How to avoid having duplicate index result?
이전 댓글 표시
I have written this code it works correct but one thing is that when I run it it gives me duplicate result how can I avoid repetion if I run it for 24 times.
x=zeros(4,4);
temp=0;
test=zeros(3,3);
b=sum(x,1);
r=randperm(4);
for i=1:4
temp=0;
for j=1:4
% r=randi([1,3]);
if temp~=r(j)
temp=r(j);
if sum(x(i,:))==0 && b(temp)==0
x(i,temp)=1;
end
end
end
b=sum(x,1);
end
x
댓글 수: 3
Jan
2019년 2월 21일
The question is not clear yet. Which "index" is duplicated? What is repeated?
The code is not clear also. Why do you create the variable test which is not used anywhere? Defining b=sum(x,1) could be simplified to: b=zeros(1, 4).
The purpose of the code is not clear also, because you do not provide meaningul comments.
Do I understand correctly, that your code creates a 4x4 matrix of zeros with one 1 in each row and column? Then this is easier:
x = zeros(4, 4);
x(sunb2ind(size(x), 1:4, randperm(4,4))) = 1;
Original (better explained) question, with simple three line answer:
@Hardi Mohammed: why do refuse to use perms, which is the best way to generate those 6/24/... permutations that you request. Please explain why perms does not work for you.
Hardi Mohammed
2019년 2월 25일
답변 (2개)
Jan
2019년 2월 21일
If I assume, that this code satisfies your needs for 1 call:
x = zeros(4, 4);
x(sub2ind(size(x), 1:4, randperm(4,4))) = 1;
I assume, that this creates all wanted results:
order = perms(1:4);
n = size(order, 1);
order = order(randperm(n, n), :); % If a random order is wanted
x = zeros(4, 4, n);
for k = 1:n
index = sub2ind([4, 4, n], 1:4, order(k, :), repmat(k, 1, 4));
x(index) = 1;
end
Now x(:, :, k) is the wanted submatrix.
Jos (10584)
2019년 2월 21일
편집: Jos (10584)
2019년 2월 21일
This also produces the N! possibilities. No loop, no sub2ind, only the outcome of perms as column indices ...
N = 4
X = eye(N) ;
X = reshape(X(:, perms(1:N).'), N, N, [])
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