130 lines
No EOL
5 KiB
Python
130 lines
No EOL
5 KiB
Python
# -*- coding: utf-8 -*-
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import numpy as np
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import matplotlib.pyplot as plt
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def cubic_poly(x, a0, a1, a2, a3):
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return a0 + a1*x + a2*x**2 + a3*x**3
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def polyfit4(x1, x2, x3, x4, y1, y2, y3, y4):
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A = np.array([[1, x1, x1**2, x1**3], [1, x2, x2**2, x2**3], [1, x3, x3**2, x3**3], [1, x4, x4**2, x4**3]])
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y = np.array([y1, y2, y3, y4])
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a = np.linalg.solve(A.transpose()@A, A.transpose()@y)
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return a
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def cubic_lagrange_minimize(f, a, b, tol=1e-6):
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''' Function to find minimum of f over interval [a, b] using cubic Lagrange polynomial interpolation.
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If the function is monotonic, then the minimum is one of the bounds of the interval, and the minimum is found in a single iteration.
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The best estimate of the minimum is returned.
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The number of function evaluations is 2 + 2*Niter.
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'''
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# initialize interval endpoints and function values
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x0, x1, x2, x3 = a, a*2./3. + b*1./3., a*1./3. + b*2./3., b
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f0, f3 = f(x0), f(x3)
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x_prev = x0
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small_coefficient = 1e-9
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delta = 1# only needed to debug print
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x_sol_1 = 1# only needed to debug print
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x_sol_2 = 1# only needed to debug print
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y_sol_1 = 1# only needed to debug print
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y_sol_2 = 1# only needed to debug print
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reduction_factor = 3# reduction factor of interval size at each iteration is 3 because we sample two points in the interval each iteration
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Niter = int(np.ceil(np.log(np.abs(b-a)/tol)/np.log(reduction_factor)))
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for i in range(Niter):
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# Compute function values at two points in the interval
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x1, x2 = x0*2./3. + x3*1./3., x0*1./3. + x3*2./3.
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f1, f2 = f(x1), f(x2)
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print('-----------------')
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print(f'Iteration {i}')
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print(x0, x1, x2, x3)
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# compute Lagrange polynomial using least-squares fit to 4 points, which is equivalent to the cubic Lagrange polynomial
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a0, a1, a2, a3 = polyfit4(x0, x1, x2, x3, f0, f1, f2, f3)
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print('p : ', a0, a1, a2, a3)
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# Solve the first derivative of the Lagrange polynomial for a zero
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if np.abs(a3) > small_coefficient:
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delta = -3*a1*a3 + a2**2
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if delta < 0:
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x_sol = x0 if f0 < f3 else x3 # just choose the interval tha contains the minimum of the linear polynomial
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y_sol = cubic_poly(x_sol, a0, a1, a2, a3)
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else: # solve for the two solutions of the quadratic equation of the first derivative of the Lagrange polynomial
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x_sol_1 = (-a2 + np.sqrt(delta))/(3*a3)
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x_sol_2 = (-a2 - np.sqrt(delta))/(3*a3)
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y_sol_1 = cubic_poly(x_sol_1, a0, a1, a2, a3)
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y_sol_2 = cubic_poly(x_sol_2, a0, a1, a2, a3)
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x_sol = x_sol_1 if y_sol_1 < y_sol_2 else x_sol_2
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y_sol = y_sol_1 if y_sol_1 < y_sol_2 else y_sol_2
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elif np.abs(a2) > small_coefficient: # if a3 is zero, then the Lagrange polynomial is a quadratic polynomial
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x_sol = -a1/(2*a2)
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y_sol = cubic_poly(x_sol, a0, a1, a2, a3)
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else: # if a3 and a2 are zero, then the Lagrange polynomial is a linear polynomial
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x_sol = x0 if f0 < f3 else x3 # just choose the interval tha contains the minimum of the linear polynomial
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y_sol = cubic_poly(x_sol, a0, a1, a2, a3)
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print(f'{delta}, f({x_sol_1}) = {y_sol_1} f({x_sol_2}) = {y_sol_2}\t{x_sol}')
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if np.abs(x_sol - x_prev) < tol:
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break
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# Determine which interval contains the minimum of the cubic polynomial
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if x_sol < x1:
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x3, f3 = x1, f1
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elif x_sol < x2:
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x0, f0 = x1, f1
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x3, f3 = x2, f2
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else:
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x0, f0 = x2, f2
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x_prev = x_sol
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# return best estimate of minimum
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if y_sol < f0 and y_sol < f3:
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return x_sol
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elif f0 < f3:
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return x0
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else:
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return x3
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def f(x):
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return x**2 + np.sin(5*x)
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# return (x-1)**2
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# return -x
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# return np.exp(x)
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# return -np.exp(-x**2)
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# return np.exp(-x**2)
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# return np.ones_like(x)
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if __name__ == '__main__':
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def test_cubic_lagrange_polynomial():
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x = np.linspace(0, 16, 100)
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x1, x2, x3, x4 = 1, 3, 7, 15
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y1, y2, y3, y4 = 1, -2, 3.5, 5
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a0, a1, a2, a3 = polyfit4(x1, x2, x3, x4, y1, y2, y3, y4)
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print(a0, a1, a2, a3)
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y = cubic_poly(x, a0, a1, a2, a3)
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# plt.plot(x, cubic_lagrange(x, x1, x2, x3, x4, y1, y2, y3, y4), 'b', label='cubic Lagrange ChatGPT one shot')
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plt.plot(x, y, 'g', label='cubic Lagrange through coeffs polyfit style baby')
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plt.plot([x1, x2, x3, x4], [y1, y2, y3, y4], 'ro')
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plt.grid()
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plt.show()
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def test_cubic_lagrange_minimize():
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x = np.linspace(-2, 2, 100)
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x_min = cubic_lagrange_minimize(f, -1.2, 1.5, tol=1e-6)
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print('Solution : ', x_min, f(x_min))
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plt.plot(x, f(x), 'b')
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plt.plot(x_min, f(x_min), 'ro')
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plt.grid()
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plt.show()
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test_cubic_lagrange_polynomial()
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test_cubic_lagrange_minimize() |