How to count alternating ones and zeros in a matrix

Lets just say that I have a matrix v as following:

v =

     0     0     0     0     0     0
     1     1     0     1     1     0
     1     0     1     0     1     0
     0     0     1     0     1     0
     0     0     0     1     1     1
     

I want to get rows where alternating ones and zeros are occuring which are 3rd and 4th rows. By alternating ones and zeros, I mean a sequence of [1 0 1 0] at least.

[Walter Roberson]

How about if a row ends with 0 1 0 1, is that also alternating 0 and 1, or does the sequence have to start with a 1?

[Awais Saeed]

Yes 0 1 0 1 would also be an alternating sequence.

Answer 1

[Devanuj Deka]

% Get the dimensions of 'v'
sz = size(v);   % Gives two numbers. sz(1) is the number of rows; sz(2) is the number of columns

% Initialize a zero vector, having as many elements as the number of rows
% in v. After the code is run, the i'th element of this vector will be 1 if
% the i'th row of 'v' has an alternating pattern
rows = zeros(1,sz(1));

% Look for the pattern
for i=1:sz(1)   % Iterating through rows
    count = 0;  % Count increments if the next element in the row is different
    for j=1:sz(2)-1 % Iterating through the elements of each row
        if v(i,j)~=v(i,j+1) % If current element and next element are unequal, increment count
            count = count+1;
        else                % If next element is equal to the current element, count is reset
            count = 0;
        end
        if count==3         % Minimum 3 changes back to back are required for your pattern (0->1->0->1, or 1->0->1->0)
            rows(j) = 1;    % If pattern detected in i'th row, row(i) is set to 1
            break;          % Since pattern detected, no need to look for more in the same row.
        end
    end
end

rows_with_pattern = find(rows); % A vector containing the rows which have the alternating pattern

find(X) gives the indices of non-zero elements in an array X, which is why we initialized rows as a vector of zeroes.

[Awais Saeed]

Thanks. That did the trick.

[Devanuj Deka]

You're welcome!

Answer 2

[Matt J]

v =[
     0     0     0     0     0     0
     1     1     0     1     1     0
     1     0     1     0     1     0
     0     0     1     0     1     0
     0     0     0     1     1     1];

A=v(:,1:end-3); B=v(:,2:end-2); C=v(:,3:end-1); D=v(:,4:end);

pattern1=A==0 & B==1 & C==0 & D==1;
pattern2=A==1 & B==0 & C==1 & D==0;

rows=find( any(pattern1|pattern2,2) )

rows = 2×1

3 4

[Stephen23]

+1 nice use of MATLAB.

Answer 3

[Stephen23]

v =[ 0     0     0     0     0     0; ...
     1     1     0     1     1     0; ...
     1     0     1     0     1     0; ...
     0     0     1     0     1     0; ...
     0     0     0     1     0     1; ... Walter Roberson's suggestion,
     1     0     1     0     0     0; ... and reversed also.
     0     1     0     1     0     1; ... Entire row alternating.
     0     0     0     1     1     1];
X = any(diff(v,3,2)>3 | diff(~v,3,2)>3, 2)

X = 8×1 logical array
   0
   0
   1
   1
   1
   1
   1
   0