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

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Kevin Bachovchin
Kevin Bachovchin 2013-2-21
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

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Azzi Abdelmalek
Azzi Abdelmalek 2013-2-21
编辑：Azzi Abdelmalek 2013-2-21
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
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Azzi Abdelmalek
Azzi Abdelmalek 2013-2-21
编辑：Azzi Abdelmalek 2013-2-21
There is another error, it's
sum(A(idx(k,:),:),1)
instead of
sum(A(idx(k,:),:)
Look at the second Edit
Kevin Bachovchin
Kevin Bachovchin 2013-2-21
编辑：Kevin Bachovchin 2013-2-21
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,:)

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更多回答（4 个）

Azzi Abdelmalek
Azzi Abdelmalek 2013-2-21
编辑：Azzi Abdelmalek 2013-2-21
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))
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Azzi Abdelmalek
Azzi Abdelmalek 2013-2-21
Ok, You did not specify that in your question. You should give all information in your question to avoid wasting of time.
Kevin Bachovchin
Kevin Bachovchin 2013-2-21
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)).

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Jan
Jan 2013-2-21
Azzi Abdelmalek
Azzi Abdelmalek 2013-2-21
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),:)
  1 个评论
Kevin Bachovchin
Kevin Bachovchin 2013-2-21
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.)

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Azzi Abdelmalek
Azzi Abdelmalek 2013-2-22
编辑：Azzi Abdelmalek 2013-2-22
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

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