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

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natalie222
natalie222 2015-4-29
评论： natalie222 2015-5-2
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
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natalie222
natalie222 2015-4-29
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
Joseph Cheng 2015-4-29
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.

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回答（2 个）

Joseph Cheng
Joseph Cheng 2015-4-29
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
pfb 2015-4-29
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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