How do I create a nested for loop to find consecutive numbers in a matrix? See example below.

A = [0 1 0 0 1 0 0

1 0 0 0 0 1 0

1 0 0 1 0 1 1

1 1 0 0 1 1 0

1 1 1 0 1 1 0

1 1 0 0 1 1 0 ];

1)

This is a 6x7 matrix. I wanted to check each row to see if it finds [0 0 0 0] and then if it finds it, then store it into a vector.

Can this be done with a nested for loop?

The way I wanted it to check for [0 0 0 0] for each row is as follows:

Row 1: A(1, 1:4), A(1,2:5), A(1,3:6), A(1,4:7)

Row 2: A(2, 1:4), A(2,2:5), A(2,3:6), A(2,4:7)

Row 3: A(3, 1:4), A(3,2:5), A(3,3:6), A(3,4:7)

etc...up until row 6.

2)

Also can this done for each column using for loops also? Check if it finds [1 1 1 1]' in each column?

Column 1: A(1:4, 1), A(2:5,1), A(3:6,1)

Column 2: A(1:4, 2), A(2:5,2), A(3:6,2)

etc...up until column 7

OR

Is there any easier method not using any fancy built in functions and just using for loops and if/else conditionals?

Answer 1

[Matt J]

A =   [0 1 0 0 1 0 0 
       1 0 0 0 0 1 0 
       1 0 0 1 0 1 1 
       1 1 0 0 1 1 0
       1 1 1 0 1 1 0
       1 1 0 0 1 1 0 ];


k=ones(1,4);

rows= find( any( conv2(~A,k,'valid')==4 ,2)  )  %containing [0,0,0,0]

rows = 2

columns= find(any( conv2(A,k','valid')==4 ,1) ) %containing [1;1;1;1]

columns = 1×2

1 6