BraninProblem.py

"""
Implementation of the Branin problem

min f(x_1, x_2) = \left( x_2 - \frac{5.1}{4\pi^2} x_1^2 + \frac{5}{\pi} x_1 - 6 \right)^2 + 10 \left(1 - \frac{1}{8\pi}\right) \cos(x_1) + 10.
s.t x_1 \in [-5, 10], \quad x_2 \in [0, 15].

It has three global minima at:
(x_1, x_2) = (-\pi, 12.275) 
             (\pi, 2.275)
             (9.42478, 2.475)
and the optimal objective 0.3979

Authors:    Tucker Hartland <hartland1@llnl.gov>
            Nai-Yuan Chiang <chiang7@llnl.gov>
"""
import numpy as np
from .problem import Problem
from numpy.random import uniform

# define the Branin problem class
class BraninProblem(Problem):
    def __init__(self):
        ndim = 2
        xlimits = np.array([[-5.0, 10], [0.0, 15]]) 
        name = 'Branin'
        super().__init__(ndim, xlimits, name=name)
            
    def _evaluate(self, x: np.ndarray) -> np.ndarray:
        
        ne, nx = x.shape
        assert nx == self.ndim
        
        y = np.zeros((ne, 1), complex)
        b = 5.1 / (4.0 * (np.pi) ** 2)
        c = 5.0 / np.pi
        r = 6.0
        s = 10.0
        t = 1.0 / (8.0 * np.pi)
        
        arg1 = (x[:,1] - b * x[:,0]**2 + c * x[:,0] - r)
        y[:,0] = arg1**2 + s * (1 - t) * np.cos(x[:,0]) + s
        
        '''
        # compute derivatives
        dy_dx0 = 2*arg1*(-2*b*x[:,0]+c) - s*(1-t)*np.sin(x[:,0])
        dy_dx1 = 2*arg1
        
        dy_dx = np.array([dy_dx0, dy_dx1])
        '''

        return y