Why doesn't this simple search program work?

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Ayob
Ayob 2014-1-22
评论： Ayob 2014-1-22
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?

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采纳的回答

Amit
Amit 2014-1-22
A simple test is type
[Dim{128}(1) Dim{128}(2)+D]
and
Dim{144}
and see if they match.
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