In a sequence of 0 and 1, how to make sure that no same values appear more than 2 times consecutively?

Question

I am generating a sequence of equal number of 0's and 1's . A = [ 0 1 1 0 1 1 1 0 0 1 0 0] My goal is to not have either 0 or 1 be repeated more than 2 times consecutively. in the above array A, you can see that 1 1 1 appears so somehow I need to switch it around so that one of them becomes 0 and if there is something like 0 0 0 then switch one of them to 1. I have tried few algorithms which work for few iterations but then I get different number of 0's and 1's. Also, the result or the output should maintain the same number of 1's and 0's i.e. six 0's and six 1's.

nums = mod( reshape(randperm(1*12), 1, 12), 2)
    for i = 1:length(nums)-2
    if nums(i+1)==nums(i) & nums(i+2) ==nums(i) 
       if nums(i) == 1
      nums(i+2) = 0;
    else
      nums(i+2) = 1;
  end
 end
end

Answer 1

This might be one way that does not require trying permutations until you find one:

A = [];
N = 100; %//number of elements you desire in A (must be even).

Initialize A to have equal number of alternating 0s and 1s:

%//initialize A = [1 0 1 0 1 0....] with equal numbers of 0 and 1.
for x=1:N/2 
    A = [A 1 0];
end

Iterate through A:

for x=3:(N-1) %//iterate through A, and decide to swap or not at each iteration

    if ( A(x) == A(x-1) && A(x) == A(x-2) ) %//must swap
        temp = A(x);
        A(x) = A(x+1);
        A(x+1) = temp;
    elseif ( A(x+1) == A(x-1) && A(x+1) == A(x-2) ) %//must not swap

    else %//can choose to swap or not
        if (randi([0 1])) %should I swap?
            temp = A(x);
            A(x) = A(x+1);
            A(x+1) = temp;
        end
    end

end

Answer 2

As your constraints are too intricated in order to solve the problem a posteriori, I'd suggest you to iterate until you find a solution. (This approach being possible considering the small length of your output vector).

The code would look like :

out=0;

while ~out

    nums = mod( reshape(randperm(1*12), 1, 12), 2);

    diff1=diff(nums);

    diff1(diff1~=0)=NaN;

    diff2=diff(diff1);

    if all(diff2~=0)

        out=1;

    end

end

It basically iterates over 2 steps :

1/ Generate a candidate :

nums = mod( reshape(randperm(1*12), 1, 12), 2);

2/ Check if the candidate respects the constraints

diff1=diff(nums);

diff1(diff1~=0)=NaN;

Here MATLAB's diff function is used in order to check for repeated numbers. A 0 means that a number is repeated 2 times. In order to check if a number is repeated 3 times we need to use this function a second time.

diff2=diff(diff1);

Now, a 0 will appear if a number has been repeated 3 times or more consecutively, thus we check the presence of 0's in diff2. If there isn't any the candidate is kept and we end the while loop.

if all(diff2~=0)

    out=1;

end

Example run :

Answer 3

You just need to make a counter.

    nums = mod( reshape(randperm(1*12), 1, 12), 2)
    idx=1;
    counter_one=0;
    counter_zero=0;
    changedIndex=0;

    while idx <= length(nums)
        if (nums(idx) == 0)
            counter_zero = counter_zero+1;
            counter_one=0;
            if (counter_zero == 3)
                nums(idx)=1;
                counter_zero=0;
                changedIndex = 1;
            end
        else
            counter_one= counter_one+1;
            counter_zero=0;
            if (counter_one== 3)
                nums(idx)=0;
                counter_one=0;
                changedIndex =1;
            end
        end

    if (changedIndex == 1)
     if (idx < length(nums))
       if (nums(idx+1)) == 1
          counter_one = 1;
          changedIndex = 0;
       else
          counter_zero=1;
          changedIndex = 0;
       end
    else
           if (nums(idx)) == 1
          counter_one = 1;
          changedIndex = 0;
       else
          counter_zero=1;
          changedIndex = 0;
       end
    end
    end
idx = idx+1;
end

I have tested it just now and it worked fine

New test, with a more difficult vector