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