How can I generate a vector of 19 numbers in such a way that all 19 numbers are repeated 10 times, but 10 consecutive numbers are not equal?

So far, the code I have is this:

nTargs = 19;
pairs = nchoosek(1:nTargs, 10);
nPairs = size(pairs, 1);
order = randperm(nPairs);
values=randsample(order,19);
targs=pairs(values,:);
Alltargs=false;
while ~Alltargs
    targs=pairs(randsample(order,19),:);
    B=[];
    for i=1:19
    G=length(find(targs==i))==10;
    B=[B G];
    end
    if sum(B)==19
        Alltargs=true;
    end
end

The computation time was a huge issue, there has to be a nicer way to do this. I also thought about generating all of the values and then doing some sort of rearrangement, but that was not fruitful. Code for that is below

N=[];
for i=1:10
    N=[N randperm(19)];
end
B=[];
 for j=1:19
    if length(unique(N(j*10-9:j*10)))<10
       B=[B 1];
    end
 end
B

Thanks for any input.

[Walter Roberson]

When you say "but 10 consecutive numbers are not equal", do you mean that no sequence of 10 digits is exactly the same as any other sequence of 10 digits, or do you mean that in every sequence of 10 digits, no digit is repeated?

[Karthik]

No digit should be repeated, i.e [1 5 19 3 2 4 5 9 12 10] is not what I want because the 5 is repeated.

Answer 1

[Niklas Nylén]

My suggestion is to start with any valid vector according to your constraints. In the example below I have chosen the most obvious one which is 1,2,3,...,18,19,1,2,...

The next step is to move around the numbers in a way that always will end up with a valid vector but still introduces some randomness. One way to do that is to take a random index and switch place with its neighbor, if the move results in a vector which fulfills the constraints. Since we know that the old vector is valid it is only necessary to check exactly +- 10 indexes from the random position.

 y = repmat(1:19,1,10);

 % Run enough iterations to get the output random enough, I selected 100000
 for ii = 1:100000
     % Select random index
     index = randi(length(y)-1);

     % Check if it is allowed to switch places
     if y(index)~=y(min(index+10, length(y))) && y(index+1)~=y(max(1,index-9))
         % Make the switch
         yTmp = y(index);
         y(index)=y(index+1);
         y(index+1)=yTmp;
     end
 end

My code runs in ~0.5 s with 100 000 iterations, of which approximately 85 000 result in a valid move.

[Karthik]

This works great!! Thanks so much. I didn't think of it this way