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

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 run into an error.

if true
  nums = mod( reshape(randperm(1*12), 1, 12), 2)
%%%%%algorithm to check for a condition repeating more than 2 times and fix it
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 end

Any help would be appreciated. Thanks

[Azzi Abdelmalek]

what if you have v=[1 1 1 1 1 1 0 1 0 0 0 0 1] ?

[Meade]

Give this a shot.

seqIn = [0 1 0 1 0 1 1 1 0 1 0 1 0 0 0 0 0 1 0 0 1 0 ];
seqOut = seqIn;

for ii = 3:length(seqOut)        % You don't care about the first 2 
    if seqOut(ii) == seqOut(ii-1) && seqOut(ii) == seqOut(ii-2)
        switch seqOut(ii)
            case 1
                seqOut(ii) = 0;
            case 0 
                seqOut(ii) = 1;
        end
    end
end
find(seqIn ~= seqOut)           % Gives the Indexes of the changed values

[Dushyant Dhundara]

@Azzi Abdelmalek One of the constraints on the generated sequence is that the number of 1's and 0's generated should be equal

[Dushyant Dhundara]

@ Meade: I tried your code for the following sequence: seqIn = [1 1 0 0 0 0 1 1 0 1 0 1]; the output came out to be: seqOut = [1 1 0 0 1 0 1 1 0 1 0 1];

As you can see, although it did do the right thing by making one of the 0's as 1 (5th index in seqOut), but it also now changed the total number of 1's to 7. As you can see in the output, there are now five 0's and seven 1's.The initial constraint on the generated sequence is that the number of 0's and 1's generated should always stay equal. Any ideas?

[Image Analyst]

"One of the constraints on the generated sequence is that the number of 1's and 0's generated should be equal" <== I don't think that's possible if you don't allow more than one 1 in a group unless your sequence is simply 0,1,0,1,0,1,0,1,0,1,......

[Dushyant Dhundara]

I mean I am able to generate the right sequence with equal number of 0's and 1's but when i go to fix it using the if statement for the repeated values, that is when the amount of 0's and 1's changes.

[Meade]

@Dushyant Dhundara Image Analyst is correct that this is not a uniquely solvable solution. For example, how would you deal with this sequence:

seqIn = [0 0 0 0 1 0 0 ]

This can't be made to satisfy your requirement of no more than 2 repeat digits & equal numbers of 1s.

[Dushyant Dhundara]

I have been able to figure out a way, it requires more computation time but does solve what I am trying to do for now. Thank you all for the help

[Image Analyst]

Are two 1's in a row allowable?

[John BG]

To Image Analyst:

from the question 'My goal is to not have either 0 or 1 be repeated more than 2 times consecutively.'

yes, you can have 010 or 0110, but you cannot have 01110 ..

Answer 1

[Jon]

Just modify your loop a little bit:

for i = 3:length(nums)
    if nums(i) == nums(i-1) && nums(i) == nums(i-2)
      if nums(i) == 1
          nums(i) = 0;
      else
          nums(i) = 1;
      end
    end
end

Oh, I see that you also require the same number of 1s and 0s. This doesn't ensure that. The easiest solution (though certainly not the fastest) is to add a check at the end that there are the same number; if not, reiterate until so.

[Dushyant Dhundara]

Same issue as I described above, doing this it fixes the sequence but it increments the number of 1's or 0's. Please test this sequence: nums = [1 1 0 0 0 0 1 1 0 1 0 1] Can you show what kind of check you mean? Do i need a counter for that?

Answer 2

[John BG]

Dushyant

the following generates sequences of random even length with your requirements of no longer than paired 1s or paired 0s bursts:

L=randi([0 10],1,1)   % choose random amount of ones

header=randi([0 1],1,2)   % choose random intitial 2 bits

S=zeros(1,2*L)    % init sequence to all zeros
S(1)=header(1);S(2)=header(2)   % cast header

k=3                 % pointer to sequence bit to decide
n1s=L;n0s=L    % meters counting how many ones and zeros left in each clip

while (n1s>0 && n0s>0)

      if (S(k-2)==0 && S(k-1)==0 && n1s>0)  % previous bits are 00
          S(k)=1
          n1s=n1s-1
          k=k+1
      end

     if (S(k-2)==1 && S(k-1)==1 && n0s>0)   % previous bits are 11
          S(k)=0
          n0s=n0s-1
          k=k+1
     end

      if (S(k-2)==0 && S(k-1)==1)                  % previous bits are 01
          S(k)=randi([0 1],1,1)
          if (S(k)==1 && n1s>0)
              n1s=n1s-1
          end
          if (S(k)==0 && n0s>0)
              n0s=n0s-1
          end

          k=k+1

      end

      if (S(k-2)==1 && S(k-1)==0)                 % previous bits are 10
          S(k)=randi([0 1],1,1)

          if (S(k)==1 && Ln1s>0)
              n1s=n1s-1
          end
          if (S(k)==0 && n0s>0)
              n0s=n0s-1
          end

          k=k+1

      end

end

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thanks in advance

John

[e-mail]

Answer 3

[Dushyant Dhundara]

I am not sure if its doing what I was looking for, but I appreciate your effort and help. I already found out a solution to the initial questions posted.

Thank you again