How can I generate a randomized vector of repeating numbers with given conditions?

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I'm trying to generate a vector of pseudo-random numbers that contains 1:11 repeated 8 times. The vector is broken up into blocks of 8. No number within each block can be repeated and each number has to occur at each position within the block once (i.e the number 4 has to occur first-eighth in eight different blocks). I've gone through a ton of iterations to do this and have attached the most recent attempt. The problem here is that by the end of the loop, the only possible numbers remaining are ones that have already been added within the current block of 8 numbers, so it becomes an endless while loop. Any help would be much appreciated!
TargTest=nan(8,11);%M by N array to indicate if Nth target was tested at Mth location
TestSessionTargs=[]; %Test Session Target order
while length(TestSessionTargs)<numel(TargTest)
TestBlock=nan(1,8);
for B=1:length(TestBlock)
PossibleTargs=find(isnan(TargTest(B,:))==1);
if length(PossibleTargs)~=1
F=randsample(PossibleTargs,1);
while ismember(F,TestBlock)
F=randsample(PossibleTargs,1);
end
else
F=PossibleTargs;
end
TestBlock(B)=F
TargTest(B,F)=1;
end
TestSessionTargs=[TestSessionTargs TestBlock];
end
  7 Comments
Karthik
Karthik on 8 May 2015
Another attempt I have made:
TargTest=nan(8,11);%M by N array to indicate if Nth target was tested at Mth location
TestSessionTargs=[]; %Test Session Target order
%While loop will go through TargTest M*N times row-wise and choose random
%index that ~isnan once it chooses that index in TargTest, places 1 as
%place holder to say location has been tested at particular time.
while length(TestSessionTargs)<numel(TargTest)
TestBlock=nan(1,8);
for B=1:size(TargTest,1)
F=randi(11,1);
while ~isnan(TargTest(B,F))||ismember(F,TestBlock)
F=randi(11,1);
end
TestBlock(B)=F;
TargTest(B,F)=1;
end
TestSessionTargs=[TestSessionTargs TestBlock];
end

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Accepted Answer

Walter Roberson
Walter Roberson on 8 May 2015
If each symbol in the current symbol set must occur exactly once in each position over the block of 8, then you should research Latin Squares of Order 8. See here
  3 Comments
Karthik
Karthik on 8 May 2015
Ah I looked at the latin square. the problem here is that from what I see, everything is in order. Another stipulation is that we do want to keep randomization, so rather than saying [1 2 3 4 5 6 7 8], it should be something more random like [7 2 3 6 1 5 8 4]
Karthik
Karthik on 8 May 2015
This works! There's a file exchange for a latin square function latsq that also provides randomized latin squares to solve the problem.
Thank you for the direction!

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