Calculate the maximum distance between two perimeter of 3d binary masks in a straight line in the z-direction.

Hello everyone,

For a project I am attempting to calculate the maximum distance between an ITV and a PTV in the x, y and z-direction. For this first I have made a code with two synthetic 3d binary masks. For the x-direction and the y-direction I was able to get a correct value. For the y-direction it is 1 and x is 2. For the z-direction for the masks I created the distance should be 3. However I am not able to figure it out. Maybe some of you can help. The entire code is down below, the z-direction that I tried is at the bottom.

%% X-direction (Loop)

% Create the two binary masks

mask1 = zeros(10,10,10);

mask1(4:6,4:6,4:6) = 1;

mask1 = bwperim(mask1);

mask2 = zeros(10,10,10);

mask2(2:7,2:8,3:9) = 1; %gaat hier dus om de rows

mask2 = bwperim(mask2);

% Initialize the maximum distances

max_distances = zeros(size(mask1,1), size(mask1,2), size(mask1,3));

% Loop over each slice in the z-direction

for k = 1:size(mask1,3)

% Find the current slices

slice1 = mask1(:,:,k);

slice2 = mask2(:,:,k);

% Find the rows in which slice1 and slice2 have non-zero elements

[row1, col1] = find(slice1);

[row2, col2] = find(slice2);

% Loop over each point in slice1 that is 1

for i = 1:numel(row1)

% Initialize the maximum distance to a point in slice2 that is 1

max_distance = -Inf;

% Loop over each point in slice2 that is 1

for j = 1:numel(row2)

% Check if the row of the current point in slice2 is the same as

% the row of the current point in slice1

if row2(j) == row1(i)

% Calculate the x-distance between the two points

x_distance = col2(j) - col1(i);

% Check if the distance is positive and less than the maximum

% distance found so far, and that there are no points of slice1

% between the two points

if x_distance > 0 && x_distance > max_distance && ~any(slice1(row1(i),col1(i)+1:col2(j)-1))

% Update the maximum distance

max_distance = x_distance;

end

end

end

% Store the maximum distance for the current point in slice1

max_distances(row1(i), col1(i), k) = max_distance;

end

end

% Find the maximum distance over all points and all slices in slice1

max_distance_x = max(max(max_distances(:,:,:)));

max_distance_value_x = max(max_distance_x);

disp(['The maximum distance between the two binary images in the x-direction is ', num2str(max_distance_value_x), '.']);

%% Y-direction (Loop)

% Create the two binary masks

mask1 = zeros(10,10,10);

mask1(4:6,4:6,4:6) = 1;

mask1 = bwperim(mask1);

mask2 = zeros(10,10,10);

mask2(2:7,2:8,3:9) = 1; %gaat hier dus om de rows

mask2 = bwperim(mask2);

% Initialize the maximum distances

max_distances = zeros(size(mask1,1), size(mask1,2), size(mask1,3));

% Loop over each slice in the y-direction

for j = 1:size(mask1,2)

% Find the current slices

slice1 = mask1(:,j,:);

slice2 = mask2(:,j,:);

% Find the columns in which slice1 and slice2 have non-zero elements

[row1, col1] = find(slice1);

[row2, col2] = find(slice2);

% Loop over each point in slice1 that is 1

for i = 1:numel(row1)

% Initialize the maximum distance to a point in slice2 that is 1

max_distance = -Inf;

% Loop over each point in slice2 that is 1

for k = 1:numel(row2)

% Check if the column of the current point in slice2 is the same as

% the column of the current point in slice1

if col2(k) == col1(i)

% Calculate the y-distance between the two points

y_distance = row2(k) - row1(i);

% Check if the distance is positive and less than the maximum

% distance found so far, and that there are no points of slice1

% between the two points

if y_distance > 0 && y_distance > max_distance && ~any(slice1(row1(i)+1:row2(k)-1,col1(i)))

% Update the maximum distance

max_distance = y_distance;

end

end

end

% Store the maximum distance for the current point in slice1

max_distances(row1(i), j) = max(max_distances(row1(i), j), max_distance);

end

end

% Find the maximum distance over all points and all slices in slice1

max_distance_y = max(max(max_distances(:)));

disp(['The maximum distance between the two binary images in the y-direction is ', num2str(max_distance_y), '.']);

%% Z-direction (loop)

% Create the two binary masks

mask1 = zeros(10,10,10);

mask1(4:6,4:6,4:6) = 1;

mask1 = bwperim(mask1);

mask2 = zeros(10,10,10);

mask2(2:7,2:8,3:9) = 1; %gaat hier dus om de rows

mask2 = bwperim(mask2);

% Initialize the maximum distances

max_distances = zeros(size(mask1,1), size(mask1,2), size(mask1,3));

% Loop over each slice in the z-direction

for j = 1:size(mask1,3)

% Find the current slices

slice1 = mask1(j,:,:);

slice2 = mask2(j,:,:);

% Find the columns in which slice1 and slice2 have non-zero elements

[row1, col1] = find(slice1);

[row2, col2] = find(slice2);

% Loop over each point in slice1 that is 1

for i = 1:numel(row1)

% Initialize the maximum distance to a point in slice2 that is 1

max_distance = -Inf;

% Loop over each point in slice2 that is 1

for k = 1:numel(row2)

% Check if the column of the current point in slice2 is the same as

% the column of the current point in slice1

if col2(k) == col1(i)

% Calculate the z-distance between the two points

z_distance = row2(k) - row1(i);

% Check if the distance is positive and less than the maximum

% distance found so far, and that there are no points of slice1

% between the two points

if z_distance > 0 && z_distance > max_distance && ~any(slice1(row1(i)+1:row2(k)-1,col1(i)))

% Update the maximum distance

max_distance = z_distance;

end

end

end

% Store the maximum distance for the current point in slice1

max_distances(row1(i), j) = max(max_distances(row1(i), j), max_distance);

end

end

% Find the maximum distance over all points and all slices in slice1

max_distance_z = max(max(max_distances(:)));

disp(['The maximum distance between the two binary images in the z-direction is ', num2str(max_distance_z), '.']);

Answer 1

[Himanshu]

Hello Sander,

As per my understanding, you are facing an error while calculating the maximum distance between an ITV (Internal Target Volume) and a PTV (Planning Target Volume) in three dimensions (x, y, and z).

The issue you are facing can be eliminated by making the following changes in the z-direction code:

1. Remove the loop for "j = 1:size(mask1,3)" as it is unnecessary.
2. Replace the existing nested loops with two new loops, one for the rows (i-direction) and one for the columns (j-direction).
3. Modify the code within the new nested loops to find the indices where "mask1" and "mask2" have non-zero elements in the current row and column.
4. Adjust the conditions in the loop and distance calculation to properly account for the z-direction.

% Loop over each row in the i-direction
for i = 1:size(mask1,1)
    % Loop over each column in the j-direction
    for j = 1:size(mask1,2)
        % Find the indices in which mask1 and mask2 have non-zero elements in the current row and column
        idx1 = find(mask1(i,j,:));
        idx2 = find(mask2(i,j,:));
        % rest of the code remains the same
    end
end

I hope this helps.