How can I find the number of specific submatrices inside a matrix?

Question

redman 2022-3-13

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https://ww2.mathworks.cn/matlabcentral/answers/1670814-how-can-i-find-the-number-of-specific-submatrices-inside-a-matrix

编辑： Matt J 2022-3-14

采纳的回答： Matt J

在 MATLAB Online 中打开

Hi, If i ran this code,

A = randi([0, 1], [10,10])

I would get something like

A =
   0     0     1     1     1     1     0     1     0
   1     0     0     1     0     0     0     0     1
   0     1     0     1     1     0     1     1     1
   0     1     0     0     1     0     1     1     0
   0     1     0     1     0     0     1     0     1
   0     0     1     1     1     1     0     0     0
   1     0     1     0     1     1     0     1     1
   0     0     1     0     1     0     0     0     0
   1     0     1     0     1     0     1     1     0
   1     0     1     1     0     1     1     1     0

In this matrix i want to find how many instances the submatrix B occurs.

B = [1 1 ; 1 1]

Does anyone know of an easy way I can achieve this?

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Torsten 2022-3-13

编辑：Torsten 2022-3-13

if-statement in for loop over the number of rows and columns of A ?

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Answer 1

Matt J 2022-3-13

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https://ww2.mathworks.cn/matlabcentral/answers/1670814-how-can-i-find-the-number-of-specific-submatrices-inside-a-matrix#answer_916754

编辑：Matt J 2022-3-14

在 MATLAB Online 中打开

    Sa=conv2(A.^2,ones(2),'valid');
    Sb=sum(B(:).^2);
    n= nnz( conv2(A, rot90(B,2),'valid').^2==Sa.*Sb );

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Image Analyst 2022-3-14

While this is correct for the given matrix of all 1's, it won't work in general for other patterns. It works by comparing the number of 1's in a sliding window over A to the number of 1's in B. It doesn't actually match the pattern. So if there are 2 or 3 in B, it would say there is a match even though the pattern is different. You want to use bwhitmiss(), which is a hit or miss transform specifically meant for this. See my answer below for the general answer that works for other patterns.

Matt J 2022-3-14

编辑：Matt J 2022-3-14

Convolution will work, but needed adjustment to deal with arbitrary patterns of integers;

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Answer 2

Image Analyst 2022-3-14

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在 MATLAB Online 中打开

You want to use the "hit or miss transform" done by bwhitmiss(). Using conv2 will not work for arbitrary patterns. See illustrations below:

A =[...
    1     0     0     1     1     1     1     0     1     0
    1     1     0     0     1     0     0     0     0     1
    1     0     1     0     1     1     0     1     1     1
    1     0     1     0     0     1     0     1     1     0
    1     0     1     0     1     0     0     1     0     1
    0     0     0     1     1     1     1     0     0     0
    0     1     0     1     0     1     1     0     1     1
    0     0     0     1     0     1     0     0     0     0
    1     1     0     1     0     1     0     1     1     0
    1     1     0     1     1     0     1     1     1     0];
B = logical([0, 1; 1, 1]);
numMatchesMJ = nnz( conv2(A,B,'valid')==nnz(B) ) % 10 is incorrect
numMatchesMJ = 10
BW = bwhitmiss(A,B,~B);
BW = BW(1:end-1, 1:end-1) % 1 if upper left corner matches.
BW = 9×9 logical array
   0   0   0   0   0   0   0   0   0
   0   0   0   0   0   0   0   0   1
   0   0   0   0   0   0   0   0   0
   0   0   0   0   0   0   0   0   0
   0   0   0   1   0   0   0   0   0
   0   0   0   0   0   0   0   0   0
   0   0   0   0   0   0   0   0   0
   0   0   0   0   0   0   0   0   0
   0   0   0   0   0   0   1   0   0
numMatches = nnz(BW) % 3 is correct.
numMatches = 3
% Another example
B = logical([1, 0; 1, 1]);
numMatchesMJ = nnz( conv2(A,B,'valid')==nnz(B) ) % 7 is incorrect
numMatchesMJ = 7
BW = bwhitmiss(A,B,~B);
BW = BW(1:end-1, 1:end-1) % 1 if upper left corner matches.
BW = 9×9 logical array
   1   0   0   0   0   0   0   0   0
   0   0   0   0   1   0   0   0   0
   0   0   0   0   0   0   0   0   0
   0   0   0   0   0   0   0   0   0
   0   0   0   0   1   0   0   0   0
   0   0   0   0   0   0   0   0   0
   0   0   0   0   0   0   0   0   0
   0   0   0   0   0   0   0   0   0
   0   0   0   1   0   0   0   0   0
numMatches = nnz(BW) % 4 is correct.
numMatches = 4