def latex_code(): r"""Return the latex code of the problem. Returns: str: Latex code """ return r'''$f(\textbf{x}) = \sum_{i = 1}^D x_i^2 + \left( \sum_{i = 1}^D 0.5 i x_i \right)^2 + \left( \sum_{i = 1}^D 0.5 i x_i \right)^4$'''
[docs] def function(self): r"""Return benchmark evaluation function. Returns: Callable[[int, Union[int, float, List[int, float], numpy.ndarray]], float]: Fitness function """ def f(D, X): r"""Fitness function. Args: D (int): Dimensionality of the problem X (Union[int, float, List[int, float], numpy.ndarray]): Solution to check. Returns: float: Fitness value for the solution. """ v1, v2 = 0.0, 0.0 for i in range(D): v1, v2 = v1 + X[i] ** 2, v2 + 0.5 * (i + 1) * X[i] return v1 + v2 ** 2 + v2 ** 4 return f
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