Steepest Descent¶

In [1]:
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.

In [2]:
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:

In [3]:
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)

Out[3]:
<mpl_toolkits.mplot3d.art3d.Poly3DCollection at 0x7f7431a9d978>
/usr/lib/python3/dist-packages/matplotlib/backends/backend_agg.py:517: DeprecationWarning: npy_PyFile_Dup is deprecated, use npy_PyFile_Dup2
filename_or_obj, self.figure.dpi)


And then as a "contour plot":

In [4]:
pt.axis("equal")
pt.contour(xmesh, ymesh, fmesh)

Out[4]:
<matplotlib.contour.QuadContourSet at 0x7f74273f6f60>

Next, initialize steepest descent with a starting guess:

In [15]:
guesses = [np.array([2, 2./5])]


Next, run Steepest Descent:

In [27]:
x = guesses[-1]
s = -df(x)

def f1d(alpha):
return f(x + alpha*s)

alpha_opt = sopt.golden(f1d)
next_guess = x + alpha_opt * s
guesses.append(next_guess)

print(next_guess)

[ 0.17558299  0.0351166 ]


Here's some plotting code to illustrate what just happened:

In [28]:
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-")

Out[28]:
[<matplotlib.lines.Line2D at 0x7f74257e43c8>]
In [ ]: