Matrix showing all combinations of multiple dice roll

I want to create a matrix showing all possible combinations of tossing n independent dice rolls, except that dice is not always 6-sided.

e.g. I want to create matrix res

 res = [1,1,1;
       1,1,2;
       1,2,1;
       1,2,2;
       2,1,1;
       2,1,2;
       2,2,1;
       2,2,2];

using a vector v = [2,2,2]. Basically vector v specifies the number and size of each dice. Is there a fast way to iterate this without using for loop? I know if the dice size is constant I can use:

    res = dec2base(0:dice_size^n-1,dice_size) - '0';
    res = res + 1;

But I am stuck when each dice has different sizes.

Answer 1

[Bruno Luong]

dice_size=[2 3 4];

n = length(dice_size);
c = arrayfun(@(n) 1:n, flip(dice_size), 'unif', 0);
[c{:}] = ndgrid(c{:});
c = cat(n+1,c{:});
c = fliplr(reshape(c,[],n))

c =

     1     1     1
     1     1     2
     1     1     3
     1     1     4
     1     2     1
     1     2     2
     1     2     3
     1     2     4
     1     3     1
     1     3     2
     1     3     3
     1     3     4
     2     1     1
     2     1     2
     2     1     3
     2     1     4
     2     2     1
     2     2     2
     2     2     3
     2     2     4
     2     3     1
     2     3     2
     2     3     3
     2     3     4

[Nicolas Tarino]

Nice! I don't completely get your code yet, but it works fast (enough) and fine.

[Bruno Luong]

If the order of the combinations is not matter for you, you can remove the FLIP and FLIPLR instructions to make it even faster.

Answer 2

[Torsten]

https://de.mathworks.com/matlabcentral/fileexchange/10064-allcomb-varargin

Best wishes

Torsten.

[Nicolas Tarino]

Not as fast as Bruno's code, but I really appreciate the fact that you can have multiple different dice entries instead of just 1:n.

Answer 3

[Image Analyst]

ndgrid or meshgrid is what experienced MATLABers would use. For beginners, a set of nested for loops would work if the number of dice was known in advance (like 3) and not a variable (like n=3).

[Nicolas Tarino]

Nope, the number of dice is not known in advance. That's what ruled out nested for loops for me. Meanwhile I might be able to do this by recursion... but it should be way too slow for large dimensions (or number of dice)