# Shuffle vector with constraints

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Pieter on 14 Jan 2014
Answered: Pieter on 14 Jan 2014
Hi,
I have an array of 50 1's, 50 2's, etc up to 50 8's. I would like to shuffle them with the constraint that two consecutive numbers cannot be the same. I tried several options but most are very cpu intensive or did not yield a good result. Are there computation efficient solutions to this problem?
Azzi Abdelmalek on 14 Jan 2014
This is not enough as information. Because you can set them for example [1 2 3 1 2 3 1 2 3 . . .]

W. Owen Brimijoin on 14 Jan 2014
A brute force method works fairly well in this scenario - 1) start with a shuffled version of the sequence, 2) find repeats, 3) swap any repeats to different locations... goto 2 (rinse and repeat). This sort of method would likely be terrible if you had fewer characters in your sequence, but for nine unique elements, it seems to halt very quickly.
seq = repmat([1:8],50,1); %make sequence
seq = [seq(:);repmat(9,30,1)]; %make sequence
Doing the reshuffling:
seq = seq(randperm(numel(seq))); %initial shuffle
old_idx= unique(find(diff(seq)==0));%find repeats
while ~isempty(old_idx), %continue until no repeats
new_idx = unique(setdiff([1:length(seq)],old_idx)); %find new spots
new_idx = new_idx((randi(length(new_idx),length(old_idx),1)))';
seq([new_idx;old_idx],:) = seq([old_idx;new_idx],:); %swap
old_idx= unique(find(diff(seq)==0));%find repeats
end

Mischa Kim on 14 Jan 2014
Edited: Mischa Kim on 14 Jan 2014
Since I do not know what techniques you have tried, here's (another) one? Start with sorting the 1's, 2's, etc. in matrix form, i.e.,
1 1 1 1...
2 2 2 2...
3 3 3 3...
and perform permutations on the individual columns (e.g using randperm). Based on the entry of the last row in the first column randomly pick a column vector with a different first row entry and make it the new second column. Using this process work your way through all the columns (you might have to do some more permutations towards the final columns). Once done, transpose and concenate all of the column vectors.

Pieter on 14 Jan 2014
Hi Mischa,
Thanks, this is a very good suggestion for one of my other experiments. This however has two issues I think:
1) It is not a constraint that every subvector of 8 numbers is a permutation. The only constraint is that no two consecutive numbers are the same.
2) This only works if all numbers occur equally. I forgot to mention that we have in fact 9 numbers. 1 to 8 occur 50 times, 9 occurs only 30 times.
Mischa Kim on 14 Jan 2014
Edited: Mischa Kim on 14 Jan 2014
Hello Pieter,
1. Understood. I used the permutation for shuffling with a certain level of randomness (while satisfying the constraint).
2. You can still use the approach outlined above by randomly tossing in the 30 9's into the vector at the end.

Pieter on 14 Jan 2014
Hi Mischa,
This looks fine. I'll try in tomorrow. Thx