Source code for NiaPy.benchmarks.rosenbrock

# encoding=utf8

"""Rosenbrock benchmark."""

import math
from NiaPy.benchmarks.benchmark import Benchmark

__all__ = ['Rosenbrock']


[docs]class Rosenbrock(Benchmark): r"""Implementation of Rosenbrock benchmark function. Date: 2018 Authors: Iztok Fister Jr. and Lucija Brezočnik License: MIT Function: **Rosenbrock function** :math:`f(\mathbf{x}) = \sum_{i=1}^{D-1} \left (100 (x_{i+1} - x_i^2)^2 + (x_i - 1)^2 \right)` **Input domain:** The function can be defined on any input domain but it is usually evaluated on the hypercube :math:`x_i ∈ [-30, 30]`, for all :math:`i = 1, 2,..., D`. **Global minimum:** :math:`f(x^*) = 0`, at :math:`x^* = (1,...,1)` LaTeX formats: Inline: $f(\mathbf{x}) = \sum_{i=1}^{D-1} (100 (x_{i+1} - x_i^2)^2 + (x_i - 1)^2)$ Equation: \begin{equation} f(\mathbf{x}) = \sum_{i=1}^{D-1} (100 (x_{i+1} - x_i^2)^2 + (x_i - 1)^2) \end{equation} Domain: $-30 \leq x_i \leq 30$ 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. """ Name = ['Rosenbrock']
[docs] def __init__(self, Lower=-30.0, Upper=30.0): r"""Initialize of Rosenbrock 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'''$f(\mathbf{x}) = \sum_{i=1}^{D-1} (100 (x_{i+1} - x_i^2)^2 + (x_i - 1)^2)$'''
[docs] def function(self): r"""Return benchmark evaluation function. Returns: Callable[[int, Union[int, float, List[int, float], numpy.ndarray]], float]: Fitness function """ def evaluate(D, sol): r"""Fitness function. Args: D (int): Dimensionality of the problem sol (Union[int, float, List[int, float], numpy.ndarray]): Solution to check. Returns: float: Fitness value for the solution. """ val = 0.0 for i in range(D - 1): val += 100.0 * math.pow(sol[i + 1] - math.pow((sol[i]), 2), 2) + math.pow((sol[i] - 1), 2) return val return evaluate