L19深度学习中的优化问题和凸性介绍

优化与深度学习

优化与估计

尽管优化方法可以最小化深度学习中的损失函数值，但本质上优化方法达到的目标与深度学习的目标并不相同。

• 优化方法目标：训练集损失函数值
• 深度学习目标：测试集损失函数值（泛化性）

%matplotlib inline

import sys

sys.path.append('/home/kesci/input')

import d2lzh1981 as d2l

from mpl_toolkits import mplot3d # 三维画图

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>

优化在深度学习中的挑战

1. 局部最小值
2. 鞍点
3. 梯度消失

局部最小值

f(x)=xcos⁡πx
f(x) = x\cos \pi x
f(x)=xcosπx

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)');

鞍点

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=[∂2f∂x12∂2f∂x1∂x2⋯∂2f∂x1∂xn∂2f∂x2∂x1∂2f∂x22⋯∂2f∂x2∂xn⋮⋮⋱⋮∂2f∂xn∂x1∂2f∂xn∂x2⋯∂2f∂xn2]
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]
A=⎣⎢⎢⎢⎢⎢⎡​∂x12​∂2f​∂x2​∂x1​∂2f​⋮∂xn​∂x1​∂2f​​∂x1​∂x2​∂2f​∂x22​∂2f​⋮∂xn​∂x2​∂2f​​⋯⋯⋱⋯​∂x1​∂xn​∂2f​∂x2​∂xn​∂2f​⋮∂xn2​∂2f​​⎦⎥⎥⎥⎥⎥⎤​

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）

基础

集合

函数

λf(x)+(1−λ)f(x′)≥f(λx+(1−λ)x′)
\lambda f(x)+(1-\lambda) f\left(x^{\prime}\right) \geq f\left(\lambda x+(1-\lambda) x^{\prime}\right)
λf(x)+(1−λ)f(x′)≥f(λx+(1−λ)x′)

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

∑iαif(xi)≥f(∑iαixi) and Ex[f(x)]≥f(Ex[x])
\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)
i∑​αi​f(xi​)≥f(i∑​αi​xi​) and Ex​[f(x)]≥f(Ex​[x])

---

性质

1. 无局部极小值
2. 与凸集的关系
3. 二阶条件

无局部最小值

证明：假设存在 x∈Xx \in Xx∈X 是局部最小值，则存在全局最小值 x′∈Xx' \in Xx′∈X, 使得 f(x)>f(x′)f(x) > f(x')f(x)>f(x′), 则对 λ∈(0,1]\lambda \in(0,1]λ∈(0,1]:

f(x)>λf(x)+(1−λ)f(x′)≥f(λx+(1−λ)x′)
f(x)>\lambda f(x)+(1-\lambda) f(x^{\prime}) \geq f(\lambda x+(1-\lambda) x^{\prime})
f(x)>λf(x)+(1−λ)f(x′)≥f(λx+(1−λ)x′)

与凸集的关系

对于凸函数 f(x)f(x)f(x)，定义集合 Sb:={x∣x∈X and f(x)≤b}S_{b}:=\{x | x \in X \text { and } f(x) \leq b\}Sb​:={x∣x∈X and f(x)≤b}，则集合 SbS_bSb​ 为凸集

证明：对于点 x,x′∈Sbx,x' \in S_bx,x′∈Sb​, 有 f(λx+(1−λ)x′)≤λf(x)+(1−λ)f(x′)≤bf\left(\lambda x+(1-\lambda) x^{\prime}\right) \leq \lambda f(x)+(1-\lambda) f\left(x^{\prime}\right) \leq bf(λx+(1−λ)x′)≤λf(x)+(1−λ)f(x′)≤b, 故 λx+(1−λ)x′∈Sb\lambda x+(1-\lambda) x^{\prime} \in S_{b}λx+(1−λ)x′∈Sb​

f(x,y)=0.5x2+cos⁡(2πy)f(x, y)=0.5 x^{2}+\cos (2 \pi y)f(x,y)=0.5x2+cos(2πy)

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])

凸函数与二阶导数

f′′(x)≥0⟺f(x)f^{''}(x) \ge 0 \Longleftrightarrow f(x)f′′(x)≥0⟺f(x) 是凸函数

必要性 (⇐\Leftarrow⇐):

对于凸函数：

12f(x+ϵ)+12f(x−ϵ)≥f(x+ϵ2+x−ϵ2)=f(x)
\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)
21​f(x+ϵ)+21​f(x−ϵ)≥f(2x+ϵ​+2x−ϵ​)=f(x)

故:

f′′(x)=lim⁡ε→0f(x+ϵ)−f(x)ϵ−f(x)−f(x−ϵ)ϵϵ
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′′(x)=ε→0lim​ϵϵf(x+ϵ)−f(x)​−ϵf(x)−f(x−ϵ)​​

f′′(x)=lim⁡ε→0f(x+ϵ)+f(x−ϵ)−2f(x)ϵ2≥0
f^{\prime \prime}(x)=\lim _{\varepsilon \rightarrow 0} \frac{f(x+\epsilon)+f(x-\epsilon)-2 f(x)}{\epsilon^{2}} \geq 0
f′′(x)=ε→0lim​ϵ2f(x+ϵ)+f(x−ϵ)−2f(x)​≥0

充分性 (⇒\Rightarrow⇒):

令 a<x<ba < x < ba<x<b 为 f(x)f(x)f(x) 上的三个点，由拉格朗日中值定理:

f(x)−f(a)=(x−a)f′(α) for some α∈[a,x] and f(b)−f(x)=(b−x)f′(β) for some β∈[x,b]
\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}
f(x)−f(a)=(x−a)f′(α) for some α∈[a,x] and f(b)−f(x)=(b−x)f′(β) for some β∈[x,b]​

根据单调性，有 f′(β)≥f′(α)f^{\prime}(\beta) \geq f^{\prime}(\alpha)f′(β)≥f′(α), 故:

f(b)−f(a)=f(b)−f(x)+f(x)−f(a)=(b−x)f′(β)+(x−a)f′(α)≥(b−a)f′(α)
\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}
f(b)−f(a)​=f(b)−f(x)+f(x)−f(a)=(b−x)f′(β)+(x−a)f′(α)≥(b−a)f′(α)​

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')

限制条件

minimize⁡xf(x) subject to ci(x)≤0 for all i∈{1,…,N}
\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}
xminimize​f(x) subject to ci​(x)≤0 for all i∈{1,…,N}​

拉格朗日乘子法

Boyd & Vandenberghe, 2004

L(x,α)=f(x)+∑iαici(x) where αi≥0
L(\mathbf{x}, \alpha)=f(\mathbf{x})+\sum_{i} \alpha_{i} c_{i}(\mathbf{x}) \text { where } \alpha_{i} \geq 0
L(x,α)=f(x)+i∑​αi​ci​(x) where αi​≥0

惩罚项

欲使 ci(x)≤0c_i(x) \leq 0ci​(x)≤0, 将项 αici(x)\alpha_ic_i(x)αi​ci​(x) 加入目标函数，如多层感知机章节中的 λ2∣∣w∣∣2\frac{\lambda}{2} ||w||^22λ​∣∣w∣∣2

投影

Proj⁡X(x)=argmin⁡x′∈X∥x−x′∥2
\operatorname{Proj}_{X}(\mathbf{x})=\underset{\mathbf{x}^{\prime} \in X}{\operatorname{argmin}}\left\|\mathbf{x}-\mathbf{x}^{\prime}\right\|_{2}
ProjX​(x)=x′∈Xargmin​∥x−x′∥2​