# Random shuffling while reduce the identical pairs

2 views (last 30 days)
J T on 3 Mar 2020
Commented: Walter Roberson on 3 Mar 2020
Hello,
I have a 50-by-1 array like this:
P = [4 5 5 5 5 5 5 5 5 9 9 9 9 9 9 9 27 27 27 35 40 40 40 40 4 5 5 5 5 5 5 5 5 5 9 9 9 9 9 9 27 27 27 35 35 40 40 40 27 9]';
P should be split into two 25-by-1 arrays A and B after shuffling, and, I want to efficiently shuffle P such that the two children arrays, A and B, consider pairing A(i) and B(i) for i = 1:25, there is the least amout of A(i) == B(i).
Could someone help me out here? Thank you very much!

Walter Roberson on 3 Mar 2020
tries = 1000;
NB = length(B);
[~,perms] = sort(rand(tries, NB),2);
matchcounts = sum(A == B(perms),2);
bestidx = find(matchcounts == min(matchcounts));
best_shuffles = B(perms(bestidx,:));
You could also sort on matchcounts and take however many that you need.
In my test, I got 11 permutations with only a single match each.
Walter Roberson on 3 Mar 2020
sP = sort(P(:).');
A = sP(1:end/2);
B = sp(end/2+1:end);
tries = 1000;
NB = length(B);
[~,perms] = sort(rand(tries, NB),2);
matchcounts = sum(A == B(perms),2);
bestidx = find(matchcounts == min(matchcounts));
best_shuffles = B(perms(bestidx,:));
61 entries with overlap 0 in my test.
The above algorithm is not fool-proof. A better algorithm would be to select the value that occurs most often and put all of the copies of it into A, and then to select the second most common value and put all the copies into B, then put all copies of the third most common into A, and so on back and forth, partitioning into non-overlapping sets as much as possible. Ideally you would get A and B that had no overlapping values, in which case every permutation of B would be a total mismatch with every permutation of A.

R2019a

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