Determine adjacent points in a logical matrix

I have a 2-D matrix of logical values, eg

000000000
010000000
000000000
000000000
000000000
000001000
000000000
000000000
100000000

How can I create a similar sized matrix with True in all the locations adjacent to the Trues in the first matrix, eg

111000000
101000000
111000000
000000000
000011100
000010100
000011100
110000000
010000000

If a location is assigned True from two or more adjacent locations, then it should be True.

[dormant]

Thanks everyone. I knew it would be trivial, but I don't have much experience with manipulating 2D arrays.

Answer 1

[John D'Errico]

Trivial. Learn to think in terms of MATLAB operations. I've added another 1 in there, just to make it clear what the problem is.

A = [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 0 1 1 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];

For example, what does this do? Does it get you close to what you want?

B = conv2(A,ones(3),'same')

B = 9×9

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

Close, but we can fix that.

B = conv2(A,[1 1 1;1 0 1;1 1 1],'same')

B = 9×9

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

Next, we need to be careful, as two elements near each other can create something bigger than 1 due to the convolution. This next will correct that.

B = conv2(A,[1 1 1;1 0 1;1 1 1],'same') ~= 0

B = 9×9 logical array
   1   1   1   0   0   0   0   0   0
   1   0   1   0   0   0   0   0   0
   1   1   1   0   0   0   0   0   0
   0   0   0   0   0   0   0   0   0
   0   0   0   0   1   1   1   1   0
   0   0   0   0   1   1   1   1   0
   0   0   0   0   1   1   1   1   0
   1   1   0   0   0   0   0   0   0
   0   1   0   0   0   0   0   0   0

The ones are adjacent to each other in the original array, and that meant that it filled in the ones in the result. We can zap them out too.

B = (conv2(A,[1 1 1;1 0 1;1 1 1],'same') ~= 0) & (~A)

B = 9×9 logical array
   1   1   1   0   0   0   0   0   0
   1   0   1   0   0   0   0   0   0
   1   1   1   0   0   0   0   0   0
   0   0   0   0   0   0   0   0   0
   0   0   0   0   1   1   1   1   0
   0   0   0   0   1   0   0   1   0
   0   0   0   0   1   1   1   1   0
   1   1   0   0   0   0   0   0   0
   0   1   0   0   0   0   0   0   0

[dormant]

This is perfect. Many thanks.

[John D'Errico]

Thank you. I hope the explanations made sense. The important point is to think of conv and conv2 when you have problems like this.

Answer 2

[DGM]

This can also be done succinctly with image processing tools:

% a logical array
A = [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 0 1 1 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];
A = logical(A);

% output is a logical array
B = imdilate(A,ones(3)) & ~A

B = 9×9 logical array
   1   1   1   0   0   0   0   0   0
   1   0   1   0   0   0   0   0   0
   1   1   1   0   0   0   0   0   0
   0   0   0   0   0   0   0   0   0
   0   0   0   0   1   1   1   1   0
   0   0   0   0   1   0   0   1   0
   0   0   0   0   1   1   1   1   0
   1   1   0   0   0   0   0   0   0
   0   1   0   0   0   0   0   0   0