How can I generate a vector of 19 numbers in such a way that all 19 numbers are repeated 10 times, but 10 consecutive numbers are not equal?

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Karthik
Karthik on 4 Feb 2014
Commented: Karthik on 4 Feb 2014
So far, the code I have is this:
nTargs = 19;
pairs = nchoosek(1:nTargs, 10);
nPairs = size(pairs, 1);
order = randperm(nPairs);
values=randsample(order,19);
targs=pairs(values,:);
Alltargs=false;
while ~Alltargs
targs=pairs(randsample(order,19),:);
B=[];
for i=1:19
G=length(find(targs==i))==10;
B=[B G];
end
if sum(B)==19
Alltargs=true;
end
end
The computation time was a huge issue, there has to be a nicer way to do this. I also thought about generating all of the values and then doing some sort of rearrangement, but that was not fruitful. Code for that is below
N=[];
for i=1:10
N=[N randperm(19)];
end
B=[];
for j=1:19
if length(unique(N(j*10-9:j*10)))<10
B=[B 1];
end
end
B
Thanks for any input.
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Accepted Answer

Niklas Nylén
Niklas Nylén on 4 Feb 2014
Edited: Niklas Nylén on 4 Feb 2014
My suggestion is to start with any valid vector according to your constraints. In the example below I have chosen the most obvious one which is 1,2,3,...,18,19,1,2,...
The next step is to move around the numbers in a way that always will end up with a valid vector but still introduces some randomness. One way to do that is to take a random index and switch place with its neighbor, if the move results in a vector which fulfills the constraints. Since we know that the old vector is valid it is only necessary to check exactly +- 10 indexes from the random position.
y = repmat(1:19,1,10);
% Run enough iterations to get the output random enough, I selected 100000
for ii = 1:100000
% Select random index
index = randi(length(y)-1);
% Check if it is allowed to switch places
if y(index)~=y(min(index+10, length(y))) && y(index+1)~=y(max(1,index-9))
% Make the switch
yTmp = y(index);
y(index)=y(index+1);
y(index+1)=yTmp;
end
end
My code runs in ~0.5 s with 100 000 iterations, of which approximately 85 000 result in a valid move.

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