permute with exact repeats

I'm trying to permute a comprehensive list of 3 elements, say [1 2 3]

over a 5-size array with at least 1 element included from the 3 element array. (I'd like it the result be general, so the variables might be 3 over 6, 4 over 7, etc.) Obviously I can do this by brute force but I'd prefer a more elegant solution.

The result would be

1 1 1 2 3

1 1 1 3 2

1 1 2 1 3

1 1 2 3 1

1 1 3 1 2

1 1 3 2 1

1 1 3 2 2

1 1 3 3 2

1 2 1 1 3

1 2 1 2 3

...

3 3 3 2 1

Answer 1

[Voss]

Here's one way:

x = [1 2 3];
n = 5;

% generate a matrix M whose rows are all possible
% sets of n elements from x:
m = numel(x);
M = x(1+dec2base(0:m^n-1,m)-'0');

% remove rows of M that don't have each element 
% of x at least once:
for ii = 1:m
    M(~any(M == x(ii),2),:) = [];
end

disp(M)

     1     1     1     2     3
     1     1     1     3     2
     1     1     2     1     3
     1     1     2     2     3
     1     1     2     3     1
     1     1     2     3     2
     1     1     2     3     3
     1     1     3     1     2
     1     1     3     2     1
     1     1     3     2     2
     1     1     3     2     3
     1     1     3     3     2
     1     2     1     1     3
     1     2     1     2     3
     1     2     1     3     1
     1     2     1     3     2
     1     2     1     3     3
     1     2     2     1     3
     1     2     2     2     3
     1     2     2     3     1
     1     2     2     3     2
     1     2     2     3     3
     1     2     3     1     1
     1     2     3     1     2
     1     2     3     1     3
     1     2     3     2     1
     1     2     3     2     2
     1     2     3     2     3
     1     2     3     3     1
     1     2     3     3     2
     1     2     3     3     3
     1     3     1     1     2
     1     3     1     2     1
     1     3     1     2     2
     1     3     1     2     3
     1     3     1     3     2
     1     3     2     1     1
     1     3     2     1     2
     1     3     2     1     3
     1     3     2     2     1
     1     3     2     2     2
     1     3     2     2     3
     1     3     2     3     1
     1     3     2     3     2
     1     3     2     3     3
     1     3     3     1     2
     1     3     3     2     1
     1     3     3     2     2
     1     3     3     2     3
     1     3     3     3     2
     2     1     1     1     3
     2     1     1     2     3
     2     1     1     3     1
     2     1     1     3     2
     2     1     1     3     3
     2     1     2     1     3
     2     1     2     2     3
     2     1     2     3     1
     2     1     2     3     2
     2     1     2     3     3
     2     1     3     1     1
     2     1     3     1     2
     2     1     3     1     3
     2     1     3     2     1
     2     1     3     2     2
     2     1     3     2     3
     2     1     3     3     1
     2     1     3     3     2
     2     1     3     3     3
     2     2     1     1     3
     2     2     1     2     3
     2     2     1     3     1
     2     2     1     3     2
     2     2     1     3     3
     2     2     2     1     3
     2     2     2     3     1
     2     2     3     1     1
     2     2     3     1     2
     2     2     3     1     3
     2     2     3     2     1
     2     2     3     3     1
     2     3     1     1     1
     2     3     1     1     2
     2     3     1     1     3
     2     3     1     2     1
     2     3     1     2     2
     2     3     1     2     3
     2     3     1     3     1
     2     3     1     3     2
     2     3     1     3     3
     2     3     2     1     1
     2     3     2     1     2
     2     3     2     1     3
     2     3     2     2     1
     2     3     2     3     1
     2     3     3     1     1
     2     3     3     1     2
     2     3     3     1     3
     2     3     3     2     1
     2     3     3     3     1
     3     1     1     1     2
     3     1     1     2     1
     3     1     1     2     2
     3     1     1     2     3
     3     1     1     3     2
     3     1     2     1     1
     3     1     2     1     2
     3     1     2     1     3
     3     1     2     2     1
     3     1     2     2     2
     3     1     2     2     3
     3     1     2     3     1
     3     1     2     3     2
     3     1     2     3     3
     3     1     3     1     2
     3     1     3     2     1
     3     1     3     2     2
     3     1     3     2     3
     3     1     3     3     2
     3     2     1     1     1
     3     2     1     1     2
     3     2     1     1     3
     3     2     1     2     1
     3     2     1     2     2
     3     2     1     2     3
     3     2     1     3     1
     3     2     1     3     2
     3     2     1     3     3
     3     2     2     1     1
     3     2     2     1     2
     3     2     2     1     3
     3     2     2     2     1
     3     2     2     3     1
     3     2     3     1     1
     3     2     3     1     2
     3     2     3     1     3
     3     2     3     2     1
     3     2     3     3     1
     3     3     1     1     2
     3     3     1     2     1
     3     3     1     2     2
     3     3     1     2     3
     3     3     1     3     2
     3     3     2     1     1
     3     3     2     1     2
     3     3     2     1     3
     3     3     2     2     1
     3     3     2     3     1
     3     3     3     1     2
     3     3     3     2     1

[Adam Danz]

I had a very similar solution I'll add here as an adapted version of @Voss' answer.

Note the final step that converts the indices back into values from the input vector. I intentionally avoided using 1:3 so that the input values are not confused with index values.

k = [4 5 6]; 
n = 5;

% generate a matrix M whose rows are all possible
% sets of n elements from k:
m = numel(k);
M = (1+dec2base(0:m^n-1,m)-'0');

% remove rows of M that don't have each element 
c = arrayfun(@(x)any(ismember(M,x),2),1:m,'UniformOutput',false);
rm = ~all([c{:}],2);
M(rm,:) = []; 

% convert indices to k values
out = k(M)

out = 150×5

4 4 4 5 6 4 4 4 6 5 4 4 5 4 6 4 4 5 5 6 4 4 5 6 4 4 4 5 6 5 4 4 5 6 6 4 4 6 4 5 4 4 6 5 4 4 4 6 5 5

[mark palmer]

Thanks both that was quick and very helpful!

[Voss]

You're welcome!

Answer 2

[Cris LaPierre]

I think this does what you are looking for.

v = 1:3;
% create all 5-digit possible combinations
T = table2array(combinations(v,v,v,v,v));

% Extract only those that contain a 3 numbers
idx = any(T==1,2) & any(T==2,2) & any(T==3,2);
T = T(idx,:)

T = 150×5

1 1 1 2 3 1 1 1 3 2 1 1 2 1 3 1 1 2 2 3 1 1 2 3 1 1 1 2 3 2 1 1 2 3 3 1 1 3 1 2 1 1 3 2 1 1 1 3 2 2

[Adam Danz]

Great idea to use combinations (R2023a and later).

I fiddled with your solution a bit to make it flexibly receive any number of combinations.

v = 1:3;
n = 5; 

% create all 5-digit possible combinations
in = repelem({v},1,n); 
A = table2array(combinations(in{:}));

% Extract only those that contain a 3 numbers
c = arrayfun(@(x)any(ismember(A,x),2),v,'UniformOutput',false);
rm = ~all([c{:}],2);
A(rm,:) = []

A = 150×5

1 1 1 2 3 1 1 1 3 2 1 1 2 1 3 1 1 2 2 3 1 1 2 3 1 1 1 2 3 2 1 1 2 3 3 1 1 3 1 2 1 1 3 2 1 1 1 3 2 2

[Dyuman Joshi]

Nice approach @Cris LaPierre and @Adam Danz, but OP is working with R2022a, so they won't be able to utilize this.