Permutation with repeating elements.

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Sourov Kumar Mondal
Sourov Kumar Mondal 2022-10-16
编辑： Bruno Luong 2022-10-16
Hello, greetings! I have the following array, A= 1:7. I want to get all the possible permutations of seven elements and I am trying to get my answer like this
ans =
1 2 3 4 5 6 7
1 2 5 3 4 7 6
2 3 4 5 1 7 6...... etc
It should be easy. But in the 'A' matrix, 1 and 2 indicates the simillar thing; 3 and 4 indicates the simillar thing; 5, 6, and 7 indicates the simillar thing. So, in the 'ans' 1 2 3 4 5 6 7 and 2 1 4 3 7 6 5 will be same. I want the code to return the values without repitating it. Is there any way to do this?
note: I want this code to do linear indexing. So, I can't replace A with any other matrix. It has to be 1:7.

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采纳的回答

Bruno Luong
Bruno Luong 2022-10-16
Ran in:
Just brute force of filter out what is considered as duplicated
g = [1 1 2 2 3 3 3];
x = 1:7;
p = perms(x);
[~,i] = unique(perms(g),'rows');
p = p(i,:);
p
p = 210×7
2 1 4 3 7 6 5 2 1 4 7 3 6 5 2 1 4 7 6 3 5 2 1 4 7 6 5 3 2 1 7 4 3 6 5 2 1 7 4 6 3 5 2 1 7 4 6 5 3 2 1 7 6 4 3 5 2 1 7 6 4 5 3 2 1 7 6 5 4 3
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Bruno Luong
Bruno Luong 2022-10-16
编辑：Bruno Luong 2022-10-16
Simplify lperms
function p = lperms(x, k)
p = nchoosek(x,k);
p = reshape(p(:,perms(1:k)),[],k);
end

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更多回答（1 个）

KSSV
KSSV 2022-10-16
Ran in:
A = {[1 2], [3 4], [5 6 7]} ;
iwant = perms(A)
iwant = 6×3 cell array
{[5 6 7]} {[ 3 4]} {[ 1 2]} {[5 6 7]} {[ 1 2]} {[ 3 4]} {[ 3 4]} {[5 6 7]} {[ 1 2]} {[ 3 4]} {[ 1 2]} {[5 6 7]} {[ 1 2]} {[5 6 7]} {[ 3 4]} {[ 1 2]} {[ 3 4]} {[5 6 7]}
  2 个评论
Sourov Kumar Mondal
Sourov Kumar Mondal 2022-10-16
Thanks for the answer. But this is keeping 1 and 2; 3 and 4; 5, 6, and 7 together always. The output I am looking for is just like normal permutations of 7 elements. but 1 and 2; 3 and 4; and 5,6 and 7 will behave like simillar elements. The no. of expected permutations would be (7! / (2! * 2! * 3!)). Is there any solution to this? thanks is advanced.
Sourov Kumar Mondal
Sourov Kumar Mondal 2022-10-16
Some expected outputs would be like this
1 3 5 7 6 4 2
1 4 3 5 6 2 7.... etc
But there won't be 2 3 5 7 6 1 because 1 and 2 are simillar

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