using break to stop nested loop

X = [1 2 3 4 5 4 3 6 7 8 9 10 11 12 11 10 9 8 7 8 9 10 9 8 7 6 5 4 3 2 1 2 4 5 6 7 8 9 11 12 11 10 9 8 7 9 10 9 8 7 6 5 4 3 2 1];

I want to take out vectors which start when value greater than 2 and end value end less than 11. The resulting vectors will be [3 4 5 4 3 6 7 8 9 10 ] and [4 5 6 7 8 9], which will be kept in cell array.

The current code I tried saved only [4 5 6 7 8 9] in fwd cell array.

Could anyone please help me to save both vectors in the cell array? thank you.

clc; clear; close all
X = [1 2 3 4 5 4 3 6 7 8 9 10 11 12 11 10 9 8 7 8 9 10 9 8 7 6 5 4 3 2 1 1 2  4 5 6 7 8 9  11 12 11 10 9 8 7  9 10 9 8 7 6 5 4 3 2 1];
start = min(X)+2;
ed = max(X) -2;
m = length(X);
f = 1; 
i = 2; 
while i < m
    k=1;
    if X(i) >= start && X(i-1) <= start && X(i+1) > X(i) 
        a(k) = X(i);
        for j = i+1:m-1
            if X(j+1) > ed && X(j+2) >= X(j+1) && X(j) < X(j+1)
            b (k) = X(j);
            fwd{k} = [X(i):X(j)];
             
            if ~isempty(b(k))                
                break;
              
            end
                                        
            end
        end        
        i =j+1;
        k =k+1; 
    else
        i = i+1;
        
         
        
    end
      
end

[Stephen23]

Yu Zhi: your problem is not sufficiently defined. For example, here are two sequences that meet your requirement "start when value greater than 2 and end value end less than 11", why are they not included in the output?:

7 8 9 10
8 7 9 10

What about ther prefectly good sequence

8

which also fulfills your conditions? Why is that sequence not in your output?

[HYZ]

Actually I am simulating a situation that an animal leaves one end and enter the other end as in the diagram.

I somehow changed the postion of k and got what I want. But honestly, I still don't really understand how putting k=1; k=k+1; can make it work.

clc; clear; close all
X = [1 2 3 4 5 4 3 6 7 8 9 10 11 12 11 10 9 8 7 8 9 10 9 8 7 6 5 4 3 2 1 1 2  4 5 6 7 8 9  11 12 11 10 9 8 7  9 10 9 8 7 6 5 4 3 2 1];

start = min(X)+10;
ed = max(X) -10;
m = length(X);
i = 2; 
k=1;
while i < m    
    if X(i) >= start && X(i-1) <= start && X(i+1) > X(i)         
        a(k) = X(i);
        for j = i+1:m-1
            if X(j+1) > ed && X(j+2) >= X(j+1) && X(j) < X(j+1)
            b (k) = X(j);           
            fwd{k} = [X(i):X(j)]; 
           
                if ~isempty(b(k)) 
                    k=k+1;  
                break; 
                
                end
                                        
            end
        end        
        i =j+1;     
       else
        i = i+1;
          
    end 
       
end
fwd;

[HYZ]

7 8 9 10 and 8 7 9 10 are not starting points of the track. the starting point is the first value which came after 2. the end is the last value which comes just before 9.