(which is just using Newton on a Lagrangian)

In [2]:
import numpy as np
import numpy.linalg as la


Here's an objective function $f$ and a constraint $g(x)=0$:

In [3]:
def f(vec):
x = vec[0]
y = vec[1]
return (x-2)**4 + 2*(y-1)**2

def g(vec):
x = vec[0]
y = vec[1]
return x + 4*y - 3


Now define the Lagrangian, its gradient, and its Hessian:

In [4]:
def L(vec):
lam = vec[2]
return f(vec) + lam * g(vec)

x = vec[0]
y = vec[1]
lam = vec[2]

return np.array([
4*(x-2)**3 + lam,
4*(y-1) + 4*lam,
x+4*y-3
])

def hess_L(vec):
x = vec[0]
y = vec[1]
lam = vec[2]

return np.array([
[12*(x-2)**2, 0, 1],
[0, 4, 4],
[1, 4, 0]
])


At this point, we only need to find an unconstrained minimum of the Lagrangian!

Let's fix a starting vector vec:

In [5]:
vec = np.zeros(3)


Implement Newton and run this cell in place a number of times (Ctrl-Enter):

In [18]:
vec = vec - la.solve(hess_L(vec), grad_L(vec))
vec

Out[18]:
array([ 1.46398989,  0.38400253,  0.61599747])

Let's first check that we satisfy the constraint:

In [21]:
g(vec)

Out[21]:
0.0

Next, let's look at a plot:

In [22]:
import matplotlib.pyplot as pt
x, y = np.mgrid[0:4:30j, -3:5:30j]
pt.contour(x, y, f(np.array([x,y])), 20)
pt.contour(x, y, g(np.array([x,y])), levels=[0])
pt.plot(vec[0], vec[1], "o")

Out[22]:
[<matplotlib.lines.Line2D at 0x7fd6b4398390>]