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

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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

##### 3 Comments

Joseph Cheng
on 29 Apr 2015

### Answers (2)

Joseph Cheng
on 29 Apr 2015

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.

pfb
on 29 Apr 2015

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

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