n! permutation matrices

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Daniel Brower
Daniel Brower el 11 de Dic. de 2019
Respondida: Bandar el 11 de Dic. de 2019
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.

Respuestas (4)

JESUS DAVID ARIZA ROYETH
JESUS DAVID ARIZA ROYETH el 11 de Dic. de 2019
identity=eye(4);
total=perms(1:4);
for k=1:24
matrixperm=identity(total(k,:),:)
end
  1 comentario
Daniel Brower
Daniel Brower el 11 de Dic. de 2019
How would I put each matrixperm in a cell or array?

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Fabio Freschi
Fabio Freschi el 11 de Dic. de 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)
  1 comentario
Daniel Brower
Daniel Brower el 11 de Dic. de 2019
Is there a way to change the cells in B into matrices? I am not familiar with using cells.

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Stephen23
Stephen23 el 11 de Dic. de 2019
Editada: Stephen23 el 11 de Dic. de 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

Bandar
Bandar el 11 de Dic. de 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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R2019b

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