Randomise vector numbers to maximum variance

2022 9 9
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I have a vector that contains numbers 1, 0 and 10. In partcular, I have 100 times the 1, 400 times the 0 and 20 times the 10.
Is there any function or any way to shuffle the numbers to the maximum variance regarding their repetition? Namely, I want to shuffle them in a way to minimise their sequential repeatibility, so a to achive the low number of times that I get 1,1 or 0,0 or 10,10 in my vector?
Thank you so much in advance!

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Torsten
Torsten 2022-9-9
编辑：Torsten 2022-9-9
0,1,0,1,...,0,1,0,1,(100 times)
0,10,0,10,...,0,10, (20 times)
0,0,0,0,....0 (280 times)
Or is it important how often the 0 is repeated ?
You should define an exact expression of what you want to minimize or maximize.
abaza
abaza 2022-9-9
Thank you fo the comment!
I mean that I have this vector: [1,1,1,1...1,0,0,0,0,...0,10,10,10....10]
And I want to shuffle the numbers so as to minimize the repetition of 1 and 0 and 10.
It is obvious that I would get repetiton of 0 since the 1 and 10 are less than 1.
Torsten
Torsten 2022-9-9
编辑：Torsten 2022-9-9
Yes, it's obvious that 1 and 10 don't need to repeat and that 0 needs to repeat 279 times. That's optimal and the solution I already posted. Or do you think there is a better one ?
With maximum variance you mean
sum(abs(x(i+1)-x(i)),i=1,...,519)
?
abaza
abaza 2022-9-9
I would like to minimise the repetition.... I cannot avoid repetition of some ZEROS since they are more than 1 and 10, but I would like a more automated way (e.g. when you have more numbers...)

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Bruno Luong
Bruno Luong 2022-9-9

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A = repmat([10 repmat([repmat(0, 1, 2), 1 repmat(0, 1, 2)], 1, 5)], 1, 20);

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abaza
abaza 2022-9-9
Thank you Bruno for your reply!
Could you please make it more general?? It works for this case, but if I had different cases of numbers and sizes?
Bruno Luong
Bruno Luong 2022-9-9
Essentially you just need put each chunk of same value regularly spaced, so only largest chunk repeat.
I let you make the code.
abaza
abaza 2022-9-9
I see, this is totally empirical...
Bruno Luong
Bruno Luong 2022-9-9
编辑：Bruno Luong 2022-9-9
Ran in:
Ran in:
Not more empirical then what you call "maximum variance", since the true variance estimated from such sequence doesn't change at all with order.
A = repmat([10 repmat([repmat(0, 1, 2), 1 repmat(0, 1, 2)], 1, 5)], 1, 20);
for k=1:10
Ak = A(randperm(length(A)));
var(Ak)
end
ans = 3.7128
ans = 3.7128
ans = 3.7128
ans = 3.7128
ans = 3.7128
ans = 3.7128
ans = 3.7128
ans = 3.7128
ans = 3.7128
ans = 3.7128
You must have different definition of variance than standard statisics variance anyone knows. But as long as you don't define it, any sequence will be empirical.
abaza
abaza 2022-9-9
编辑：abaza 2022-9-9
Thank you! Probably variability or maximum randomisation should be more correct!
Torsten
Torsten 2022-9-9
Probably variability or bst randomisation should be more correct!
And what's the meaning of those ? You operate with vague associations, not with precise definitions.
abaza
abaza 2022-9-9
Minimum number of pairs...
Thanks! Or
if D= my_vector
MINIMUM number of:
unique(find(diff(D)==0))
Torsten
Torsten 2022-9-9
编辑：Torsten 2022-9-9
Ran in:
Ran in:
D = repmat([10 repmat([repmat(0, 1, 2), 1 repmat(0, 1, 2)], 1, 5)], 1, 20);
nnz(diff(D)==0)
ans = 280
D = [repmat([0,1],1,100),repmat([0,10],1,20),zeros(1,280)];
nnz(diff(D)==0)
ans = 279
Bruno Luong
Bruno Luong 2022-9-9
Ran in:
Ran in:
This is a method for generic case
v=[0 1 10];
n=[400 100 20];
A = shuffle(v, n)
A = 1×520
0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1
function A = shuffle(v, n)
[m,i] = maxk(n,2);
mm = min(m);
if mm > 0
n(i) = n(i)-mm;
A = [repmat(v(i), 1, mm), shuffle(v, n)];
else
[mm,j] = max(m);
A = repmat(v(i(j)), 1, mm);
end
end
abaza
abaza 2022-9-9
That is amazing!!! Thank you Bruno!

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