I want to apply a for loop to an array (say [1 2 3]) so that the output gives every possible combination of that array (i.e. [123], [132], [213], [231], [312], [321])

%this is the code i have present%
%Selection Sort - Iterating Arrays%
%1. Define input array and variables%
a = input('Please enter a list of numbers: ');
n=length(a);
%2. For loop - finding min index in list%
for i=1: n-1
  imin=i;
    for j=i+1:n
      if a(j)~=a(imin)
      imin=j;    
                %3. swapping elements - forming temp variable%
    if imin~=i
      temp=a(i);
      a(i)=a(imin);
      a(imin)=temp
            end
  end
        end
end
%4. Displaying results%
disp(a)
%But it only gives 3 of the possible combinations%
%any help would be much appreciated%

Answer 1

[Jan]

result = perms(1:3)

This replies a matrix, which contains the permutations as rows. If you really want the number 123 instead of the vector [1,2,3], use:

result = perms(1:3) * [100; 10; 1]

[Stephen23]

+2 I wish I could give two votes for this. Brilliant.

[Jan]

@Stephen: Thanks. I hope, I n more can offer even more sophisticated methods than a matrix multiplication ;-)

Answer 2

[Akira Agata]

If the input is 1-by-3 array with different digits, like [1 2 3], the following code generates all possible combination.

a = [1 2 3];
c = cellstr(num2str(a'))';

% List of all possible combination
[x,y,z] = meshgrid(c,c,c);
list = strcat(x(:), y(:), z(:));

[Jos (10584)]

why convert to strings???

[Stephen23]

@Jos (10584): I guess it was a literal interpretation of what is shown in the question title: i.e. [123] rather than [1,2,3] . Using chars makes sense for that case, but judging by the OP's code I doubt that this was their intention.