use "for loop" for n times but n is not fixed

hi

i have a problrm with "for loop"

consider that i only have 5 integers, say 1 2 3 4 5

if the input n=1

i will expect i will get a set of data [1] [2] [3] [4] [5]

if n=2

i will get

[1 1] [1 2] [1 3] [1 4] [1 5]

[2 1] [2 2] [2 3] [2 4] [2 5]

[3 1] [3 2] [3 3] [3 4] [3 5]

[4 1] [4 2] [4 3] [4 4] [4 5]

[5 1] [5 2] [5 3] [5 4] [5 5]

actually, i want to get all the permutations of a n-digit number which can take the value of 1 2 3 4 5 (in this case)

of course if i know the number of digits (say n=2), i can write a for loop like this

for i = 1:5

for j = 1:5

    A = [i,j]

end

end

however, i would like to input n for any integers and then matlab will generate all the possible cases autometically

for example, if i input n = 2, i will get 25 results listed above

if i input n = 3, i will get 125 results...

thanks for the help

Answer 1

[Jan]

If you want to do this fast: FEX: VChooseKRO.

If you want to find out, how to solve such question in general:

n = 5;
k = 3;
m = zeros(n^k, k);
v = ones(1, 3);
c = 0;
ready = false;
while ~ready
  % Store current value in the output matrix:
  c       = c + 1;
  m(c, :) = v;

    % Increase the current value, care for overflow:
    q = k;
    while ~ready
      v(q) = v(q) + 1; 
      if v(q) <= n
        break;        % Index inside limit, proceed with outer loop
      end
      v(q) = 1;       % Reset current index to 1
      q    = q - 1;   % Go to previous index
      if q == 0       % All indices are exhausted
        ready = true; % Stop both loops
      end
    end
  end

By this way, you do not increase a scalar in a FOR loop, but the elements of a vector in a WHILE loop.

There are more efficient methods to create permutations. Exploiting the obvious structure of the output would be more efficient.

Answer 2

[Daniel Shub]

I wouldn't do it this way, but recursion is a useful technique when you have an unknown number of loops.

function x = looper(n)
  if n > 1
    temp = looper(n-1);
    x = [];
    for ii = 1:5
      x = [x; repmat(ii, length(temp), 1), temp];
    end
  else
    x = (1:5)';
  end
end

Answer 3

[Andrei Bobrov]

v = 1:5;
n = 3;

a = cell(1,n);
[a{:}] = ndgrid(v);
out = cell2mat(cellfun(@(x)x(:),a,'un',0));