Reconstruction from Partial Tomography Measurements

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This numerical tour explores the reconstruction from tomographic measurement with TV regularization.

In [2]:

Width $n$ of the image.

In [3]:
n = 128;

We first load an image.

In [4]:
name = 'phantom';
f0 = load_image(name, n);

Rotation over the Spacial Domain

Given an image $f$, its rotation is defined by rotating the pixels locations $$ R_\th(f) = f(R_{-\th}(x) $$ where we have use the same notation $R_\th$ for the 2-D linear operator defined by the matrix $$ R_\th = \begin{pmatrix}$ \cos(t) & -\sin(t) \\ \sin(t) & \cos(t) \end{pmatrix}$ $$

Computing rotation $R_\th(f)$ over a discrete grid requires interpolation.

In [5]:
t = linspace(-1,1,n);
[Y,X] = meshgrid(t,t);
rotation = @(f,t)interp2(Y,X,f, sin(t)*X + cos(t)*Y, cos(t)*X - sin(t)*Y, 'cubic', 0);

Rotation Over the Fourier Domain

A difficulty with spine-based rotation is that the corresponding operator is not exactly orthogonal.

In particular, one does not have $R_{-\th} = R_{\th}^{-1}$.

In [6]:

To circunvent this problem, one can define the rotation using a Fourier-domain interpolation (a.k.a. Shannon). This is easily performed by decomposing a rotation as a series of shears which can themselves be implemented as diagonal operator over the Fourier domain.

X-aligned and Y-aligned shear matrices are defined as $$ S_a^x = \begin{pmatrix}$ 1 & a \\ 0 & 1 \end{pmatrix}$ \qandq S_a^y = \begin{pmatrix}$ 1 & 0 \\ a & 1 \end{pmatrix}$ $$

One has the following decomposition of a rotation matrix as a product of shears matrices : $$ R_\th = S_{-\tan(t/2)}^x \circ S_{\sin(t)}^y \circ S_{-\tan(t/2)}^y. $$

If we consider continuous functions defined over the plane $x \in \RR^2$ and define the Fourier transform along the X-direction as $$ \hat f(\om,x_2) = \int f(x_1,x_2) e^{-i \om x_1} d x_2 $$ one can compute the shear transformed function $$ f_a(x) = f( S_{-a}(x) ) $$ as a diagonal operator over the Fourier domain $$ \hat f_a(\om,x_2) = e^{i \om x_2} \hat f(\om,x_2). $$

By analogy, we define the shear transform of a discrete image as a diagonal operator over the Fourier domain. Remark: to ensure that the transform is reald-valued, one should be careful about how to handle the highest frequency $n/2$, otherwise the transform would be complex-valued. Note also that this is an orthogonal transform (the eigenvalues have unit modulus).

In [7]:
t = [0:n/2-1 0 -n/2+1:-1]';
[Y,X] = meshgrid(t,t);
shearx = @(f,a)real( ifft( fft(f) .* exp(-a*2i*pi*X.*Y/n) ) );
shearx = @(f,a)fftshift(shearx(fftshift(f),a));
sheary = @(f,a)shearx(f',a)';

Exercise 1

Display the effect of shearing $S_a^x$ for several values of shear $a$.

In [8]:
In [9]:
%% Insert your code here.

By analogy to the continuous setting, we define a discrete rotation by decomposing it using a series of shear.

In [10]:
rotation = @(f,t)shearx( sheary( shearx(f,-tan(t/2)) ,sin(t)) ,-tan(t/2));

Exercise 2

Display the effect of rotations $R_\th$ for several values of angles $\th$.

In [11]:
In [12]:
%% Insert your code here.

To avoid issues when using large angles, we replace a rotation of angle $\th$ by 4 rotations of angle $\th/4$. Note that this makes computation slower.

In [13]:
rotation = @(f,t)rotation(rotation(f,t/2),t/2);
rotation = @(f,t)rotation(rotation(f,t/2),t/2);

Exercise 3

Display the effect of rotations $R_\th$ for several values of angles $\th$.

In [14]:
In [15]:
%% Insert your code here.

We check that this discrete rotation operator is exactly invertible, $R_{\th}^{-1} = R_{-\th}$.

In [16]:
theta = .2;
e = norm( rotation(rotation(f0,theta),-theta)-f0 );
disp(['Error (should be 0): ' num2str(e, 2) '.']);
Error (should be 0): 1.3e-14.

We check that the inverse is also the adjoint operator, $R_{\th}^{-1} = R_{\th}^* = R_{-\th}$.

In [17]:
theta = .1;
a = randn(n); b = randn(n);
dotp = @(a,b)sum(a(:).*b(:));
e = dotp(a,rotation(b,theta)) - dotp(b,rotation(a,-theta));
disp(['Error (should be 0): ' num2str(e, 2) '.']);
Error (should be 0): -1.3e-12.

Partial Radon Operator

The Radon transform is defined, for an angle $\th \in [-\pi/2,\pi/2)$ as an integration along all lines of angle $\th$. It can thus be defined equivalently by 1-D vertical integration of the image rotated by an angle $\th$. We use here a simple sum to replace the continuous integration: $$ \Phi_\th(f)(x_1) = \sum_{x_2} R_\th(f)(x_1,x_2). $$

The radon transform is defined as the collection of projections for several angles, $\{ \th_i \}_{i=1}^{m}$, usually equi-spaced in $[-\pi/2,\pi/2)$. $$ \Phi(f) = \{ \Phi_{\th_i}(f) \}_{i=1}^m \in \RR^{n \times m}. $$

For $y = \{y_{i}\}_{i=1}^m \in \RR^{n \times m}$, where $y_i \in \RR^n$, the adjoint of the Radon transform is $$ \Phi^*(y) = \sum_i R_{-\th_i}(U(y_i)), $$ where $ U : \RR^n \rightarrow \RR^{n \times n} $ replicates the vector along the lines, $U(v)(x_1,x_2)=v(x_1)$.

Number of angles for the partial tomography.

In [18]:
m = 64;

List of angles.

In [19]:
tlist = linspace(-pi/2,pi/2, m+1); tlist(end) = [];

Define the operators $\Phi$ and $\Phi^*$

In [20]:
Phi  = @(f)perform_radon_transform(f, tlist, +1, rotation);
PhiS = @(R)perform_radon_transform(R, tlist, -1, rotation);

Display a Radon transform.

In [21]:

Check that $\Phi^*$ is indeed the adjoint of $\Phi$.

In [22]:
a = randn(n,m); b = randn(n);
e = dotp(a,Phi(b)) - dotp(PhiS(a),b);
disp(['Error (should be 0): ' num2str(e, 2) '.']);
Error (should be 0): 3.5e-12.

Partial Tomography Pseudo-Inversion

We consider here a coarse sub-sampling of the Radon transform.

In [23]:
m = 16;
tlist = linspace(-pi/2,pi/2, m+1); tlist(end) = [];
Phi  = @(f)perform_radon_transform(f, tlist, +1, rotation);
PhiS = @(R)perform_radon_transform(R, tlist, -1, rotation);

We use noiseless measurements $y=\Phi f_0$.

In [24]:
y = Phi(f0);

Display the corresponing 1-D projections.

In [25]:
plot(y); axis tight;

The pseudo inverse reconstruction is defined as $$ \Phi^+ y = \Phi^* ( \Phi\Phi^* )^{-1} y. $$ It is the minimum $L^2$ norm reconstruction $$ \Phi^+ y = \uargmin{ f} \norm{y-\Phi f}^2. $$

Remark: When $m$ tends to infinity (continuous set of angles), the Radon transform is not sub-sampled and $ \Phi \Phi^* $ is a filtering. For $y(r,\th)$ a continuous set of Radon coefficients, $$ (\Phi\Phi^*) y(r,\th) = \int y(s,\th) h(r-s) d s \qwhereq \hat h(\om) = 1/\abs{\om}. $$ When $m$ is large enough, it is common to approximate $\Phi^+$ using $$ \Phi^+ R = \Phi^*( g \star y ) $$ where $g$ is a 1-D filter with $\hat g(\om) = \abs{\om}$ and $\star$ is the convolution along the first coordinate of the Radon coefficients.

This is not the case in our setting, so that $ \Phi \Phi^* $ is actually difficult to invert, and it is thus computaitonaly difficult to compute the exact pseudo inverse $\Phi^+ = \Phi^* (\Phi \Phi^*)^{-1} $.

Exercise 4

Compute the pseudo inverse reconstruction using a conjugate gradient descent (function |cgs|).

In [26]:
In [27]:
%% Insert your code here.

Partial Tomography Total Variation Inversion

We consider the following reconstruction $$ \umin{\Phi f = y} J(f) \qwhereq J(f)=\norm{\nabla f}_1$$ where the $\ell^1$ norm of a vector field $(u_i)_i$ is defined as $$ J(f) = \sum_i \norm{u_i}. $$


In [28]:

This problem is rewritten as $$ \umin{f} F(K(f)) + G(f) $$ where $G=0$, $Kf = (\Phi f, \nabla f)$ and $F(u,v)=i_{\{y\}}(u) + \norm{v}_1.$

The primal-dual algorithm reads: $$ g_{k+1} = \text{Prox}_{\sigma F^*}( g_k + \sigma K(\tilde f_k) $$ $$ f_{k+1} = \text{Prox}_{\tau G}( f_k-\tau K^*(g_k) ) $$ $$ \tilde f_{k+1} = f_{k+1} + \theta (f_{k+1} - f_k) $$

One has $$ \text{Prox}_{\la F}(u,v) = \pa{ y, \text{Prox}_{\la \norm{\cdot}_1}(v) } $$ where $$ \text{Prox}_{\la \norm{\cdot}_1}(v)_k = \max\pa{0,1-\frac{\la}{\norm{v_k}}} v_k $$

In [29]:
Amplitude = @(u)sqrt(sum(u.^2,3));
F1 = @(u)sum(sum(Amplitude(u)));
u = @(z)reshape(z(1:n*m),n,m);
v = @(z)reshape(z(n*m+1:end),n,n,2);
In [30]:
ProxF1 = @(u,lambda)max(0,1-lambda./repmat(Amplitude(u), [1 1 2])).*u;
ProxF = @(z,lambda)[y(:); reshape(ProxF1(v(z),lambda),n*n*2,1) ];
ProxFS = @(y,sigma)y-sigma*ProxF(y/sigma,1/sigma);
ProxG = @(x,lambda)x;
K  = @(f)[reshape(Phi(f),n*m,1); reshape(grad(f), n*n*2,1)];
KS = @(z)PhiS(u(z)) - div(v(z));

We set parameters for the algorithm. Note that in our case, $L=\norm{K}^2=n m$. One should has $L \sigma \tau < 1$.

In [31]:
L = n*m;
sigma = 10;
tau = .9/(L*sigma);
theta = 1;
In [32]:
f = fL2;


In [ ]:
g = K(f)*0;
f1 = f;
In [ ]:
niter = 100;
E = []; C = []; F = [];
for i=1:niter    
    % update
    fold = f;
    g = ProxFS( g+sigma*K(f1), sigma);
    f = ProxG(  f-tau*KS(g), tau);
    f1 = f + theta * (f-fold);
    % monitor the decay of the energy
    E(i) = F1(grad(f));
    F(i) = norm(y-Phi(f), 'fro');
    C(i) = snr(f0,f);
h = plot(E);
set(h, 'LineWidth', 2);