n! permutation matrices
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I need to generate 24 (4!) distict permutation 4x4 matrices. How would I do that?
The first one would need to be the identity matrix =
[1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1]
Each other would be variations of the identity matrix having different order of rows of the identity matrix. There would be exactly 24 (4!) different possible distict variations of the identity matrix, counting the identity matrix.
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Antworten (4)
JESUS DAVID ARIZA ROYETH
am 11 Dez. 2019
identity=eye(4);
total=perms(1:4);
for k=1:24
matrixperm=identity(total(k,:),:)
end
Fabio Freschi
am 11 Dez. 2019
Is this what you wish?
% identity matrix
A = eye(4);
% permuatations
idx = perms(1:4);
% all matrices in a cell array
B = arrayfun(@(i)A(idx(i,:),:),1:24,'UniformOutput',false)
Stephen23
am 11 Dez. 2019
Bearbeitet: Stephen23
am 11 Dez. 2019
A purely numeric solution without loops:
>> I = eye(4);
>> M = reshape(I(:,flipud(perms(1:4)).'),4,4,24)
M =
ans(:,:,1) =
1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1
ans(:,:,2) =
1 0 0 0
0 1 0 0
0 0 0 1
0 0 1 0
ans(:,:,3) =
1 0 0 0
0 0 1 0
0 1 0 0
0 0 0 1
ans(:,:,4) =
1 0 0 0
0 0 0 1
0 1 0 0
0 0 1 0
ans(:,:,5) =
1 0 0 0
0 0 1 0
0 0 0 1
0 1 0 0
ans(:,:,6) =
1 0 0 0
0 0 0 1
0 0 1 0
0 1 0 0
... more here
ans(:,:,21) =
0 0 1 0
0 1 0 0
0 0 0 1
1 0 0 0
ans(:,:,22) =
0 0 0 1
0 1 0 0
0 0 1 0
1 0 0 0
ans(:,:,23) =
0 0 1 0
0 0 0 1
0 1 0 0
1 0 0 0
ans(:,:,24) =
0 0 0 1
0 0 1 0
0 1 0 0
1 0 0 0
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Bandar
am 11 Dez. 2019
You may consider creating multidemintional matrix as follows:
I=eye(4);
pr=perms(1:4);
A=zeros(4,4,24);
for i=1:24
A(:,:,i) = I(pr(i,:),:);
end
A
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