How to get all possible permutations in binary format?

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lilly lord 2022-4-29

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https://ww2.mathworks.cn/matlabcentral/answers/1708485-how-to-get-all-possible-permutations-in-binary-format

评论： lilly lord 2022-5-2

Hello, I want to get possible permutation of the vector v which actually contains a boolean values. Suppose the first permutation is v=[b2 b3 b1] . So it put the second column first then third column and finally the first colum and convert it into decimals

B=[0,1,0;0,0,1;1,1,0;0,1,1;1,0,1;1,0,0];
b1=B(:,1);
b2=B(:,2);
b3=B(:,3);
v=[b1 b2 b3];
c=perms(v);
final =bi2de(c)

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Voss 2022-4-29

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You say v is a vector, but in the code v is a matrix. And it's not clear why in the code you split the columns of matrix B into three separate variables b1, b2, and b3, only to recombine them into matrix v, so that v is the same as B.

B=[0,1,0;0,0,1;1,1,0;0,1,1;1,0,1;1,0,0];
b1=B(:,1);
b2=B(:,2);
b3=B(:,3);
v=[b1 b2 b3];
isequal(B,v)
ans = logical
   1

Can you please clarify what exactly you want to do? You want all permutations of what exactly?

lilly lord 2022-4-29

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v=[b1 b2 b3] I want to permute v ,for exapmle v=[b1,b3 b2] then v=[b3 ,b2,b1] and by doing so it actually takes the valurs for b1=B(:,1) etc. I need different decimal values after each permutation

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Answer 1

DGM 2022-4-29

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https://ww2.mathworks.cn/matlabcentral/answers/1708485-how-to-get-all-possible-permutations-in-binary-format#answer_954180

编辑：DGM 2022-4-30

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I'm not sure, but are you thinking of something like this?

B = [0,1,0;0,0,1;1,1,0;0,1,1;1,0,1;1,0,0];
% all permutations of the columns of B
v = B(:,perms(1:size(B,2)));
v = reshape(v.',3,[]).'
v = 36×3
     0     0     1
     1     0     0
     1     0     0
     0     0     1
     0     1     0
     0     1     0
     1     1     0
     0     0     0
     0     0     1
     0     1     0
% convert to dec
c = bi2de(v,'left-msb')
c = 36×1
     1
     4
     4
     1
     2
     2
     6
     0
     1
     2

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DGM 2022-4-30

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If you want to organize the results in something other than one long vector, you can represent the results for each permutation as columns of a numeric array:

B = [0,1,0;0,0,1;1,1,0;0,1,1;1,0,1;1,0,0];
v = B(:,perms(1:size(B,2)));
v = reshape(v.',3,[]).';
c = bi2de(v,'left-msb');
c = reshape(c,size(B,1),[]) % one column for each perm
c = 6×6
     1     6     1     7     6     0
     4     0     7     4     3     3
     4     1     6     5     3     2
     1     2     5     3     6     4
     2     0     7     2     5     5
     2     5     2     7     5     0

If you're thinking of making a bunch of dynamically-created variables for each vector, just don't. It would only make things slower and more problematic. For 8 columns, there are only 107520 permutations, so it's still manageable. You may find that the vast majority are nonunique.

lilly lord 2022-5-2