Source code for NiaPy.benchmarks.weierstrass
# encoding=utf8
"""Implementations of Weierstrass functions."""
from math import pi, cos
from NiaPy.benchmarks.benchmark import Benchmark
__all__ = ['Weierstrass']
[docs]class Weierstrass(Benchmark):
r"""Implementations of Weierstrass functions.
Date: 2018
Author: Klemen Berkovič
License: MIT
Function:
**Weierstass Function**
:math:`f(\textbf{x}) = \sum_{i=1}^D \left( \sum_{k=0}^{k_{max}} a^k \cos\left( 2 \pi b^k ( x_i + 0.5) \right) \right) - D \sum_{k=0}^{k_{max}} a^k \cos \left( 2 \pi b^k \cdot 0.5 \right)`
**Input domain:**
The function can be defined on any input domain but it is usually
evaluated on the hypercube :math:`x_i ∈ [-100, 100]`, for all :math:`i = 1, 2,..., D`.
Default value of a = 0.5, b = 3 and k_max = 20.
**Global minimum:** :math:`f(x^*) = 0`, at :math:`x^* = (420.968746,...,420.968746)`
LaTeX formats:
Inline:
$$f(\textbf{x}) = \sum_{i=1}^D \left( \sum_{k=0}^{k_{max}} a^k \cos\left( 2 \pi b^k ( x_i + 0.5) \right) \right) - D \sum_{k=0}^{k_{max}} a^k \cos \left( 2 \pi b^k \cdot 0.5 \right)
Equation:
\begin{equation} f(\textbf{x}) = \sum_{i=1}^D \left( \sum_{k=0}^{k_{max}} a^k \cos\left( 2 \pi b^k ( x_i + 0.5) \right) \right) - D \sum_{k=0}^{k_{max}} a^k \cos \left( 2 \pi b^k \cdot 0.5 \right) \end{equation}
Domain:
$-100 \leq x_i \leq 100$
Reference:
http://www5.zzu.edu.cn/__local/A/69/BC/D3B5DFE94CD2574B38AD7CD1D12_C802DAFE_BC0C0.pdf
"""
Name = ['Weierstrass']
a, b, k_max = 0.5, 3, 20
[docs] def __init__(self, Lower=-100.0, Upper=100.0, a=0.5, b=3, k_max=20):
r"""Initialize of Bent Cigar benchmark.
Args:
Lower (Optional[float]): Lower bound of problem.
Upper (Optional[float]): Upper bound of problem.
a (Optional[float]): TODO
b (Optional[float]): TODO
k (Optional[float]): TODO
See Also:
:func:`NiaPy.benchmarks.Benchmark.__init__`
"""
Benchmark.__init__(self, Lower, Upper)
Weierstrass.a, Weierstrass.b, Weierstrass.k_max = a, b, k_max
[docs] @staticmethod
def latex_code():
r"""Return the latex code of the problem.
Returns:
str: Latex code
"""
return r'''$f(\textbf{x}) = \sum_{i=1}^D \left( \sum_{k=0}^{k_{max}} a^k \cos\left( 2 \pi b^k ( x_i + 0.5) \right) \right) - D \sum_{k=0}^{k_{max}} a^k \cos \left( 2 \pi b^k \cdot 0.5 \right)$'''
[docs] def function(self):
r"""Return benchmark evaluation function.
Returns:
Callable[[int, Union[int, float, List[int, float], numpy.ndarray]], float]: Fitness function
"""
a, b, k_max = self.a, self.b, self.k_max
def f(D, sol, a=a, b=b, k_max=k_max):
r"""Fitness function.
Args:
D (int): Dimensionality of the problem
sol (Union[int, float, List[int, float], numpy.ndarray]): Solution to check.
a (Optional[float]): TODO
b (Optional[float]): TODO
k (Optional[float]): TODO
Returns:
float: Fitness value for the solution.
"""
val1 = 0.0
for i in range(D):
val = 0.0
for k in range(k_max): val += a ** k * cos(2 * pi * b ** k * (sol[i] + 0.5))
val1 += val
val2 = 0.0
for k in range(k_max): val2 += a ** k * cos(2 * pi * b ** k * 0.5)
return val1 - D * val2
return f
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