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  1. 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 $p$th 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.

  2. 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

  3. 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)

  4. Solve the Hobbit Village problem. Open In Colab

  5. Solve the problem of constrained optimization using projected gradient descent Open In Colab.