查找用户定义函数的局部最大值和最小值

Question

What I want

I want to find a list of the stationary points, of their values and locations, and of whether they are minima or maxima.

My function looks like:

import numpy as np

def func(x,y):
  return (np.cos(x*10))**2 + (np.sin(y*10))**2

Methods

Here are methods that I am thinking of using:

1. I am actually already doing something like this on Mathematica. I differentiate the function once and then twice. I look at the points where the first derivative is 0, compute their values and their locations. Then I take the second derivative at those locations and check whether they are minima or maxima.

2. I am also wondering whether just making a 2D array of the values of the function in x and y, and find the maximum and minimum values of that array. But this requires me to know how finely to define the x and y meshgrids to reliably capture the behaviour of the function

For the latter case I already found some ways like this one .

I just wanted to know, which method makes more sense in terms of efficiency, speed, accuracy or even elegance in Python?

Answer 1

find a list of the stationary points, of their values and locations, and of whether they are minima or maxima.

This is in general an unsolvable problem. Method 1 (symbolic) is appropriate for that, but for complicated functions there is no symbolic solution for stationary points (there is no method for solving a general system of two equations symbolically).

Symbolic solution with SymPy

For simple functions like your example, SymPy will work fine. Here is a complete example of finding the stationary points and classifying them by the eigenvalues of the Hessian.

import sympy as sym
x, y = sym.symbols("x y")
f = sym.cos(x*10)**2 + sym.sin(y*10)**2
gradient = sym.derive_by_array(f, (x, y))
hessian = sym.Matrix(2, 2, sym.derive_by_array(gradient, (x, y)))

So far, Hessian is a symbolic matrix 2 by 2: [[200*sin(10*x)**2 - 200*cos(10*x)**2, 0], [0, -200*sin(10*y)**2 + 200*cos(10*y)**2]] . Next, we find the stationary points by equating gradient to zero, and plug them into Hessian one by one.

stationary_points = sym.solve(gradient, (x, y))
for p in stationary_points:
    value = f.subs({x: p[0], y: p[1]})
    hess = hessian.subs({x: p[0], y: p[1]})
    eigenvals = hess.eigenvals()
    if all(ev > 0 for ev in eigenvals):
        print("Local minimum at {} with value {}".format(p, value))
    elif all(ev < 0 for ev in eigenvals):
        print("Local maximum at {} with value {}".format(p, value))
    elif any(ev > 0 for ev in eigenvals) and any(ev < 0 for ev in eigenvals):
        print("Saddle point at {} with value {}".format(p, value))
    else:
        print("Could not classify the stationary point at {} with value {}".format(p, value))

The last clause is necessary because when the Hessian is only semi definite, we cannot tell what kind of stationary point is that ( x**2 + y**4 and x**2 - y**4 have the same Hessian at (0, 0) but different behavior). The output:

Saddle point at (0, 0) with value 1
Local maximum at (0, pi/20) with value 2
Saddle point at (0, pi/10) with value 1
Local maximum at (0, 3*pi/20) with value 2
Local minimum at (pi/20, 0) with value 0
Saddle point at (pi/20, pi/20) with value 1
Local minimum at (pi/20, pi/10) with value 0
Saddle point at (pi/20, 3*pi/20) with value 1
Saddle point at (pi/10, 0) with value 1
Local maximum at (pi/10, pi/20) with value 2
Saddle point at (pi/10, pi/10) with value 1
Local maximum at (pi/10, 3*pi/20) with value 2
Local minimum at (3*pi/20, 0) with value 0
Saddle point at (3*pi/20, pi/20) with value 1
Local minimum at (3*pi/20, pi/10) with value 0
Saddle point at (3*pi/20, 3*pi/20) with value 1

Obviously, solve did not find all solutions (there are infinitely many of those). Consider solve vs solveset but in any case, handling infinitely many solutions is hard.

Numeric optimization with SciPy

SciPy offers a lot of numerical minimization routines , including brute force (which is your method 2; generally it's very very slow). These are powerful methods, but consider these points.

1. Each run will only find one minimum.
2. Replacing f with -f you can also find a maximum.
3. Changing the starting point of the search (argument x0 of minimize ) may yield another maximum or minimum. Still, you'll never know if there are other extrema that you have not seen yet.
4. None of this will find saddle points.

Mixed strategy

Using lambdify one can turn a symbolic expression into a Python function that can be passed to SciPy numeric solvers.

from scipy.optimize import fsolve
grad = sym.lambdify((x, y), gradient)
fsolve(lambda v: grad(v[0], v[1]), (1, 2))

This returns some stationary point, [0.9424778 , 2.04203522] in this example. Which point it is depends on the initial guess, which was (1, 2). Typically (but not always) you'll get a solution that's near the initial guess.

This has an advantage over the direct minimization approach in that saddle points can be detected as well. Still, it is going to be hard to find all solutions, as each run of fsolve brings up only one.