# Generate a matrix of combinations without repetition (array exceeds maximum array size preference)

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Reinforcement Learning on 25 Oct 2021
Edited: Jan on 25 Oct 2021
Hey,
I am trying to generate a matrix, that has all unique combinations of [0 0 1 1], I wrote this code for this:
v1 = [0 0 1 1];
M1 = unique(perms([0 0 1 1]),'rows');
This isn't ideal, because perms() is seeing each vector element as unique and doing 4! = 4 * 3 * 2 * 1 = 24 combinations.
With unique() I tried to delete all the repetitive entries so I end up with the combination matrix M1 -> only [4!/ 2! * (4-2)!] = 6 combinations!
Now, when I try to do something very simple like:
n = 15;
i = 1;
v1 = [zeros(1,n-i) ones(1,i)];
M = unique(perms(vec_1),'rows');
Instead of getting [15!/ 1! * (15-1)!] = 15 combinations, the perms() function is trying to do 15! = 1.3077e+12 combinations and it's interrupted.
How would you go about doing in a much better way? Thanks in advance!

Jan on 25 Oct 2021
Edited: Jan on 25 Oct 2021
Instead of permuting [0,0,1,1] and removing the duplicates, you can obtain the list of indices of the ones (or zeros):
M1 = unique(perms([0 0 1 1]), 'rows')
M1 = 6×4
0 0 1 1 0 1 0 1 0 1 1 0 1 0 0 1 1 0 1 0 1 1 0 0
nDigit = 4;
nOnes = 2;
index = nchoosek(1:nDigit, nOnes) % Indices of 2 ones in 4 elements
index = 6×2
1 2 1 3 1 4 2 3 2 4 3 4
n = size(index, 1);
v = (1:n).' + n * (index - 1);
M2 = zeros(n, nDigit);
M2(v(:)) = 1
M2 = 6×4
1 1 0 0 1 0 1 0 1 0 0 1 0 1 1 0 0 1 0 1 0 0 1 1
The sorting order differs from the unique(perms()) solution.
This method works well with nDigits=14 and nOnes=1

R2020b

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