Why doesn't this simple search program work?

I wrote a program to give me some nodes as shown in the figure

For each node like i, I used 'Dim{i}=[something something]' in my programming which introduces the position of each node.

Each node has contact which some other nodes like what is shown for i=128

.

For each node like i, I introduce position as given below (from 1 to 13) :

.

For example:

NODZ{i}(1)=127

NODZ{i}(2)=144

...

NODZ{i}(13)=128

Differences between nodes is a constant value and is called D in my programming.

In part of my program, I used a kind of search program to give me neighborhood of each node like i for all p nodes:

for i=1:p
  NODZ{i}=zeros(1,13);
  for j=1:p
      if Dim{j}==[Dim{i}(1)-D Dim{i}(2)]
          NODZ{i}(1)=j;
      end 
      if Dim{j}==[Dim{i}(1) Dim{i}(2)+D]
          NODZ{i}(2)=j;
      end
      if Dim{j}==[Dim{i}(1)+D Dim{i}(2)]
          NODZ{i}(3)=j;
      end
      if Dim{j}==[Dim{i}(1) Dim{i}(2)-D]
          NODZ{i}(4)=j;
      end
      if Dim{j}==[Dim{i}(1)-D Dim{i}(2)+D]
          NODZ{i}(5)=j;
      end
      if Dim{j}==[Dim{i}(1)+D Dim{i}(2)+D]
          NODZ{i}(6)=j; 
      end
      if Dim{j}==[Dim{i}(1)+D Dim{i}(2)-D]
          NODZ{i}(7)=j;
      end
      if Dim{j}==[Dim{i}(1)-D Dim{i}(2)-D]
          NODZ{i}(8)=j;
      end
      if Dim{j}==[Dim{i}(1)-2*D Dim{i}(2)]
          NODZ{i}(9)=j;
      end
      if Dim{j}==[Dim{i}(1) Dim{i}(2)+2*D]
          NODZ{i}(10)=j;
      end
      if Dim{j}==[Dim{i}(1)+2*D Dim{i}(2)]
          NODZ{i}(11)=j;
      end
      if Dim{j}==[Dim{i}(1) Dim{i}(2)-2*D]
          NODZ{i}(12)=j;
      end
      if Dim{j}==[Dim{i}(1) Dim{i}(2)]
          NODZ{i}(13)=j;
      end
  end
end

BUT, when I run the program and ask it something like NODZ{128} it gives me:

>> NODZ{128}
ans =
 127     0   129   113     0     0   114   112   126   160   130     0   128

This not correct because real value of NODZ{128} should be :

NODZ{128}=[127 160 129 113 143 145 114 112 126 160 130 99 128]

Why some members becomes zero, How can I solve it?

Answer 1

[Amit]

A simple test is type

[Dim{128}(1) Dim{128}(2)+D]

and

Dim{144}

and see if they match.

[Ayob]

Actually my problem is that they match in the value but "IF orders" I've used sometimes doesn't work . I'm confused about it an can not find the problem in this simple codes.

for example:

Dim{160}=[Dim{128}(1) Dim{128}(2)+D] in Matlab check but for i=128 and j=160 in the if order

      if Dim{j}==[Dim{i}(1) Dim{i}(2)+D]
          NODZ{i}(2)=j;
      end

if order doesn't work!!?

[Amit]

what is if orders ?

[Ayob]

    CODE:
for i=1:p
  NODZ{i}=zeros(1,13);
  for j=1:p
      if Dim{j}==[Dim{i}(1)-D Dim{i}(2)] * _%These are if orders_ *
          NODZ{i}(1)=j;
      end 
      if Dim{j}==[Dim{i}(1) Dim{i}(2)+D] 
          NODZ{i}(2)=j;
      end
      if Dim{j}==[Dim{i}(1)+D Dim{i}(2)]
          NODZ{i}(3)=j;
      end
      if Dim{j}==[Dim{i}(1) Dim{i}(2)-D]
          NODZ{i}(4)=j;
      end
      if Dim{j}==[Dim{i}(1)-D Dim{i}(2)+D]
          NODZ{i}(5)=j;
      end
      if Dim{j}==[Dim{i}(1)+D Dim{i}(2)+D]
          NODZ{i}(6)=j; 
      end
      if Dim{j}==[Dim{i}(1)+D Dim{i}(2)-D]
          NODZ{i}(7)=j;
      end
      if Dim{j}==[Dim{i}(1)-D Dim{i}(2)-D]
          NODZ{i}(8)=j;
      end
      if Dim{j}==[Dim{i}(1)-2*D Dim{i}(2)]
          NODZ{i}(9)=j;
      end
      if Dim{j}==[Dim{i}(1) Dim{i}(2)+2*D]
          NODZ{i}(10)=j;
      end
      if Dim{j}==[Dim{i}(1)+2*D Dim{i}(2)]
          NODZ{i}(11)=j;
      end
      if Dim{j}==[Dim{i}(1) Dim{i}(2)-2*D]
          NODZ{i}(12)=j;
      end
      if Dim{j}==[Dim{i}(1) Dim{i}(2)]
          NODZ{i}(13)=j;
      end
  end
end

[Amit]

try

if isequal(Dim{j},[Dim{i}(1) Dim{i}(2)+D])

and see if it makes a difference.

[Ayob]

OK, I will check it, but maybe this is a MATlAB weakness point? And it should be checked.

[Amit]

The thing is, what you're trying to do can be done more simply without going through so many loop.

Like for example, if you store your original node information in a matrix, you don need a cell to store that information.

[Ayob]

*It still doesn't work!* Is there any other way to set NODZ{i} matrices, I'm disappointed of Matlab in this type of code writing.

[Amit]

Don't be dissapointed with MATLAB. There must be something happening. Can you save the workspace with value for Dim and upload? If I have Dim values, I might be able to help you better.

[Ayob]

Yes, I will upload my code.m.

[Amit]

not your code. I need Dim values.

Once you've generated Dim cell, do this:

save('ToUpload.mat',Dim)

This will create ToUpload.mat which you should upload.

[Ayob]

I've uploaded it.I'm sure about 'Dim' because it gives me true virtual plot shown in untitled.fig . I think all the problem is with IF orders.

[Amit]

From your file: Dim{144} = [-0.0400 -0.0640]; Dim{128} = [-0.0360 -0.0680];

From you if statement: Dim{144} == [Dim{128}(1) Dim{128}(2)+D]

But Dim{144}(1) does not matches with Dim{128}(1), Then how can you expect MATLAB to say that they are equal.

[Ayob]

No, you made a mistake. As you see in 'untitled.fig' i=144 is in the left-top of i=128 so its position is 5 so this If order should be activated for it:

       if Dim{j}==[Dim{i}(1)-D Dim{i}(2)+D]
          NODZ{i}(5)=j;
      end

Dim{144}==[Dim{128}(1)-D Dim{128}(2)+D]

(In this example D=0.04)

.

[Amit]

I was looking at the figure you posted in your original post.

[Amit]

Here is the thing. Sometimes in floating point numbers, you get very small decimal deference. In which case they both will not be equal. for example 0.01 and 0.01000000001.

you should try something like this

tol = 1e-6; % Define Tolerance
for ... % Rest of the code
  for ..
      if norm(Dim{j}-[Dim{i}(1)-D Dim{i}(2)+D]) < tol

This takes care of small differences in floating point numbers.

[Ayob]

It's probably right.I will check it.

[Ayob]

IT WORKS., THANKS GOD., THANKS Amit.