# n! permutation matrices

조회 수: 4(최근 30일)
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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### 답변 수 (4)

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

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

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Fabio Freschi 11 Dec 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)

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Daniel Brower 11 Dec 2019
Is there a way to change the cells in B into matrices? I am not familiar with using cells.

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Stephen Cobeldick 11 Dec 2019
Stephen Cobeldick 님이 편집함. 11 Dec 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 Aldhafeeri 11 Dec 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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