what function for computing all subsets given an array of consecutive number?

Hi I'm a little lost looking up the function call for the task: computing all subset of this array, which has 12 consecutive numbers So, like given we have

a = [1:12]; 
  % need to get all 2^12 subsets [1, 2], [1, 2, 3]....

I'm trying but I don't think combntns() could work

[Bhanu Pratap Yadav]

can you give a MATLAB code so that I can generate all subsets of given number of elements

Answer 1

[John D'Errico]

S = dec2bin(0:2^12-1) - '0';

The presence of a one in each row will indicate the elements included in the corresponding subset. There are 2^12 rows, this will be a list of all possible subsets.

Answer 2

[Jan]

a = 1:12;
R = cell(1, numel(a) + 1);
R{1} = [];
for k = 1:numel(a)
  R{k + 1} = nchoosek(a, k);
end

Now R{k} contains the matrix with k-1 elements choosen from the input a. Or perhaps you want:

RR    = cell(1, numel(a) + 1);
RR{1} = {[]};
for k = 1:numel(a)
  RR{k + 1} = num2cell(nchoosek(a, k), 2);
end
R = cat(1, RR{:});

Answer 3

[Charles Kluepfel]

a = [1:12]; 
idxs = dec2bin(0:2^length(a)-1) - '0';
for i=1:length(idxs)
    idx=idxs(i,:);
    subset=a(find(idx));
    disp(subset);
end

12

11

11 12

10

10 12

10 11

10 11 12

9

9 12

9 11

9 11 12

9 10

9 10 12

9 10 11

9 10 11 12

8

8 12

8 11

8 11 12

8 10

8 10 12

8 10 11

8 10 11 12

8 9

8 9 12

8 9 11

. . .

1 2 3 4 5 6 7 9 11 12

1 2 3 4 5 6 7 9 10

1 2 3 4 5 6 7 9 10 12

1 2 3 4 5 6 7 9 10 11

1 2 3 4 5 6 7 9 10 11 12

1 2 3 4 5 6 7 8

1 2 3 4 5 6 7 8 12

1 2 3 4 5 6 7 8 11

1 2 3 4 5 6 7 8 11 12

1 2 3 4 5 6 7 8 10

1 2 3 4 5 6 7 8 10 12

1 2 3 4 5 6 7 8 10 11

1 2 3 4 5 6 7 8 10 11 12

1 2 3 4 5 6 7 8 9

1 2 3 4 5 6 7 8 9 12

1 2 3 4 5 6 7 8 9 11

1 2 3 4 5 6 7 8 9 11 12

1 2 3 4 5 6 7 8 9 10

1 2 3 4 5 6 7 8 9 10 12

1 2 3 4 5 6 7 8 9 10 11

1 2 3 4 5 6 7 8 9 10 11 12

[Walter Roberson]

You can do slightly better,

a = [1:12]; 
idxs = logical(dec2bin(0:2^length(a)-1) - '0');
for i=1:length(idxs)
    subset = a(idxs(i,:));
    disp(subset);
end