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

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Trung Ngo
Trung Ngo 2017-4-17
评论： Walter Roberson 2024-8-9
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

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回答（3 个）

John D'Errico
John D'Errico 2017-4-17
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.
Jan
Jan 2017-4-17
编辑：Jan 2017-4-17
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{:});
Charles Kluepfel
Charles Kluepfel 2024-8-9
编辑：Walter Roberson 2024-8-9
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
  1 个评论
Walter Roberson
Walter Roberson 2024-8-9
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

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