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How to get all possible permutations in binary format?

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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)
  2 Comments
Voss
Voss on 29 Apr 2022
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
lilly lord on 29 Apr 2022
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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Accepted Answer

DGM
DGM on 29 Apr 2022
Edited: DGM on 30 Apr 2022
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
  3 Comments
DGM
DGM on 30 Apr 2022
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
lilly lord on 2 May 2022
Thank you sir. YOur answer clearify the problem I am facing.

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