generate all possible upper triangular matricies with variables

I am trying to find a way of generating all possible matricies with different combinations of elements in a list.

The base matrix is an upper triangular matrix:

g = [1,a,b;

0,1,c;

0,0,1]

In this case, the values for a,b, and c are in the list [0,1,2]. I want to generate every possible g based on the combination of the options for a,b, and c ie:

g = [1,0,0;0,1,0;0,0,1] , g = [1,1,0;0,1,0;0,0,1] , g = [1,1,1;0,1,0;0,0,1] , ...

Is there a good way to do this in Matlab?

[Torsten]

Hint:

Use "ndgrid" to create all possible triple combinations of {0,1,2} and insert these combinations in the matrix for a, b and c.

Answer 1

[Voss]

g = [1,NaN,NaN; 0,1,NaN; 0,0,1];
v = [0,1,2];

n_values = numel(v);

slots = find(isnan(g));
n_slots = numel(slots);

n_combos = n_values^n_slots;

M = v(1+dec2base(0:n_combos-1,n_values)-'0');
idx = (slots+numel(g)*(0:n_combos-1)).';
g_all = repmat(g,[1,1,n_combos]);
g_all(idx) = M;


format compact
disp(g_all)

(:,:,1) =
     1     0     0
     0     1     0
     0     0     1
(:,:,2) =
     1     0     0
     0     1     1
     0     0     1
(:,:,3) =
     1     0     0
     0     1     2
     0     0     1
(:,:,4) =
     1     0     1
     0     1     0
     0     0     1
(:,:,5) =
     1     0     1
     0     1     1
     0     0     1
(:,:,6) =
     1     0     1
     0     1     2
     0     0     1
(:,:,7) =
     1     0     2
     0     1     0
     0     0     1
(:,:,8) =
     1     0     2
     0     1     1
     0     0     1
(:,:,9) =
     1     0     2
     0     1     2
     0     0     1
(:,:,10) =
     1     1     0
     0     1     0
     0     0     1
(:,:,11) =
     1     1     0
     0     1     1
     0     0     1
(:,:,12) =
     1     1     0
     0     1     2
     0     0     1
(:,:,13) =
     1     1     1
     0     1     0
     0     0     1
(:,:,14) =
     1     1     1
     0     1     1
     0     0     1
(:,:,15) =
     1     1     1
     0     1     2
     0     0     1
(:,:,16) =
     1     1     2
     0     1     0
     0     0     1
(:,:,17) =
     1     1     2
     0     1     1
     0     0     1
(:,:,18) =
     1     1     2
     0     1     2
     0     0     1
(:,:,19) =
     1     2     0
     0     1     0
     0     0     1
(:,:,20) =
     1     2     0
     0     1     1
     0     0     1
(:,:,21) =
     1     2     0
     0     1     2
     0     0     1
(:,:,22) =
     1     2     1
     0     1     0
     0     0     1
(:,:,23) =
     1     2     1
     0     1     1
     0     0     1
(:,:,24) =
     1     2     1
     0     1     2
     0     0     1
(:,:,25) =
     1     2     2
     0     1     0
     0     0     1
(:,:,26) =
     1     2     2
     0     1     1
     0     0     1
(:,:,27) =
     1     2     2
     0     1     2
     0     0     1

Answer 2

[Mann Baidi]

Hi Alec,

I understand you would like to generate all the possible combinations matrices. I would like to suggest you the below function from the File Exchange which will help you to find all the possible permutations of "a","b" and "c". You can download the zipped folder, extract it and add the path to MATLAB.

After that you can get the permutions by the following code:

r1 = permn([0 1 2],3)

This will give you all the possible values for "a","b" and "c" respectively.

After that you can simply make a loop and put the values of "a","b" and "c" in the matrices.

Here is the link of the tool

http://www..mathworks.com/matlabcentral/fileexchange/7147-permn/

Hoping that this helps!

[Dyuman Joshi]

Why should OP download a function file, when there are options/methods available utilizing built-in functions?

Answer 3

[Steven Lord]

Since you're using release R2023b, you can use the combinations function introduced in release R2023a.

values = [0, 1, 2];
g = @(a, b, c) [1,a,b; 0,1,c; 0,0,1];
V = combinations(values, values, values);
M = rowfun(g, V, 'OutputFormat', 'cell')

M = 27×1 cell array
    {3×3 double}
    {3×3 double}
    {3×3 double}
    {3×3 double}
    {3×3 double}
    {3×3 double}
    {3×3 double}
    {3×3 double}
    {3×3 double}
    {3×3 double}
    {3×3 double}
    {3×3 double}
    {3×3 double}
    {3×3 double}
    {3×3 double}
    {3×3 double}

Let's look at one of the values to check it satisfies your requirements.

M{17}

ans = 3×3

1 1 2 0 1 1 0 0 1

[Dyuman Joshi]

@Steven Lord, a nice idea, but it feels (to me) like brute force, as values has to be supplied manually n times, where n is the number of elements in values.

What if there are more elements say 6 or 8 or a bigger array with 15 such placeholders?