How can stop "while" loop

Hi, I have a while loop, my code is inside the loop. I want to stop the loop when the same number (must be non zero)created in the matrix from the first row to the last row.

ex.

this matrix is inside the while loop:

U3(:,:,1) = [

0 , 0 , 0 , 1 , 0;

0 , 1 , 0 , 0 , 0;

0 , 1 , 0 , 0 , 1;

0 , 0 , 0 , 0 , 0;

0 , 0 , 0 , 0 , 0

]

U3(:,:,2) = [

0 , 0 , 0 , 0 , 0;

0 , 0 , 0 , 0 , 0;

0 , 1 , 0 , 0 , 1;

0 , 1 , 0 , 0 , 0;

0 , 1 , 0 , 0 , 0

]

U3(:,:,3) = [

0 , 0 , 0 , 0 , 0;

0 , 0 , 0 , 0 , 0;

0 , 1 , 0 , 0 , 1;

0 , 0 , 0 , 0 , 0;

0 , 0 , 0 , 0 , 0

]

Now, the loop will continue but if U3(1,2, 1) becomes equal to 1, the loop will stop. How can I do it?

Note:

The bath from the first row to the last row can go in any direction (x,y,z) but it must pass from the topped row to the bottom row similar to the above example. The bath from the topped row to the bottom row is as:

U3(1,2,1), U3(2,2,1), U3(3,2,1), U3(3,2,2), U3(4,2,2), U3(5,2,2) and U3(3,2,3) ONLY

Answer 1

[Hisham]

But how can I do it for 3D matrix?

Note: I might have more than an island, but I want to stop the while loop when any of the islands created a path from the topped row to the bottom row.

Answer 2

[José-Luis]

Your condition should be

[x y z] = size(U3);
condZ = sum(sum(sum(U3,3) == z)) > 0;
condY = sum(sum(sum(U3,2) == y)) > 0;
condX = sum(sum(sum(U3) == x)) > 0;
stopBool = condX || condY || condZ;

Cheers!

[Hisham]

Sorry, what do you mean by "data" and numbers "3 and 2" Thanks!:)

[José-Luis]

By data I meant U3., sorry. The numbers indicate along which dimension you do the sum. If the sum is equal to the matrix size along that dimension -> success. Any success-> Great success!

[Hisham]

Actually, I've tried to apply it but it shows me error. Here is the code and the error message.

________

The Code

________

[x y z] = size(U3);

condX = sum(sum(U3,3) == z) > 0;

condY = sum(sum(U3,2) == y) > 0;

condZ = sum(sum(U3) == x) > 0;

if ((sum(sum(U3,3) == z) > 0) (sum(sum(U3,2) == y) > 0) (sum(sum(U3) == x) > 0))

disp('Bath has been created')

break;

else

disp('the bath has not been created yet')

end

_______________

The output

_______________

Operands to the and && operators must be convertible to logical scalar values.

[José-Luis]

The operator for the while loop is stopBool

while (~stopBool)
%do your stuff
end

[Hisham]

I did this code but there is an error):

________

while (~stopBool)

[x y z] = size(U3);

condX = sum(sum(U3,3) == z) > 0;

condY = sum(sum(U3,2) == y) > 0;

condZ = sum(sum(U3) == x) > 0;

if ((sum(sum(U3,3) == z) > 0) || (sum(sum(U3,2) == y) > 0) || (sum(sum(U3) == x) > 0))

disp('Bath has been created')

break;

else

disp('the bath has not been created yet')

end
end

How can I defin stopBool?

Undefined function or variable 'stopBool'.

Also,I definded stopBool as equal to 0. The out put still the same: Operands to the and && operators must be convertible to logical scalar values.

[José-Luis]

Look at my original answer...

[Hisham]

I did as you said but the error stil the same.

Here is the code:
   while(~stopBool)
  [x y z] = size(U3);
  condX = sum(sum(U3,3) == z) > 0;
  condY = sum(sum(U3,2) == y) > 0;
  condZ = sum(sum(U3) == x) > 0;
  stopBool = condX || condY || condZ;
  end

Operands to the and && operators must be convertible to logical scalar values.

[José-Luis]

You are totally right, i forgot that you need to aggregate along two dimensions. It should be sum(sum(sum....)). Modified the original posting.

[Hisham]

I think the code that you provided is not true. The reason is the value of "stopBool", either the bath created or not created, is equal to 1. Thus, the code should has something to distinguish whether the bath has been created or not?

Here is the code:

while(~stopBool)
  [x y z] = size(U3);
  condX = sum(sum(U3,3) == z) > 0;
  condY = sum(sum(U3,2) == y) > 0;
  condZ = sum(sum(U3) == x) > 0;
  stopBool = condX || condY || condZ;
  end

Just to let you know, the value of "stopBool" must be defined before the "while" loop, I defined it as 0.

Here is an example when the bath is created in the 3D matrix.

U3(:,:,1) = [

0 , 1 , 0 , 1 , 0;

0 , 1 , 0 , 0 , 0;

0 , 1 , 0 , 0 , 1;

0 , 0 , 0 , 0 , 0;

0 , 0 , 0 , 0 , 0

]

U3(:,:,2) = [

0 , 0 , 0 , 0 , 0;

0 , 0 , 0 , 0 , 0;

0 , 1 , 0 , 0 , 1;

0 , 1 , 0 , 0 , 0;

0 , 1 , 0 , 0 , 0

]

U3(:,:,3) = [

0 , 0 , 0 , 0 , 0;

0 , 0 , 0 , 0 , 0;

0 , 1 , 0 , 0 , 1;

0 , 0 , 0 , 0 , 0;

0 , 0 , 0 , 0 , 0

]

However, here is the 3D matrix but the bath has not been created yet.

U3(:,:,1) = [

0 , 0 , 0 , 1 , 0;

0 , 1 , 0 , 0 , 0;

0 , 1 , 0 , 0 , 1;

0 , 0 , 0 , 0 , 0;

0 , 0 , 0 , 0 , 0

]

U3(:,:,2) = [

0 , 0 , 0 , 0 , 0;

0 , 0 , 0 , 0 , 0;

0 , 1 , 0 , 0 , 1;

0 , 1 , 0 , 0 , 0;

0 , 1 , 0 , 0 , 0

]

U3(:,:,3) = [

0 , 0 , 0 , 0 , 0;

0 , 0 , 0 , 0 , 0;

0 , 1 , 0 , 0 , 1;

0 , 0 , 0 , 0 , 0;

0 , 0 , 0 , 0 , 0

]

[José-Luis]

I think i understand what you mean. I have a question. Should the "islands" be in contact to form a path. That is, do U(1,2,1)=1 together with U(2,2,3)=1 form a path segment or not? The solution will be different whether they do or not.

[Hisham]

Yes the "islands" will be in contact to form a path.

[José-Luis]

Use bwconncomp, if you have the image processing toolbox. If i understand correctly you want connectivity 26. Look at all the connected components and find whether any of them goes from one side to the other.

[Hisham]

Here is the problem in "bwconncomp".

See the whole example please.

    U3(:,:,1) = [
0 , 0 , 0 , 0 , 0;
0 , 0 , 0 , 0 , 0;
0 , 1 , 0 , 0 , 1;
0 , 0 , 0 , 0 , 1;
0 , 0 , 0 , 0 , 0
];
U3(:,:,2) = [
0 , 0 , 0 , 0 , 0;
0 , 0 , 0 , 0 , 0;
0 , 1 , 0 , 0 , 1;
0 , 1 , 0 , 0 , 0;
0 , 1 , 0 , 0 , 0
];
U3(:,:,3) = [
0 , 0 , 0 , 0 , 0;
0 , 0 , 0 , 0 , 0;
0 , 1 , 0 , 0 , 0;
0 , 0 , 0 , 0 , 1;
0 , 0 , 0 , 0 , 1
];

Now I will explain the all chain(s):

The first chain is: U3(3,2,1) and U3(3,2,2)and U3(4,2,2) and U3(5,2,2)and U3(3,2,3)

The second chain is: U3(3,5,1) and U3(4,5,1) and U3(3,5,2)

The third chain is: U3(4,5,3) and U3(5,5,3)

to give each chain a similar individual number

(i.e).

U3(3,2,1) and U3(3,2,2)and U3(4,2,2) and U3(5,2,2)and U3(3,2,3)=2

U3(3,5,1) and U3(4,5,1) and U3(3,5,2)=3

U3(4,5,3) and U3(5,5,3)=4

One of the suggested solutions is to use :

    CC = bwconncomp(U3);
    labelmatrix(CC)

But this function is not valid for my case because it gives the first chain individual number (which is fine), and combined the second and third chains in one number (which is not fine)

See the output (which is wrong not as I want).

CC = bwconncomp(U3); labelmatrix(CC)

ans(:,:,1) =

    0    0    0    0    0
    0    0    0    0    0
    0    1    0    0    2
    0    0    0    0    2
    0    0    0    0    0
ans(:,:,2) =

    0    0    0    0    0
    0    0    0    0    0
    0    1    0    0    2
    0    1    0    0    0
    0    1    0    0    0
ans(:,:,3) =

    0    0    0    0    0
    0    0    0    0    0
    0    1    0    0    0
    0    0    0    0    2
    0    0    0    0    2

[Hisham]

I did as you suggested but the output still the same. Only two chains but in the reality there are three chains.

 CC = bwconncomp(U3,26)

CC =

    Connectivity: 26
       ImageSize: [5 5 3]
      NumObjects: 2
    PixelIdxList: {[5x1 double]  [5x1 double]}

labelmatrix(CC)

ans(:,:,1) =

    0    0    0    0    0
    0    0    0    0    0
    0    1    0    0    2
    0    0    0    0    2
    0    0    0    0    0

ans(:,:,2) =

    0    0    0    0    0
    0    0    0    0    0
    0    1    0    0    2
    0    1    0    0    0
    0    1    0    0    0

ans(:,:,3) =

    0    0    0    0    0
    0    0    0    0    0
    0    1    0    0    0
    0    0    0    0    2
    0    0    0    0    2

You can see there is no connection between ans(3,5,2) and ans(4,5,3)

[José-Luis]

I think i finally got it:

conn = zeros(3,3,3);
conn(2,2,:) = 1;
conn(:,2,2) = 1;
conn(2,:,2) = 1;
CC = bwconncomp(U3,conn); labelmatrix(CC)
ans(:,:,1) =

    0    0    0    0    0
    0    0    0    0    0
    0    1    0    0    2
    0    0    0    0    2
    0    0    0    0    0

ans(:,:,2) =

    0    0    0    0    0
    0    0    0    0    0
    0    1    0    0    2
    0    1    0    0    0
    0    1    0    0    0

ans(:,:,3) =

    0    0    0    0    0
    0    0    0    0    0
    0    1    0    0    0
    0    0    0    0    3
    0    0    0    0    3

[Hisham]

Great...Real you are expert and intelligent. I appreciate your helps.... Thanks so much for your helps.

[José-Luis]

I'm no expert nor intelligent, just stubborn. I learned something new today. I'm liking Matlab answers.

[Hisham]

So, Now how can I stop the "while loop"? Is it by using the same code

while(~stopBool)
  [x y z] = size(U3);
  condX = sum(sum(U3,3) == z) > 0;
  condY = sum(sum(U3,2) == y) > 0;
  condZ = sum(sum(U3) == x) > 0;
  stopBool = condX || condY || condZ;
  end

Or by applying another thing?

Also, if I want to use the same code, I need to defined "stopBool " befor the while loop. What shall I define it please?

[Hisham]

If I use this example:

U3(:,:,1) = [

0 , 1 , 0 , 1 , 0;

0 , 1 , 0 , 0 , 0;

0 , 1 , 0 , 0 , 1;

0 , 0 , 0 , 0 , 0;

0 , 0 , 0 , 0 , 0

]

U3(:,:,2) = [

0 , 0 , 0 , 0 , 0;

0 , 0 , 0 , 0 , 0;

0 , 1 , 0 , 0 , 1;

0 , 1 , 0 , 0 , 0;

0 , 1 , 0 , 0 , 0

]

U3(:,:,3) = [

0 , 0 , 0 , 0 , 0;

0 , 0 , 0 , 0 , 0;

0 , 1 , 0 , 0 , 1;

0 , 0 , 0 , 0 , 0;

0 , 0 , 0 , 0 , 0

]

Then put your code:

conn = zeros(3,3,3);
conn(:,:,2) = 1;
conn(:,2,:) = 1;
conn(2,:,:) = 1;
CC = bwconncomp(U3,conn); 
[x y z] = size(U3);
isConnected = false;
while (~isConnected)
    %Do your stuff here
    for i = CC.PixelIdxList
        [sub_x sub_y sub_z] = ind2sub([x y z],cell2mat(i));
        if (sum(sub_x==1) && (sub_x==x))
            isConnected = true;
        end       
      end
  end

The output is error:

Operands to the || and && operators must be convertible to logical scalar values.

[Hisham]

I think you missed some bracts. The correct is: if ((sum(sub_x==1)) && ((sub_x==x)))

But I have a small question pleas.. If I use more than layer in depth (i.e. U3(:,:,4) , and U3(:,:,5).....U3(:,:,i)

Do I need to make any change in this code:

conn = zeros(3,3,3);
conn(:,:,2) = 1;
conn(:,2,:) = 1;
conn(2,:,:) = 1;

[José-Luis]

conn = zeros(3,3,3);
conn(:,:,2) = 1;
conn(:,2,:) = 1;
conn(2,:,:) = 1;
CC = bwconncomp(U3,conn); 

[x y z] = size(U3);

isConnected = false;
while (~isConnected)
    %Do your stuff here
    for i = CC.PixelIdxList
        [sub_x sub_y sub_z] = ind2sub([x y z],cell2mat(i));
        if (~isempty(find(sub_x == 1)) && ~isempty(find(sub_x == x)) )
            isConnected = true;
        end       

      end
  end

[José-Luis]

Nope, no changes, unless you want a different connectivity pattern. You can read the documentation:

doc bwconncomp;

[Hisham]

I would like to thank you for your helps.

I left the last part to explain it to you. If I imagine that the "U3" is represented as a box, and I want to plot the number ones in U3 matrix to see the created path. How can I do it in 3D matrix?

[José-Luis]

http://www.mathworks.com/matlabcentral/fileexchange/28497-plot-a-3d-array-using-patch

Glad it worked.

[Hisham]

I downloaded the file from the above link. Then I used this code:

load exampleA.mat 
figure 
hpat = PATCH_3Darray(gridINPUT,gridX,gridY,gridZ);

But the issue is about how can I put U3 matrix in the code.

Do I need to change "gridINPUT" to "U3"? or how to plot the U£ matrix?

[José-Luis]

Hisham, I would strongly recommend reading the getting started part of the documentation. grinInput, gridX, gridY, and gridZ are just placeholder variable names that are passed to the function PATCH_3Darray. You can name them whatever you want. In this case, yes, gridINPUT would be U3. Copy/pasting from the function description:

gridX (optional) - A 1xP array - List of the X axis coordinates.
gridY (optional) - A 1xQ array - List of the Y axis coordinates.
gridZ (optional) - A 1xR array - List of the Z axis coordinates.

That means that passing PATCH_3Darray(U3) should be enough to produce some results.

[Hisham]

I tried the above function but it is suitable for my case. Do you know any function or way to plot similar to the link below?

I did it for 2D by using the following code:

   [r,c]=size(U3);

figure

for cIdx=1:c,
    y=find(U3(:,cIdx)==1);

      if length(y)>1
          y=(r+1)-y;
          x=ones(size(y))*cIdx;
          for idx=1:length(x)-1
              if abs( y(idx) - y(idx+1))<=1
                  line(x(idx:idx+1),y(idx:idx+1),'LineWidth',4)
              end
          end
          hold on
      end

end

for rIdx=1:r,
    x=find(U3(rIdx,:)==1);

      if length(x)>1
          y=(r+1)-ones(size(x))*rIdx;
          for idx=1:length(x)-1
              if abs( x(idx) - x(idx+1))<=1
                  line(x(idx:idx+1),y(idx:idx+1),'LineWidth',4)
              end
          end
          hold on
      end

end

It is very complicated to do it for 3D. Do you know any function able to do it?

Thanks!

Answer 3

[Azzi Abdelmalek]

add a condition to your while loop

test=1
while condition1 & test==1
%your program
%if you want exit a loop change the value of test for example
test=0, %any value different from 1

[Hisham]

It is very nice, but my problem is how can I build the condition that should be valid for a 3D matrix.

can you help me more...please