Splitting an array in every possible combination

Question

Selva Kumar 2022-7-14

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1759910-splitting-an-array-in-every-possible-combination

评论： Selva Kumar 2022-7-16

采纳的回答： Voss

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.

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Jonas 2022-7-14

编辑：Jonas 2022-7-14

在 MATLAB Online 中打开

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

Jonas 2022-7-14

编辑：Jonas 2022-7-14

在 MATLAB Online 中打开

@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

Jonas 2022-7-14

@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 2022-7-14

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

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Answer 1

Voss 2022-7-15

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1759910-splitting-an-array-in-every-possible-combination#answer_1007795

编辑：Voss 2022-7-15

在 MATLAB Online 中打开

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

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Selva Kumar 2022-7-16

This is exactly what I wished for !!!

Thanks a lot Voss ...

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

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Splitting an array in every possible combination

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