How to avoid having duplicate index result?
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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
3 Kommentare
Jan
am 21 Feb. 2019
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
am 25 Feb. 2019
Antworten (2)
Jan
am 21 Feb. 2019
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.
2 Kommentare
Jos (10584)
am 21 Feb. 2019
you can leave out the sub2ind ...
x = eye(4) ;
x = x(randperm(4),:)
Jos (10584)
am 21 Feb. 2019
Bearbeitet: Jos (10584)
am 21 Feb. 2019
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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