Fastest way to find a cross in a matrix - matlab

Question

Definition:
A(i, j) = 1 is a midpoint of a cross if the elements
A(i-1, j) = 1
A(i+1, j) = 1
A(i, j+1) = 1
A(i, j-1) = 1.
Together the elements and the midpoint form a cross in a matrix A, where A is at least a 3-by-3 matrix and i, j ∈ ℕ\{0}.
Suppose the image above is the 8-by-8 matrix A with natural numbers 1, 2, 3 ... as elements. From this definition the matrix has a total of 3 crosses. The crosses have their midpoints on A(2,2), A(5, 4) and A(5, 5).
What I want to do is write a function that finds the number of crosses in the matrix A. I have an idea but I'm not sure it's the most optimal one. Here's the pseudocode for it:
ITERATE FROM row 2 TO row 7
ITERATE FROM column 1 TO column 8
IF current element contains 1
INCREMENT xcount by 1
IF xcount >= 3
CHECK IF counted 1:s is part of a cross
ELSE IF xcount IS NOT 0
SET xcount to 0
The idea is to iterate through every column from row 2 to row 7. If I find 3 consecutive 1:s on the same row I immediately check if the 1:s belongs to a cross. This should work, but imagine having a very large matrix A - how efficient would this code be in that situation? Couldn't this problem be solved using vector notation?
Any answer is very much appreciated. Thanks in advance!

Answer 1

Not near matlab at the moment, but this is what I'd do. Assuming A is binary (has only 0'a and 1's):
crs=[0 1 0 ; 1 1 1 ; 0 1 0]; % a minimal "cross" filter
C=conv2(A,crs./sum(crs(:)),'same'); % convolve A with it
[x y]=find(C>0.9); % find x,y positions of the crosses by looking
% for peak values of C
so you basically convolve with a "minimal" (normalized) cross (crs) and look for peaks using max. x and y are the coordinates of your cross positions. No need to use for loops, just the built in (and pretty fast) 2d convolution, and the max function.
The threshold condition C>0.9, is just to illustrate that there's need to be a threshold that is weighted by intensity of crs. In this case I have normalized crs in the colvolution line (crs/sum(crs(:))) so if A is a binary matrix as in the example, you find that the convolution of the minimal normalized cross will leave the value of the pixel where the cross is at 1, whereas other pixels will be less than 1 (that's why I arbitrarily chose 0.9) . So you can replace the threshold to C==1, if it's always a binary.
Another way to visulize the position of the cross is just to look at C.*(C==1). This will generate a matrix the size of A with 1s only where the crosses were...
EDIT:
For maximal speed, you may consider writing it as a one liner, for example:
[x y]=find(conv2(A,[0 1 0 ; 1 1 1 ; 0 1 0]./5,'same')==1);

Answer 2

Using bit masks:
ux = [false(size(A,1),1) (A(:,3:end) & A(:,2:end-1) & A(:,1:end-2)) false(size(A,1),1)]
uy = [false(1,size(A,2)); (A(3:end,:) & A(2:end-1,:) & A(1:end-2,:)); false(1, size(A,2))]
u = ux & uy
[x y] = find(u)