Borrar filtros
Borrar filtros

All possible permutation of a given vector.

6 visualizaciones (últimos 30 días)
luca
luca el 23 de Jul. de 2019
Editada: Stephen23 el 23 de Jul. de 2019
Given a vector, e.g [1 2 3 4 1], I would like to obtain all the possible permutations with NO REPETITION and NO ROTATION.
With
v = [1 2 3 4 1];
P = unique(perms(v),'rows');
I can avoid repetition but no rotation.
With rotation I mean that [1 2 1 4 3] and [2 1 4 3 1] are the same vector because the sequence is the same, just view from a different initial position.
I would like to be able to obtain all the possible combinations that avoid also this fact (rotation).
Thanks
  1 comentario
Rik
Rik el 23 de Jul. de 2019
I suspect there are two options:
  1. looping through your array (from end to beginning) and remove all rotations (use circshift)
  2. design your own implementation of perms that avoids rotations
Neither is likely to be fast, and the first will also require the generation of a very large array for slightly larger vectors.

Iniciar sesión para comentar.

Respuesta aceptada

Stephen23
Stephen23 el 23 de Jul. de 2019
Editada: Stephen23 el 23 de Jul. de 2019
V = [1,2,3,4,1];
N = numel(V);
% Permutations:
P = perms(V(2:N));
P(:,N) = V(1);
P = unique(P,'rows')
% Rotations:
R = P; % just to allow comparison.
X = hankel(2:N,[N,1:N-1]);
for k = size(R,1):-1:1
T = R(k,:);
if any(ismember(T(X),R,'rows'))
R(k,:) = [];
end
end
Giving a 12x5 matrix:
R =
1 2 3 4 1
1 2 4 3 1
1 3 2 4 1
1 3 4 2 1
1 4 2 3 1
1 4 3 2 1
2 1 3 4 1
2 1 4 3 1
2 3 1 4 1
2 4 1 3 1
3 1 4 2 1
3 2 1 4 1
See also:

Más respuestas (0)

Categorías

Más información sobre Dates and Time en Help Center y File Exchange.

Productos


Versión

R2019a

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by