How to find the positions of pixels with 4 neighborhood in a binary pixel grid
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Hi @Harrison,
There are several issues that need to be addressed to ensure the code functions correctly. Below, I will outline the necessary corrections and provide a complete, updated version of the code.
Boundary Conditions: The code does not adequately handle boundary conditions, which can lead to indexing errors when accessing pixels outside the 3x3 grid.
Variable Initialization: The variable fourneighbors is used but not defined in the provided code. We need to ensure it is initialized properly.
Logical Structure: The nested if statements can be simplified for better readability and efficiency.
Here is the revised version of your MATLAB code, which includes boundary checks and proper initialization of variables:
% Initialize a sample 3x3 image matrix image = [1 0 1; 1 1 1; 0 1 0];
% Initialize the value to assign to neighboring pixels fourneighbors = 2; % Example value to represent neighboring pixels
% Loop through the 3x3 grid for i = 1:3 for j = 1:3 % Check the upper neighbor if i > 1 && image(i-1,j) == 1 && image(i,j) == 1 image(i,j) = fourneighbors; end
% Check the left neighbor if j > 1 && image(i,j-1) == 1 && image(i,j) == 1 image(i,j) = fourneighbors; end
% Check the lower neighbor if i < 3 && image(i+1,j) == 1 && image(i,j) == 1 image(i,j) = fourneighbors; end
% Check the right neighbor if j < 3 && image(i,j+1) == 1 && image(i,j) == 1 image(i,j) = fourneighbors; end end end
% Display the final image matrix disp('Updated Image Matrix:'); disp(image);
Please see attached.
So, in the above code snippet,image matrix is initialized with sample values. The variable fourneighbors is set to 2, which will be used to mark pixels that have neighboring pixels with the same brightness. The outer loop iterates over the rows, while the inner loop iterates over the columns of the 3x3 grid. Each if statement includes conditions to ensure that the code does not attempt to access indices outside the bounds of the matrix. For example, if i > 1 checks that the upper neighbor exists, and if j < 3 checks that the right neighbor exists. If the current pixel and its neighbor share the same brightness value (both equal to 1), the current pixel is updated to the value of fourneighbors. Finally, the updated image matrix is displayed using disp.
This updated code should now function correctly, checking neighboring pixels in the 3x3 grid while avoiding any indexing errors. You can modify the image matrix to test different scenarios.
Hope this helps.
If you have any further questions or need additional modifications, feel free to ask!
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