First order methods

A function is said to belong to the class f \in C^{k,p}_L (Q) if it k times is continuously differentiable on Q and the pth derivative has a Lipschitz constant L.

\|\nabla^p f(x) - \nabla^p f(y)\| \leq L \|x-y\|, \qquad \forall x,y \in Q

The most commonly used C_L^{1,1}, C_L^{2,2} on \mathbb{R}^n. Note that:
- p \leq k
- If q \geq k, then C_L^{q,p} \subseteq C_L^{k,p}. The higher the order of the derivative, the stronger the constraint (fewer functions belong to the class)
Prove that the function belongs to the class C_L^{2,1} \subseteq C_L^{1,1} if and only if \forall x \in \mathbb{R}^n:

\||\nabla^2 f(x)\| \leq L

Prove also that the last condition can be rewritten, without generality restriction, as follows:

-L I_n \preceq \nabla^2 f(x) \preceq L I_n

Note: by default the Euclidean norm is used for vectors and the spectral norm is used for matrices.
Покажите, что с помощью следующих стратегий подбора шага в градиентному спуске:
- Постоянный шаг h_k = \dfrac{1}{L}
- Убывающая последовательность h_k = \dfrac{\alpha_k}{L}, \quad \alpha_k \to 0
можно получить оценку убывания функции на итерации вида:

f(x_k) - f(x_{k+1}) \geq \dfrac{\omega}{L}\|\nabla f(x_k)\|^2

\omega > 0 - некоторая константа, L - константа Липщица градиента функции

Рассмотрим функцию двух переменных:

f(x_1, x_2) = x_1^2 + k x_2^2,

где k - некоторый параметр. Постройте график количества итераций, необходимых для сходимости алгоритма наискорейшего спуска (до выполнения условия \|\nabla f(x_k)\| \leq \varepsilon = 10^{-7}) в зависимости от значения k. Рассмотрите интервал k \in [10^{-3}; 10^3] (будет удобно использовать функцию ks = np.logspace(-3,3)) и строить график по оси абсцисс в логарифмическом масштабе plt.semilogx() или plt.loglog() для двойного лог. масштаба.

Сделайте те же графики для функции:

f(x) = \ln(1 + e^{x^\top A x}) + \mathbf{1}^\top x

Объясните полученную зависимость.

Для наглядности можете пользоваться кодом отрисовки картинок:

def f_6(x, *f_params):
    if len(f_params) == 0:
        k = 2
    else:
        k = float(f_params[0])
    x_1, x_2 = x
    return x_1**2 + k*x_2**2

def df_6(x, *f_params):
    if len(f_params) == 0:
        k = 2
    else:
        k = float(f_params[0])
    return np.array([2*x[0], 2*k*x[1]])

%matplotlib inline
from mpl_toolkits.mplot3d import Axes3D
from matplotlib import cm
from matplotlib.ticker import LinearLocator, FormatStrFormatter
import numpy as np

def plot_3d_function(x1, x2, f, title, *f_params, minima = None, iterations = None):
    '''
    '''
    low_lim_1 = x1.min()
    low_lim_2 = x2.min()
    up_lim_1  = x1.max()
    up_lim_2  = x2.max()

    X1,X2 = np.meshgrid(x1, x2) # grid of point
    Z = f((X1, X2), *f_params) # evaluation of the function on the grid

    # set up a figure twice as wide as it is tall
    fig = plt.figure(figsize=(16,7))
    fig.suptitle(title)

    #===============
    #  First subplot
    #===============
    # set up the axes for the first plot
    ax = fig.add_subplot(1, 2, 1, projection='3d')

    # plot a 3D surface like in the example mplot3d/surface3d_demo
    surf = ax.plot_surface(X1, X2, Z, rstride=1, cstride=1, 
                        cmap=cm.RdBu,linewidth=0, antialiased=False)

    ax.zaxis.set_major_locator(LinearLocator(10))
    ax.zaxis.set_major_formatter(FormatStrFormatter('%.02f'))
    if minima is not None:
        minima_ = np.array(minima).reshape(-1, 1)
        ax.plot(*minima_, f(minima_), 'r*', markersize=10)



    #===============
    # Second subplot
    #===============
    # set up the axes for the second plot
    ax = fig.add_subplot(1, 2, 2)

    # plot a 3D wireframe like in the example mplot3d/wire3d_demo
    im = ax.imshow(Z,cmap=plt.cm.RdBu,  extent=[low_lim_1, up_lim_1, low_lim_2, up_lim_2])
    cset = ax.contour(x1, x2,Z,linewidths=2,cmap=plt.cm.Set2)
    ax.clabel(cset,inline=True,fmt='%1.1f',fontsize=10)
    fig.colorbar(im)
    ax.set_xlabel(f'$x_1$')
    ax.set_ylabel(f'$x_2$')

    if minima is not None:
        minima_ = np.array(minima).reshape(-1, 1)
        ax.plot(*minima_, 'r*', markersize=10)

    if iterations is not None:
        for point in iterations:
            ax.plot(*point, 'go', markersize=3)
        iterations = np.array(iterations).T
        ax.quiver(iterations[0,:-1], iterations[1,:-1], iterations[0,1:]-iterations[0,:-1], iterations[1,1:]-iterations[1,:-1], scale_units='xy', angles='xy', scale=1, color='blue')

    plt.show()

up_lim  = 4
low_lim = -up_lim
x1 = np.arange(low_lim, up_lim, 0.1)
x2 = np.arange(low_lim, up_lim, 0.1)
k=0.5
title = f'$f(x_1, x_2) = x_1^2 + k x_2^2, k = {k}$'

plot_3d_function(x1, x2, f_6, title, k, minima=[0,0])

from scipy.optimize import minimize_scalar

def steepest_descent(x_0, f, df, *f_params, df_eps = 1e-2, max_iter = 1000):
    iterations = []
    x = np.array(x_0)
    iterations.append(x)
    while np.linalg.norm(df(x, *f_params)) > df_eps and len(iterations) <= max_iter:
        res = minimize_scalar(lambda alpha: f(x - alpha * df(x, *f_params), *f_params))
        alpha_opt = res.x
        x = x - alpha_opt * df(x, *f_params)
        iterations.append(x)
    print(f'Finished with {len(iterations)} iterations')
    return iterations

x_0 = [10,1]
k = 30
iterations = steepest_descent(x_0, f_6, df_6, k, df_eps = 1e-9)
title = f'$f(x_1, x_2) = x_1^2 + k x_2^2, k = {k}$'

plot_3d_function(x1, x2, f_6, title, k, minima=[0,0], iterations = iterations)

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