# Source code for niapy.problems.weierstrass

# encoding=utf8
"""Implementations of Weierstrass functions."""

import numpy as np
from niapy.problems.problem import Problem

__all__ = ['Weierstrass']

[docs]class Weierstrass(Problem):
r"""Implementations of Weierstrass functions.

Date: 2018

Author: Klemen Berkovič

Function:
**Weierstrass 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:
$$f(\textbf{x}) = \sum_{i=1}^D \left( \sum_{k=0}^{k_{max}} a^k \cos\left( 2 \pi b^k ( x_i + 0.5) \right) \right) - D \sum_{k=0}^{k_{max}} a^k \cos \left( 2 \pi b^k \cdot 0.5 \right)$$

Domain:
$-100 \leq x_i \leq 100$

Reference:

"""

[docs]    def __init__(self, dimension=4, lower=-100.0, upper=100.0, a=0.5, b=3, k_max=20, *args, **kwargs):
r"""Initialize Bent Cigar problem..

Args:
dimension (Optional[int]): Dimension of the problem.
lower (Optional[Union[float, Iterable[float]]]): Lower bounds of the problem.
upper (Optional[Union[float, Iterable[float]]]): Upper bounds of the problem.
a (Optional[float]): The a parameter.
b (Optional[float]): The b parameter.
k_max (Optional[int]): Number of elements of the series to compute.

:func:niapy.problems.Problem.__init__

"""
super().__init__(dimension, lower, upper, *args, **kwargs)
self.a = a
self.b = b
self.k_max = k_max

[docs]    @staticmethod
def latex_code():
r"""Return the latex code of the problem.

Returns:
str: Latex code.

"""
return r'''$f(\textbf{x}) = \sum_{i=1}^D \left( \sum_{k=0}^{k_{max}} a^k \cos\left( 2 \pi b^k ( x_i + 0.5) \right) \right) - D \sum_{k=0}^{k_{max}} a^k \cos \left( 2 \pi b^k \cdot 0.5 \right)$'''

def _evaluate(self, x):
val1 = 0.0
for i in range(self.dimension):
val = 0.0
for k in range(self.k_max):
val += self.a ** k * np.cos(2.0 * np.pi * self.b ** k * (x[i] + 0.5))
val1 += val
val2 = 0.0
for k in range(self.k_max):
val2 += self.a ** k * np.cos(2 * np.pi * self.b ** k * 0.5)
return val1 - self.dimension * val2