Find all possible alternative arrays
in [Matlab]
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From: Vinh Le on 20 Mar 2010 20:55 Almost get it for n = 1, try A = [ 0 0 0 0 0 0 0; 0 0 0 1 1 1 1; 0 0 1 0 0 1 1; 0 0 1 1 1 0 0; 0 1 0 0 1 0 1; 0 1 0 1 0 1 0; 0 1 1 0 1 1 0; 0 1 1 1 0 0 1; 1 0 0 0 1 1 0; 1 0 0 1 0 0 1; 1 0 1 0 1 0 1; 1 0 1 1 0 1 0; 1 1 0 0 0 1 1; 1 1 0 1 1 0 0; 1 1 1 0 0 0 0; 1 1 1 1 1 1 1] [r, c] = size(A); resultMatrix = []; for i=1:r row = A(i,:); resultMatrix =[resultMatrix;row]; for j=0:r-1 saveRow = row; try if row(1,c-j) == 1 row(1,c-j) = 0; else row(1,c-j) = 1; end addedRow = row; row = saveRow; resultMatrix = [resultMatrix;addedRow]; catch end end end =====================
From: Matt J on 21 Mar 2010 08:52 "Vinh Le" <lekhanhvinh(a)gmail.com> wrote in message <ho3qp9$mqn$1(a)fred.mathworks.com>... > Almost get it for n = 1, try > > A = [ 0 0 0 0 0 0 0; > 0 0 0 1 1 1 1; > 0 0 1 0 0 1 1; > 0 0 1 1 1 0 0; =================== Here's an automatic way to compute A for arbitrary number of colums and arbitrary n m=6; n=3; idxc=nchoosek(1:m,n); [p,q]=size(idxc); idxr=repmat((1:p).',1,q); A=false(p,m); A( sub2ind([p,m],idxr,idxc) )=true,
From: Jos (10584) on 21 Mar 2010 16:41
"Vinh Le" <lekhanhvinh(a)gmail.com> wrote in message <ho3k5d$9g8$1(a)fred.mathworks.com>... > I have a binary matrix A (3x7): I am not sure I understand your problem completely, but perhaps PERMPOS might be useful here: http://www.mathworks.com/matlabcentral/fileexchange/11216-permpos A small example: permpos(2,4) % 1 1 0 0 % 1 0 1 0 % 1 0 0 1 % 0 1 1 0 % 0 1 0 1 % 0 0 1 1 hth Jos |