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This tour explores the use of gradient descent method for unconstrained and constrained optimization of a smooth function
addpath('toolbox_signal')
addpath('toolbox_general')
addpath('solutions/optim_1_gradient_descent')
We consider the problem of finding a minimum of a function $f$, hence solving $$ \umin{x \in \RR^d} f(x) $$ where $f : \RR^d \rightarrow \RR$ is a smooth function.
Note that the minimum is not necessarily unique. In the general case, $f$ might exhibit local minima, in which case the proposed algorithms is not expected to find a global minimizer of the problem. In this tour, we restrict our attention to convex function, so that the methods will converge to a global minimizer.
The simplest method is the gradient descent, that computes $$ x^{(k+1)} = x^{(k)} - \tau_k \nabla f(x^{(k)}), $$ where $\tau_k>0$ is a step size, and $\nabla f(x) \in \RR^d$ is the gradient of $f$ at the point $x$, and $x^{(0)} \in \RR^d$ is any initial point.
In the convex case, if $f$ is of class $C^2$, in order to ensure convergence, the step size should satisfy $$ 0 < \tau_k < \frac{2}{ \sup_x \norm{Hf(x)} } $$ 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).
We consider a simple problem, corresponding to the minimization of a 2-D 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.
Anisotropy parameter $\eta$.
eta = 10;
Function $f$.
f = @(x)( x(1)^2 + eta*x(2)^2 ) /2;
Background image of the function.
t = linspace(-.7,.7,101);
[u,v] = meshgrid(t,t);
F = ( u.^2 + eta*v.^2 )/2 ;
Display the function as a 2-D image.
clf; hold on;
imagesc(t,t,F); colormap jet(256);
contour(t,t,F, 20, 'k');
axis off; axis equal;
Gradient.
Gradf = @(x)[x(1); eta*x(2)];
The step size should satisfy $\tau_k < 2/\eta$. We use here a constrant step size.
tau = 1.8/eta;
Exercise 1
Perform the gradient descent using a fixed step size $\tau_k=\tau$. Display the decay of the energy $f(x^{(k)})$ through the iteration. Save the iterates so that |X(:,k)| corresponds to $x^{(k)}$.
exo1()
%% Insert your code here.
Display the iterations.
clf; hold on;
imagesc(t,t,F); colormap jet(256);
contour(t,t,F, 20, 'k');
h = plot(X(1,:), X(2,:), 'k.-');
set(h, 'LineWidth', 2);
set(h, 'MarkerSize', 15);
axis off; axis equal;
Exercise 2
Display the iteration for several different step sizes.
exo2()
%% Insert your code here.
Local differential operators like gradient, divergence and laplacian are the building blocks for variational image processing.
Load an image $x_0 \in \RR^N$ of $N=n \times n$ pixels.
n = 256;
x0 = rescale( load_image('lena',n) );
Display it.
clf;
imageplot(x0);
For a continuous function $g$, the gradient reads $$ \nabla g(s) = \pa{ \pd{g(s)}{s_1}, \pd{g(s)}{s_2} } \in \RR^2. $$ (note that here, the variable $s$ denotes the 2-D spacial position).
We discretize this differential operator on a discrete image $x \in \RR^N$ using first order finite differences. $$ (\nabla x)_i = ( x_{i_1,i_2}-x_{i_1-1,i_2}, x_{i_1,i_2}-x_{i_1,i_2-1} ) \in \RR^2. $$ Note that for simplity we use periodic boundary conditions.
Compute its gradient, using finite differences.
grad = @(x)cat(3, x-x([end 1:end-1],:), x-x(:,[end 1:end-1]));
One thus has $ \nabla : \RR^N \mapsto \RR^{N \times 2}. $
v = grad(x0);
One can display each of its components.
clf;
imageplot(v(:,:,1), 'd/dx', 1,2,1);
imageplot(v(:,:,2), 'd/dy', 1,2,2);
One can also display it using a color image.
clf;
imageplot(v);
One can display its magnitude $\norm{(\nabla x)_i}$, which is large near edges.
clf;
imageplot( sqrt( sum3(v.^2,3) ) );
The divergence operator maps vector field to images. For continuous vector fields $v(s) \in \RR^2$, it is defined as $$ \text{div}(v)(s) = \pd{v_1(s)}{s_1} + \pd{v_2(s)}{s_2} \in \RR. $$ (note that here, the variable $s$ denotes the 2-D spacial position). It is minus the adjoint of the gadient, i.e. $\text{div} = - \nabla^*$.
It is discretized, for $v=(v^1,v^2)$ as $$ \text{div}(v)_i = v^1_{i_1+1,i_2} - v^1_{i_1,i_2} + v^2_{i_1,i_2+1} - v^2_{i_1,i_2} . $$
div = @(v)v([2:end 1],:,1)-v(:,:,1) + v(:,[2:end 1],2)-v(:,:,2);
The Laplacian operatore is defined as $\Delta=\text{div} \circ \nabla = -\nabla^* \circ \nabla$. It is thus a negative symmetric operator.
delta = @(x)div(grad(x));
Display $\Delta x_0$.
clf;
imageplot(delta(x0));
Check that the relation $ \norm{\nabla x} = - \dotp{\Delta x}{x}. $
dotp = @(a,b)sum(a(:).*b(:));
fprintf('Should be 0: %.3i\n', dotp(grad(x0), grad(x0)) + dotp(delta(x0),x0) );
Should be 0: 000
We consider now the problem of denoising an image $y \in \RR^d$ where $d = n \times n$ is the number of pixels ($n$ being the number of rows/columns in the image).
Add noise to the original image, to simulate a noisy image.
sigma = .1;
y = x0 + randn(n)*sigma;
Display the noisy image $y$.
clf;
imageplot(clamp(y));
Denoising is obtained by minimizing the following functional $$ \umin{x \in \RR^d} f(x) = \frac{1}{2} \norm{y-x}^2 + \la J_\epsilon(x) $$ where $J_\epsilon(x)$ is a smoothed total variation of the image. $$ J_\epsilon(x) = \sum_i \norm{ (G x)_i }_{\epsilon} $$ where $ (Gx)_i \in \RR^2 $ is an approximation of the gradient of $x$ at pixel $i$ and for $u \in \RR^2$, we use the following smoothing of the $L^2$ norm in $\RR^2$ $$ \norm{u}_\epsilon = \sqrt{ \epsilon^2 + \norm{u}^2 }, $$ for a small value of $\epsilon>0$.
The gradient of the functional read $$ \nabla f(x) = x-y + \lambda \nabla J_\epsilon(x) $$ where the gradient of the smoothed TV norm is $$ \nabla J_\epsilon(x)_i = G^*( u ) \qwhereq u_i = \frac{ (G x)_i }{\norm{ (G x)_i }_\epsilon} $$ where $G^*$ is the adjoint operator of $G$ which corresponds to minus a discretized divergence.
Value for $\lambda$.
lambda = .3/5;
Value for $\epsilon$.
epsilon = 1e-3;
TV norm.
NormEps = @(u,epsilon)sqrt(epsilon^2 + sum(u.^2,3));
J = @(x,epsilon)sum(sum(NormEps(grad(x),epsilon)));
Function $f$ to minimize.
f = @(y,x,epsilon)1/2*norm(x-y,'fro')^2 + lambda*J(x,epsilon);
Gradient of $J_\epsilon$. Note that |div| implement $-G^*$.
Normalize = @(u,epsilon)u./repmat(NormEps(u,epsilon), [1 1 2]);
GradJ = @(x,epsilon)-div( Normalize(grad(x),epsilon) );
Gradient of the functional.
Gradf = @(y,x,epsilon)x-y+lambda*GradJ(x,epsilon);
The step size should satisfy $$ 0 < \tau_k < \frac{2}{ 1 + 4 \lambda / \epsilon }. $$ Here we use a slightly larger step size, which still work in practice.
tau = 1.8/( 1 + lambda*8/epsilon );
tau = tau*4;
Exercise 3
Implement the gradient descent. Monitor the decay of $f$ through the iterations.
exo3()
%% Insert your code here.
Display the resulting denoised image.
clf;
imageplot(clamp(x));
We consider a linear imaging operator $\Phi : x \mapsto \Phi(x)$ that maps high resolution images to low dimensional observations. Here we consider a pixel masking operator, that is diagonal over the spacial domain.
To emphasis the effect of the TV functional, we use a simple geometric image.
n = 64;
name = 'square';
x0 = load_image(name,n);
We consider here the inpainting problem. This simply corresponds to a masking operator. Here we remove the central part of the image.
a = 4;
Lambda = ones(n);
Lambda(end/2-a:end/2+a,:) = 0;
Masking operator $ \Phi $. Note that it is symmetric, i.e. $\Phi^*=\Phi$
Phi = @(x)x.*Lambda;
PhiS = @(x)Phi(x);
Noiseless observations $y=\Phi x_0$.
y = Phi(x0);
Display.
clf;
imageplot(x0, 'Original', 1,2,1);
imageplot(y, 'Damaged', 1,2,2);
We want to solve the noiseless inverse problem $y=\Phi f$ using a total variation regularization: $$ \umin{ y=\Phi x } J_\epsilon(x). $$ We use the following projected gradient descent $$ x^{(k+1)} = \text{Proj}_{\Hh}( x^{(k)} - \tau_k \nabla J_{\epsilon}(x^{(k)}) ) $$ where $\text{Proj}_{\Hh}$ is the orthogonal projection on the set of linear constraint $\Phi x = y$, and is easy to compute for inpainting
ProjH = @(x,y) x + PhiS( y - Phi(x) );
Exercise 4
Display the evolution of the inpainting process.
exo4()
%% Insert your code here.
Exercise 5
Try with several values of $\epsilon$. au = tau * 100;
exo5()
%% Insert your code here.