Notebook

Gradient Descent¶

In [1]:

from __future__ import division
%pylab inline
%load_ext autoreload
%autoreload 2

Populating the interactive namespace from numpy and matplotlib

$\newcommand{\dotp}[2]{\langle #1, #2 \rangle}$ $\newcommand{\umin}[1]{\underset{#1}{\min}\;}$ $\newcommand{\RR}{\mathbb{R}}$ $\newcommand{\pd}[2]{ \frac{ \partial #1}{\partial #2} }$ $\newcommand{\norm}[1]{\|#1\|}$ $\newcommand{\pa}[1]{\left(#1\right)}$

We consider the problem of finding a minimizer of a convex smooth function $\;f : \RR^d \rightarrow \RR$; that is, we want to solve $$\umin{x \in \RR^d} f(x)$$

Note that the minimizer is not necessarily unique.

The simplest method is gradient descent, which iteratively computes $$ x^{(k+1)} = x^{(k)} - \tau \nabla f(x^{(k)}), $$ where $\nabla f(x) \in \RR^d$ is the gradient of $f$ at $x$, $x^{(0)} \in \RR^d$ is an arbitrary initial point, and the stepsize $\tau$ must satisfy $$0<\tau<2/\beta,$$ to have convergence, where $\beta$ is a Lipschitz constant of $\nabla f$; that is, $$ \forall (x,x') \in \RR^N \times \RR^N, \quad \norm{\nabla f(x) - \nabla f(x')} \leq \beta \norm{x-x'}. $$

For instance, if $f$ is of class $C^2$, $$\beta= \sup_x \norm{Hf(x)},$$ where $Hf(x) \in \RR^{d \times d}$ is the Hessian of $f$ at $x$ and $\norm{\cdot}$ is the spectral operator norm (largest eigenvalue).

Gradient descent in 2-D¶

We consider a simple problem, corresponding to the minimization of a 2-D ($d=2$) quadratic form $$ f(x) = \frac{1}{2} \pa{ x_1^2 + \eta x_2^2 } ,$$ where $\eta>0$ controls the anisotropy, and hence the difficulty, of the problem.

We can note that minimizing a strongly convex quadratic function is equivalent to solving a positive definite linear system. Gradient descent is then equivalent to the Richardson iteration applied to this linear system.

We define the anisotropy parameter $\eta$:

In [2]:

eta = 8

We define the function $f$:

In [3]:

f = lambda x : (x[0]**2 + eta*x[1]**2) / 2

Note: a 'lambda' function is a one-line function definition; in Matlab, this would be f=@(x)(x(1)^2+eta*x(2)^2)/2;

An example of function evaluation:

In [4]:

f([2,3])

Out[4]:

38.0

We display the function using a contourplot:

In [5]:

figsize(6,6)
t = linspace(-.7,.7,101)
[u,v] = meshgrid(t,t)
F = (u**2 + eta*v**2) / 2
contourf(t,t,F,35)

Out[5]:

<matplotlib.contour.QuadContourSet instance at 0x10853c9e0>

We define the gradient of $f$:

In [6]:

Grad_f = lambda x : array([x[0], eta*x[1]])

An example of evaluation:

In [7]:

Grad_f([1,2])

Out[7]:

array([ 1, 16])

Since $f$ is quadratic, its Hessian is the constant matrix $\left(\begin{array}{cc}1&0\\0&\eta\end{array}\right)$. Its spectral norm, which is the Lipschitz constant of Grad_f, is $\beta=\max(1,\eta)$.

The stepsize $\tau$ must satisfy $0< \tau < 2/\beta$.

In [8]:

tau = 1.8/max(eta,1); tau

Out[8]:

0.225

Now we implement the gradient descent method: given the initial estimate $x^{(0)}$ of the solution, the stepsize $\tau$ and the number $k$ of iterations, we compute $x^{(k)}$:

In [9]:

nbiter = 10
x = [1,1]  # initial estimate of the solution
for iter in range(nbiter):  # iter goes from 0 to nbiter-1
    x = x - tau*Grad_f(x)
x   # to display x, like in Matlab. Use print(x) if this is not the last command of the cell, else nothing is displayed

Out[9]:

array([ 0.07816584,  0.10737418])

Note: there is no 'end' for the loops in Python; instead, the content of the loop is determined by the indentation: here four blank spaces before x=...

In [10]:

f(_) # _ is the current variable, like Matlab's ans

Out[10]:

0.049171809817584886

We encapsulate the code in a function called GradientDescent.

In [11]:

def GradDescent(Grad_f, x0, nbiter, tau):
    x = x0;
    for iter in range(nbiter):  # iter goes from 0 to nbiter-1
        x = x - tau*Grad_f(x)   # x has type 'array'. Like in C, one can also write x-=tau*Grad_f(x)
    return x

In [12]:

GradDescent(Grad_f,[1,1],10,tau)

Out[12]:

array([ 0.07816584,  0.10737418])

We define a function called GradDescentArray which puts the iterates in a 'matrix' (a 2-D array in fact); they are first put in a list and the list is converted to an array at the end (the + operation on lists concatenates them, whereas on arrays this is the classical elementwise addition).

In [13]:

def GradDescentArray(Grad_f, x0, nbiter, tau):
    x = x0;
    xlist=[list(x0)];  # the cast is just in case x0 would be an array
    for iter in range(nbiter):  
        x = x - tau*Grad_f(x)     
        xlist = xlist + [list(x)]
    return array(xlist)

In [14]:

xarray=GradDescentArray(Grad_f,[0.6,0.6],10,0.225)
xarray

Out[14]:

array([[ 0.6       ,  0.6       ],
       [ 0.465     , -0.48      ],
       [ 0.360375  ,  0.384     ],
       [ 0.27929063, -0.3072    ],
       [ 0.21645023,  0.24576   ],
       [ 0.16774893, -0.196608  ],
       [ 0.13000542,  0.1572864 ],
       [ 0.1007542 , -0.12582912],
       [ 0.07808451,  0.1006633 ],
       [ 0.06051549, -0.08053064],
       [ 0.04689951,  0.06442451]])

We can access the elements of an array like in Matlab. Be careful: like in C, the first element has index 0! Some examples:

In [15]:

xarray[1,0]

Out[15]:

0.46499999999999997

In [16]:

xarray[1,:]

Out[16]:

array([ 0.465, -0.48 ])

In [17]:

xarray[:,0]

Out[17]:

array([ 0.6       ,  0.465     ,  0.360375  ,  0.27929063,  0.21645023,
        0.16774893,  0.13000542,  0.1007542 ,  0.07808451,  0.06051549,
        0.04689951])

We can apply the function $f$ to every iterate $x^{(k)}$ of the list xarray, using the command map:

In [18]:

map(f,xarray)

Out[18]:

[1.6199999999999999,
 1.0297125000000003,
 0.65475907031250036,
 0.41648898660644562,
 0.26501726238049639,
 0.16868867468928564,
 0.10740675137733219,
 0.06840757437694138,
 0.043580991731946392,
 0.027771796276949725,
 0.017701851534330564]

We plot the cost function $f(x^{(k)})$ as a function of $k$, in log-scale:

In [19]:

plot(range(len(xarray)),log10(map(f,xarray)),'o-')

Out[19]:

[<matplotlib.lines.Line2D at 0x1086b7b50>]

We remark that because the function $f$ is strongly convex, the convergence is linear. Also, we can note that gradient descent is monotonic: $f(x^{(k)})$ is decreased at every iteration.

We plot the iterates above the contourplot of $f$:

In [20]:

figsize(8,8)
contourf(t,t,F,35)
xarray=GradDescentArray(Grad_f,[0.6,0.6],30,0.225)
plot(xarray[:,0], xarray[:,1], 'w.-')

Out[20]:

[<matplotlib.lines.Line2D at 0x1062a1b90>]

Exercise: re-run the previous cells with different values of tau (try 0.25,0.225,0.2,0.125,0.05).

Also try different initial estimates.

Also try different values of the anisotropy eta.

Gradient descent in large dimension: image restoration¶

Local differential operators like the discrete gradient are fundamental for variational image processing.

We load the image Lena in the array $x^\sharp$ (we choose the symbol # - 'sharp', because the image is clean, not degraded).

In [27]:

from scipy import misc
xsharp = misc.lena()
print(xsharp.shape) # like Matlab's size(xsharp). Given as a tuple.
print("The size of the image is %s x %s." % (xsharp.shape[0],xsharp.shape[1]))
print("The range of the pixel values is [%s,%s]." % (xsharp.min(),xsharp.max()))
xsharp = xsharp.astype(float32)  # like Matlab's xsharp=double(xsharp) so that the pixel values are floating point numbers

(512, 512)
The size of the image is 512 x 512.
The range of the pixel values is [25,245].

We display the image.

In [28]:

figsize(11,11)
imshow(xsharp, interpolation='nearest', cmap=cm.gray, vmin=0, vmax=255)
# Without specifying vmin and vmax, imshow auto-adjusts its range so that black and white are
# the min and max of the data, respectively, like Matlab's imagesc.
colorbar()       # displays the color bar close to the image
#axis('off')     # uncomment to remove the axes
subplots_adjust(top=0.75)
title('This is Lena')

Out[28]:

<matplotlib.text.Text at 0x10958b1d0>

Gradient Descent¶

Gradient descent in 2-D¶

__Exercise: re-run the previous cells with different values of tau (try 0.25,0.225,0.2,0.125,0.05).__

__Also try different initial estimates.__

__Also try different values of the anisotropy eta__.

Gradient descent in large dimension: image restoration¶

Exercise: re-run the previous cells with different values of tau (try 0.25,0.225,0.2,0.125,0.05).

Also try different initial estimates.

Also try different values of the anisotropy eta.