problem in perimeter

Question

Mohammad Golam Kibria 2011-11-30

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https://ww2.mathworks.cn/matlabcentral/answers/22642-problem-in-perimeter

hi,

I have a matrix as follows:

I =

   0     0     0     0     0
   0     1     1     0     0
   1     1     1     1     0
   1     1     1     1     0
   0     1     1     0     0
   0     0     0     0     0

I want to have another matrix as follows

I =

   2     2     2     2     0
   2     1     1     2     2
   1     1     1     1     2
   1     1     1     1     2
   2     1     1     2     2
   2     2     2     2     0

i.e. replace every position with 2 within distance 1 from the perimeter of region 1. Thanks

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Answer 1

Chandra Kurniawan 2011-11-30

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https://ww2.mathworks.cn/matlabcentral/answers/22642-problem-in-perimeter#answer_29782

Sorry. Here I modified my code

clear all; clc  
I =[0 0 1 1 0 0;
      0 1 1 1 1 0;
      0 1 1 1 1 0;
      0 0 1 1 0 0;
      0 0 0 0 0 0];
J = zeros(size(I,1)+2, size(I,2)+2);
K = zeros(size(I,1)+2, size(I,2)+2);
J(2:size(J,1)-1, 2:size(J,2)-1) = I;
for x = 2 : size(J,1)-1
    for y = 2 : size(J,2)-1
        neighbour = [J(x-1,y-1) J(x-1,y) J(x-1,y+1) ...
                               J(x,y-1) J(x,y+1) ...
                               J(x+1,y-1) J(x+1,y) J(x+1,y+1)];
        if (find(neighbour))
            K(x,y) = 2;
        end            
    end
end
L = K - J;
L(1,:) = [];     L(end,:) = [];
L(:,1) = [];     L(:,end) = []

And the result :

L =
       2     2     1     1     2     2
       2     1     1     1     1     2
       2     1     1     1     1     2
       2     2     1     1     2     2
       0     2     2     2     2     0

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Mohammad Golam Kibria 2011-11-30

Thanks

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Answer 2

Chandra Kurniawan 2011-11-30

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https://ww2.mathworks.cn/matlabcentral/answers/22642-problem-in-perimeter#answer_29780

clear all; clc;
I =[0 0 0 0 0 0;
      0 0 1 1 0 0;
      0 1 1 1 1 0;
      0 1 1 1 1 0;
      0 0 1 1 0 0;
      0 0 0 0 0 0];
J = zeros(size(I,1)+2, size(I,2)+2);
K = zeros(size(I,1)+2, size(I,2)+2);
J(2:7,2:7) = I;
for x = 2 : 7
    for y = 2 : 7
        neighbour = [J(x-1,y-1) J(x-1,y) J(x-1,y+1) ...
                               J(x,y-1) J(x,y+1) ...
                               J(x+1,y-1) J(x+1,y) J(x+1,y+1)];
        if (find(neighbour))
            K(x,y) = 2;
        end            
    end
end
L = K - J;
L(1,:) = [];     L(end,:) = [];
L(:,1) = [];     L(:,end) = []

And you will get L =

L =
       0     2     2     2     2     0
       2     2     1     1     2     2
       2     1     1     1     1     2
       2     1     1     1     1     2
       2     2     1     1     2     2
       0     2     2     2     2     0

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Mohammad Golam Kibria 2011-11-30

but if my matrix is as follows:

I =[0 0 1 1 0 0;

0 1 1 1 1 0;

0 0 1 1 0 0;

0 0 0 0 0 0];

It does not work, how to extend your idea for any matrix I give, I am not good in matlab. could u help me in this regard.

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Answer 3

Andrei Bobrov 2011-11-30

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https://ww2.mathworks.cn/matlabcentral/answers/22642-problem-in-perimeter#answer_29794

variant use conv2 without Image Processing Toolbox

t = conv2(I,ones(3),'same')>0 
out = t + 0 
out(t>0&t~=I) = 2

or

out = 2*(conv2(I,ones(3),'same')>0+0)-I

variant use with function imdilate by Image Processing Toolbox

out = imdilate(I,ones(3))
out(out~=I) = 2

or

out = 2*imdilate(I,ones(3))-I

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Mohammad Golam Kibria 2011-12-1

It also works fine

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Answer 4

Image Analyst 2011-11-30

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https://ww2.mathworks.cn/matlabcentral/answers/22642-problem-in-perimeter#answer_29809

If you have the Image Processing Toolbox you can call imdilate() and then bwperim() and then combine the perimeter image with the original by multiplying the perimeter image by 2 and adding to the original image.