# Program to generate permutations in a certain order

조회 수: 3(최근 30일)
Ronald Niwamanya 2021년 5월 24일
댓글: David Hill 2021년 7월 24일
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)
##### 댓글 수: 1표시숨기기 없음
Torsten 2021년 5월 24일
The order of your permutations is not clear to me.
Maybe you want to use Matlab's "perms" to generate all n! permutations.

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### 채택된 답변

David Hill 2021년 5월 24일
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,[])';
##### 댓글 수: 3표시숨기기 이전 댓글 수: 2
David Hill 2021년 7월 24일
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(:)';

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### 추가 답변(1개)

Bruno Luong 2021년 5월 24일
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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