randperm in while loop

Hi, I want to achieve the following with my code:

For A(1) I can chose randomly from 1, 2 or 3 - but if A(1) say is equal to 1, then A(2) should be randomly chosen from either 2 or 3, and if A(2) would be say 3, then A(3) = 2. And with A(4) the game starts again. So I wrote the following code - and run it with RandomizeTrials(3,12) - why is it not running?

function [Array] = RandomizeTrials(numberofSeq, lengthofArray)
  Array = zeros(1,lengthofArray);
  k = 1;
  while k < lengthofArray-1
        m = k+numberofSeq-1;
        Array(k:m) = randperm(numberofSeq,numberofSeq,false);
        k+numberofSeq;
  end
  end

[Adam]

What do you mean by "not running"?

When I run it I get a clear error message which tells me exactly what the problem is which I can consult the randperm help page for to confirm. I assume you can do likewise if you are getting the same error.

Answer 1

[pfb]

If I get it right, you want a vector A that is "divided" into bunches of 3 consecutive elements. Each of these "bunches" should be a random permutation of the integers 1, 2, 3.

This should be it

 A = zeros(3*N);
 for k=1:N
  A((k-1)*3+(1:3))=randperm(3);
 end

You can also create a Nx3 matrix of random permutations, and then possibly "unwind" it to a single vector.

 A = zeros(N,3);
 for k =1:N
  A(k,:)=randperm(3);
 end
 A = A';   % transpose so that the permutations are along columns
 A = A(:); % the columns of A are concatenated into a single column vector

[pfb]

of course, if the sequence length is not 3 but a different number L, you just write L instead of 3.

[MiauMiau]

Hi, thanks, but what does A((k-1)*3+(1:3)) exactly do? I mean we would end up with A((1:3)) or then A((1:3)+3) etc. - what does eg. A((1:3)+3) or A((1:3)+6) exaclty do? I hadn't encounter this syntax. But I see that if we would use A((1:3)+N)) = randperm(3), we would not neet a loop altogether, right?

[pfb]

Hi

 (k-1)*3+(1:3)

is the same as

 ((k-1)*3+1):(k*3)

i.e. the indices

   k*3-2, k*3-1, k*3

if k=1 these are 1,2,3; if k=2 these are 4,5,6, and so on... This particular choice is related to the fact that indices start from 1 in matlab.

The second choice above is more or less what Thorsten suggested below, although he did that in two lines, probably for clarity reason.

I thought that

 (1:3)+(k-1)*3

was sufficiently clear:

 (1:3)

is the vector [1 2 3]. Then I add (k-1)*3, and I use the result as the index for A.

Answer 2

[Thorsten]

In this solution you have a growing array in a loop, which is not the fasted way, but you do not have to fiddle with indices:

 N = length_of_array/number_of_sequence; 
 x = []; 
 for i=1:N
   x = [x randperm(number_of_sequence)]; 
 end

If you insist on preallocation and indices

 x = zeros(1, length_of_array);
 for i=1:N
   i1 = (i-1)*number_of_sequence;
   x(i1+1:i1+number_of_sequence) = randperm(number_of_sequence); 
 end