Logo Questions Linux Laravel Mysql Ubuntu Git Menu
 

Rigidly register a 2D image to a 3D volume with good initial guess for affine transformation

I have a 3D volume and a 2D image and an approximate mapping (affine transformation with no skwewing, known scaling, rotation and translation approximately known and need fitting) between the two. Because there is an error in this mapping and I would like to further register the 2D image to the 3D volume. I have not written code for registration purposes before, but because I can't find any programs or code to solve this I would like to try and do this. I believe the standard for registration is to optimize mutual information. I think this would also be suitable here, because the intensities are not equal between the two images. So I think I should make a function for the transformation, a function for the mutual information and a function for optimization.

I did find some Matlab code on a mathworks thread from two years ago, based on an article. The OP reports that she managed to get the code to work, but I'm not getting how she did that exactly. Also in the IP package for matlab there is an implementation, but I dont have that package and there does not seem to be an equivalent for octave. SPM is a program that uses matlab and has registration implemented, but does not cope with 2d to 3d registration. On the file exchange there is a brute force method that registers two 2D images using mutual information.

What she does is pass a multi planar reconstruction function and an similarity/error function into a minimization algorithm. But the details I don't quite understand. Maybe it would be better to start fresh:

load mri; volume = squeeze(D);

phi = 3; theta = 2; psi = 5; %some small angles
tx = 1; ty = 1; tz = 1; % some small translation
dx = 0.25, dy = 0.25, dz = 2; %different scales
t = [tx; ty; tz];
r = [phi, theta, psi]; r = r*(pi/180);
dims = size(volume);
p0 = [round(dims(1)/2);round(dims(2)/2);round(dims(3)/2)]; %image center
S = eye(4); S(1,1) = dx; S(2,2) = dy; S(3,3) = dz;

Rx=[1 0 0 0;
    0 cos(r(1)) sin(r(1)) 0;
    0 -sin(r(1)) cos(r(1)) 0;
    0 0 0 1];
Ry=[cos(r(2)) 0 -sin(r(2)) 0;
    0 1 0 0;
    sin(r(2)) 0 cos(r(2)) 0;
    0 0 0 1];
Rz=[cos(r(3)) sin(r(3)) 0 0;
    -sin(r(3)) cos(r(3)) 0 0;
    0 0 1 0;
    0 0 0 1];
R = S*Rz*Ry*Rx;
%make affine matrix to rotate about center of image
T1 = ( eye(3)-R(1:3,1:3) ) * p0(1:3);
T = T1 + t; %add translation
A = R;
A(1:3,4) = T;
Rold2new = A;
Rnew2old = inv(Rold2new);

%the transformation
[xx yy zz] = meshgrid(1:dims(1),1:dims(2),1:1);
coordinates_axes_new = [xx(:)';yy(:)';zz(:)'; ones(size(zz(:)))'];
coordinates_axes_old = Rnew2old*coordinates_axes_new;
Xcoordinates = reshape(coordinates_axes_old(1,:), dims(1), dims(2), dims(3));
Ycoordinates = reshape(coordinates_axes_old(2,:), dims(1), dims(2), dims(3));
Zcoordinates = reshape(coordinates_axes_old(3,:), dims(1), dims(2), dims(3));

%interpolation/reslicing
method = 'cubic'; 
slice= interp3(volume, Xcoordinates, Ycoordinates, Zcoordinates, method);
%so now I have my slice for which I would like to find the correct position

% first guess for A
A0 = eye(4); A0(1:3,4) = T1; A0(1,1) = dx; A0(2,2) = dy; A0(3,3) = dz; 
% this is pretty close to A
% now how would I fit the slice to the volume by changing A0 and examining some similarity measure?
% probably maximize mutual information?
% http://www.mathworks.com/matlabcentral/fileexchange/14888-mutual-information-computation/content//mi/mutualinfo.m
like image 205
Leo Avatar asked Oct 31 '22 11:10

Leo


1 Answers

Ok I was hoping for someone else's approach, that would probably have been better than mine as I have never done any optimization or registration before. So I waited for Knedlsepps bounty to almost finish. But I do have some code thats working now. It will find a local optimum so the initial guess must be good. I wrote some functions myself, took some functions from the file exchange as is and I extensively edited some other functions from the file exchange. Now that I put all the code together to work as an example here, the rotations are off, will try and correct that. Im not sure where the difference in code is between the example and my original code, must have made a typo in replacing some variables and data loading scheme.

What I do is I take the starting affine transformation matrix, decompose it to an orthogonal matrix and an upper triangular matrix. I then assume the orthogonal matrix is my rotation matrix so I calculate the euler angles from that. I directly take the translation from the affine matrix and as stated in the problem I assume I know the scaling matrix and there is no shearing. So then I have all degrees of freedom for the affine transformation, which my optimisation function changes and constructs a new affine matrix from, applies it to the volume and calculates the mutual information. The matlab optimisation function patternsearch then minimises 1-MI/MI_max.

What I noticed when using it on my real data which are multimodal brain images is that it works much better on brain extracted images, so with the skull and tissue outside of the skull removed.

%data
load mri; volume = double(squeeze(D));

%transformation parameters
phi = 3; theta = 1; psi = 5; %some small angles
tx = 1; ty = 1; tz = 3; % some small translation
dx = 0.25; dy = 0.25; dz = 2; %different scales
t = [tx; ty; tz];
r = [phi, theta, psi]; r = r*(pi/180);
%image center and size
dims = size(volume);
p0 = [round(dims(1)/2);round(dims(2)/2);round(dims(3)/2)]; 
%slice coordinate ranges
range_x = 1:dims(1)/dx;
range_y = 1:dims(2)/dy;
range_z = 1;

%rotation 
R = dofaffine([0;0;0], r, [1,1,1]);
T1 = ( eye(3)-R(1:3,1:3) ) * p0(1:3); %rotate about p0
%scaling 
S = eye(4); S(1,1) = dx; S(2,2) = dy; S(3,3) = dz;
%translation
T = [[eye(3), T1 + t]; [0 0 0 1]];
%affine 
A = T*R*S;

% first guess for A
r00 = [1,1,1]*pi/180;
R00 = dofaffine([0;0;0], r00, [1 1 1]);
t00 = T1 + t + ( eye(3) - R00(1:3,1:3) ) * p0;
A0 = dofaffine( t00, r00, [dx, dy, dz] );
[ t0, r0, s0 ] = dofaffine( A0 );
x0 = [ t0.', r0, s0 ];

%the transformation
slice = affine3d(volume, A, range_x, range_y, range_z, 'cubic');
guess = affine3d(volume, A0, range_x, range_y, range_z, 'cubic');

%initialisation
Dt = [1; 1; 1]; 
Dr = [2 2 2].*pi/180;
Ds = [0 0 0];
Dx = [Dt', Dr, Ds];
%limits
LB = x0-Dx;
UB = x0+Dx;
%other inputs
ref_levels = length(unique(slice));
Qref = imquantize(slice, ref_levels);
MI_max = MI_GG(Qref, Qref);
%patternsearch options
options = psoptimset('InitialMeshSize',0.03,'MaxIter',20,'TolCon',1e-5,'TolMesh',5e-5,'TolX',1e-6,'PlotFcns',{@psplotbestf,@psplotbestx});

%optimise
[x2, MI_norm_neg, exitflag_len] = patternsearch(@(x) AffRegOptFunc(x, slice, volume, MI_max, x0), x0,[],[],[],[],LB(:),UB(:),options);

%check
p0 = [round(size(volume)/2).'];
R0 = dofaffine([0;0;0], x2(4:6)-x0(4:6), [1 1 1]);
t1 = ( eye(3) - R0(1:3,1:3) ) * p0;
A2 = dofaffine( x2(1:3).'+t1, x2(4:6), x2(7:9) ) ;
fitted = affine3d(volume, A2, range_x, range_y, range_z, 'cubic');
overlay1 = imfuse(slice, guess);
overlay2 = imfuse(slice, fitted);
figure(101); 
ax(1) = subplot(1,2,1); imshow(overlay1, []); title('pre-reg')
ax(2) = subplot(1,2,2); imshow(overlay2, []); title('post-reg');
linkaxes(ax);

function normed_score = AffRegOptFunc( x, ref_im, reg_im, MI_max, x0 )
    t = x(1:3).';
    r = x(4:6);
    s = x(7:9);

    rangx = 1:size(ref_im,1);
    rangy = 1:size(ref_im,2);
    rangz = 1:size(ref_im,3);

    ref_levels =  length(unique(ref_im));
    reg_levels =  length(unique(reg_im));

    t0 = x0(1:3).';
    r0 = x0(4:6);
    s0 = x0(7:9);
    p0 = [round(size(reg_im)/2).'];
    R = dofaffine([0;0;0], r-r0, [1 1 1]);
    t1 = ( eye(3) - R(1:3,1:3) ) * p0;
    t = t + t1;
    Ap = dofaffine( t, r, s );

    reg_im_t = affine3d(reg_im, A, rangx, rangy, rangz, 'cubic');

    Qref = imquantize(ref_im, ref_levels);
    Qreg = imquantize(reg_im_t, reg_levels);

    MI = MI_GG(Qref, Qreg);

    normed_score = 1-MI/MI_max;
end

function [ varargout ] = dofaffine( varargin )
% [ t, r, s ] = dofaffine( A )
% [ A ] = dofaffine( t, r, s )
if nargin == 1
    %affine to degrees of freedom (no shear)
    A = varargin{1};

    [T, R, S] = decompaffine(A);
    r = GetEulerAngles(R(1:3,1:3));
    s = [S(1,1), S(2,2), S(3,3)];
    t = T(1:3,4);

    varargout{1} = t;
    varargout{2} = r;
    varargout{3} = s;
elseif nargin == 3
    %degrees of freedom to affine (no shear)
    t = varargin{1};
    r = varargin{2};
    s = varargin{3};

    R = GetEulerAngles(r); R(4,4) = 1;
    S(1,1) = s(1); S(2,2) = s(2); S(3,3) = s(3); S(4,4) = 1;
    T = eye(4); T(1,4) = t(1); T(2,4) = t(2); T(3,4) = t(3);
    A = T*R*S;

    varargout{1} = A;
else
    error('incorrect number of input arguments');
end
end

function [ T, R, S ] = decompaffine( A )
    %I assume A = T * R * S
    T = eye(4);
    R = eye(4);
    S = eye(4);
    %decompose in orthogonal matrix q and upper triangular matrix r
    %I assume q is a rotation matrix and r is a scale and shear matrix
    %matlab 2014 can force real solution
    [q r] = qr(A(1:3,1:3));
    R(1:3,1:3) = q;
    S(1:3,1:3) = r;

    % A*S^-1*R^-1 = T*R*S*S^-1*R^-1 = T*R*I*R^-1 = T*R*R^-1 = T*I = T
    T = A*inv(S)*inv(R);
    t = T(1:3,4);
    T = [eye(4) + [[0 0 0;0 0 0;0 0 0;0 0 0],[t;0]]];
end

function [varargout]= GetEulerAngles(R)

assert(length(R)==3)

dims = size(R);

    if min(dims)==1
        rx = R(1); ry = R(2); rz = R(3);
        R = [[                           cos(ry)*cos(rz),                          -cos(ry)*sin(rz),          sin(ry)];...
             [ cos(rx)*sin(rz) + cos(rz)*sin(rx)*sin(ry), cos(rx)*cos(rz) - sin(rx)*sin(ry)*sin(rz), -cos(ry)*sin(rx)];...
             [ sin(rx)*sin(rz) - cos(rx)*cos(rz)*sin(ry), cos(rz)*sin(rx) + cos(rx)*sin(ry)*sin(rz),  cos(rx)*cos(ry)]];
        varargout{1} = R;
    else
        ry=asin(R(1,3));
        rz=acos(R(1,1)/cos(ry));
        rx=acos(R(3,3)/cos(ry));

        if nargout > 1 && nargout < 4
            varargout{1} = rx;
            varargout{2} = ry;
            varargout{3} = rz;
        elseif nargout == 1
            varargout{1} = [rx ry rz];
        else
            error('wrong number of output arguments');
        end
    end
end
like image 138
Leo Avatar answered Nov 15 '22 10:11

Leo