Splitting an array in every possible combination

Hello,

I am attempting to split an array in all possible combination. I have illustrated it in the figure given below.

Depending on the number of rows, I wish to split the given array in all possible combinations.

One of the condition is Number of rows <= Number of elements in given array

After toiling for 3 days, I am posting this here.

Kindly guide me on how to accomplish this task.

Thanks for your time and valuable guidance.

[Jonas]

are there restriction on the number of columns? or is the number given?

rows with 0 only are not allowed?

[Selva Kumar]

Thank you for the swift response Jonas.

Yes, rows with zeros are not allowed.

Each row must have atleast 1 non zero element.

[Jonas]

maybe it helps you to see how such a matrix is generated. You can run this script multiple times to see different variants

you did still not tell us how many columns are allowed

numElements=randi([2 10]); % number of elements shall be random between 2 and 10

elements=1:numElements; % define elements

numOfRows=randi([2 numElements]); % random number of rows

elInRowNrX=zeros(numOfRows,1); % vector containing the number of elements per row

createdMatrix=[]; 

for rowNr=1:numOfRows
    if rowNr<numOfRows
        elInRowNrX(rowNr)=randi(numElements-sum(elInRowNrX)-(numOfRows-rowNr)); % we cannot enter more values than total number of values minus already given values minus 1 for each remaining row
        createdMatrix(rowNr,1:elInRowNrX(rowNr))=elements(1:elInRowNrX(rowNr));
        elements(1:elInRowNrX(rowNr))=[];
    else
        elInRowNrX(rowNr)=numElements-sum(elInRowNrX); % last row has to contian the remaining number of elements
        createdMatrix(rowNr,1:elInRowNrX(rowNr))=elements;
    end
end

disp(createdMatrix)

     1     0     0     0     0
     2     3     4     5     6
     7     8     9    10     0

[Steven Lord]

Those aren't "every possible combination". What about:

[x6  0  0  0  0  0;
 x1 x2 x3 x4 x5  0]

What about:

[ 0  0  0  0  0  x1;
 x2 x3 x4 x5 x6   0]

or:

[x1  0  0  0  0;
 x2 x3 x4 x5 x6]

or

[x1 x2 x3;
 x4 x5 x6]

What rules haven't you stated that disallow those options?

[Jonas]

@Steven Lord to my understanding we have to fill the rows from the left and put in the elements one by one

your last case depends on a definition of number of columns, which was also not given. but the requested output can be generated by my script

numElements=6; % number of elements shall be random between 2 and 10
elements=1:numElements; % define elements
numOfRows=2; % random number of rows
elInRowNrX=zeros(numOfRows,1); % vector containing the number of elements per row
createdMatrix=[]; 
for rowNr=1:numOfRows
    if rowNr<numOfRows
        elInRowNrX(rowNr)=randi(numElements-sum(elInRowNrX)-(numOfRows-rowNr)); % we cannot enter more values than total number of values minus already given values minus 1 for each remaining row
        createdMatrix(rowNr,1:elInRowNrX(rowNr))=elements(1:elInRowNrX(rowNr));
        elements(1:elInRowNrX(rowNr))=[];
    else
        elInRowNrX(rowNr)=numElements-sum(elInRowNrX); % last row has to contian the remaining number of elements
        createdMatrix(rowNr,1:elInRowNrX(rowNr))=elements;
    end
end
disp(createdMatrix)

     1     2     3
     4     5     6

[Selva Kumar]

Jonas,

Your script gives one of the correct combination while I require all possible combinations.

For example, if number of elements in array is 6 and number of rows is 2, then there will be a total of 5 possible combinations. I have illustrated all the possible 5 combinations in the figure attached.

If number of elements in array is 6 and number of rows is 3, then there will be a total of 10 possible combinations. I have illustrated all the possible 5 combinations in the figure attached.

In order to make my question clearer, I am listing few points

1. Presence of zeroes at the end is allowed.
2. 1<=Number of columns<=Number of elements in the array.
3. Each row shall atleast one element.

[Dyuman Joshi]

Is the number of rows allowed only 2 and 3?

[Selva Kumar]

Both the combinations given below are same.

[ 0 0 0 0 0 x1;

x2 x3 x4 x5 x6 0]

and

[x1 0 0 0 0;

x2 x3 x4 x5 x6]

[Selva Kumar]

Dyuman Joshi

I have stated the number of rows as an example.

I wish to change the number of rows as per my requirement.

The number of rows shall be less than equal to number of elements in the array

[Dyuman Joshi]

This looks like a herculean task. I'll give it a try and update you.

[Jonas]

@Selva Kumar, i know that it gives only one solution, i also wrote that it gives one at a time. but all of the solutions are at least valid. it should just be a help how the problem could finally be solved, that's why the code is a comment and not in the answer section

[Jonas]

one 'solution' could be to run the script multiple times and collect unique solutions. but we still do not know restrictions on the number of columns

Answer 1

[Voss]

I believe this will do the required "splitting" of the array:

format compact
x = 1:6;
disp(split_x(x,1));

     1     2     3     4     5     6

disp(split_x(x,2));

(:,:,1) =
     1     0     0     0     0
     2     3     4     5     6
(:,:,2) =
     1     2     0     0     0
     3     4     5     6     0
(:,:,3) =
     1     2     3     0     0
     4     5     6     0     0
(:,:,4) =
     1     2     3     4     0
     5     6     0     0     0
(:,:,5) =
     1     2     3     4     5
     6     0     0     0     0

disp(split_x(x,3)); % scroll down to see them all

(:,:,1) =
     1     0     0     0
     2     0     0     0
     3     4     5     6
(:,:,2) =
     1     0     0     0
     2     3     0     0
     4     5     6     0
(:,:,3) =
     1     0     0     0
     2     3     4     0
     5     6     0     0
(:,:,4) =
     1     0     0     0
     2     3     4     5
     6     0     0     0
(:,:,5) =
     1     2     0     0
     3     0     0     0
     4     5     6     0
(:,:,6) =
     1     2     0     0
     3     4     0     0
     5     6     0     0
(:,:,7) =
     1     2     0     0
     3     4     5     0
     6     0     0     0
(:,:,8) =
     1     2     3     0
     4     0     0     0
     5     6     0     0
(:,:,9) =
     1     2     3     0
     4     5     0     0
     6     0     0     0
(:,:,10) =
     1     2     3     4
     5     0     0     0
     6     0     0     0

disp(split_x(x,4)); % scroll down to see them all

(:,:,1) =
     1     0     0
     2     0     0
     3     0     0
     4     5     6
(:,:,2) =
     1     0     0
     2     0     0
     3     4     0
     5     6     0
(:,:,3) =
     1     0     0
     2     0     0
     3     4     5
     6     0     0
(:,:,4) =
     1     0     0
     2     3     0
     4     0     0
     5     6     0
(:,:,5) =
     1     0     0
     2     3     0
     4     5     0
     6     0     0
(:,:,6) =
     1     0     0
     2     3     4
     5     0     0
     6     0     0
(:,:,7) =
     1     2     0
     3     0     0
     4     0     0
     5     6     0
(:,:,8) =
     1     2     0
     3     0     0
     4     5     0
     6     0     0
(:,:,9) =
     1     2     0
     3     4     0
     5     0     0
     6     0     0
(:,:,10) =
     1     2     3
     4     0     0
     5     0     0
     6     0     0

disp(split_x(x,5)); % scroll down to see them all

(:,:,1) =
     1     0
     2     0
     3     0
     4     0
     5     6
(:,:,2) =
     1     0
     2     0
     3     0
     4     5
     6     0
(:,:,3) =
     1     0
     2     0
     3     4
     5     0
     6     0
(:,:,4) =
     1     0
     2     3
     4     0
     5     0
     6     0
(:,:,5) =
     1     2
     3     0
     4     0
     5     0
     6     0

disp(split_x(x,6));

     1
     2
     3
     4
     5
     6

function x_split = split_x(x,n_rows)

N = numel(x);
n_cols = N-n_rows+1;

split_idx = nchoosek(1:N-1,n_rows-1);
n_combos = size(split_idx,1);
split_idx = [zeros(n_combos,1) split_idx N*ones(n_combos,1)];
n_split = diff(split_idx,1,2);

x_split = zeros(n_rows,n_cols,n_combos);
for ii = 1:n_combos
    for jj = 1:n_rows
        x_split(jj,1:n_split(ii,jj),ii) = x(split_idx(ii,jj)+(1:n_split(ii,jj)));
    end
end

end

[Selva Kumar]

This is exactly what I wished for !!!

Thanks a lot Voss ...

Thanks to Jonas, Dyuman Joshi, Steven Lord for your time and effort :)