how to write such a loop?

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Shuoguo
Shuoguo 2011-4-12
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
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Shuoguo
Shuoguo 2011-4-12
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
Shuoguo 2011-4-12
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.

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回答（3 个）

the cyclist
the cyclist 2011-4-12
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
Shuoguo 2011-4-12
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
  2 个评论
Matt Fig
Matt Fig 2011-4-12
With that P, your code simply returns....
Shuoguo
Shuoguo 2011-4-12
get a wrong matrix...fixed

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Walter Roberson
Walter Roberson 2011-4-12
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.

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