how to write such a loop?

hello,

if i have a n*1 vector. every element can interact with each other.an 1 at position (i,j) means member i will change its state with member j.an zero means they do not change together. so all the interactions between any two elements can be expressed in a n*n matrix (diagnal is 0 means i^th element do not interact with itself):

     0     1     0     1     0     0     0     0
     1     0     0     0     0     0     0     0
     0     0     0     0     0     0     0     0
     1     0     0     0     1     0     0     0
     0     0     0     1     0     0     0     0
     0     0     0     0     0     0     0     0
     0     0     0     0     0     0     0     0
     0     0     0     0     0     0     0     0

now my question is: i want to begin with a random place, and find all that co-change with it (all the friends), then go on and find all that co-change with its friends (new friends), and go on until no new friends can be find.

eg.

1. i begin with element 1.

2. i go to first row, find element 2 and 4 are friends

3. then i go to row2 find 1 (not new friends); go to row 4 find 1 (again) and 5 (new friends);

4. then continue to row 5, find 4 (not new friends) 5. mark 1,2,4,5 as friends. 6. go back to step 1, find a new element (other than 1,2,4,5) to start with.

i have some bigger matrix to work with (100*100)...and this has been puzzled me for a while, please help me out.

thanks a lot.

Shuoguo

[Matt Fig]

You show the matrix, but where is the vector?

[Sean de Wolski]

I think I smell accumarray...

[Shuoguo]

oh i skipped something...lets say the so called vector is, for this case, 8 random numbers like V=[1 2 3 4 8 7 6 5].each element of V, Vi has a spin variable either 0 or 1 (only two "states", 0 or 1 for simplicity. It can be 10 states numbered from 1~10).

the interaction between any two elements forms a 8*8 matrix P, if Vi and Vj have the same spin, then P_ij=1, else 0. so by looking at P, an 1 simply means there is an interaction between two elements and 0 means no interaction.

if Vi interacts with Vj, and Vj interacts with Vk, then Vi interacts with Vk, and so on.

so my question is if i begin with any Vi, i want to find all that can interact with it directly (like Vi-Vj) or indirectly (Vi-Vk).

thanks

[Shuoguo]

by the way the so called "spin" variable will be a separate vector that defines some property of Vi...i gave random numbers to vector V since the actual value is not critical when we are talking about spin interactions.

Answer 1

[the cyclist]

Here is a solid kick in the right direction. The while-loop only tests if you have found all possible friends (based on the size of the original array). It does not test whether you have reached a "steady-state" because the network does not extend to all possible people. Should be easy to add that, if you want.

A = [0     1     0     1     0     0     0     0
     1     0     0     0     0     0     0     0
     0     0     0     0     0     0     0     0
     1     0     0     0     1     0     0     0
     0     0     0     1     0     0     0     0
     0     0     0     0     0     0     0     0
     0     0     0     0     0     0     0     0
     0     0     0     0     0     0     0     0];

N = size(A,1); % Assume symmetric, square array A 

allFriends = 1:N;

oldFriends = 1
pause(1)

while not(numel(oldFriends)==numel(allFriends))

      for nf = oldFriends
          newFriends = find(A(nf,:));
          oldFriends = union(oldFriends,newFriends); % Automatically sorted by MATLAB, which is important for while test.
      end

      oldFriends
      pause(1)

end

[Shuoguo]

when i run this i think matlab stuck somewhere?

Answer 2

[Shuoguo]

thanks the cyclist...i also just worked out a code (bigger than yours)and so far i am not eyeballing any bugs.

P =
   0     1     0     1     0     0     0     0
   1     0     0     0     0     0     0     0
   0     0     0     0     0     0     0     0
   1     0     0     0     1     0     0     0
   0     0     0     1     0     0     0     0
   0     0     0     0     0     0     0     0
   0     0     0     0     0     0     0     0
   0     0     0     0     0     0     0     0 
while ~isempty(P>0)
[rs,~]=find(P>0);

if isempty(rs)
    return;
else
    rs=rs(1);
end

func = @(x) find(P(x,:)>0);
c = rs;
oldc = c;
while true
   c=oldc;
   [r c] = func(c);
   c=sort(unique(c))';
   if isequal(c,oldc)
     break;
   end
    oldc=[oldc c];
    oldc=sort(unique(oldc));
end
P(c,:)=0;
end

[Matt Fig]

With that P, your code simply returns....

[Shuoguo]

get a wrong matrix...fixed

Answer 3

[Walter Roberson]

I didn't catch the bit about the spin variable, but the other part sound like standard matrix traversal that is implemented by taking successive matrix powers.