《動手學深度學習》task6_2 凸優化

系統學習《動手學深度學習》點擊下面這個鏈接，有全目錄哦~
https://blog.csdn.net/Shine_rise/article/details/104754764

優化與深度學習

優化與估計

儘管優化方法可以最小化深度學習中的損失函數值，但本質上優化方法達到的目標與深度學習的目標並不相同。

• 優化方法目標：訓練集損失函數值
• 深度學習目標：測試集損失函數值（泛化性）

%matplotlib inline
import sys
sys.path.append('/home/kesci/input')
import d2lzh1981 as d2l
from mpl_toolkits import mplot3d # 三維畫圖
import numpy as np

def f(x): return x * np.cos(np.pi * x)
def g(x): return f(x) + 0.2 * np.cos(5 * np.pi * x)

d2l.set_figsize((5, 3))
x = np.arange(0.5, 1.5, 0.01)
fig_f, = d2l.plt.plot(x, f(x),label="train error")
fig_g, = d2l.plt.plot(x, g(x),'--', c='purple', label="test error")
fig_f.axes.annotate('empirical risk', (1.0, -1.2), (0.5, -1.1),arrowprops=dict(arrowstyle='->'))
fig_g.axes.annotate('expected risk', (1.1, -1.05), (0.95, -0.5),arrowprops=dict(arrowstyle='->'))
d2l.plt.xlabel('x')
d2l.plt.ylabel('risk')
d2l.plt.legend(loc="upper right")

<matplotlib.legend.Legend at 0x7fc092436080>

優化在深度學習中的挑戰

1. 局部最小值
2. 鞍點
3. 梯度消失

局部最小值

$f(x) = x\cos \pi x$

def f(x):
    return x * np.cos(np.pi * x)

d2l.set_figsize((4.5, 2.5))
x = np.arange(-1.0, 2.0, 0.1)
fig,  = d2l.plt.plot(x, f(x))
fig.axes.annotate('local minimum', xy=(-0.3, -0.25), xytext=(-0.77, -1.0),
                  arrowprops=dict(arrowstyle='->'))
fig.axes.annotate('global minimum', xy=(1.1, -0.95), xytext=(0.6, 0.8),
                  arrowprops=dict(arrowstyle='->'))
d2l.plt.xlabel('x')
d2l.plt.ylabel('f(x)');

鞍點

x = np.arange(-2.0, 2.0, 0.1)
fig, = d2l.plt.plot(x, x**3)
fig.axes.annotate('saddle point', xy=(0, -0.2), xytext=(-0.52, -5.0),
                  arrowprops=dict(arrowstyle='->'))
d2l.plt.xlabel('x')
d2l.plt.ylabel('f(x)');

$A=\left[\begin{array}{cccc}{\frac{\partial^{2} f}{\partial x_{1}^{2}}} & {\frac{\partial^{2} f}{\partial x_{1} \partial x_{2}}} & {\cdots} & {\frac{\partial^{2} f}{\partial x_{1} \partial x_{n}}} \\ {\frac{\partial^{2} f}{\partial x_{2} \partial x_{1}}} & {\frac{\partial^{2} f}{\partial x_{2}^{2}}} & {\cdots} & {\frac{\partial^{2} f}{\partial x_{2} \partial x_{n}}} \\ {\vdots} & {\vdots} & {\ddots} & {\vdots} \\ {\frac{\partial^{2} f}{\partial x_{n} \partial x_{1}}} & {\frac{\partial^{2} f}{\partial x_{n} \partial x_{2}}} & {\cdots} & {\frac{\partial^{2} f}{\partial x_{n}^{2}}}\end{array}\right]$
e.g.

x, y = np.mgrid[-1: 1: 31j, -1: 1: 31j]
z = x**2 - y**2

d2l.set_figsize((6, 4))
ax = d2l.plt.figure().add_subplot(111, projection='3d')
ax.plot_wireframe(x, y, z, **{'rstride': 2, 'cstride': 2})
ax.plot([0], [0], [0], 'ro', markersize=10)
ticks = [-1,  0, 1]
d2l.plt.xticks(ticks)
d2l.plt.yticks(ticks)
ax.set_zticks(ticks)
d2l.plt.xlabel('x')
d2l.plt.ylabel('y');

梯度消失

x = np.arange(-2.0, 5.0, 0.01)
fig, = d2l.plt.plot(x, np.tanh(x))
d2l.plt.xlabel('x')
d2l.plt.ylabel('f(x)')
fig.axes.annotate('vanishing gradient', (4, 1), (2, 0.0) ,arrowprops=dict(arrowstyle='->'))

Text(2, 0.0, 'vanishing gradient')

凸性 （Convexity）

基礎

集合

函數

$\lambda f(x)+(1-\lambda) f\left(x^{\prime}\right) \geq f\left(\lambda x+(1-\lambda) x^{\prime}\right)$

def f(x):
    return 0.5 * x**2  # Convex

def g(x):
    return np.cos(np.pi * x)  # Nonconvex

def h(x):
    return np.exp(0.5 * x)  # Convex

x, segment = np.arange(-2, 2, 0.01), np.array([-1.5, 1])
d2l.use_svg_display()
_, axes = d2l.plt.subplots(1, 3, figsize=(9, 3))

for ax, func in zip(axes, [f, g, h]):
    ax.plot(x, func(x))
    ax.plot(segment, func(segment),'--', color="purple")
    # d2l.plt.plot([x, segment], [func(x), func(segment)], axes=ax)

Jensen 不等式

$\sum_{i} \alpha_{i} f\left(x_{i}\right) \geq f\left(\sum_{i} \alpha_{i} x_{i}\right) \text { and } E_{x}[f(x)] \geq f\left(E_{x}[x]\right)$

---

性質

1. 無局部極小值
2. 與凸集的關係
3. 二階條件

無局部最小值

證明：假設存在 $x \in X$ 是局部最小值，則存在全局最小值 $x' \in X$, 使得 $f(x) > f(x')$, 則對 $\lambda \in(0,1]$:
$f(x)>\lambda f(x)+(1-\lambda) f(x^{\prime}) \geq f(\lambda x+(1-\lambda) x^{\prime})$

與凸集的關係

對於凸函數 $f(x)$，定義集合 $S_{b}:=\{x | x \in X \text { and } f(x) \leq b\}$，則集合 $S_b$ 爲凸集

證明：對於點 $x,x' \in S_b$, 有 $f\left(\lambda x+(1-\lambda) x^{\prime}\right) \leq \lambda f(x)+(1-\lambda) f\left(x^{\prime}\right) \leq b$, 故 $\lambda x+(1-\lambda) x^{\prime} \in S_{b}$

$f(x, y)=0.5 x^{2}+\cos (2 \pi y)$

x, y = np.meshgrid(np.linspace(-1, 1, 101), np.linspace(-1, 1, 101),
                   indexing='ij')

z = x**2 + 0.5 * np.cos(2 * np.pi * y)

# Plot the 3D surface
d2l.set_figsize((6, 4))
ax = d2l.plt.figure().add_subplot(111, projection='3d')
ax.plot_wireframe(x, y, z, **{'rstride': 10, 'cstride': 10})
ax.contour(x, y, z, offset=-1)
ax.set_zlim(-1, 1.5)

# Adjust labels
for func in [d2l.plt.xticks, d2l.plt.yticks, ax.set_zticks]:
    func([-1, 0, 1])

凸函數與二階導數

$f^{''}(x) \ge 0 \Longleftrightarrow f(x)$ 是凸函數

必要性 ($\Leftarrow$):

對於凸函數：
$\frac{1}{2} f(x+\epsilon)+\frac{1}{2} f(x-\epsilon) \geq f\left(\frac{x+\epsilon}{2}+\frac{x-\epsilon}{2}\right)=f(x)$
故:
$f^{\prime \prime}(x)=\lim _{\varepsilon \rightarrow 0} \frac{\frac{f(x+\epsilon) - f(x)}{\epsilon}-\frac{f(x) - f(x-\epsilon)}{\epsilon}}{\epsilon}$

$f^{\prime \prime}(x)=\lim _{\varepsilon \rightarrow 0} \frac{f(x+\epsilon)+f(x-\epsilon)-2 f(x)}{\epsilon^{2}} \geq 0$

充分性 ($\Rightarrow$):

令 $a < x < b$ 爲 $f(x)$ 上的三個點，由拉格朗日中值定理:
$\begin{array}{l}{f(x)-f(a)=(x-a) f^{\prime}(\alpha) \text { for some } \alpha \in[a, x] \text { and }} \\ {f(b)-f(x)=(b-x) f^{\prime}(\beta) \text { for some } \beta \in[x, b]}\end{array}$
根據單調性，有 $f^{\prime}(\beta) \geq f^{\prime}(\alpha)$, 故:
$\begin{aligned} f(b)-f(a) &=f(b)-f(x)+f(x)-f(a) \\ &=(b-x) f^{\prime}(\beta)+(x-a) f^{\prime}(\alpha) \\ & \geq(b-a) f^{\prime}(\alpha) \end{aligned}$

def f(x):
    return 0.5 * x**2

x = np.arange(-2, 2, 0.01)
axb, ab = np.array([-1.5, -0.5, 1]), np.array([-1.5, 1])

d2l.set_figsize((3.5, 2.5))
fig_x, = d2l.plt.plot(x, f(x))
fig_axb, = d2l.plt.plot(axb, f(axb), '-.',color="purple")
fig_ab, = d2l.plt.plot(ab, f(ab),'g-.')

fig_x.axes.annotate('a', (-1.5, f(-1.5)), (-1.5, 1.5),arrowprops=dict(arrowstyle='->'))
fig_x.axes.annotate('b', (1, f(1)), (1, 1.5),arrowprops=dict(arrowstyle='->'))
fig_x.axes.annotate('x', (-0.5, f(-0.5)), (-1.5, f(-0.5)),arrowprops=dict(arrowstyle='->'))

Text(-1.5, 0.125, 'x')

限制條件

$\begin{array}{l}{\underset{\mathbf{x}}{\operatorname{minimize}} f(\mathbf{x})} \\ {\text { subject to } c_{i}(\mathbf{x}) \leq 0 \text { for all } i \in\{1, \ldots, N\}}\end{array}$

拉格朗日乘子法

Boyd & Vandenberghe, 2004
$L(\mathbf{x}, \alpha)=f(\mathbf{x})+\sum_{i} \alpha_{i} c_{i}(\mathbf{x}) \text { where } \alpha_{i} \geq 0$

懲罰項

欲使 $c_i(x) \leq 0$, 將項 $\alpha_ic_i(x)$ 加入目標函數，如多層感知機章節中的 $\frac{\lambda}{2} ||w||^2$

投影

$\operatorname{Proj}_{X}(\mathbf{x})=\underset{\mathbf{x}^{\prime} \in X}{\operatorname{argmin}}\left\|\mathbf{x}-\mathbf{x}^{\prime}\right\|_{2}$