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.

Réponses (4)

JESUS DAVID ARIZA ROYETH
JESUS DAVID ARIZA ROYETH le 11 Déc 2019

0 votes

identity=eye(4);
total=perms(1:4);
for k=1:24
matrixperm=identity(total(k,:),:)
end

1 commentaire

Daniel Brower
Daniel Brower le 11 Déc 2019
How would I put each matrixperm in a cell or array?

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Fabio Freschi
Fabio Freschi le 11 Déc 2019

0 votes

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 commentaire

Daniel Brower
Daniel Brower le 11 Déc 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 le 11 Déc 2019
Modifié(e) : Stephen23 le 11 Déc 2019

0 votes

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 le 11 Déc 2019

0 votes

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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