How to place the minimum value of a 3*3 matrix at center of the matrix.?

Suppose a matrix x= 512*512 elements. How to select 3*3 matrix and the smallest element should be at the center position. i.e suppose the matrix is x1=

56 57 90

98 79 49

20 30 39

In the above 3*3 matrix as 20 is smaller compared to all so 20 and 79 position should be interchanged. let the above matrix is first 3*3 sub matrix of 512*512 matrix, then how to select the next 3*3 matrix and so on..

Answer 1

[Andrei Bobrov]

Let A - your array and we use Image Processing Toolbox :

out = imerode(A,ones(3));

add

[ii,jj] = ndgrid(ceil((1:size(A,1))/3),ceil((1:size(A,2))/3));
[~,~,c] = unique([ii(:),jj(:)],'rows');
for kk = 1:max(c)
    t = kk == c;
    k = A(t);
    [~,n] = min(k(:));
    nn = min(numel(k),5);
    k([nn,n]) = k([n,nn]);
    A(t) = k;
end

[aditya sahu]

Not working..

[Andrei Bobrov]

Please see add part in my answer. [EDIT answer 2017-09-08 01:44EDT]

[aditya sahu]

How to find the smallest value for the subsequent 3*3 matrix. You process is for only first 3*3 matrix of 512*512 matrix.

[Image Analyst]

No - ii and jj scan over the entire large image.

What is this swapping algorithm for? I've never heard about it.

By the way, a neat trick for swapping numbers is

[b, a] = deal(a, b); % Swap a and b.

Also you need to specify whether you want the code to be recursive (which I think Andrei's code is), or non-recursive (which I think is usually wanted in spatial filters.) In other words, swapping something to the middle or bottom row will affect later positions of the window since it changes the main "A" input matrix.

[aditya sahu]

Actually it is non recursive..and is applied to an spatial domian.

Answer 2

[Stephen23]

This is easy with linear indexing:

>> X = [56,57,90;98,79,49;20,30,39]
X =
 56   57   90
 98   79   49
 20   30   39
>> [~,idx] = min(X(:));
>> X([idx,5]) = X([5,idx])
X =
 56   57   90
 98   20   49
 79   30   39

Put this inside a function, and then use blockproc to call it for each 3x3 submatrix inside matrix A (untested):

function X = swapmin(S)
X = S.data;
[~,idx] = min(X(:));
X([idx,5]) = X([5,idx]);
end

and calling:

B = blockproc(A,[3,3],@swapmin)

[aditya sahu]

Why you have done X = S.data; and what is the value of 'data' here. and in B = blockproc(A,[3,3],@swapmin) what is 'A' refers to.

[aditya sahu]

and it is not working too.

[Stephen23]

"Why you have done X = S.data; and what is the value of 'data' here. and in B = blockproc(A,[3,3],@swapmin)"

All of these questions are clearly answered in the blockproc documentation. Did you read it?

"what is 'A' refers to."

The input matrix.

[Stephen23]

"and it is not working too."

It works for me:

>> A = randi(9,6,6)
A =
     8     3     9     8     7     7
     9     5     5     9     7     1
     2     9     8     6     7     3
     9     9     2     1     4     1
     6     2     4     8     6     1
     1     9     9     9     2     8
>> blockproc(A,[3,3],@swapmin)
ans =
     8     3     9     8     7     7
     9     2     5     9     1     7
     5     9     8     6     7     3
     9     9     2     6     4     1
     6     1     4     8     1     1
     2     9     9     9     2     8

As you can see, the center of each 3x3 block is now the minimum value, which is what your question requested.