Creating random order of trials (numbers in a matrix) with restrictions

Hi! I am new to MATLAB and I am struggling with creating a matrix with numbers which would indicate the different trials of an experiment. The experiment consists of 5 blocks with 50 trials in each block (10 different trials, each repeated 5 times per block). I would like to always start with a different first trial (so not start for instance with trial index 3 in two blocks) and not to repeat the same trial more than once in any given run (so for instance 5 6 2 2 9 ... is ok, but not 5 6 2 2 2 9). I only have the core of the structure - just the randomization, but not the two constraints.

Many thanks!

A=zeros(5,50);

for i=1:5,
    p=randi(10,1,50);

        if p(1)==A(1:5,1)

???????? How to make the array re-do itself so that the first number will definitely be different?

        else

        end

    A(i,:)=p;
end

[Joseph Cheng]

Just to get a confirmation that i'm reading your problem correctly. you'll end up with a 5x50 matrix (A) where any element can be a number between 1 and 10. The condition is that no number can be present more than 2 times per column and in column 1 no number can be repeated.

[natalie222]

Indeed, dimensions are correct and like you say - no number in the five rows of column 1 can be repeated. Secondly, no number can be present more than twice consequently in any of the five rows - they can (and indeed will be repeated), but should not be repeated more than twice one after another (e.g. 3 3 3). Is this clear?

[Joseph Cheng]

yes... the math didn't add up on the repetition portion but in your clarification you were talking about consecutive trials. I'll come back if i think of anything that doesn't require lots of loops.

Answer 1

[Joseph Cheng]

Well initial thoughts are to generate the initial matrix using

for ind = 1:5
    A(ind,:)=mod(randperm(49,49),10)+1;
end

to generate columns 2 to 50. The chances of 3 repeats are low but not infeasible. Then attach column 1 using the randperm(10,5) to get non repeating numbers.

Then go through and detect of groups of consecutive numbers. This can be done through something like this

 consecutive=diff(diff(A(ind,:))==0)

and then look for the pattern of [1 0 -1] which would mean there would be 3 consecutive same numbers. more consecutive numbers would be more zeros between the 1 and -1. Then knowing the index locations substitute these numbers with another random set of numbers and recheck/repeat till no consecutive numbers are repeated >2 times in a row.

[natalie222]

Thanks a lot! :)

Answer 2

[pfb]

So you need a 5x50 matrix of numbers between 1 and 10, such that:

- all the numbers in the first column are different - no number can appear more than two consecutive times in a row

I think I have it. Since it is not a lot of numbers, I did this in a perhaps trivial way, using 2 loops and some parsing

A = zeros(5,50);
pf=0; % previous element in the 1st column. It is 0 at the beginning
for r =1:5
    % first trial
    t=randi(10);
    % this takes care of the constraint in 1st column
    while(t==pf)
      t=randi(10);        
    end
    A(r,1)=t;
    pf=t;
    cons=1;
    for c=2:50
       t=randi(10);
       % this should avoid that more than 2 consecutive elements appear in
       % each row
       cons=cons+(t==A(r,c-1));
       if(cons==3)
         t=randi(10);
         while(t==A(r,c-1))  
              t=randi(10);
         end
         cons=1;
       end
       A(r,c)=t;
    end
end

[natalie222]

Thank you! I made it work.