Preventing repeated sequences in a matrix

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Lauren Shone
Lauren Shone el 21 de Sept. de 2012
I have a 192 x 3 matrix, order(192 x 3):
order(:, 1) and order(:, 2) both contain repeating values of 1 - 16, and order(:, 3) contains repeating values of 1 and 2. I need to shuffle the matrix, while preventing any repeats of more than three of the same value in the last column, so order(:, 3) should never show more than 3 repeats of 1 or 2.
This is what I have, which worked for a smaller version of the matrix just fine, but seems to get stuck with a slightly larger matrix:
not_good = true;
while not_good
not_good = false;
order = Shuffle(order);
% returns an array of 1s and 0s indexing the position of the values for 1 and 2
R1 = order(:, 3) == 1;
R2 = order(:, 3) == 2;
% checks for repeats, returns 1 if repeats are present
rep_test1 = any(diff([1; find(R1)])>3);
rep_test2 = any(diff([1; find(R2)])>3);
if rep_test1 > 0 || rep_test2 > 0
not_good = true;
end
end
Any comments much appreciated. Thanks.
  1 comentario
Matt Fig
Matt Fig el 21 de Sept. de 2012
Editada: Matt Fig el 21 de Sept. de 2012
What is Shuffle?
>> help Shuffle
Shuffle not found.
And give the matrix order. Is this a good approximation of order?
order = cell2mat(arrayfun(@(x) randperm(x).',16*ones(12,2),'Un',0));
order = [order cell2mat(arrayfun(@(x) randperm(x).',2*ones(96,1),'Un',0))];

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Matt Fig
Matt Fig el 21 de Sept. de 2012
Editada: Matt Fig el 21 de Sept. de 2012
I think I figured out that Shuffle is on the FEX.
Part of the problem is that you are just going to have more chance to get 3+ of a kind in a row the more numbers you add. This is faster, however. For this order array (or the more structured one I show above), it takes less than a minute.
order = [mod(randperm(192),16);mod(randperm(192),16);...
mod(randperm(192),2)].'+1;
T = true(1,4); % 3 in a row is o.k.
while true
order = Shuffle(order);
if isempty(strfind(order(:,3).'==2,T))
if isempty(strfind(order(:,3).'==1,T))
break
end
end
end
Another note: Your code will let pass an order that has 4 or more in a row as the end of the third column...
  2 comentarios
Matt Fig
Matt Fig el 21 de Sept. de 2012
I noticed the first code I posted used
T = ones(1,3);
When it should have been
T = ones(1,4);
Now fixed.
Lauren Shone
Lauren Shone el 25 de Sept. de 2012
Thank you! This works.

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Más respuestas (1)

Jan
Jan el 21 de Sept. de 2012
Editada: Jan el 21 de Sept. de 2012
It will be much cheaper to shuffle the index vector only and use an UINT8 as long as less than 255 elements are processed (based on Matt's code):
index = uint8(1):uint8(192);
data = [mod(randperm(192), 16); mod(randperm(192), 16)].';
crit = mod(randperm(192), 2) + 1; % Avoid repeated transposition
T = true(1,4);
while 1 % Cheaper than the function(!) TRUE
index = Shuffle(index);
test = crit(index);
if isempty(strfind(test == 2, T))
if isempty(strfind(test == 1, T))
break
end
end
end
% Create the full data once only after the calculations:
result = cat(2, data(index, :), double(crit(index).'));

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