Newton's method in $n$ dimensions¶
Copyright (C) 2020 Andreas Kloeckner
MIT License
Permission is hereby granted, free of charge, to any person obtaining a copy of this software and associated documentation files (the "Software"), to deal in the Software without restriction, including without limitation the rights to use, copy, modify, merge, publish, distribute, sublicense, and/or sell copies of the Software, and to permit persons to whom the Software is furnished to do so, subject to the following conditions:The above copyright notice and this permission notice shall be included in all copies or substantial portions of the Software.
THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, EXPRESS OR IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY, FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT. IN NO EVENT SHALL THE AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING FROM, OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN THE SOFTWARE.
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 are two functions. The first one is an oblong "bowl-shaped" one made of quadratic functions.
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]])
def ddf(x):
return np.array([
[1,0],
[0,5]
])
The second one is a challenge problem for optimization algorithms known as Rosenbrock's banana function.
def f(X):
x = X[0]
y = X[1]
val = 100.0 * (y - x**2)**2 + (1.0 - x)**2
return val
def df(X):
x = X[0]
y = X[1]
val1 = 400.0 * (y - x**2) * x - 2 * x
val2 = 200.0 * (y - x**2)
return np.array([val1, val2])
def ddf(X):
x = X[0]
y = X[1]
val11 = 400.0 * (y - x**2) - 800.0 * x**2 - 2
val12 = 400.0
val21 = -400.0 * x
val22 = 200.0
return np.array([[val11, val12], [val21, val22]])
Let's take a look at these functions. First in 3D:
ax = pt.figure().add_subplot(projection="3d")
xmesh, ymesh = np.mgrid[-2:2:50j,-2:2:50j]
fmesh = f(np.array([xmesh, ymesh]))
ax.plot_surface(xmesh, ymesh, fmesh,
alpha=0.3, cmap=pt.cm.coolwarm, rstride=3, cstride=3)
<mpl_toolkits.mplot3d.art3d.Poly3DCollection at 0x7f04fcb90160>
Then as a "contour plot":
pt.axis("equal")
pt.contour(xmesh, ymesh, fmesh, 50)
<matplotlib.contour.QuadContourSet at 0x7f04fdd7c780>
- You may need to add contours to seee more detail.
- The function is not symmetric about the y axis!
Newton¶
First, initialize:
guesses = [np.array([2, 2./5])]
Then evaluate this cell lots of times:
x = guesses[-1]
s = la.solve(ddf(x), df(x))
next_guess = x - s
print(f(next_guess), next_guess)
guesses.append(next_guess)
0.0 [ 0. 0.]
Here's some plotting code to see what's going on:
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-")
[<matplotlib.lines.Line2D at 0x7f04fc7b6dd8>]
For comparison: Conjugate Gradients ("CG") -- later in the class¶
Initialize the method:
x0 = np.array([2, 2./5])
#x0 = np.array([2, 1])
iterates = [x0]
gradients = [df(x0)]
directions = [-df(x0)]
Evaluate this cell many times in-place:
# Evaluate this cell many times in-place
x = iterates[-1]
s = directions[-1]
def f1d(alpha):
return f(x + alpha*s)
alpha_opt = sopt.golden(f1d)
next_x = x + alpha_opt*s
g = df(next_x)
last_g = gradients[-1]
gradients.append(g)
beta = np.dot(g, g)/np.dot(last_g, last_g)
directions.append(-g + beta*directions[-1])
print(f(next_x))
iterates.append(next_x)
# plot function and iterates
pt.axis("equal")
pt.contour(xmesh, ymesh, fmesh, 50)
it_array = np.array(iterates)
pt.plot(it_array.T[0], it_array.T[1], "x-")
7.81186907775e-19
[<matplotlib.lines.Line2D at 0x7f04fc66e6d8>]