Finding all possible row combinations of a matrix that add to zero

Hello,

I'm looking for a general way to find all possible row combinations of a matrix that add to zero.

For instance, for the matrix

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

the following 6 row combinations would all sum to zero

A(1,:)+A(2,:)
A(1,:)+A(3,:)+A(4,:)+A(6,:)
A(1,:)+A(3,:)+A(5,:)+A(6,:)
-A(2,:)+A(3,:)+A(4,:)+A(6,:)
-A(2,:)+A(3,:)+A(5,:)+A(6,:)
-A(4,:)+A(5,:)

Does MATLAB have any built-in functions that can help me do this? Generating all possible row combinations and testing to see which ones sum to zero seems like it would be extremely computationally intensive.

Thanks,

Kevin

Answer 1

[Azzi Abdelmalek]

Edit2

A = [-1 0 0 ; 1 0 0 ; 1 -1 0 ; 0 1 -1 ; 0 1 -1; 0 0 1];
n=size(A,1);
idx=logical(npermutek([0 1],n));
p=size(idx,1);
out=cell(p,1);
for k=1:p
  out{k}=sum(A(idx(k,:),:),1);
end

You can get you sum in a matrix 647x3

M=cell2mat(out)

npermutek is a Matt Fig submission at http://www.mathworks.com/matlabcentral/fileexchange/11462-npermutek/content/npermutek.m

[Kevin Bachovchin]

I'm confused by your answer. I just ran this code. What is supposed to be the variable here that stores all possible row combinations summing to zero? I don't see any variable like that after running the code.

[Azzi Abdelmalek]

 out{1} is the first sum
out{2} is the second sum 
.
.
.
out{64} is the last sum

[Azzi Abdelmalek]

Sorry there is typo in

out{k}=sum(A(idx(k,:),:));

Look at the edited answer

[Azzi Abdelmalek]

There is another error, it's

sum(A(idx(k,:),:),1)

instead of

sum(A(idx(k,:),:)

Look at the second Edit

[Kevin Bachovchin]

Ok, it works now, but the code doesn't seem to pick up on solutions that involve subtraction.

The following should also be solutions -A(2,:)+A(3,:)+A(4,:)+A(6,:)

-A(2,:)+A(3,:)+A(5,:)+A(6,:)

-A(4,:)+A(5,:)

Answer 2

[Azzi Abdelmalek]

Ok try this

Edit

A = [-1 0 0 ;1 0 0; 1 -1 0 ; 0 1 -1 ; 0 1 -1; 0 0 1];
n=size(A,1)
idx1=npermutek([0 -1 1],n);
p=size(idx1,1);
out=cell(p,1);
for k=1:p
  out{k}=sum(bsxfun(@times,A,idx1(k,:)'),1);
end
M=cell2mat(out)
find(~any(M,2))

[Kevin Bachovchin]

I'm not sure what the code is doing wrong, but out has 17 rows that are [0 0 0]. It should only have the 6 solutions I mentioned above.

[Azzi Abdelmalek]

Are you sur of your result?

[Azzi Abdelmalek]

Look at the different combinaisons that give 0;

idx1(find(~any(M,2)),:)

each row corresponds to one combinaison

[Kevin Bachovchin]

I suppose it could be looked at that since

A(1,:)+A(2,:) = 0

-A(4,:)+A(5,:) = 0

then A(1,:)+A(2,:)-A(4,:)+A(5,:)=0 for example.

However, for my specific purpose, I would not like sums of individual solutions to also be solutions.

In the same regard, I would also not like solutions that are negatives of each other for my specific purpose.

[Azzi Abdelmalek]

There is just one individual which is equal to zero, the first one.

[Kevin Bachovchin]

I'm not sure what you mean. When you add the first and second rows together for instance, that forms [0 0 0].

[Azzi Abdelmalek]

Yes I know, what do you mean?

[Kevin Bachovchin]

I mean that since A1+A2 and -A4+A5 are solutions, it is undesirable for my specific purpose to also have -Al-A2, A4-A5, A1+A2-A4+A5, -Al-A2-A4+A5, A1+A2+A4-A5, -Al-A2+A4-A5 returned as solutions as well.

[Azzi Abdelmalek]

Ok, You did not specify that in your question. You should give all information in your question to avoid wasting of time.

[Kevin Bachovchin]

Sorry, the issue with sums of solutions didn't occur to me until I saw it. Seems like there is no difference in the result of the code from before though. Still 17 elements in find(~any(M,2)).

Answer 3

[Jan]

FEX: Permutations with sum criteria

Answer 4

[Azzi Abdelmalek]

Try this

A = [-1 0 0 ;1 0 0; 1 -1 0 ; 0 1 -1 ; 0 1 -1; 0 0 1];
n=size(A,1)
idx1=npermutek([0 -1 1],n);
p=size(idx1,1);
out=cell(p,1);
for k=1:p
  out{k}=sum(bsxfun(@times,A,idx1(k,:)'),1);
end
M=cell2mat(out)
ii=find(~any(M,2))
idx2=idx1(ii,:)
for k=1:size(idx2,1)
  jj{k}=find(idx2(k,:))
end
for k=1:numel(jj)
  iddx{k}=find(cellfun(@(x) isequal(x,jj{k}),jj))
  ee(k)=iddx{k}(1)
end
result=idx2(unique(ee),:)

[Kevin Bachovchin]

This fixed the problem of both positives and negatives being solutions (ie, A1+A2 and -A1-A2). One thing it did not fix is sums of individual solutions being solutions (for instance, since -A1-A2 and -A4+A5 are both solutions, then -A1-A2-A4+A5 is also given as a solution but this solution is unwanted for my specific purpose because it corresponds to a non-physical solution for my application.)

Answer 5

[Azzi Abdelmalek]

Try this code

 A = [-1 0 0 ;1 0 0; 1 -1 0 ; 0 1 -1 ; 0 1 -1; 0 0 1];
n=size(A,1)
idx1=npermutek([0 -1 1],n);
p=size(idx1,1);
out=cell(p,1);
for k=1:p
  out{k}=sum(bsxfun(@times,A,idx1(k,:)'),1);
end
M=cell2mat(out);
ii=find(~any(M,2));
idx2=idx1(ii,:);
for k=1:size(idx2,1);
  jj{k}=find(idx2(k,:));
end
for k=1:numel(jj);
  iddx{k}=find(cellfun(@(x) isequal(x,jj{k}),jj));
  ee(k)=iddx{k}(1);
end
result=idx2(unique(ee),:);
n=size(result,1);
test=0;
ii=1;
while test==0
     ii=ii+1;
     a=find(result(ii,:));
     c=result(ii,a);
     e=result(:,a);
     f=find(ismember(e,c,'rows'));
     f(1)=[];
     if ~isempty(f);
     result(f,:)=[];
     n=n-1;
     end
     if ii==n-1
         test=1;
     end
end
result

[Kevin Bachovchin]

Thanks. That works perfectly for what I wanted.