Source code for niapy.benchmarks.perm

# encoding=utf8

"""Implementations of Perm function."""

from niapy.benchmarks.benchmark import Benchmark

__all__ = ['Perm']


[docs]class Perm(Benchmark): r"""Implementations of Perm functions. Date: 2018 Author: Klemen Berkovič License: MIT Function: **Perm Function** :math:`f(\textbf{x}) = \sum_{i = 1}^D \left( \sum_{j = 1}^D (j - \beta) \left( x_j^i - \frac{1}{j^i} \right) \right)^2` **Input domain:** The function can be defined on any input domain but it is usually evaluated on the hypercube :math:`x_i ∈ [-D, D]`, for all :math:`i = 1, 2,..., D`. **Global minimum:** :math:`f(\textbf{x}^*) = 0` at :math:`\textbf{x}^* = (1, \frac{1}{2}, \cdots , \frac{1}{i} , \cdots , \frac{1}{D})` LaTeX formats: Inline: $f(\textbf{x}) = \sum_{i = 1}^D \left( \sum_{j = 1}^D (j - \beta) \left( x_j^i - \frac{1}{j^i} \right) \right)^2$ Equation: \begin{equation} f(\textbf{x}) = \sum_{i = 1}^D \left( \sum_{j = 1}^D (j - \beta) \left( x_j^i - \frac{1}{j^i} \right) \right)^2 \end{equation} Domain: $-D \leq x_i \leq D$ Reference: https://www.sfu.ca/~ssurjano/perm0db.html """ Name = ['Perm']
[docs] def __init__(self, dimension=10.0, beta=.5): r"""Initialize of Bent Cigar benchmark. Args: dimension (float): Dimension of the problem. -dimension will be the lower bound, dimension will be the upper bound. beta (float): Beta parameter. See Also: :func:`niapy.benchmarks.Benchmark.__init__` """ super().__init__(-dimension, dimension) self.beta = beta
[docs] @staticmethod def latex_code(): r"""Return the latex code of the problem. Returns: str: Latex code. """ return r'''$f(\textbf{x}) = \sum_{i = 1}^D \left( \sum_{j = 1}^D (j - \beta) \left( x_j^i - \frac{1}{j^i} \right) \right)^2$'''
[docs] def function(self): r"""Return benchmark evaluation function. Returns: Callable[[int, Union[int, float, List[int, float], numpy.ndarray]], float]: Fitness function. """ def f(dimension, x): r"""Fitness function. Args: dimension (int): Dimensionality of the problem x (Union[int, float, List[int, float], numpy.ndarray]): Solution to check. Returns: float: Fitness value for the solution. """ v = .0 for i in range(1, dimension + 1): vv = .0 for j in range(1, dimension + 1): vv += (j + self.beta) * (x[j - 1] ** i - 1 / j ** i) v += vv ** 2 return v return f
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