please help in this program. is to find the index of zeros which is in consecutive order which should more or equal to 5 time.

a=[0,0,0,0,0,0,0,0,9,8,5,6,0,0,0,0,0,0,3,4,6,8,0,0,9,8,4,0,0,7,8,9,5,0,0,0,0,0,8,9,0,5,8,7,0,0,0,0,0]; [x,y]=size(a);

for i=0:y i+1; k=1; l=0; n=i; count=0;

    while (a==0)
           count+1;
           break;  
           n+1;
    end
           if(count>=5)
               v([]);
               for l=k:l<n
                   v(m)=l+1;
                   m+1;
               end    
           end    
        count=1;
        i=n;
end    
for i=o:i<m
    i+1;
fprintf('index of continous zero more than 5 or equal=%d',v(i));
end

Answer 1

[Image Analyst]

It's trivial if you have the Image Processing Toolbox:

a = [0,0,0,0,0,0,0,0,9,8,5,6,0,0,0,0,0,0,3,4,6,8,0,0,9,8,4,0,0,7,8,9,5,0,0,0,0,0,8,9,0,5,8,7,0,0,0,0,0]; 
% Measure lengths of all "0" regions.
measurements = regionprops(a == 0, 'Area', 'PixelIdxList');
% Get indexes of those regions that are >= 5 in length.
fiveLongRegions = find([measurements.Area] >= 5)
theIndexes = vertcat(measurements(fiveLongRegions).PixelIdxList)

% Get a logical output vector which is true if it's in a >=5 region.
output = false(1, length(a)); % Initialize
% Set elements = 1 if it's in a >=5 region.
output(theIndexes) = true

[Y.L.K KHUMAN]

thank you...

Answer 2

[María]

I don't know if this is what you men:

a=[0,0,0,0,0,0,0,0,9,8,5,6,0,0,0,0,0,0,3,4,6,8,0,0,9,8,4,0,0,7,8,9,5,0,0,0,0,0,8,9,0,5,8,7,0,0,0,0,0]; [x,y]=size(a); cont=0; v=[];

for i=1:y

    if a(i)==0
        cont=cont+1;
    else
        if cont>=5
            v=[v,i-cont]
        end
        cont=0;
    end
end   
if cont>=5
   v=[v,i-cont]
end
for i=1:length(v)

fprintf('index of continous zero more than 5 or equal=%d',v(i)); end

[Y.L.K KHUMAN]

it's for finding the index of zero's element, only those which are in consecutive order of greater than 5.