Copyright (C) 2020 Andreas Kloeckner
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import numpy as np
import numpy.linalg as la
import scipy.optimize as sopt
import matplotlib.pyplot as pt
from mpl_toolkits.mplot3d import axes3d
Here's a function. It's an oblong bowl made of two quadratic functions.
This is pretty much the easiest 2D optimization job out there.
def f(x):
return 0.5*x[0]**2 + 2.5*x[1]**2
def df(x):
return np.array([x[0], 5*x[1]])
Let's take a look at the function. First in 3D:
fig = pt.figure()
ax = fig.gca(projection="3d")
xmesh, ymesh = np.mgrid[-2:2:50j,-2:2:50j]
fmesh = f(np.array([xmesh, ymesh]))
ax.plot_surface(xmesh, ymesh, fmesh)
And then as a "contour plot":
pt.axis("equal")
pt.contour(xmesh, ymesh, fmesh)
Next, initialize steepest descent with a starting guess:
guesses = [np.array([2, 2./5])]
Next, run Steepest Descent:
Here's some plotting code to illustrate what just happened:
pt.figure(figsize=(9, 9))
pt.axis("equal")
pt.contour(xmesh, ymesh, fmesh, 50)
it_array = np.array(guesses)
pt.plot(it_array.T[0], it_array.T[1], "x-")
for i, guess in enumerate(guesses):
print(i, la.norm(guess, 2))
Steepest descent with added "momentum" term:
$$x_{k+1} = x_k - \alpha \nabla f(x_k) \color{red}{+ \beta (x_{k}-x_{k-1})}$$guesses = [np.array([2, 2./5])]
# beta = 0.01
beta = 0.1
# beta = 0.5
# beta = 1
Explore different choices of the "momentum parameter" $\beta$ above.
pt.figure(figsize=(9, 9))
pt.axis("equal")
pt.contour(xmesh, ymesh, fmesh, 50)
it_array = np.array(guesses)
pt.plot(it_array.T[0], it_array.T[1], "x-")
for i, guess in enumerate(guesses):
print(i, la.norm(guess, 2))