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generate all possible upper triangular matricies with variables

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Alec Bledsoe
Alec Bledsoe 2023-10-25
评论： Dyuman Joshi 2023-10-25
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?
  1 个评论
Torsten
Torsten 2023-10-25
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.

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采纳的回答

Voss
Voss 2023-10-25
编辑：Voss 2023-10-25
Ran in:
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

更多回答（2 个）

Mann Baidi
Mann Baidi 2023-10-25
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!
  1 个评论
Dyuman Joshi
Dyuman Joshi 2023-10-25
Why should OP download a function file, when there are options/methods available utilizing built-in functions?

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Steven Lord
Steven Lord 2023-10-25
Ran in:
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
  1 个评论
Dyuman Joshi
Dyuman Joshi 2023-10-25
@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?

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