Program to generate permutations in a certain order

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Greetings.
I need to write a program that can generate m permutations from a possible number of permutation sequences n!. (m<n!).
Below is a description. Thank you very much and be blessed.
X=[X1 X2 X3 X4 X5 X6];
%Where
X1=[1 1 0 1];
X2=[1 0 1 0];
X3=[1 0 1 1];
X4=[1 1 1 0];
X5=[0 1 0 1];
X6=[0 0 0 1];
% generate a matrix that contains k permutations out of a possible number
% % of 6!=720. Forexample k=4.
%generate the first 4 permutations (1-4)
P=X1 X2 X3 X4 X5 X6; X1 X2 X3 X4 X6 X5
X1 X2 X3 X5 X4 X6; X1 X2 X3 X5 X6 X4
%Which will be P=[1 1 0 1 1 0 1 0 1 0 1 1 1 1 1 0 0 1 0 1 0 0 0 1;
......
....1 1 0 1 1 0 1 0 0 1 0 1 1 1 1 0 0 0 0 1 1 1 1 0]
% generate last 4 permutations (717-720)
P=X6 X5 X4 X2 X3 X1; X6 X5 X4 X3 X1 X2
X6 X5 X4 X3 X2 X1; X1 X2 X3 X4 X5 X6
% generate the 4 permutations e,g 520-523
% generate the 4 permuations randomly (e.g 230, 310, 320, 640)
%generate the 4 permuations one after the other (e.g 1, 3 , 5, 7 or 210,
%212, 214, 216)
  1 Commento
Torsten
Torsten il 24 Mag 2021
Modificato: Torsten il 24 Mag 2021
The order of your permutations is not clear to me.
Maybe you want to use Matlab's "perms" to generate all n! permutations.

Accedi per commentare.

Risposta accettata

David Hill
David Hill il 24 Mag 2021
k=4;
X=[1 1 0 1;1 0 1 0;1 0 1 1;1 1 1 0;0 1 0 1;0 0 0 1]';
p=perms(6:-1:1);
m1=reshape(X(:,p(1:k,:)'),24,[])';
m2=reshape(X(:,p(end-k+1:end,:)'),24,[])';
m3=reshape(X(:,p(520:523,:)'),24,[])';
m4=reshape(X(:,p(randi(720,1,4),:)'),24,[])';
m5=reshape(X(:,p(1:2:7,:)'),24,[])';
  3 Commenti
Ronald Niwamanya
Ronald Niwamanya il 24 Lug 2021
@David Hill Just a quick one, Considering the permutations from the above e.g from m4, how can you reverse any of the permutations to obtain the original sequence X i.e X=[1 1 0 1 1 0 1 0 1 0 1 1 1 1 1 0 0 1 0 1 0 0 0 1]?
Thank you very much.
David Hill
David Hill il 24 Lug 2021
You have to know the order of one of the permutations of the rows of m. For example is you know that the first row of m4 permuted x by [5 3 6 4 1 2], then it would be easy to generate x.
m=reshape(m4(1,:),4,[]);%generate the row of m4 for the known permutation
idx=[5 3 6 4 1 2];%known permutation of x for the 1st row of m4
x=x(:,idx);
x=x(:)';

Accedi per commentare.

Più risposte (1)

Bruno Luong
Bruno Luong il 24 Mag 2021
clear X
X{1}=[1 1 0 1];
X{2}=[1 0 1 0];
X{3}=[1 0 1 1];
X{4}=[1 1 1 0];
X{5}=[0 1 0 1];
X{6}=[0 0 0 1];
n=length(X);
P=flipud(perms(1:n));
c=arrayfun(@(r) cat(2,X{P(r,:)}), 1:4,'Unif',0);
P=cat(1,c{:})
P = 4×24
1 1 0 1 1 0 1 0 1 0 1 1 1 1 1 0 0 1 0 1 0 0 0 1 1 1 0 1 1 0 1 0 1 0 1 1 1 1 1 0 0 0 0 1 0 1 0 1 1 1 0 1 1 0 1 0 1 0 1 1 0 1 0 1 1 1 1 0 0 0 0 1 1 1 0 1 1 0 1 0 1 0 1 1 0 1 0 1 0 0 0 1 1 1 1 0

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